240511165 Instructor S Manual 3363v

Business Logistics/ Supply Chain Management Planning, Organizing, and Controlling the Supply Chain Fifth Edition

Instructor’s Manual

Ronald H. Ballou Weatherhead School of Management Case Western Reserve University

CONTENTS Preface Chapter 1 2 3 4 5 6 7

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10 11 12 13

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Business Logistics/Supply ChainA Vital Subject……… Logistics/Supply Chain Strategy and Planning…………… The Logistics/Supply Chain Product...…………………… Logistics/Supply Chain Customer Service…..…………… Order Processing and Information Systems………………. Transport Fundamentals………………………………….. Transport Decisions………………………………………. Fowler Distributing Company………………………….. Metrohealth Medical Center……………………………. Orion Foods, Inc………………………………............... R & T Wholesalers……………………………………... Forecasting Supply Chain Requirements…………….…… World Oil……………………………………………….. Metro Hospital …………………………………………. Inventory Policy Decisions……………………………….. Complete Hardware Supply, Inc….…………………….. American Lighting Products……………………………. American Red Cross: Blood Services………………….. Purchasing and Supply Scheduling Decisions……………. Industrial Distributors, Inc……………………………… The Storage and Handling System………………………... Storage and Handling Decisions………………………….. Facility Location Decisions………………………………. Superior Medical Equipment Company……………….... Ohio Auto & Driver’s License Bureau…………………. Southern Brewery ……………………………………… The Logistics Planning Process…………………………... Usemore Soap Company………………….…………….. Essen USA………………….…………….. Logistics/Supply Chain Organization……………………. Logistics/Supply Chain Control……………..…………….

iii 1 2 4 9 13 14 17 35 41 48 52 65 84 88 94 121 124 131 134 144 147 148 162 186 190 198 204 208 217 229 230

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PREFACE This instructor's guide provides answers to the more quantitatively oriented problems at the end of the textbook chapters. If the questions or problems are for discussion or they involve a substantial amount of individual judgment, they have not been included. Solutions to the cases and exercises in the text are also included. These generally require computer assistance for solution. With the text, you are provided with a collection of software programs, called LOGWARE, that assist in the solution of the problems, cases, and exercises in the text. The LOGWARE software along with a ’s manual is available for ing from the Prentice Hall website or this book. The ’s manual is in Microsoft Word or Acrobat .pdf formats. This software, along with the ’s manual, may be freely reproduced and distributed to your classes without requiring permission from the copyright holder. This permission is granted as long as the use of the software is for educational purposes. If you encounter difficulty with the software, direct questions to Professor Ronald H. Ballou Weatherhead School of Management Case Western Reserve University Cleveland, Ohio 44106 Tel: (216) 368-3808 Fax: (216) 368-6250 E-mail: [email protected] Web site: www.prenhall.com/ballou

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CHAPTER 1 BUSINESS LOGISTICS/SUPPLY CHAINA VITAL SUBJECT 12 (a) This problem introduces the student to the evaluation of alternate channels of production and distribution. To know whether domestic or foreign production is least expensive, the total of production and distribution costs must be computed from the source point to the marketplace. Two alternatives are suggested, and they can be compared as follows. Production at Houston: Total cost = Production cost at Houston + Transportation and storage costs = $8/shirt100,000 shirts + $5/cwt. 1,000 cwt. = $805,000/year Production at Taiwan: Total cost = Production cost in Taiwan + Transportation and storage costs from Taiwan to Chicago + Import duty + Raw material transportation cost from Houston to Taiwan = $4/shirt100,000 shirts + $6/cwt. 1,000 cwt. + $0.5/shirt100,000 shirts + $2/cwt. 1,000 cwt. = $458,000/year Producing in Taiwan would appear to be the least expensive. (b) Other factors to consider before a final decision is made might be: (i) How reliable would international transportation be compared with domestic transportation? (ii) What is the business climate in Taiwan such that costs might change in favor of Houston as a production point? (iii) How likely is it that the needed transportation and storage will be available? (iv) If the market were to expand, would there be adequate production capacity available to the increased demand?

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CHAPTER 2 LOGISTICS/SUPPLY CHAIN STRATEGY AND PLANNING 13 The purpose of this exercise is to allow the student, in an elementary way, to examine the tradeoffs between transportation and inventory-related costs when an incentive transportation rate is offered. Whether the incentive rate should be implemented depends on the shipment size corresponding to the minimum of the sum of transportation, inventory, and order processing costs. These costs are determined for various shipping quantities that might be selected to cover the range of shipment sizes implied in the problem. Table 2-1 gives a summary of the costs to Monarch for various shipment sizes. From Monarch's point of view, the incentive rate would be beneficial. Shipment sizes should be approximately doubled so that the 40,000 lb. minimum is achieved. It is important to note that the individual cost elements are not necessarily at a minimum at low shipment sizes, whereas order-processing costs are low at high shipment sizes. They are in cost conflict with each other. Transportation costs are low at high shipment sizes, but exact costs depend on the minimum volume for which the rate is quoted. In preparation for a broader planning perspective to be considered later in the text, the student might be asked what the place of the supplier is in this decision. How does he affect the decision, and how is he affected by it? This will focus the student's attention on the broader issues of the physical distribution channel.

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TABLE 2-1

Evaluation of Alternative Shipment Sizes for the Monarch Electric Company Current Proposed 57 motors 114 motors 171 motors 228 motors 285 motors or or or or or 10,000 lb. 20,000 lb. 30,000 lb. 40,000 lb. 50,000 lb. Type of cost Transportation 38,750 98,750 58,750 58,750 38,750 = $26,250 RD = $78,750 = $43,750 = $43,750 = $26,250a b Inventory carrying 0.25200114/2 0.25200171/2 0.25200228/2 0.25200285/2 0.2520057/2 = $2,850 = $4,275 = $5,700 = $7,125 ICQ/2 = $1,425a c Order processing 5,00015/57 5,00015/114 5,00015/171 5,00015/228 5,00015/285 DS/Q = $1,316 = $658 = $439 = $329 = $263a Handling 0.308,750 0.308,750 0.308,750 0.308,750 0.308,750 HD = $2,625 = $2,625 = $2,625 = $2,625 = $2,625 Total $84,116 $49,883 $51,089 $36,263 $34,904a a

Minimum values. Students should be informed that average inventory can be approximated by one half the shipment size. c Demand D has been converted to units per year. LEGEND R = transportation rate, $/cwt. D = annual demand, cwt. I = inventory carrying cost, %/year. C = cost of a motor, $/motor. Q = shipment size in motors, where Q/2 represents the average number of motors maintained in inventory. S = order processing costs, $/order. H = handling costs, $/cwt. b

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CHAPTER 3 THE LOGISTICS/SUPPLY CHAIN PRODUCT 3 The 80-20 principle applies to sales and items where 80 percent of the dollar volume is generated from 20 percent of the product items. While this ratio rarely holds exactly in practice, the concept does. We can apply it to these data by ranking the products by sales, and the percentage that the cumulative sales represent of the total. The following table shows the calculations.

Product code 08776 12121 10732 11693 10614 12077 07071 10542 06692 09721 14217 11007 Total

Dollar sales $71,000 63,000 56,000 51,000 46,000 27,000 22,000 18,000 14,000 10,000 9,000 4,000 $391,000

Cumulative sales $ 71,000 134,000 190,000 241,000 287,000 314,000 336,000 336,000 354,000 368,000 378,000 391,000

Cumulative sales as % of total 18.2 34.3 48.6 61.6 73.4 80.3 85.9 90.5 94.1 96.7 98.9 100.0

Cumulative items as % of total 8.3 16.7 25.0 33.3 41.7 50.0 58.3 66.7 75.0 83.3 91.7 100.0

The 80-20 rule cannot be applied exactly, since the cumulative percent of items does not break at precisely 20 percent. However, we might decide that only products 08776 and 12121 should be ordered directly from vendors. The important principle derived from the 80-20 rule is that not every item is of equal importance to the firm, and that different channels of distribution can be used to handle them. The 80-20 rule gives some rational basis for deciding which products should be shipped directly from vendors and which are more economically handled through a system of warehouses. 6 (a) Reading the ground transport rates for the appropriate zone as determined by zip code and the weight of 27 lb. (rounding upward of 26.5 lb.) gives the following total cost table for the four shipments.

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To zip code 11107 42117 74001 59615 a

Catalog price $99.95 99.95 99.95 99.95

UPS zone 2 5 6 8

Transport costa $ 7.37 10.46 13.17 18.29

Total cost $107.32 110.41 113.12 118.24

Use 27 lb.

(b) The transport rate structure is reasonably fair, since ground rates generally follow distance and size of shipment. These are the factors most directly affecting transport costs. They are not fair in the sense that customers within a zone are all charged the same rate, regardless of their distance from the shipment origin point. However, all customers may benefit from lower overall rates due to this simplified zone-rate structure. 10 (a) This is a delivered pricing scheme where the seller includes the transport charges in the product price. The seller makes the transport arrangements. (b) The seller prices the product at the origin, but prepays any freight charges; however, the buyer owns the goods in transit. (c) This is a delivered pricing scheme where the freight charges are included in the product price, however the freight charges are then deducted from the invoice, and the seller owns the goods in transit. (d) The seller initially pays the freight charges, but they are then collected from the buyer by adding them to the invoice. The buyer owns the goods in transit, since the pricing is f.o.b. origin. (e) The price is f.o.b. origin. The buyer pays the freight charges and owns the goods in transit. Regardless of the price policy, the customer will ultimately pay all costs. If a firm does not consider outbound freight charges, the design of the distribution system will be different than if it does. Since pricing policy is an arbitrary decision, it can be argued that transport charges should be considered in decision making, whether the supplying firm directly incurs them or not. 11 This shows how Pareto's law (80-20 principle) is useful in estimating inventory levels when a portion of the product line is to be held in inventory. An empirical function that approximates the 80-20 curve is used to estimate the level of sales for each product to be held in inventory. According to Equation 3-2, the constant A is determined as follows.

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A

X (1  Y ) 0.25(1.75)   0125 . YX 0.75  0.25

The 80-20 type curve according to Equation 3-1 is: Y

(1  A) (1  0125 . )X  A X 0125 . X

This formula can be used to estimate the cumulative sales from the cumulative item proportion. For example, item 1 is 0.05 of the total number of items (20) so that:

Y

(1  0125 . )( 0.05)  0.321 0125 .  0.05

Of the $2,600,000 in total annual warehouse sales, item 1 should for 0.3212,600,000 = $835,714. By applying this formula to all items, the following inventory investment table can be developed which shows sales by item. The average inventory investment by item is found by dividing the turnover ratio into the item sales. The sum of the average inventory value for each item gives a total projected inventory of $380,000. Inventory Investment Table Product 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

A

B

C

Cumulative item proportion, X 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50 0.55 0.60 0.65 0.70 0.75 0.80 0.85 0.90 0.95 1.00

Cumulative sales, Y $ 835,714 1,300,000 1,595,454 1,800,000 1,950,000 2,064,705 2,155,263 2,228,571 2,289,130 2,340,000 2,383,333 2,420,689 2,453,226 2,481,818 2,507,142 2,529,719 2,550,000 2,568,293 2,584,884 2,600,000

Projected item sales $ 835,714 464,286 295,454 204,546 150,000 114,706 90,558 73,308 60,559 50,870 43,333 37,356 32,537 28,592 25,324 22,587 20,271 18,293 16,591 15,116

Turnover ratio 8 8 8 8 6 6 6 6 6 6 4 4 4 4 4 4 4 4 4 4 Total

Average inventory value $104,464 58,036 36,932 25,568 25,000 19,118 15,093 12,218 10,093 8,478 10,833 9,339 8,134 7,148 6,331 5,647 5,068 4,473 4,148 3,779 $380,000

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12 This problem involves the application of Equations 3-1 and 3-2. We can develop an 8020 curve based on 30 percent of the items ing for 70 percent of sales. That is,

X (1  Y ) 0.30(1  0.70)   0.225 0.70  0.30 YX

A

Therefore, the sales estimating equation is: Y

(1  0.225) X 0.225  X

By applying this estimating curve, we can find the sales of A and B items. For example, 20 percent of the items, or 0.220 = 4 items, will be A items with a cumulative proportion of sales of: YA 

(1  0.225)( 0.20)  0.5765 0.225  0.20

and 3,000,0000.5765 = 1,729,412. The A+B item proportion will be: YA B 

(1  0.225)( 0.50)  0.8448 0.225  0.50

and 3,000,0000.8448 = 2,534,400. The product group B sales will A+B sales less A sales, or 2,534,400  1,729,412 = $804,988. The product group C will be the remaining sales, but these are not of particular interest in this problem. The average inventories for A and B products are found by dividing the estimated sales by the turnover ratio. That is, A: B:

1,729,412/9 804,988/5 Total inventory

= 192,157 = 160,988 353,155 cases

The total cubic footage required for this inventory would be 353,1551.5 = 529,732 cu. ft. The total square footage for products A and B is divided by the stacking height. That is, 529,731/16 = 33,108 sq. ft.

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13 This problem is an application of Equations 3-1 and 3-2. We first determine the constant A. That is,

A

X (1  Y ) 0.20(1  0.65)   0156 . 0.65  0.20 YX

and 0.75 

(1  0156 . )X 0156 . X

Solving algebraically for X, we have: X

AxY 0156 . x 0.75   0.288 1  A  Y 1  0156 .  0.75

That is, about 29% of the items (0.2885,000 = 1,440 items) produce 75% of the sales. 14 The price would be the sum of all costs plus an increment for profit to place the automotive component in the hands of the customer. This would be 25+10+5+8+5+transportation cost, or 53+T. Based on the varying transportation cost, the following price schedule can be developed.

Quantity 1 to 1,000 units 1,001 to 2,000 units >2,000 units a

Price per unit 53+5=$58 53+4.00=57 53+3.00=56

Discount 0 1.7%a 3.5%

[(58 - 57)/58][100]=1.7%

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CHAPTER 4 LOGISTICS/SUPPLY CHAIN CUSTOMER SERVICE 6 (a) This company is fortunate to be able to estimate the sales level that can be achieved at various levels of distribution service. Because of this, the company should seek to maximize the difference between sales and costs. These differences are summarized as follows.

Percent of orders delivered within 1 day Contribution to 50 60 70 80 90 95 100 profit -1.8 2.0 3.5 4.0 3.4 2.8 -2.0 The company should strive to make deliveries within 1 day 80 percent of the time for a maximum contribution to profit. (b) If a competing company sets its delivery time so that more than 80 percent of the orders are delivered in 1 day and all other factors that attract customers are the same, the company will lose customers to its competitor, as the sales curve will have shifted downward. Cleanco should adjust its service level once again to the point where the profit contribution is maximized. Of course, there is no guarantee that the previous level of profits can be achieved unless the costs of supplying the service can correspondingly be reduced. 7 (a) This problem solution requires some understanding of experimental design and statistical inference, which are not specifically discussed in the text. Alert the students to this. The first task is to determine the increase in sales that can be attributed to the change in the service policy. To determine if there is a significant change in the control group, we set up the following hypothesis test. z

X 2  X1 s22 s2  1 N 2 N1



224  185 612 79 2  102 102



39 .  394 36.48  6118 .

Now, referring to a normal distribution table in Appendix A of the text, there is a significant difference at the 0.01 level in the sales associated with the control group. That is, some factors other than the service policy alone are causing sales to increase. Next, we analyze the test group in the same manner.

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z

2,295  1,342 576 2 3352  56 56



953  10.7 5,924  2,004

This change is also significant at the 0.01 level. The average increase in sales for the control group is 224/185 = 1.21, or 21%. The average sales increase in the test group is 2295/1342 = 1.71, or 71%. If we believe that 21% of the 71% increase in the test group is due to factors other than service policy, then 71  21 = 50% was the true service effect. Therefore, for each sales unit, an incremental increase in profit of (0.4095)(0.50) = $19 can be realized. Since the cost of the service improvement is $2, the benefit exceeds the cost. The service improvement should be continued. Note: If the students are not well versed in statistical methodology, you may wish to instruct them to consider the before and after differences in the mean values of both groups as significant. The solution will be the same. (b) The use of the before-after-with-control-group experimental design is a methodology that has been used for some time, especially in marketing research studies. The outstanding feature of the design is that the use of the control group helps to isolate the effect of the single service variable. On the other hand, there are a number of potential problems with the methodology:  The sales distributions may not be normal.  The time that it takes for diffusing the information that a service change has taken place may distort the results.  The products in the control group may not be mutually exclusive from those in the test group.  The method only shows the effect of a single step change in service and does not develop a sales-service relationship.  It may not always be practical to introduce service changes into on-going operations to test the effect. 8 (a) The optimum service level is set at that point where the change in gross profit equals the change in cost.

The change in gross profit: P = Trading margin  Sales response rate  Annual sales = 1.000.0015100,000 = $150 per year per 1% change in the service level The change in cost: C = Annual carrying cost  Standard product cost  z 10

 Demand standard deviation for order cycle = 0.3010.00400z Now, set P = C and solve for z. 150 = 1200z z = 0.125 From the tabulated changes in service level with those changes in z, the service level should be set between 96-97%. (b) The weakest link in this analysis is estimating the effect that a change in service will have on revenue. This implies that a sales-service relationship is known. 9 The methodology is essentially the same as that in question 7, except that we are asked to find X instead of Y. That is,

P = 0.750.001580,000 = 90 and C = 0.251,000500z = 1250z Then, P= C 90 = 1250z z = 0.072 From the normal distribution (see Appendix A), the z for an area under the curve of 93% is 1.48, and for 92%, z is 1.41. Since the difference of 1.48  1.41 = 0.07, we can conclude that the in-stock probability should be set at 92-93%. Of course, the change in z is found by taking the difference in z values for 1% differences in the area values under the normal distribution curve for a wide range of area percentages. 10 Apply Taguchi’s concept of the loss function. First, estimate the loss per item if the target level of service is not met. We know the profit per item as follows.

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Sales price Cost of item Other costs Profit per item

$5.95 -4.25 -0.30 $1.40

Since one-half of the sales are lost, the opportunity loss per item would be Profit per item

Sales lost $1.40  (1/2)(880) Opportunity loss   $0.70/item 880 Current sales

Next, find k in the loss function. L  k ( y  m) 2 0.70  k (10  5)2 0.70  k ( 25) k  0.03

out-of-stock % at point where ½ sales are lost Target %

Finally, the point where the marginal supply cost equals the marginal sales loss is ( y  5) 

B 0.10   1.67% 2k 2(0.03)

y  1.67  5  6.67% The retailer should not allow the out-of-stock percentage to deviate more than 1.67%, and should not allow the out-of-stock level to fall below 1.67 + 5 = 6.67%.

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CHAPTER 5 ORDER PROCESSING AND INFORMATION SYSTEMS

All questions in this chapter require individual judgment and response. No answers are offered.

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CHAPTER 6 TRANSPORT FUNDAMENTALS 14 The maximum that the power company can pay for coal at its power plant location in Missouri is dictated by competition. Therefore, the landed cost at the power plant of coal production costs plus transportation costs cannot exceed $20 per ton. Since western coal costs $17 per ton at the mine, the maximum worth of transportation is $20  $17 = $3 per ton. However, if the grade of coal is equal to the coal from the western mines, eastern coal can be landed in Missouri for $18 per ton. In light of this competitive source, transportation from the western mines is worth only $18  $17 = $1 per ton. 15 Prior to transport deregulation, it was illegal for a carrier to charge shippers less for the longer haul than for the shorter haul under similar conditions when the shorter haul was contained within the longer one. To be fair, the practice probably should be continued. If competitive conditions do not permit an increase in the rate to Z, then all rates that exceed $1 per cwt. on a line between X and Z should not exceed $1 per cwt. Therefore, the rate to Z is blanketed back to Y so that the rate to Y is $1 per cwt. By blanketing the rate to Z on intervening points, no intervening point is discriminated against in of rates. 16 (a) From text Table 6-4, the item number for place mats is 4745-00. For 2,500 lb., the classification is 100 since 2,500 lb. is less than the minimum weight of 20,000 lb. for a truckload shipment. From text Table 6-5, the rate for a shipment 2,000 lb. is 8727¢/cwt. The shipping charges are $87.27  25 cwt. = $2,181.75.

(b) This is an LTL shipment with a classification of 100, item number 4980-00 in text Table 6-4. From Table 6-5, the minimum charge is 9351¢ and the rate for a <500 lb. shipment is 5401¢/cwt. Check the charges using the <500 lb. rate and compare it to the minimum charge. That is, $54.01  1.5 cwt. = $81.02 Since this is less than the minimum charge of $93.51, pay the minimum charge. (c) From Table 6-4, the item number is 2055-00 with a classification of 55 for LTL and 37.5 for TL at a minimum weight of 36,000 lb. There are three possibilities that need to be examined: (1) Ship LTL at class 55 and 27,000 lb. shipment. (2) Ship at class 55 and 30,000 lb. rate. (3) Ship at class 37.5 and 36,000 lb. rate.

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Try (1): Rate is $5.65/cwt. 5.65  270 = $1,525.50. Try (2): Rate is $3.87/cwt. 3.87  300 = $1,161.00

Lowest cost

Try (3): Rate is $3.70/cwt. 3.70  360 = $1,332.00 (d) The shipment is a truckload classification (2070-00) of 65. The rate at 30,000 lb. is $4.21/cwt. The charges are 4.21  300 = $1,263.00. (e) Classification of this product is 55 (4860-00) for a truckload of 24,000 lb. Check the break weight according to Equation 6-1. Break weight =

3.87  30,000  20,549 lb. 5.65

Since current shipping weight of 24,000 lb. exceeds the break weight, ship as if 30,000 lb. Hence, 3.87  300 = $1,161.00. Now, discount the charges by 40 percent. That is, $1,161  (1  0.40) = $696.60 21 The question involves evaluating two alternatives. The first is to compute the transport charges as if there are three separate shipments. The next is to see if a stop-off privilege offers any cost reduction. The comparison is shown below. Separate shipments

Loading/unloading 22,000 3,000 15,000

Route A to D A to C B to C

Rate, Stop-off $/cwt. charge Charges $3.20 --$704.00 2.50 --75.00 1.50 --225.00 Total charges $1,004.00

With stop-off Ship direct to B and split deliver thereafter.

Rate, Stop-off Loading/unloading Route $/cwt. charge Charges 25,000 A to B $1.20 $ 300.00 40,000 B to D 2.20 880.00 Stop-off @ C $25.00 25.00 Stop-off @ D 25.00 25.00 Total charges $1,230.00

Direct shipment

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Split deliver at all stops. Rate, Stop-off Loading/unloading Route $/cwt charge Charges 40,000 A to D 3.20 1,280.00 Stop-off @ B 25.00 25.00 Stop-off @ C 25.00 25.00 Stop-off @ D 25.00 25.00 Total charges $1,335.00 Other combinations may be tried. In this case, there appears to be no advantage to using the stop-off privilege.

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CHAPTER 7 TRANSPORT DECISIONS 1 Selecting a mode of transportation requires balancing the direct cost of transportation with the indirect costs of both vendor and buyer inventories plus the in-transit inventory costs. The differences in transport mode performance affect these inventory levels, and, therefore, the costs for maintaining them, as well as affect the time that the goods are in transit. We wish to compare these four cost factors for each mode choice as shown in Table 7-1 of the manual. The symbols used are:

R = transportation rate, $/unit D = annual demand, units C = item value at buyer's inventory, $ C' = item value at vendor's inventory, $ T = time in transit, days Q = Shipping quantity, units Rail has the lowest total cost. TABLE 7-1

An Evaluation of the Transport Alternatives for the Wagner Company

Cost type Transport

Method RD

Rail 2550,000 = $1,250,000 In-transit 0.2547550,000 inventorya IC’Dt/365 (16/365) = $260,274 Wager’s 0.25475(10,000/2) inventorya IC’Q/2 = $593,750 Electronic’s 0.25500(10,00/2) inventory ICQ/2 = $625,000 Total $2,729,024 a C’ refers to price less transport cost per unit.

Piggyback 4450,000 = $2,200,000 0.2545650,000 (10/365) = $156,164 0.25456(7,000/2) = $399,000 0.25500(7,000/2) = $437,500 $3,192,664

Truck 8850,000 = $4,400,000 0.2541250,000 (4/365) = $56,438 0.25412(5,000/2) = $257,500 0.25500(5,000/2) = $312,500 $5,026,438

2 As in question 1, this problem is one of balancing transport costs with the indirect costs associated with inventories. However, in this case we must for the variability in transit time as it affects the warehouse inventories. We can develop the following decision table.

Service type

17

Cost type

Method

Transport

RD

In-transit inventory Plant inventory

ICDt/365

Warehouse inventory Total

ICQ*/2

IC’Q*/2 + IC’r

A 129,600 = $115,200 0.20509,600 (4/365) = $1,052 0.3050(321.8/2) = $2,684 0.3062(321.3/2) + 0.306250.5 = $3,927 $122,863

B 11.809,600 =$114,048 0.20509,600 (5/365) = $1,315 0.3050(357.8/2) = $2,684 0.3061.80(321.8/2) + 0.3061.8060.6 = $4,107 $122,154

Recall that Q*  2 DS / IC  2(9,600 )(100 ) / 0.3(50 )  357.8 cwt. for the plant, assuming the order cost is the same at plant and warehouse. However, for the warehouse, we must for safety stock (r) and for the transportation cost in the value of the product. Therefore, For A: Q *  2 DS / IC  2(9,600)(100) / 0.3( 62)  3213 . cwt.

and for z = 1.28 for an area under the normal distribution of 0.90, the safety stock is: r  zs LT ( d )  1.28  1.5  (9,600 / 365)  50.5 cwt.

For B: Q *  2(9,600 )(100 ) / 0.3( 6180 . )  3218 . cwt.

and

r  1.28  1.8  (9,600 / 365)  60.6 cwt. Service B appears to be slightly less expensive. 3 The shortest route method can be applied to this problem. The computational table is shown in Table 7-2. The shortest route is defined by tracing the links from the destination node. They are shown in Table 7-2 as A D F G for a total distance of 980 miles. TABLE 7-2

Tabulation of Computational Steps for the Shortest Route Method Applied to Transcontinental Trucking Company Problem

18

Solved nodes directly Its closest connected to connected unsolved unsolved Step nodes node 1 A B A D 2 A D B C 3 B C D C D F 4 C E C F D F 5 C F E G D F 6 E G F G a Asterisk indicates the shortest route

Total time involved 186 mi. 276 276 186+110= 296 186+110= 296 276+ 58= 334 276+300= 576 296+241= 537 296+350= 646 276+300= 576 296+350= 646 537+479=1016 276+300= 576 537+479=1016 576+404= 980

nth nearest node B

Its minimum time 186 mi.

Its last connectiona AB

D

276

AD*

C

296

BC

E

537

CE

F

576

DF*

G

980

FG*

4 In this actual problem, the U.S. Army used the transportation method of linear programming to solve its allocation problem. The problem can be set up in matrix form as follows:

Origin Destination Letterkenny Fort Hood Fort Riley

Cleveland 150 150 325 50 275 100 375

South Charleston 100 150 350

San Jose 800

325

350

Demand 300

300 50

100 400

275

Fort Carson

100

300

100

100

450

250

Fort Benning

100

0

Supply

400

150

100 150

The cell values shown in bold represent the number of personnel carriers to be moved between origin and destination points for minimum transportation costs of $153,750. An alternative solution at the same cost would be:

19

Origin Cleveland S. Charleston Cleveland San Jose Cleveland San Jose Cleveland

Destination Letterkenny Letterkenny Fort Hood Fort Hood Fort Riley Fort Carson Fort Benning

Number of carriers 150 150 50 50 100 100 100

5 This problem can be used effectively as an in-class exercise. Although the problem might be solved using a combination of the shortest route method to find the optimum path between stops and then a traveling salesman method to sequence the stops, it is intended that students will use their cognitive skills to find a good solution. The class should be divided into teams and given a limited amount of time to find a solution. They should be provided with a transparency of the map and asked to draw their solution on it. The instructor can then show the class each solution with the total distance achieved. From the least-distance solutions, the instructor may ask the teams to explain the logic of their solution process. Finally, the instructor may explore with the class how this and similar problems might be treated with the aid of a computer. Although the question asks the student to use cognitive skills to find a good route, a route can be found with the aid of the ROUTER software in LOGWARE. The general approach is to first find the route in ROUTER without regard to the rectilinear distances of the road network. Because this may produce an infeasible solution, specific travel distances are added to the database to represent actual distances traveled or to block infeasible paths from occurring. A reasonable routing plan is shown in Figure 7-1 and the ROUTER database that generates it is given in Figure 7-2. The total distance for the route is 9.05 miles and at a speed of 20 miles per hour, the route time is approximately 30 minutes.

0.5

1.0

1.5

2.0

0 19

20

21

0

17

20

FIGURE 7-2 Input Data for ROUTER for School Bus Routing Problem —PARAMETERS AND LABELS— Problem label – School Bus Routing Exercise Grid corner with 0,0 coordinates (NW, SW, SE, or NE) - NW DEPOT DATA Depot description - Atlanta Located in zone - 0 Horizontal coordinate – 0.14 Vertical coordinate – 0.45 Earliest starting time (min) - 0 Latest return time (min) - 9999 Default vehicle speed (miles per hour) - 20 After how many clock hours will overtime begin - 9999 GENERAL DATA Percent of vehicle in use before allowing pickups - 0 Horizontal scaling factor - 1 Vertical scaling factor - 1 Maximum TIME allowed on a route (hours) - 9999 Maximum DISTANCE allowed on a route (miles) - 9999 LOAD/UNLOAD TIME FORMULA Fixed time per stop - 0 Variable time per stop by weight - 0 By cube - 0 BREAK TIMES Duration of 1st break (minutes) - 0 To begin after - 9999 Duration of 2nd break (minutes) - 0 To begin after - 9999 Duration of 3rd break (minutes) - 0 To begin after - 9999 Duration of 4th break (minutes) - 0 To begin after - 9999

21

--STOP DATA— NO 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21

STOP LOAD VOL. LOAD DESCRIPTION TY WGHT CUBE HCRD VCRD ZN TIME BEG1 END1 Stop 1 D 1 0 0.14 0.80 0 0 0 9999 Stop 2 D 1 0 0.14 1.14 0 0 0 9999 Stop 3 D 1 0 0.14 1.31 0 0 0 9999 Stop 4 D 1 0 0.35 1.31 0 0 0 9999 Stop 5/22 D 1 0 0.52 0.61 0 0 0 9999 Stop 6 D 1 0 0.58 1.31 0 0 0 9999 Stop 7 D 1 0 0.80 1.31 0 0 0 9999 Stop 8 D 1 0 1.03 0.61 0 0 0 9999 Stop 9 D 1 0 1.03 0.96 0 0 0 9999 Stop 10 D 1 0 1.03 1.31 0 0 0 9999 Stop 11 D 1 0 1.36 1.31 0 0 0 9999 Stop 12 D 1 0 1.48 1.31 0 0 0 9999 Stop 13 D 1 0 1.80 1.31 0 0 0 9999 Stop 14 D 1 0 1.87 1.31 0 0 0 9999 Stop 15 D 1 0 1.84 0.61 0 0 0 9999 Stop 16 D 1 0 1.95 0.61 0 0 0 9999 Stop 17 D 1 0 1.29 0.10 0 0 0 9999 Stop 18 D 1 0 1.26 0.61 0 0 0 9999 Stop 19 D 1 0 1.15 0.10 0 0 0 9999 Stop 20 D 1 0 0.69 0.23 0 0 0 9999 Stop 21 D 1 0 0.14 0.26 0 0 0 9999

BEG2 9999 9999 9999 9999 9999 9999 9999 9999 9999 9999 9999 9999 9999 9999 9999 9999 9999 9999 9999 9999 9999

END2 9999 9999 9999 9999 9999 9999 9999 9999 9999 9999 9999 9999 9999 9999 9999 9999 9999 9999 9999 9999 9999

—VEHICLE DATA— --CAPACITY--

NO. 1

VEHICLE DESCRIPTION Bus

TP 1

NO 1

WGHT 9999

CUBE 9999

--VEHICLE-FIXED COST 0

PER MI COST 0

--DRIVER-FIXED COST 0

PER HR COST 0

OVER TIME COST 0

—SPECIFIED STOP-TO-STOP DISTANCES—

NO 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19

STOP NO. 14 14 15 16 18 19 19 19 19 9 9 9 5/22 5/22 5/22 5/22 20 20 20

STOP DESCRIPTION Stop 14 Stop 14 Stop 15 Stop 16 Stop 18 Stop 19 Stop 19 Stop 19 Stop 19 Stop 9 Stop 9 Stop 9 Stops 5&22 Stops 5&22 Stops 5&22 Stops 5&22 Stop 20 Stop 20 Stop 20

STOP NO. 16 15 17 17 9 8 20 5/22 18 20 19 21 1 21 20 9 21 0 5/22

STOP DISTANCE DESCRIPTION IN MILES Stop 16 0.78 Stop 15 0.90 Stop 17 1.06 Stop 17 1.18 Stop 9 0.58 Stop 8 0.76 Stop 20 0.59 Stops5&22 1.14 Stop 18 0.53 Stop 20 1.08 Stop 19 1.11 Stop 21 1.69 Stop 1 0.56 Stop 21 1.05 Stop 20 1.14 Stop 9 0.97 Stop 21 0.84 School 1.03 Stops 5&22 0.55

22

20 21 22

17 0 2

Stop 17 School Stop 2

0 School 5/22 Stops 5&22 5/22 Stops 5&22

2.43 1.37 1.03

6 Strategy 1 is to stay at motel M2 and serve the two routes on separate days. Using the ROUTESEQ module in LOGWARE gives us the sequence of stops and the coordinate distance. The routes originating at M2 would be:

Distancea 95.55 mi. 86.45 182.00 mi.

Route Stop sequence 1 8,6,1,4,2,3,5,7,9 2 10,13,14,17,18,16,12,15,11 a

Includes map scaling factor

The total cost of this strategy would be: Motel 3 nights @ 49.00 Travel 182 miles @ $.30/mi. Total

$147.00 54.60 $201.60

Strategy 2 is a mixed strategy involving staying at motels closest to the center of the stops clusters. The route sequences from different motels are: Route Stop sequence Distance 1 4,2,3,5,7,9,8,6,2 98.50 mi. 2 18,17,13,14,10,11,15,12,16 80.30 178.80 mi. The total cost of this strategy is: Motel M1 1st night $ 40.00 M1 2nd night 40.00 rd M1 3 night 45.00 Travela 214.80 mi. @ 0.30/mi. 64.44 Total $189.44 a

178.80 + 36 = 214.80

Strategy 2 appears to be most economical. 7 (a) Since distances are asymmetrical, we cannot use the geographically based traveling salesman method in LOGWARE. Rather, we use a similar module in STORM that allows such asymmetrical matrices, or the problem is small enough to be solved by inspection. For this problem, the minimal cost stop sequence would be:

23

BakeryStop 5Stop 3Stop 4Stop 2Stop 1Bakery with a tour time of 130 minutes. (b) Loading/unloading times may be added to the travel times to a stop. The problem may then be solved as in part a. (c) The travel times between stop 3 and all other nodes are increased by 50%. The remaining times are left unchanged. Optimizing on this matrix shows no change in the stop sequence. However, the tour time increases to 147.50 minutes. 8 This may be solved by using the ROUTER module in LOGWARE. The screen set up for this is as follows.

24

FIGURE 7-3 Input Data for ROUTER for Sima Donuts --PARAMETERS AND LABELS— Problem label - Sima Donuts Grid corner with 0,0 coordinates (NW, SW, SE, or NE) - NE DEPOT DATA Depot description - Atlanta Located in zone - 0 Horizontal coordinate - 2084 Vertical coordinate - 7260 Earliest starting time (min) - 180 Latest return time (min) - 9999 Default vehicle speed (miles per hour) - 45 After how many clock hours will overtime begin - 168 GENERAL DATA Percent of vehicle in use before allowing pickups - 0 Horizontal scaling factor - 0.363 Vertical scaling factor - 0.363 Maximum TIME allowed on a route (hours) - 40 Maximum DISTANCE allowed on a route (miles) - 1400 LOAD/UNLOAD TIME FORMULA Fixed time per stop - 0 Variable time per stop by weight - 0 By cube - 0 BREAK TIMES Duration of 1st break (minutes) - 60 To begin after - 720 Duration of 2nd break (minutes) - 60 To begin after - 1200 Duration of 3rd break (minutes) - 60 To begin after - 2160 Duration of 4th break (minutes) - 60 To begin after - 2640 --STOP DATA— STOP NO DESCRIPTION 1 Tampa FL 2 Clearwater FL 3 Daytona Beach F 4 Ft Lauderdale FL 5 N Miami FL 6 Oakland Park FL 7 Orlando FL 8 St Petersburg FL 9 Tallahassee FL 10 W Palm Beach F 11 Puerto Rico

TY D P D D D P D P D D D

LOAD WGHT 20 14 18 3 5 4 3 3 3 3 4

VOL. CUBE 0 0 0 0 0 0 0 0 0 0 0

HCRD VCRD ZN 1147 8197 0 1206 8203 0 1052 7791 0 557 8282 0 527 8341 0 565 8273 0 1031 7954 0 1159 8224 0 1716 7877 0 607 8166 0 527 8351 0

LOAD TIME 15 45 45 45 45 45 45 45 15 45 45

BEG1 360 360 360 180 360 180 180 180 600 360 360

END1 BEG2 1440 1800 1440 1800 1440 1800 1440 1800 1440 1800 1440 1800 1440 1800 1440 1800 1440 1800 1440 1800 1440 1800

END2 2880 2880 2880 2880 2880 2880 2880 2880 2880 2880 2880

--VEHICLE DATA— -CAPACITY--

NO. 1 2 3

VEHICLE DESCRIPTION Truck #1-20 Truck #2-25 Truck #3-30

TP 1 2 3

NO 3 1 1

WGHT 20 25 30

CUBE 9999 9999 9999

--VEHICLE-FIXED COST 0 0 0

PER MI COST 1.30 1.30 1.30

--DRIVER-FIXED COST 0 0 0

PER HR COST 0 0 0

OVER TIME COST 0 0 0

Making a run with ROUTER will give the route design.

25

Pickup

Pickup

FIGURE 7-4 Graphical Display of Route Design for Sima Donuts

The route design involves 3 routes for a total distance of 3,830 miles, a cost of $4,978.71, and a total time of 100.4 hours. The route details are as follows: Route #1 with 20-pallet truck Depot Start time  3:00AM of day 1 Daytona Beach Deliver 18 pallets Clearwater Pickup 14 pallets Depot Return time  5:48AM of day 2 Route #2 with 20-pallet truck Depot Start time  3:00AM of day 1 Orlando Deliver 3 pallets W Palm Beach Deliver 3 pallets Ft Lauderdale Deliver 3 pallets N Miami Deliver 5 pallets Miami-Puerto R. Deliver 4 pallets Depot Return time  4:43PM of day 2

26

Route #3 with 30-pallet truck Depot Start time  4:13AM of day 1 Tallahassee Deliver 3 pallets Tampa Deliver 20 pallets St Petersburg Pickup 3 pallets Oakland Park Pickup 4 pallets Depot Return time  4:03PM of day 2 9 Given sailing times and dates when deliveries are to be made, loadings need to be accomplished no later than the following dates:

To: A B C D From: 1 16 40 1 2 69 25 5 The problem can be expressed as a transportation problem of linear programming. There will be 6 initial states [(1,1), (2,5), (1,16), (2,25), (1,40), and (2,69)] and 6 terminal states [(D,10), (C,15), (A,36), (B,39), (C,52), and (A,86)]. The linear program is structured as shown in Figure 7-4. Using a transportation solution method, we determine one of the optimum solutions. There are several. The solution is read by starting with the slack on initial loading state 1. This tells us to next select the cell of terminal state 1. In turn, this defines initial state 3 and hence terminal state 3. And so it goes until we reach the terminal state slack column. This procedure is repeated until all initial state slacks are exhausted. Our solution shows two routings. The first is (1,1)(D,10)(1,16)(A,36)(2,69)(A,86). The second is (2,5)(C,15)(2,25)(B,39)(1,40)(C,52). Two ships are needed.

27

1 1

Load date Discharge date 100

100

1

XXa 100

C 15

2 69

XX

A 36

Rim restriction

XX

B 39

1

10

100

100

1

1

10

1

1

10

1

10

100

10

1

XX XX

XX 100

XX 100

XX

XX 10

1

1 100

XX

1

1

XX 100

10

1

XX 100

XX 10

1

1

100

XX

1

1

XX 100

10

1

100

XX

A 86

1

XX 100

100

10

100

XX

C 52

1

XX 100

100

1

1

XX 100

100

1

XX 100

100

a

1 40

Slack

D 10

Slack Rim restriction

Loading points and dates 1 2 16 25

2 5

XX 10

XX 10

1

1

1

1

1

1

4

6

10

1 1

1

1

6

XX inissible cells given a high cost

FIGURE 7-5 Transportation Matrix Setup and Solution for the Queens Lines Tanker Scheduling Problem 10 This is a problem of freight consolidation brought about by holding orders so they can be shipped with orders from subsequent periods. The penalty associated with holding the orders is a lost sales cost.

(i) Orders shipped as received Weight  Rate a Hays 10,000  0.0519 Manhattan 14,000  0.0519 Salina 13,000  0.0408 Great Bend 10,000  0.0498 Transportation Lost sales Total a

= = = = =

Cost $519.00 726.00 530.00 498.00 $2,274.00 .00 $2,274.00

Ship 8,000 lb. as if 10,000 lb.

Average period cost is $2,274.00

28

(ii) Consolidate first period orders with second period orders. Weight  Rate Cost = Hays 16,000  0.0519 = $830.40 a 40,000  0.0222 = 888.00 Manhattan Salina 26,000  0.0342 = 889.20 Great Bend 10,000  0.0498 = 498.00 Transportation $3,105.60 Lost sales 1,050.00 Total $4,155.60 a

Ship 28,000 lb. as if 40,000 lb.

The lost sales cost is 1,000 cases  $1.05 = $1,050.00 to hold one group of orders for 2 weeks. Average cost per period is $4,155.60/2 = $2,077.80. (iii) Hold all orders until the third period. Weight  Rate Hays 24,000  0.0426 Manhattan 42,000  0.0222 40,000  0.0246 Salinaa Great Bend 15,000  0.0498 Transportation Lost sales Total a

= = = = =

Cost $1,022.40 932.40 984.00 747.00 $3,685.80 3,150.00 $6,835.80

Ship as if 40,000 lb.

Lost sales Hold 1st period orders for 2 periods 1,0001.05.2 = $ 2,100 Hold 2nd period sales for 1 period 1,0001.05 = 1,050 $ 3,150 Average period cost is $6,835.80/3 = $2,278.60 Summary Ship immediately $ 2,274.00 Hold orders 1 period 2,077.80 Hold orders 2 periods 2,278.60

Optimum

11 Routes are built by placing the trips end-to-end throughout the day from 4AM until 11PM, respecting the times that a warehouse can receive a shipment. This is a 19-hour block of time per day, or there are 95 hours per week per truck in which a truck may

29

operate. If there were no delivery time restrictions on warehouses and trips could be placed end-to-end for a truck without any slack at the end of the day, the absolute minimum number of trucks can be found multiplying the number of trips by the route time and then dividing the total by the 95 hours allowed per week. That is, Warehouse location Flint Alpena Saginaw Lansing Mt. Pleasant W. Branch Pontiac Traverse City Petoskey

(1) Number of trips 43 5 8 21 12 5 43 6 5

(2) Total time per trip, hr. 1.25 10.50 2.25 3.75 5.50 6.00 2.75 10.50 11.75 Total

(3)=(1)(2) Total time, hr. 53.75 52.50 18.00 78.75 66.00 30.00 118.25 63.00 58.75 539.00

For 539 trip hours, 539/95 = 5.67 rounded to six trucks needed per week. Now, it is necessary to adjust for the problem constraints. A good schedule can be found by following a few simple rules that can be developed by examining the data. First, begin the day with a trip where the driving time to a warehouse is just long enough for the truck to arrive at the warehouse just after it opens. One-half the driving time should exceed 6:30  4:00 = 2:30, or 2½ hr. Trips to Alpena, Traverse City, and Petoskey qualify. Second, use the short trips at the end of the day to avoid slack time. Third, allocate the trips to the days using the longest ones first. Make sure that the total trip time for a day does not exceed 19 hours. For a minimum of six trucks, the following feasible schedule can be developed by inspection. Truck 1

Truck 2

Truck 3

Truck 4

Truck 5

Truck 6

Day 1 Petoskey 11.75 W Branch 6.00 Flint 1.25 Total =19.00 hr. T. City 10.50 2 Lansing 7.50 Total =18.00 hr. T. City 10.50 2 Lansing 7.50 Total =18.00 hr. Alpena 10.50 Lansing 3.75 3 Flint 3.75 Total =18.00 hr. M Pleasant 5.50 4 Pontiac 11.00 2 Flint 2.50 Total =19.00 hr. M Pleasant 5.50 4 Pontiac 11.00 2 Flint 2.50 Total =19.00 hr.

Day 2 Petoskey 11.75 W Branch 6.00 Flint 1.25 Total =19.00 hr. T. City 10.50 2 Lansing 7.50 Total =18.00 hr. Alpena 10.50 2 Lansing 7.50 Total =18.00 hr. M Pleasant 5.50 4 Pontiac 11.00 2 Flint 2.50 Total =19.00 hr. M Pleasant 5.50 4 Pontiac 11.00 2 Flint 2.50 Total =19.00 hr. M Pleasant 5.50 3 Pontiac 8.25 2 Flint 2.50 Total =16.25 hr.

Day 3 Petoskey 11.75 W Branch 6.00 Flint 1.25 Total =19.00 T. City 10.50 2 Lansing 7.50 Total =18.00 Alpena 10.50 2 Lansing 7.50 Total =18.00 hr. M Pleasant 5.50 4 Pontiac 11.00 2 Flint 2.50 Total =19.00 hr. M Pleasant 5.50 4 Pontiac 11.00 Flint 1.25 Total =17.75 M Pleasant 5.50 6 Saginaw 13.50

Day 4 Petoskey 11.75 5 Flint 6.25

Day 5 Petoskey 11.75 5 Flint 6.25

Total =18.00 hr. T. City 10.50 2 Lansing 7.50 Total =18.00 hr. Alpena 10.50 2 Lansing 7.50 Total =18.00 hr. M Pleasant 5.50 4 Pontiac 11.00 2 Flint 2.50 Total =19.00 hr. M Pleasant 5.50 4 Pontiac 11.00

Total =18.00 hr. T. City 10.50 2 Lansing 7.50 Total =18.00 hr. Alpena 10.50 2 Lansing 7.50 Total =18.00 hr. M Pleasant 5.50 4 Pontiac 11.00 2 Flint 2.50 Total =19.00 M Pleasant 5.50 4 Pontiac 11.00

Total =16.50 hr. W Branch 6.00 2 Saginaw 4.50

Total = 16.50 W Branch 6.00 10 Flint 12.50

Total =19.00 hr.

Total =10.50 hr.

Total =18.50 hr.

30

Although this schedule meets the requirements of the problem, it might be improved by better balancing the workload across the trucks and the days. 12 (a) A sweep method solution is shown on the following figure. Five trucks are needed with a total route distance of (30+29+39+44+19.5)10 = 1,615 miles.

20

2

18 Route #1 Load 19

Route #5 Load 9

3

2 5

4 3

3

14

4

Miles 12 x 10 10

3

4

Route #4 Load `8

2 3 Warehouse

8

5

5 4

6 4

1

4

3

16

3

6

Route #2 Load 20

5

Route #3 Load 17

3

2

2 4

0 0

2

4

6

8

10 12 14 Miles x 10

16

18

20

22

24

26

(b) The sweep method is a fast and relatively simple method for finding a solution to rather complex vehicle routing problems. Solutions can be found graphically without the aid of a computer. However, there are some limitations. Namely,   

The method is heuristic and has an average error of about 10 to 15 percent. This error is likely to be low if the problem contains many points and the weight of each point is small relative to the capacity of the vehicle. The method does not handle timing issues well, such as time windows. Too many trucks may be used in the route design.

13 This problem may be solved with the aid of ROUTER in LOGWARE. The model input data may be formatted as shown in Figure 7-6.

31

(a) The solution from ROUTER shows that four routes are needed with a minimum total distance of 492 miles. The route design is shown graphically in Figure 7-7. A summary for these routes is given in following partial output report. Route Route no 1 2 3 4 Total

Run Stop Brk Stem time, time, time, time, time, Start Return No of Route hr hr hr hr hr time time stops dist,Mi 1.2 1.0 .3 .0 .4 08:59AM 10:12AM 3 29 8.9 6.6 1.3 1.0 1.1 08:32AM 05:25PM 19 199 6.2 3.7 1.4 1.0 .9 08:42AM 02:54PM 14 112 7.5 5.1 1.5 1.0 1.4 08:30AM 04:02PM 12 152 23.8 16.4 4.4 3.0 3.8 48 492

Route cost,$ .00 .00 .00 .00 .00

(b) Note that route #1 is short and that a driver and a station wagon would be used for a route that takes 1.2 hours to complete. By “attaching” route #1 to route #3, the same driver and station wagon may be used, and the constraints of the problems are still met. The refilled station wagon can leave the depot by 3:30-3:45PM and still meet the customers’ time windows and return to the depot by 6PM. Thus, only three drivers and station wagons are actually needed for this problem. FIGURE 7-6 Input Data for ROUTER for Medic Drugs --PARAMETERS AND LABELS— Problem label - Medic Drugs Grid corner with 0,0 coordinates (NW, SW, SE, or NE) - SW DEPOT DATA Depot description - Pharmacy Located in zone - 0 Horizontal coordinate - 13.7 Vertical coordinate - 21.2 Earliest starting time (min) - 480 Latest return time (min) - 9999 Default vehicle speed (miles per hour) - 30 After how many clock hours will overtime begin - 168 GENERAL DATA Percent of vehicle in use before allowing pickups - 0 Horizontal scaling factor - 4.6 Vertical scaling factor - 4.6 Maximum TIME allowed on a route (hours) - 168 Maximum DISTANCE allowed on a route (miles) - 9999 LOAD/UNLOAD TIME FORMULA Fixed time per stop - 0 Variable time per stop by weight - 0 By cube - 0 BREAK TIMES Duration of 1st break (minutes) - 60 To begin after - 720 Duration of 2nd break (minutes) - 0 To begin after - 9999 Duration of 3rd break (minutes) - 0 To begin after - 9999 Duration of 4th break (minutes) - 0 To begin after - 9999 --STOP DATA— STOP NO DESCRIPTION 1 Covington House 2 Cuyahoga Falls 3 Elyria   

TY D D D

LOAD WGHT 1 9 1

VOL. CUBE 0 0 0

HCRD VCRD ZN 23.40 12.90 0 13.40 13.40 0 6.30 16.80 0

LOAD TIME 2 18 5

BEG1 540 540 540

END1 BEG2 1020 9999 1020 9999 1020 9999

END2 9999 9999 9999

32

46 47 48

Westbay Westhaven Broadfiels Mnr

D D D

6 2 6

0 0 0

8.40 8.50 18.20

18.00 18.10 22.90

0 0 0

10 5 2

630 540 540

690 1020 1020

9999 9999 9999

9999 9999 9999

--VEHICLE DATA— -CAPACITY--

NO. 1

VEHICLE DESCRIPTION Station wagon

TP 1

NO 50

WGHT 63

CUBE 9999

--VEHICLE-FIXED COST 0

--DRIVER--

PER MI COST 0

FIXED COST 0

PER HR COST 0

OVER TIME COST 0

FIGURE 7-7 Graphical Solution for Medic Drugs

14 There is no exact answer to this problem nor is one intended. Several approaches might be taken to this problem. We could apply the savings method or the sweep method to solve the routing problem for each day of the week, given the current demand patterns. However, we can see that there is much overlap in the locations of the customers by delivery day of the week. We might encourage orders to be placed so that deliveries form tight clusters by working with the sales department and the customers. Perhaps some incentives could be provided to help discipline the order patterns. The orders should form a general pattern as shown below. Currently, the volume for Thursday exceeds the available truck capacity of 45 caskets. Maybe the farthest stops could be handled by a for-hire service rather than acquiring another truck for such little usage.

33

Monday

Tuesday

Friday Depot

Wednesday

Thursday

It appears that the truck capacity is about right, given that some slack capacity is likely to be needed. Once the pattern orders are established, either as currently given or as may be revised, apply principles numbers 1, 3, 4, 5, and 7.

34

FOWLER DISTRIBUTING COMPANY Teaching Note

Fowler Distributing Company involves a problem of routing and scheduling a fleet of trucks to serve customers for beer and wine products on a daily basis. Determining an efficient design for making daily deliveries when stop sequence, number of trucks and their sizes, warehouse location, and time window restrictions are variables is the objective of this exercise. For the most part, students should consider that they are to prepare a consulting report to management about this problem. The several questions provided at the end of the case study should help direct the analysis. The ROUTER model is used to cost out and to optimize the various design scenarios. Run results are tabulated in Table 1. Q1. The current design may not be as efficient as it might be. Therefore, our first task is establish a profile of the current design and then to plan the routes so as to minimize the total miles driven and the number of trucks needed to serve the customers, subject to the restrictions of truck capacity, time windows, total time on the route, etc. Running the current route design in ROUTER establishes a benchmark as shown in Figure 1. The design requires 334 miles of travel with 5 trucks. The daily routing cost is $764.62. Next, optimizing this problem, given no change in data or restrictions, gives the revised benchmark design as shown in Figure 2. Now, costs are reduced to $731.31 per day for a total routing distance of 311 miles. Six trucks are needed. There appears to be a substantial change in the design of the routes that can achieve a ((334  311)/334)100 = 7% reduction in the total miles traveled to make the deliveries. The saving in cost is ($764.62  731.31)250 = $8,327.50 per year. This is not a large savings, but it can be achieved without changing the restriction on deliveries or incurring additional investment. Q2. Time window restrictions often force route design into stop sequences that are not very economical, that is, routes cross themselves. This is the case for the revised benchmark design as shown in Figure 2. To determine the impact of the time windows, we can make an optimizing ROUTER run where there are no time window restrictions. The daily time windows are all set 8:00AM to 5:00PM. The optimized routes are shown in Figure 3. The route mileage is 270 miles per day for five trucks. The total cost is $673.64 per day. From the revised benchmark, this is an additional cost reduction of ($731.31  673.64) 250 = $14,417.50 per year.

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FIGURE 1 Current Route Design

The questions for management are: How restrictive are the time windows? Can deliveries be made outside of the 's "open hours", such as by giving the driver the key to a safe storage area? Can management offer a small incentive to widen the time window when it otherwise would not be convenient for the ? Relaxing such time windows is often one of the important sources for cost savings in routing problems. Q3. When there are no truck capacity restrictions on a routing problem, the most efficient route design would be to use one large truck to serve all s. Therefore, we would expect that trucks of larger capacity would reduce the total distance traveled. A ROUTER optimizing run was made with 600 case capacity trucks to find out. All other conditions were set at the current design.

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FIGURE 2 Optimized Current Route Design

Compared with the optimized current design, the potential savings is ($731.31  722.98)  250 = $2,082.50 per year. Two 600-case trucks would be needed, along with 3 of the smaller trucks. Although there is a positive savings associated with using larger trucks, the savings seems quite small and perhaps not worth switching to some larger trucks at this time. We should note that these savings are a result of comparing full costs, which include truck depreciation. Although a present value analysis would be appropriate here, we would need some additional information which might include (1) the proportion of truck costs allocated to equipment depreciation, (2) the life of the trucks, (3) the estimated salvage value of the trucks, and (4) and the required rate of return on investments of this type.

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FIGURE 3 Route Design with Relaxed Time Window Constraints

Q4. From the incremental costs in the detailed report of the optimized current design, we can see that 14 costs $39.05 and 19 costs $35.04 to make the deliveries. Since the volume of 19 is 90 cases and exceeds the 50 case capacity of an outside transport service, 19 cannot be considered for alternate delivery. If it only costs $35.00 for an outside delivery service to handle 14, then a cost savings can be realized. Dropping this and redeg the routes shows that only 5 trucks are needed for a route cost of $690.17 and route miles of 286. The total cost for handling all s would be $690.17 + 35.00 = $725.17. Compared with the optimized current design, this is an annual savings of ($731.31  725.17)  250 = $1,535.00. Economically, Roy should use the outside transportation. However, losing direct control over the deliveries and possible adverse reactions from route salesmen may give him second thoughts about it. Q5. Two ROUTER runs were made here in order to determine the effect of a shorter workday before overtime begins. If no overtime is allowed, then six trucks are required for a total daily routing cost of $754.77. If some overtime is permitted, then the route cost can be reduced to $733.28 with 6 trucks required. Compared with the optimized current design, Roy Fowler would seem to be putting himself at a disadvantage by not wanting to pay overtime. Q6. Moving the warehouse to a more central location would have the appeal of being closer to all stops and would result in shorter routes. Testing this shows that total route distance can be reduced to 305 miles with a daily cost of $716.91. However, in order to meet the time window and other constraints, an extra truck is needed. There is a potential annual savings of ($731.31  716.91)  250 = $3,600.00. Since it costs $15,000 to make

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the move, a simple return on investment would be (3,600/15,000)  100 = 24%. At this ROI, the move should be seriously considered. Q7. Routing and scheduling beer trucks from a central depot is quite similar to delivery problems found in manufacturing, retail, and service industries. Several examples are listed below.       

Serving branch banks by making pickups and deliveries of canceled checks, supplies, and other paper work from a central domicile point for the trucks Making Federal Express pickups and deliveries around an airport Dispatching school buses around a school for student pick up and drop off Dispatching service personnel for various service companies such as electric, water, gas, and telephone Making retail deliveries from a warehouse Making deliveries from a plant location to warehouses and then picking up supplies from vendors on the return trip. Making deliveries (and pickups) of medical records to doctor's offices and other locations from a central record-keeping location such as a hospital

Summary Roy Fowler should consider optimizing his route design. This can be done at no investment to him, and he can gain about a 7% reduction in operating costs on the average. A substantial benefit can be realized by widening the time windows. He should especially see if those time windows that cause routes to overlap themselves can be widened. The potential for doing this is up to $14,418 per year. He should explore the possibility and reasonableness of serving s 14 and 19 by an outside transport service. Finally, he should consider relaxing his rigid policy of not wanting to pay overtime, especially if the union is successful in negotiating a 7½-hour workday. There seems to be little opportunity for reducing cost by increasing truck capacity or relocating the warehouse. Although there are operating cost reductions available, they do not seem sufficient to justify a change in truck size unless Roy can make good use of the trucks in other ways.

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TABLE 1 Results of Various ROUTER Runs Run type Miles Cost Trucks Comments Current design 334 $764.62 5 Cost out current design Improves on current Optimized current 311 731.31 6 route design No time windows restrictions 270 673.64 5 Time windows opened Trucks at 600-case Two routes require larger capacity 299 722.98 5 trucks 7½ hr. work day - no Route time allowed to be overtime 332 754.77 6 no longer than 8.0 hours 7½ hr. work day Route time allowed to with overtime 311 733.28 6 exceed 8.0 hours New warehouse Warehouse moved to location 305 716.91 7 more central location

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METROHEALTH MEDICAL CENTER Teaching Note Strategy MetroHealth Medical Center’s mission is to provide transportation to and from its facility for those patients that cannot afford the expense or are unable to use other modes of transportation. MMC currently has a fleet of vans and drivers to serve this patient population. In addition, some transportation can be provided as part of the hospital’s contract ambulance service. The primary service provided by Transportation Services is to pick patients up and bring them to MMC for their scheduled appointments and then to return them to their origin points. Careful management of the fleet is needed through vehicle assignment to patients, routing the vans, and asg some of the patients to the ambulance service for transportation. It is the purpose of this case to show how a basic routing and scheduling model can be used to assist in answering the typical questions surrounding fleet utilization in a service setting. The ROUTER module in LOGWARE is used as the analytical tool. This case does require some insight on the part of the student to deal with the complexities of this problem. Considering the amount of data provided and the richness of the issues involved, this case is probably best assigned as a project rather than as a homework exercise. Answer to Questions (1) Which vans from the current fleet should be used? subcontracting be used?

To what extent should

This problem is complex, considering that it is quite dynamic. Although the patient pick up list appears to be fixed for a particular day, changes take place in the form of cancellations and occasional additions. Patient appointment times are met a high percentage of the time, but not always. Patients to be returned to their origins may find alternate means of transportation so that 100 percent of the patients picked up may not have to be returned. For simplicity, it will be assumed that the patient pick up list is fixed for a particular day and appointment times are rigid. In addition, the returns are not directly considered in deg the routes. Rather, they are expected to be seeded into the pick up routes. This need not be a serious limitation as long as van capacity (number of pick-up patients to the number of available seats) is not highly utilized on a pickup route and van utilization for pick ups drops as appointments are concentrated in the morning hours and return times are shifted toward the afternoon hours, as shown in Figure 1.

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Appointment times

Return times

FIGURE 1 Appointment Time and Return Time Distributions

A typical day’s patient list can be submitted to ROUTER. Coordinate points for zip code areas are scaled from the map in Figure 2 of the case study. Appointment times are represented as time windows with a 2-hour gap for pick up. A large number of 15enger and 6-enger vans are assumed available to start the ROUTER model. Since ROUTER is a basic vehicle routing and scheduling model, it simply assigns patients to vans and gives a sequence to pick them up. It assumes that a single van leaves and returns to MMC only once in a day. Analysis must for this, since vans make several out and back trips per day. It is not possible to accurately recreate the current routing patterns for the vans from the data given. However, optimized routing and scheduling can be found through the application of ROUTER. First, the typical daily appointment list is solved with ROUTER using a 2-hour time window where the ending window time is the appointment time. An average speed of 27.5 miles per hour, variable costs of $0.11 per mile1, and a large number of 15-enger and 6-enger vans are used. A complete database is shown in the appendix of this note. Using the appointment list for the typical day as representative with all 56 patients, ROUTER shows that 6 vans are needed, where one is a 15-enger type (route 1). This is routing where no patient is transported by the contract transport service. The routing represents a total of 329 miles driven at a variable cost of 329 mi.  $0.11/mi. = $36.19. The routes are placed end-to-end to cover the full day and then ranked, first by the number of patients transported throughout the day on a van and second by the total miles driven, as shown in Table 1. A reasonable procedure for allocating routes to vans is to assign routes with the largest number of patients first to the largest vans. This is because 1

Per mile variable cost is determined as repair cost/30,000 mi. per yr. = $1,000/30,000 = $0.03 per mi. plus fuel cost at $1.00 per gallon/average miles per gallon = 1/13 = $0.08 per mi. for a total of 0.08 + 0.03 = $0.11 per mile.

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the ambulance service charges on a per patient basis, and these routes would be most costly to subcontract. If the number of patients is tied for a van, rank first the van with the shortest total route distance. TABLE 1 Route Design for Representative Patient Appointments with No Transport Subcontracting Van no. 1 1 1 1 1 2 2 3 3 4 4 5 6 1

Start from MMC 6:07AM 6:52AM 8:55AM 11:04AM 11:41AM 7:24AM 10:59AM 7:41AM 10:45AM 7:37AM 11:52AM 7:24AM 7:43AM

Return to MMC 6:28AM 8:28AM 10:30AM 11:31AM 1:27PM 9:05AM 11:57AM 9:16AM 12:27PM 9:12AM 1:34PM 8:44AM 9:03AM

Patients on route 1 81 4 1 5 5 5 5 5 5 5 4 3 56

Route distance 7 mi. 22 33 9 35 32 13 29 33 30 33 25 28 329 mi.

Route time 0.3 hr. 1.6 1.6 0.4 1.8 1.7 1.0 1.6 1.7 1.6 1.7 1.3 1.6 17.6 hr.

A 15-enger van needed.

The cost of this initial route configuration can be summarized as follows: Annual cost of 1 15-enger van Annual cost of 5 6-enger vans Annual repair cost of 6 vans Annual cost for 6 drivers Variable cost Total cost

$ 5,750 1 28,500 2 6,000 141,000 8,686 3 $189,936

1

$23,000/4 yr. = $5,750 ($19,000/4 yr.)  6 = $28,500 3 329 mi. x $0.11/mi.  240 days/yr. = $8,686 2

The question now is whether substituting contract carriage for some of the routes will significantly lower vehicle and driver costs while only slightly increasing variable costs. Since the routes are ranked in Table 1, a reasonable rule for dropping routes would be to start from the bottom of the route list in Table 1 and work upward. This will determine the number of vans needed and the patients to be transported by each mode. These results are shown in Table 2. An unlimited number of subcontracting trips is assumed.

TABLE 2 Optimal Number of Vans

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Vans 6 5 4 3 2 1 0 1

Driver $141,000 117,500 94,000 70,500 47,000 23,500 0

Vehicle $34,250 29,500 20,000 15,250 10,500 5,750 0

Variable (gas + repair) $14,686 12,946 11,286 8,623 5,986 3,798 0

Subcontracting $ 0 6,235 1 14,549 35,333 56,117 76,901 116,390

Total $189,936 166,181 139,835 129,706 119,603 109,949 116,390

Annual subcontracting cost is $8.66 per round trip  240 days per year  3 patients = $6,235.20

(2) How many subcontracted trips should MMC negotiate and at what price? One 15-enger van appears to be the optimal number, but this requires 37 patients  20 days per month = 740 subcontracting trips per month. This is more than the 500 allowed. If the 500-trip limit (500/20 days per mo. = 25 patients per day) is to be respected, approximately one additional van will be needed. On the other hand, the cost of the additional van is the annual cost of a van + the annual driver’s salary + the gas and repair cost = $19,000/4 + 23,500 + 1,1882 = $29,438/yr. to transport 10 patients per day. At this rate, MMC could afford to pay $29,438/240/10 = $12.27 per round trip (find 10 patients for van 2 in Table 1). Possibly the subcontractor would be willing to offer this additional transport for a price between the $8.66 and $12.27 per trip. It would be in Mac’s interest to do this. Renegotiating all trips at the 750-trip level is another possibility. MMC can afford to pay a subcontracting cost of up to $189,936 - 31,0483 = $158,888/yr. This would require transporting 37 patients per day at a maximal average trip cost of $158,888/240/37 = $17.89. Since $17.89 is the optimal worth of subcontracting, MMC probably can negotiate a rate much less than this while still encouraging the subcontractor to provide service at the higher level. (3) MMC is considering using all 6-enger vans. Would this be a good decision? Developing daily routes and asg them to 6-enger vans is the same as for Table 1. This initial allocation without subcontracting is shown in Table 3. Considering the subcontracting trip limit of 500 patients per month (25 patients per day), two vans are needed in the MMC fleet. This number of vans is determined by counting the patients from the bottom of Table 3 until approximately 25 patients are found, considering increments of full vans and noting the number of vans remaining at the top of the list. (Assume 27, an extra 2 patients can be accommodated by the subcontractor at no additional cost.) The variable cost for two vans is shown below.

Variable cost type 2 3

Two 6-enger

One 15- & one 6-enger van

45 miles x $0.11/mile  240 days/yr. = $1,188. Driver + vehicle +variable costs for one van, or 19,000/4 + 23,500 + 2,798 = $31,048.

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vans Annual cost of 2 drivers Vehicle mileage + repair cost Subcontracting cost Total 1 2

47,000 5,9601 56,1172 $109,077

47,000 5,986 56,117 $109,103

150 mi.  $0.11/mi.  240 days/yr. + 2,000 = $3,960. 27 patients  $8.66/trip  240 days/yr. = $56,117.

In contrast, the 15-enger and 6-enger van configuration compared with the two 6-enger van configuration saves $109,103  109,077 = $26 per year. The initial outlay is less for the smaller van by $4,000. From purely an economic standpoint, using just 6-enger vans would appear to be the best choice, however there may be customer service reasons for having the larger van available to meet peak demand needs. TABLE 3 Route Design for 6-enger Vans Only Van no. 1 1 1 1 1 2 2 2 3 3 4 4 5 6

Start from MMC 6:07AM 6:52AM 7:37AM 11:04AM 11:52AM 7:22AM 8:55AM 10:59AM 7:41AM 10:45AM 7:24AM 11:41AM 7:24AM 7:43AM

Return to MMC 6:28AM 7:37AM 9:12AM 11:31AM 1:34PM 8:19AM 10:30AM 11:57AM 9:16AM 12:27PM 9:05AM 1:27PM 8:44AM 9:03AM

Patients on route 1 3 5 1 5 5 4 5 5 5 5 5 4 3 56

Route distance 7 mi. 12 30 9 33 13 33 13 29 33 32 35 25 28 332 mi.

Route time 0.3 hr. 0.8 1.6 0.4 1.7 1.0 1.6 1.0 1.6 1.7 1.7 1.8 1.3 1.3 17.8 hr.

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APPENDIX A Database for ROUTER --PARAMETERS AND LABELS-Problem label - METROHEALTH MEDICAL HOSPITAL Grid corner with 0,0 coordinates (NW, SW, SE, or NE) - SW DEPOT DATA Depot description - Medical Center Located in zone - 0 Horizontal coordinate 7.40 Vertical coordinate 6.13 Earliest starting time (min) - 360 Latest return time (min) - 960 Default vehicle speed (miles per hour) 28 After how many clock hours will overtime begin 68.0 GENERAL DATA Percent of vehicle in use before allowing pickups - 100 Horizontal scaling factor 3.102 Vertical scaling factor 2.820 Maximum TIME allowed on a route (hours) 2.0 Maximum DISTANCE allowed on a route (miles) - 9999.0 LOAD/UNLOAD TIME FORMULA Fixed time per stop 6.00 Variable time per stop by weight 0.00 By cube 0.00 BREAK TIMES Duration of 1st break (minutes) 0 To begin after - 9999 Duration of 2nd break (minutes) 0 To begin after - 9999 Duration of 3rd break (minutes) 0 To begin after - 9999 Duration of 4th break (minutes) 0 To begin after - 9999 --STOP DATA-NO. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44

STOP DESCRIPTION---- TY Baker P Boyd P Carver P Ivey P Rashed P Walsh P Johnson P Burgess P Delgado P Fairrow P Middlebrooks P Suech P Lawson P Reed P Bongiovanni P Miller P Talley P Williams P Dumas P Taylor P Barker P Lhota P Manco P Webb P Wilson P Arrington P Staunton P Wall P Williams P Caruso P West P Amaro P Brown P Ciesicki P Pinkevich P Staufer P Winterich P Brown P Ball P Lanza P Mayernik P Suech P Heffner P Jarrell P

LOAD VOLUME WGHT CUBE 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0

-COORDINATESLOAD ---TIME WINDOWS---HCRD VCRD ZN TIME BEG1 END1 BEG2 END2 6.50 8.90 0 0 390 510 9999 9999 6.20 9.00 0 0 390 510 9999 9999 11.00 5.00 0 0 420 540 9999 9999 10.00 6.00 0 0 420 540 9999 9999 10.00 10.00 0 0 420 540 9999 9999 3.50 5.00 0 0 420 540 9999 9999 8.00 5.50 0 0 450 570 9999 9999 9.00 7.50 0 0 465 585 9999 9999 8.20 7.20 0 0 465 585 9999 9999 8.60 7.80 0 0 480 600 9999 9999 8.00 5.40 0 0 480 600 9999 9999 5.00 6.50 0 0 480 600 9999 9999 6.30 8.80 0 0 510 630 9999 9999 9.05 7.30 0 0 510 630 9999 9999 9.00 5.80 0 0 525 645 9999 9999 7.80 5.40 0 0 540 660 9999 9999 6.50 10.20 0 0 570 690 9999 9999 7.50 7.00 0 0 585 705 9999 9999 7.20 5.90 0 0 630 750 9999 9999 9.80 7.00 0 0 645 765 9999 9999 7.20 6.00 0 0 660 780 9999 9999 8.00 6.20 0 0 660 780 9999 9999 10.10 9.00 0 0 660 780 9999 9999 9.70 7.00 0 0 660 780 9999 9999 9.00 5.80 0 0 660 780 9999 9999 10.20 5.90 0 0 720 840 9999 9999 6.40 8.80 0 0 720 840 9999 9999 9.50 5.10 0 0 720 840 9999 9999 8.20 7.20 0 0 720 840 9999 9999 7.00 5.00 0 0 375 495 9999 9999 5.60 6.00 0 0 390 510 9999 9999 6.00 6.60 0 0 420 540 9999 9999 7.20 4.20 0 0 420 540 9999 9999 6.50 3.00 0 0 420 540 9999 9999 6.60 5.10 0 0 420 540 9999 9999 5.80 6.00 0 0 420 540 9999 9999 7.30 4.80 0 0 420 540 9999 9999 4.00 5.00 0 0 435 555 9999 9999 6.20 6.30 0 0 450 610 9999 9999 6.20 5.90 0 0 450 610 9999 9999 7.10 7.00 0 0 450 610 9999 9999 4.00 6.50 0 0 465 625 9999 9999 5.20 4.50 0 0 480 640 9999 9999 4.20 6.80 0 0 480 640 9999 9999

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45 46 47 48 49 50 51 52 53 54 55 56

Piatak Swaysland Baer Wills Fauber Mullins Pack Westerfield Lisiewski Mherson Mykytuk Gutschmidt

VEHICLE NO. DESCRIPTION---1 15-pasngr vehcl 2 6-pasngr vehcl

P P P P P P P P P P P P

TP 1 2

1 1 1 1 1 1 1 1 1 1 1 1

0 0 0 0 0 0 0 0 0 0 0 0

7.00 6.10 3.80 4.70 5.20 7.20 6.20 5.90 5.00 4.20 6.00 5.50

3.50 6.70 4.90 6.10 6.80 6.30 5.00 6.30 5.50 6.20 6.50 5.90

0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0

480 480 540 540 660 660 660 675 690 720 750 780

640 640 700 700 780 780 780 795 810 840 870 900

9999 9999 9999 9999 9999 9999 9999 9999 9999 9999 9999 9999

9999 9999 9999 9999 9999 9999 9999 9999 9999 9999 9999 9999

--VEHICLE DATA----VEHICLE--- ----DRIVER---OVER --CAPACITY--FIXED PER MI FIXED PER HR TIME NO WEIGHT --CUBE ---COST --COST ---COST --COST --COST 20 15 99 0.00 0.11 0.00 0.00 0.00 20 6 99 0.00 0.11 0.00 0.00 0.00

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ORION FOODS, INC. Teaching Note Strategy The purpose of this case study is to allow students to analyze a distribution problem where determining the optimal paths through a network is central to the problem solution. The shortest route methodology applies, and the ROUTE module in the LOGWARE software can effectively be used. This module permits students to quickly find the shortest distance paths from any starting point to all other points in the network. A prepared data file of the network is available for use under the ROUTE module. The case has several dimensions that allow it to be used as a homework assignment, a short case study project, or as a basis for classroom discussion. It is a simple problem in network design highlighting transportation routing issues, however additional factors such as inventory consolidation and investment costs allow for an enriched analysis. Note: A database for this problem is available in the ROUTE module of the LOGWARE software. Answers to Questions (1) Can Anita improve upon the current distribution operations?

It should be obvious to Anita that the current distribution system may not be performing at optimum. Allowing carriers to decide the routes to use when they are being paid on a mileage basis is like asking a fox to watch the chicken coop. She really cannot expect that carriers will be motivated to seek out optimal routing patterns. Therefore, she should determine the best routes between regional and field warehouses and insist that carriers invoice according to the mileages along these specified routes. Using the map provided in Figure 1 of the case study and the current assignments of field warehouses to regional warehouses, she can develop a database for the ROUTE module in LOGWARE. The database is shown in the Supplement to this note. Running ROUTE will give the optimal routes from which she can develop transportation costs, as shown in Table 1 below. Compared with the current cost level, a savings of $652,274  630,140 = $22,134 per year can be realized. This savings does not require any investment; however, it will be magnified by the growth in demand in the next five years.

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TABLE 1 Optimal Transportation Costs Under Current Distribution System Design Regional warehouse Fresno Fresno Fresno Fresno Burns Burns Burns Totals

Field warehouse Los Angeles Phoenix Salt Lake City San Francisco Portland Butte Seattle

Optimal route miles 219 588 815 183 293 676 467 3,241

Average number of tripsa 366.7 200.0 116.7 280.0 143.3 16.7 186.7

Transport cost, $ 104,399b 152,880 123,644 66,612 54,583 14,676 113,346 630,140

a

Determined from the warehouse throughput divided by the average shipment size. e.g., 110,000 cwt./300 cwt. = 366.7 trips. b $1.30 per mi.  219 mi.  366.7 trips = $104,399.

(2) Is there any benefit to expanding the warehouse at Burns, OR? Finding the optimal distances when serving all field warehouses from Burns or Fresno gives a slightly different allocation of field warehouses to regional warehouses than is currently the case. That is,

Field warehouse Los Angeles Phoenix Salt Lake City San Francisco Portland Butte Seattle

If served from Burns 806 mi. 973 536* 555 293* 676* 467*

If served from Fresno 219 mi.* 588* 815 183* 757 1,120 925

*

Indicates optimal assignment.

This shows that Salt Lake City would be better served out of Burns rather than Fresno. Burns currently is near its capacity limit, so to assign Salt Lake City's volume to it would require expansion. In the short term, 35,000/8 = 4,375 cwt. of inventory capacity is needed. However, (43,000 + 5,000 + 56,000)/8 = 13,000 cwt. of the 15,000 cwt. of available capacity is currently being used. At minimum, an additional increment of capacity is required at a cost of $300,000. Reassignment of Salt Lake City to Burns would save 815  536 = 279 miles per trip. On 116.7 trips per year, the annual savings would be 1.30  279  116.7 = $42,327. The simple return on investment (ROI) would be: ROI 

$42,327  100  141% . $300,000

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If the anticipated growth in demand is realized, the number trips to Salt Lake City would increase to 56,000/300 = 186.7. The projected savings in the fifth year would be 1.30  279  186.7 = $67,716. The average annual savings would be (42,327 + 67,716)/2 = $55,022. The average annual ROI is: ROI 

$55,022  100  18.3% $300,000

Anita must now compare this return to other worthy investments in the firm to see if this opportunity is worth the risk. (3) Is there any merit to consolidating the regional warehousing operation at Reno, NV? If Reno were to replace the Burns and Fresno warehouses, an initial cost of $2,000,000 would be incurred to establish the new location and shut down the existing warehouses. Against this cost would be a savings of 40 percent of the inventory in the two regional warehouses. That is, the total inventory is 393,000/8 = 49,125 cwt. now and 528,000/8 = 66,000 cwt. in five years. The inventory cost savings now would be 0.40  0.35  60  49,125 = $412,650 and in five years 0.40 0.35  60  66,000 = $554,400. However, transportation costs will increase compared with the two-warehouse distribution system. If Reno is used, the transportation costs would be: Optimal Field warehouse route served from Reno miles Los Angeles 472 Phoenix 732 Salt Lake City 520 San Francisco 228 Portland 542 Butte 823 Seattle 716 Totals 4,033 a 472  336.7  1.30 = $225,007

Current no. of trips 366.7 200.0 116.7 280.0 143.3 16.7 186.7

Current transport cost, $ 225,007a 190,320 78,889 82,992 100,969 17,867 173,780 869,824

5th-year no. of trips 440.0 280.0 186.7 350.0 190.0 50.0 263.3

5th-year transport costs, $ 269,984 266,448 126,209 103,740 133,874 53,495 245,080 1,198,830

We cannot make a fair comparison with the current system design since there is inadequate capacity at Burns to handle the growth in volume. Two additional units of capacity will be needed for a total of $600,000. Then, the net investment attributable to Reno is 2,000,000  600,000 = $1,400,000. The revised transportation cost for the fifth year is as follows.

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Regional warehouse Fresno Fresno Burns Fresno Burns Burns Burns Totals

Field warehouse Los Angeles Phoenix Salt Lake City San Francisco Portland Butte Seattle

Optimal route miles 219 588 536 183 293 676 467 2,962

5th-year number of trips 440.0 280.0 186.7 350.0 190.0 50.0 263.3

Transport cost, $ 125,268 214,032 130,092 83,265 72,371 43,940 159,849 828,817

Now,

Transport costs Transport costs Net increase Less inventory savings Net cost savings

Current year $587,813 869,824 $282,011

5th year $828,817 1,198,830 $370,013

(412,650) $130,639

(554,400) $184,387

Burns/Fresno Reno

The average annual savings for Reno now is (130,639 + 184,387)/2 = $157,513. The relevant return on investment is ROI 

$157,513  100  1125% . $1,400,000

This is probably not a sufficient return to justify the warehouse at Reno. Rather, if Orion wishes to serve the increasing demandthere is no requirement to do sothen the better strategy would be to expand Burns by 20,000 cwt. and serve Salt Lake City from this location. An interesting question to pose to students is: What does it mean to only serve demand up to the limits of capacity? Orion could serve only the more profitable demand and avoid the risks of expansion.

51

R & T WHOLESALERS Teaching Note4

The objective of this assignment is to minimize the total monthly delivery costs for R&T Wholesalers, a company distributing general products throughout India. The focus is on one warehouse acting as a truck depot that delivers merchandise to retailers located in surrounding towns. Delivery expenses are minimized through the optimal utilization of the trucks, crews, and related expenses. The constraints and other considerations listed below were used when deg the solution methodology and identifying the optimal solution. Constraints  Operating Schedule - Trucks make deliveries every day of the week except Saturday or Sunday - Normal operation is for trucks to be loaded overnight and leave from the warehouse in the morning - Trucks make deliveries within towns from 9 a.m. to 6 p.m. - Earliest start time for trucks is 12 a.m. in the morning of delivery - Trucks returned to the depot require 2 hours for reloading and subsequent sameday delivery  Visits Per Month - Every town has a known number of visits per month  Truck Capacity - T407 trucks have a capacity of Rs500,000 - T310 trucks have a capacity of Rs350,000  Truck Operating Costs - Trucks operate at an average speed of 40 km/hr - T407 trucks have an operating cost of Rs13,500 - T310 trucks have an operating cost of Rs7,000 - T407 trucks have a running cost of Rs5 per kilometer - T310 trucks have a running cost of Rs3 per kilometer - Each truck has a crew of two, a driver and a helper - The driver is paid Rs2,200 per month - The helper is paid Rs1,400 per month - Each crewmember receives Rs60 per day for meals and other expenses while on the road  Working Schedule - Flexibly planned breaks for crew are at approximately 6 a.m., 12 p.m. and 6 p.m. - Breakfast and lunchtime breaks are 30 minutes each and dinner is 60 minutes

4

The solution to this exercise is provided by Nutthapol Dussadeenoad, Inderjot Gandhi, Earle Keith, Lisa Kuta, Jan Shahan, and Piyanuch Vichitakul who were students in the MBA program of the Weatherhead School of Management.

52

- Crew are allowed at least an 8-hour overnight break before starting a route on the following day - No overtime is paid and the company policy is to return the crews to the warehouse each day rather than plan for overnight layovers Other Considerations  Assumptions - Delivery demand is fixed by quantity and location, and city demand cannot be subdivided  Opportunities - Subcontracting is available at an estimated rate of Rs15 per km one way to the town. Methodology The methodological options leading to a good solution are outlined in Figure 1. START

1. Route 43 cities altogether

Cost = Rs116,678

2. Route all 4-visit cities and outsource all 2-visit cities

Cost = Rs135,254

3. Route 2 sets of data: all 4-visit cities in one seat and 2-visit cities in another set

4. Route 2 sets of data: mix between 4-visit cities and 2-visit cities

Cost = Rs109,092

Cost = Rs78,328

5. Outsource all cities whose demand is > Rs350,000 and route 2 sets of data: mix between 4-visit cities and 2-visit cities

N

Cost = Rs65,780

Satisfy? Y STOP

FIGURE 1 Computational Options to Find a Good Solution

53

Detail on each methodological step is offered as follows. 1. Route all 43 cities altogether Weeks 1&3: visit all cities = 14 routes/week Weeks 2&4: visit only 4-visit cities = 7 routes/week Results: - No. of truck used: T401 = 2, T301 = 2 - Cost = Rs116,678 2. Route all 4-visit cities, outsource all 2-visit cities All week = 7 routes/week Weeks 1 & 3: visit half of 2-visit cities. Weeks 2 & 4: visit another half of 2-visit cities. Results: - No. of truck used: T401 = 1, T301 = 1 - Cost = Rs135,254 3. Route 2 set of data, separate between 4-visit cities and 2-visit cities 4-visit cities: all weeks = 9 routes/week 2-visit cities: - Week 1&3: visit half of 2-visit cities = 4 routes/ week - Week 2&4 :visit another half of 2-visit cities = 3 routes/week Result - No of truck used: T401=1, T301 =2 - Cost = Rs109,092 4. Route 2 sets of data, mix between 4-visit cities and 2-visit cities 4-visit cities + half of 2-visit cities: visit in weeks 1&3 = 10 routes/week 4-visit cities + another half of 2-visit cities: visit in week 2&4 = 9 routes/week Note: Arrange 2-visit cities into 2 sets by following the below steps - Separate those 2-visit cities into 2 sets by area, east/west. - Sort cities in each set by sales per visit. - Set all cities in odd row as set 1, and all cities in even row as set 2 - The logic behind this method is that each set should be balanced in of demand and area. Result - No of truck used: T401 = 1, T301 = 1 - Cost = Rs78,328 5. Route 2 sets of data, mix 4-visit cities and 2-visit cities, and outsource all cities whose demand greater than Rs350,000 4-visit cities + half of 2-visit cities: visit in weeks 1&3 = 10 routes/week 4-visit cities + another half of 2-visit cities: visit in weeks 2&4 = 10 routes/week Note:

54

1) Arrange 2-visit cities into 2 sets by following the below steps - Separate those 2-visit cities into 2 sets by sales per visits, low demand and high demand - Sort cities in each set by distance from the central warehouse - Set all cities in odd row as set 1, and all cities in even row as set 2 - The logic behind this method is each set should be balanced in of demand and distance from the central warehouse. Using distance should reflect the better route than roughly separating cities by east/west area like in method 4. 2) Outsource cities whose demand is greater than 350,000 so that only a T310 truck is required - Guntur and Rajahmundry are the only 2 cities whose demand is greater than 350,000. Both of them are weekly visit cities, therefore, outsource to these two cities every week. Result - No. of truck used: T301 = 2 - Outsource 2 cities for all weeks: Guntur and Rajahmundry - Cost = Rs65,780, which is the best solution Best Solution The best solution per method 5 completes all deliveries for a total monthly cost of Rs65,780 using only 2 of T301 trucks together with 2 sets of crew . The cost calculation is shown in Figure 2. The daily schedule shown in Figure 3 represents the number of trucks needed, the truck routes with stop sequence, the schedule of truck usage through the month, and the schedule for using the crews. The Gantt chart of truck deliveries throughout the month is shown in Figure 4. See the router solution reports generated from LOGWARE, ROUTER module, for individual route detail. The reports divide the month into weeks 1 & 3 (Figure 5) and weeks 2 & 4 (Figure 6).

55

FIGURE 2 Cost Calculation Sheet for Best Solution Total monthly cost 2 2 2

fixed cost

allowance

Week 1&3 Week 2&4

Running cost

Common Carrier

Total

of of of

truck 2 (T310) driver helper 4 2 2

people people people

Week 1 & 3 Week 2 & 4 To Guntur To Rajahmundry

= = = x x x = = = =

2 weeks 2 weeks 2 weeks

x x x

3,146 km 2,414 km 35 76

km km

2 2 2 5 days 5 days 4 days

x x x

x x x

7,000 2,200 1,400 Rs 60 Rs 60 Rs 60

= = =

14,000 4,400 2,800

= = =

2,400 1,200 960

x x

3 Rs/km 3 Rs/km

x x

2 weeks 2 weeks

= =

18,876 14,484

x x

15 Rs/km 15 Rs/km

x x

4 weeks 4 weeks

= =

2,100 4,560

65,780

56

FIGURE 3 Daily Schedule

Daily Schedule Summary WEEK 1 & 3

Truck # DAY 1

I

II

DAY 2

I

II

DAY 3

I

II

DAY 4

I

II

Route 1 Depot Podili Kondulur Tanguturu Depot 25 Depot Jangareddygudem Kakinada Depot

Start Time

Crew # Driver I, Helper I

7:04 PM 5.46 AM

405 339

Driver II, Helper II

5:45 PM 6:39 AM

7 Depot Tadikonda Tenali Chirala Vuyyuru Depot

7:59 AM

8 Depot Sattenapalle Ongole Depot

Route Miles

4:27 AM

5 Depot Chilakalurupet Narasaraopet Macheria Depot

13 Depot Kaikalur Bhimavaram Tadepallegudem Depot

End Time

Driver I, Helper I

8:18 PM

306 Driver II, Helper II

8:47 PM

232

6:47 AM

Driver I, Helper I

7:07 PM

253

7:01 AM

Driver II, Helper II

8:12 PM

19 Depot Narasapur Mandapeta Depot

5:19 AM

11 Depot Pamarru Machilipatnam Palakolu Depot

7:19 AM

307 Driver I, Helper I

7:03 PM

349 Driver II, Helper II

5:19 PM

240

57

DAY 5

I

24 Depot Eluru Chintalapuidi Depot

II

2 Depot Kani Giri Bestavaipetta Giddalur Markapur Depot

6:56 AM

Driver I, Helper I

3:11 PM

190

3:44 AM

Driver II, Helper II

11:21 PM

525

End Time

Route Miles

Guntur Rajahmundry

Outsource

WEEK 2 & 4

DAY 1

Truck #

Route

Start Time

I

3 Depot Ongole Depot

5:01 AM

II

DAY 2

I

II

Driver I, Helper I 3:28 PM

18 Depot Jaggayyapeta Eluru Depot

6:34 AM

13 Depot Vuyyuru Machilipatnam Gudivada Depot

7:41 AM

19 Depot Hunuman Junction Chirala Bapatia Repalie Depot

8:26 AM

Crew#

278 Driver II, Helper II

5:00 PM

276 Driver I, Helper I

3:30 PM

153 Driver II, Helper II

7:49 PM

175

58

DAY 3

I

II

DAY 4

I

II

DAY 5

I

II Outsource

14 Depot Tenali Palakolu Depot

7:35 AM

24 Depot Tanuku Nidadvole Kovvur Depot

5:12 AM

16 Depot Nuzvid Tadepallegudem Bhimavaram Depot

7:27 AM

29 Depot Piduguralia Addanki Vinukonda Narasaraopet Depot

6:23 AM

23 Depot Amaiapuram Kakinada Depot

4:23 AM

Driver I, Helper I

4:40 PM

243 Driver II, Helper II

5:04 PM

315 Driver I, Helper I

5:51 PM

236

Driver II, Helper II

7:49 PM

318 Driver I, Helper I

7:52 PM

420

Open Guntur Rajahmundry

Note: Truck # represents the number of type T301 truck.

59

FIGURE 4 Gantt Chart of Truck Deliveries Throughout the Month

60

FIGURE 5 ROUTER Solution Report, Weeks 1 and 3 Label- Untitled Date- 11/23/2002 Time- 2:47:17 AM *** SUMMARY REPORT *** TIME/DISTANCE/COST INFORMATION

Route no 1 2

Route Run Stop Brk Stem time, time, time, time, time, Start Return No of Route hr hr hr hr hr time time stops dist,Mi 14.6 10.1 2.5 2.0 7.8 04:27AM 07:04PM 3 405 19.6 13.1 4.5 2.0 9.4 03:44AM 11:21PM 4 525

Route cost,$ 1215.00 1575.00

5

13.6

7.6

4.0

2.0

5.4 06:39AM 08:18PM

3

306

918.00

7 8

12.8 13.2

5.8 7.7

5.0 3.5

2.0 2.0

1.4 07:59AM 08:47PM 5.0 07:01AM 08:12PM

4 2

232 307

696.00 921.00

11

10.0

6.0

3.0

1.0

3.3 07:19AM 05:19PM

3

240

720.00

13

12.3

6.3

4.0

2.0

4.5 06:47AM 07:07PM

3

253

759.00

19

13.7

8.7

3.0

2.0

7.2 05:19AM 07:03PM

2

349

1047.00

24 25

8.2 12.0

4.8 8.5

2.5 2.5

1.0 1.0

3.6 06:56AM 03:11PM 7.7 05:46AM 05:45PM

2 2

190 339

570.00 1017.00

VEHICLE INFORMATION Route Veh Weight Delvry Pickup Weight Cube Delvry Pickup no typ capcty weight weight util capcty cube cube 1 1 350 180 0 51.4% 9999 0 0 2 1 350 134 0 38.3% 9999 0 0

Cube util .0% .0%

Vehicle description T310 T310

5

1

350

342

0

97.7%

9999

0

0

.0% T310

7 8

1 1

350 350

337 350

0 96.3% 0 100.0%

9999 9999

0 0

0 0

.0% T310 .0% T310

11

1

350

350

0 100.0%

9999

0

0

.0% T310

13

1

350

326

0

93.1%

9999

0

0

.0% T310

19

1

350

330

0

94.3%

9999

0

0

.0% T310

24 25

1 1

350 350

266 296

0 0

76.0% 84.6%

9999 9999

0 0

0 0

.0% T310 .0% T310

*** DETAIL REPORT ON ROUTE NUMBER 1 *** A T310 leaves at 4:27AM on day 1 from the depot at Vijayawada

Stop No description

Stop Drive Distance Time Arrive Depart time to stop to stop wind time Day time Day Min Min Miles met? -->Break 30 minutes

61

2 Podili 6 Kondulur

09:00AM 1 09:30AM 11:28AM 1 12:28PM -->Break 30 minutes 01:19PM 1 02:19PM -->Break 60 minutes 07:04PM 1 -------

1 Tanguturu Depot Stop No description

Stop volume Weight Cube

2 Podili 6 Kondulur 1 Tanguturu Totals Weight: Del = 180

24 90 66 Pickups

Route time: Driving Load/unload Break Total

10.1 hr 2.5 2.0 14.6 hr

Max allowed

21.0 hr

Route costs: Driver (reg time) Driver (over time) Vehicle (mileage) Fixed Total

0 0 0 =

1 1

30 60

243.0 118.5

162 79

YES YES

1

60

21.0

14

YES

--

---

225.0

150

Inc cost to serve stop In $ In $/unit

Capacity in use Weight Cube 51.4% .0% 183.00 7.6 44.6 .0 51.00 .6 18.9 .0 -48.00 -.7 .0 .0 0 Cube: Del = 0 Pickups = 0

Distance: To 1st stop From last stop On route Total Max allowed

162 mi 150 93 405 mi 9999 mi

$.00 .00 1215.00 .00 $1215.00

*** DETAIL REPORT ON ROUTE NUMBER 2 *** A T310 leaves at 3:44AM on day 1 from the depot at Vijayawada

Stop No description 5 Kani Giri 9 Bestavaipetta 7 Giddalur 4 Markapur Depot Stop No description

Arrive time -->Break 09:00AM -->Break 01:21PM 02:41PM 05:12PM -->Break 11:21PM

Stop Drive Distance Time Depart time to stop to stop wind Day time Day Min Min Miles met? 30 minutes 1 11:30AM 1 150 286.5 191 YES 30 minutes 1 01:51PM 1 30 81.0 54 YES 1 03:41PM 1 60 49.5 33 YES 1 05:42PM 1 30 91.5 61 YES 60 minutes 1 ------- -- --279.0 186

Stop volume Weight Cube

5 Kani Giri 9 Bestavaipetta 7 Giddalur 4 Markapur Totals Weight: Del = 134

24 25 25 60 Pickups

Route time: Driving Load/unload Break Total

13.1 hr 4.5 2.0 19.6 hr

Max allowed

24.0 hr

0 0 0 0 =

Inc cost to serve stop In $ In $/unit

Capacity in use Weight Cube 38.3% .0% 87.00 3.6 31.4 .0 15.00 .6 24.3 .0 210.00 8.4 17.1 .0 -21.00 -.4 .0 .0 0 Cube: Del = 0 Pickups = 0

Distance: To 1st stop From last stop On route Total Max allowed

191 mi 186 148 525 mi 9999 mi

62

Route costs: Driver (reg time) Driver (over time) Vehicle (mileage) Fixed Total

$.00 .00 1575.00 .00 $1575.00

*** DETAIL REPORT ON ROUTE NUMBER 5 *** A T310 leaves at 6:39AM on day 1 from the depot at Vijayawada

Stop No description

Arrive time -->Break 11 Chilakalurupet 09:00AM 12 Narasarapet 10:31AM -->Break 42 Macheria 01:46PM -->Break Depot 08:18PM Stop No description

Stop Drive Distance Time Depart time to stop to stop wind Day time Day Min Min Miles met? 30 minutes 1 10:00AM 1 60 111.0 74 YES 1 11:31AM 1 60 31.5 21 YES 30 minutes 1 03:46PM 1 120 105.0 70 YES 60 minutes 1 ------- -- --211.5 141

Stop volume Weight Cube

Inc cost to serve stop In $ In $/unit

Capacity in use Weight Cube 97.7% .0% 57.00 .6 71.4 .0 .00 .0 42.9 .0 405.00 2.7 .0 .0 0 Cube: Del = 0 Pickups = 0

11 Chilakalurupet 92 12 Narasarapet 100 42 Macheria 150 Totals Weight: Del = 342 Pickups

0 0 0 =

Route time: Driving Load/unload Break Total

7.6 hr 4.0 2.0 13.6 hr

Distance: To 1st stop From last stop On route Total

Max allowed

21.0 hr

Route costs: Driver (reg time) Driver (over time) Vehicle (mileage) Fixed Total

Max allowed

74 mi 141 91 306 mi 9999 mi

$.00 .00 918.00 .00 $918.00

*** DETAIL REPORT ON ROUTE NUMBER 7 *** A T310 leaves at 7:59AM on day 1 from the depot at Vijayawada

Stop No description 14 Tadikonda 19 Tenali 8 Chirala 18 Vuyyuru Depot Stop No description

Arrive time -->Break 09:00AM 10:58AM -->Break 02:00PM 05:57PM -->Break 08:47PM

Stop Drive Distance Time Depart time to stop to stop wind Day time Day Min Min Miles met? 30 minutes 1 10:00AM 1 60 31.5 21 YES 1 11:58AM 1 60 58.5 39 YES 30 minutes 1 04:00PM 1 120 91.5 61 YES 1 06:57PM 1 60 117.0 78 YES 60 minutes 1 ------- -- --49.5 33

Stop volume Weight Cube

Inc cost to serve stop In $ In $/unit

Capacity in use Weight Cube

63

14 Tadikonda 60 19 Tenali 140 8 Chirala 98 18 Vuyyuru 39 Totals Weight: Del = 337 Pickups

96.3% 0 72.00 1.2 79.1 0 33.00 .2 39.1 0 219.00 2.2 11.1 0 66.00 1.7 .0 = 0 Cube: Del = 0 Pickups = 0

Route time: Driving Load/unload Break Total

5.8 hr 5.0 2.0 12.8 hr

Distance: To 1st stop From last stop On route Total

Max allowed

21.0 hr

Route costs: Driver (reg time) Driver (over time) Vehicle (mileage) Fixed Total

Max allowed

.0% .0 .0 .0 .0

21 mi 33 178 232 mi 9999 mi

$.00 .00 696.00 .00 $696.00

*** DETAIL REPORT ON ROUTE NUMBER 8 *** A T310 leaves at 7:01AM on day 1 from the depot at Vijayawada

Stop No description 15 Sattenapalle 3 Ongole Depot

Arrive time -->Break 09:00AM -->Break 01:13PM -->Break 08:12PM

Stop No description

Stop Drive Distance Time Depart time to stop to stop wind Day time Day Min Min Miles met? 30 minutes 1 10:00AM 1 60 88.5 59 YES 30 minutes 1 03:43PM 1 150 163.5 109 YES 60 minutes 1 ------- -- --208.5 139

Stop volume Weight Cube

Capacity in use Weight Cube 100.0% .0% 15 Sattenapalle 45 0 87.00 1.9 87.1 .0 3 Ongole 305 0 567.00 1.9 .0 .0 Totals Weight: Del = 350 Pickups = 0 Cube: Del = 0 Pickups = 0 Route time: Driving Load/unload Break Total

7.7 hr 3.5 2.0 13.2 hr

Max allowed

21.0 hr

Route costs: Driver (reg time) Driver (over time) Vehicle (mileage) Fixed Total

Inc cost to serve stop In $ In $/unit

Distance: To 1st stop From last stop On route Total Max allowed

59 mi 139 109 307 mi 9999 mi

$.00 .00 921.00 .00 $921.00

*** DETAIL REPORT ON ROUTE NUMBER 11 *** A T310 leaves at 7:19AM on day 1 from the depot at Vijayawada Stop

Drive Distance Time

64

Stop No description 20 Pamarru 22 Machilipatnam 38 Palakolu Depot

Arrive Depart time to stop time Day time Day Min Min -->Break 30 minutes 09:00AM 1 10:00AM 1 60 70.5 10:36AM 1 11:36AM 1 60 36.0 -->Break 30 minutes 02:12PM 1 03:12PM 1 60 126.0 05:19PM 1 ------- -- --127.5

Stop No description

Stop volume Weight Cube

47 24

YES YES

84 85

YES

Inc cost to serve stop In $ In $/unit

Capacity in use Weight Cube 100.0% .0% -9.00 -.1 82.3 .0 126.00 1.2 51.4 .0 285.00 1.6 .0 .0 0 Cube: Del = 0 Pickups = 0

20 Pamarru 62 22 Machilipatnam 108 38 Palakolu 180 Totals Weight: Del = 350 Pickups

0 0 0 =

Route time: Driving Load/unload Break Total

6.0 hr 3.0 1.0 10.0 hr

Distance: To 1st stop From last stop On route Total

Max allowed

21.0 hr

Route costs: Driver (reg time) Driver (over time) Vehicle (mileage) Fixed Total

to stop wind Miles met?

Max allowed

47 mi 85 108 240 mi 9999 mi

$.00 .00 720.00 .00 $720.00

*** DETAIL REPORT ON ROUTE NUMBER 13 *** A T310 leaves at 6:47AM on day 1 from the depot at Vijayawada

Stop No description

Arrive time -->Break 23 Kaikalur 09:00AM 39 Bhimavaram 10:54AM -->Break 36 Tadepallegudem 01:48PM -->Break Depot 07:07PM Stop No description

Stop Drive Distance Time Depart time to stop to stop wind Day time Day Min Min Miles met? 30 minutes 1 10:00AM 1 60 102.0 68 YES 1 12:24PM 1 90 54.0 36 YES 30 minutes 1 03:18PM 1 90 54.0 36 YES 60 minutes 1 ------- -- --169.5 113

Stop volume Weight Cube

Inc cost to serve stop In $ In $/unit

Capacity in use Weight Cube 93.1% .0% -12.00 -.2 79.4 .0 -39.00 -.3 37.1 .0 123.00 .9 .0 .0 0 Cube: Del = 0 Pickups = 0

23 Kaikalur 48 39 Bhimavaram 148 36 Tadepallegudem 130 Totals Weight: Del = 326 Pickups

0 0 0 =

Route time: Driving Load/unload Break Total

6.3 hr 4.0 2.0 12.3 hr

Distance: To 1st stop From last stop On route Total

Max allowed

21.0 hr

Max allowed

68 mi 113 72 253 mi 9999 mi

65

Route costs: Driver (reg time) Driver (over time) Vehicle (mileage) Fixed Total

$.00 .00 759.00 .00 $759.00

*** DETAIL REPORT ON ROUTE NUMBER 19 *** A T310 leaves at 5:19AM on day 1 from the depot at Vijayawada

Stop No description 30 Narasapur 29 Mandapeta

Depot

Arrive time -->Break 09:00AM 11:30AM -->Break -->Break 07:03PM

Stop No description

Stop Drive Distance Time Depart time to stop to stop wind Day time Day Min Min Miles met? 30 minutes 1 10:00AM 1 60 190.5 127 YES 1 01:30PM 1 120 90.0 60 YES 30 minutes 60 minutes 1 ------- -- --243.0 162

Stop volume Weight Cube

Capacity in use Weight Cube 94.3% .0% 30 Narasapur 160 0 75.00 .5 48.6 .0 29 Mandapeta 170 0 285.00 1.7 .0 .0 Totals Weight: Del = 330 Pickups = 0 Cube: Del = 0 Pickups = 0 Route time: Driving Load/unload Break Total

8.7 hr 3.0 2.0 13.7 hr

Max allowed

21.0 hr

Route costs: Driver (reg time) Driver (over time) Vehicle (mileage) Fixed Total

Inc cost to serve stop In $ In $/unit

Distance: To 1st stop From last stop On route Total Max allowed

127 mi 162 60 349 mi 9999 mi

$.00 .00 1047.00 .00 $1047.00

*** DETAIL REPORT ON ROUTE NUMBER 24 *** A T310 leaves at 6:56AM on day 1 from the depot at Vijayawada

Stop No description 37 Eluru 41 Chintalapuidi Depot Stop No description

Stop Drive Distance Time Arrive Depart time to stop to stop wind time Day time Day Min Min Miles met? -->Break 30 minutes 09:00AM 1 11:00AM 1 120 94.5 63 YES -->Break 30 minutes 12:41PM 1 01:11PM 1 30 70.5 47 YES 03:11PM 1 ------- -- --120.0 80 Stop volume Weight Cube

Inc cost to serve stop In $ In $/unit

Capacity in use Weight Cube 76.0% .0% 37 Eluru 198 0 90.00 .5 19.4 .0 41 Chintalapuidi 68 0 192.00 2.8 .0 .0 Totals Weight: Del = 266 Pickups = 0 Cube: Del = 0 Pickups = 0

66

Route time: Driving Load/unload Break Total

4.8 hr 2.5 1.0 8.2 hr

Max allowed

21.0 hr

Route costs: Driver (reg time) Driver (over time) Vehicle (mileage) Fixed Total

Distance: To 1st stop From last stop On route Total Max allowed

63 mi 80 47 190 mi 9999 mi

$.00 .00 570.00 .00 $570.00

*** DETAIL REPORT ON ROUTE NUMBER 25 *** A T310 leaves at 5:46AM on day 1 from the depot at Vijayawada Stop Drive Distance Time Arrive Depart time to stop to stop wind time Day time Day Min Min Miles met? -->Break 30 minutes 40 Jangareddygudem 09:00AM 1 09:30AM 1 30 163.5 109 YES 32 Kakinada 10:15AM 1 12:15PM 1 120 45.0 30 YES -->Break 30 minutes Depot 05:45PM 1 ------- -- --300.0 200 Stop No description

Stop No description

Stop volume Weight Cube

Capacity in use Weight Cube 84.6% .0% 40 Jangareddygudem 68 0 -183.00 -2.7 65.1 .0 32 Kakinada 228 0 363.00 1.6 .0 .0 Totals Weight: Del = 296 Pickups = 0 Cube: Del = 0 Pickups = 0 Route time: Driving Load/unload Break Total

8.5 hr 2.5 1.0 12.0 hr

Max allowed

21.0 hr

Route costs: Driver (reg time) Driver (over time) Vehicle (mileage) Fixed Total

Inc cost to serve stop In $ In $/unit

Distance: To 1st stop From last stop On route Total Max allowed

109 mi 200 30 339 mi 9999 mi

$.00 .00 1017.00 .00 $1017.00

67

FIGURE 6 ROUTER Solution Report, Weeks 2 and 4 Label- Untitled Date- 11/23/2002 Time- 4:21:32 AM *** SUMMARY REPORT *** TIME/DISTANCE/COST INFORMATION

Route no 3

Route Run Stop Brk Stem time, time, time, time, time, Start Return No of Route hr hr hr hr hr time time stops dist,Mi 10.4 7.0 2.5 1.0 7.0 05:01AM 03:28PM 1 278

Route cost,$ 834.00

13 14

7.8 9.1

3.8 6.1

3.0 2.0

1.0 1.0

2.0 07:41AM 03:30PM 3.0 07:35AM 04:40PM

3 2

153 243

459.00 729.00

16

10.4

5.9

3.5

1.0

3.8 07:27AM 05:51PM

3

236

708.00

18 19

4.8 11.4

3.8 4.4

.5 5.0

.5 2.0

3.8 06:34AM 5:00PM 1.7 08:26AM 07:49PM

2 4

276 175

828.00 525.00

23 24

15.5 11.9

10.5 7.9

3.0 3.0

2.0 1.0

9.1 04:23AM 07:52PM 7.0 05:12AM 05:04PM

2 3

420 315

1260.00 945.00

29

13.4

8.0

3.5

2.0

4.0 06:23AM 07:49PM

4

318

954.00

VEHICLE INFORMATION Route Veh Weight Delvry Pickup Weight Cube Delvry Pickup no typ capcty weight weight util capcty cube cube

Cube Vehicle util description

3

1

350

305

0

87.1%

9999

0

0

.0% T310

13 14

1 1

350 350

327 320

0 0

93.4% 91.4%

9999 9999

0 0

0 0

.0% T310 .0% T310

16

1

350

315

0

90.0%

9999

0

0

.0% T310

18 19

1 1

350 350

235 280

0 0

67.1% 80.0%

9999 9999

0 0

0 0

.0% T310 .0% T310

23 24

1 1

350 350

318 229

0 0

90.9% 65.4%

9999 9999

0 0

0 0

.0% T310 .0% T310

29

1

350

305

0

87.1%

9999

0

0

.0% T310

*** DETAIL REPORT ON ROUTE NUMBER 3 *** A T310 leaves at 5:01AM on day 1 from the depot at Vijayawada

Stop No description 3 Ongole Depot Stop No description

Stop Drive Distance Time Arrive Depart time to stop to stop wind time Day time Day Min Min Miles met? -->Break 30 minutes 09:00AM 1 11:30AM 1 150 208.5 139 YES -->Break 30 minutes 03:28PM 1 ------- -- --208.5 139 Stop volume Weight Cube

Inc cost to serve stop In $ In $/unit

Capacity in use Weight Cube 87.1% .0%

68

3 Ongole 305 0 834.00 2.7 Totals Weight: Del = 305 Pickups = 0 Cube: Del = 0 Pickups = 0 Route time: Driving Load/unload Break Total

7.0 hr 2.5 1.0 10.4 hr

Max allowed

21.0 hr

Route costs: Driver (reg time) Driver (over time) Vehicle (mileage) Fixed Total

Distance: To 1st stop From last stop On route Total Max allowed

.0

.0

139 mi 139 0 278 mi 9999 mi

$.00 .00 834.00 .00 $834.00

*** DETAIL REPORT ON ROUTE NUMBER 13 *** A T310 leaves at 7:41AM on day 1 from the depot at Vijayawada

Stop No description 18 Vuyyuru 22 Machilipatnam 26 Gudivada Depot

Stop Drive Distance Time Arrive Depart time to stop to stop wind time Day time Day Min Min Miles met? -->Break 30 minutes 09:00AM 1 10:00AM 1 60 49.5 33 YES 10:58AM 1 11:58AM 1 60 58.5 39 YES -->Break 30 minutes 01:20PM 1 02:20PM 1 60 51.0 34 YES 03:30PM 1 ------- -- --70.5 47

Stop No description

Stop volume Weight Cube

Inc cost to serve stop In $ In $/unit

Capacity in use Weight Cube 93.4% .0% -6.00 -.2 82.3 .0 153.00 1.4 51.4 .0 21.00 .1 .0 .0 0 Cube: Del = 0 Pickups = 0

18 Vuyyuru 39 22 Machilipatnam 108 26 Gudivada 180 Totals Weight: Del = 327 Pickups

0 0 0 =

Route time: Driving Load/unload Break Total

3.8 hr 3.0 1.0 7.8 hr

Distance: To 1st stop From last stop On route Total

Max allowed

21.0 hr

Route costs: Driver (reg time) Driver (over time) Vehicle (mileage) Fixed Total

Max allowed

33 mi 47 73 153 mi 9999 mi

$.00 .00 459.00 .00 $459.00

*** DETAIL REPORT ON ROUTE NUMBER 14 *** A T310 leaves at 7:35AM on day 1 from the depot at Vijayawada

Stop No description

Stop Drive Distance Time Arrive Depart time to stop to stop wind time Day time Day Min Min Miles met? -->Break 30 minutes

69

19 Tenali

09:00AM 1 10:00AM -->Break 30 minutes 01:33PM 1 02:33PM 04:40PM 1 -------

38 Palakolu Depot Stop No description

Stop volume Weight Cube

1

60

54.0

36

YES

1 --

60 ---

183.0 127.5

122 85

YES

Capacity in use Weight Cube 91.4% .0% 19 Tenali 140 0 219.00 1.6 51.4 .0 38 Palakolu 180 0 513.00 2.8 .0 .0 Totals Weight: Del = 320 Pickups = 0 Cube: Del = 0 Pickups = 0 Route time: Driving Load/unload Break Total

6.1 hr 2.0 1.0 9.1 hr

Max allowed

21.0 hr

Route costs: Driver (reg time) Driver (over time) Vehicle (mileage) Fixed Total

Inc cost to serve stop In $ In $/unit

Distance: To 1st stop From last stop On route Total Max allowed

36 mi 85 122 243 mi 9999 mi

$.00 .00 729.00 .00 $729.00

*** DETAIL REPORT ON ROUTE NUMBER 16 *** A T310 leaves at 7:27AM on day 1 from the depot at Vijayawada Stop Drive Distance Time Arrive Depart time to stop to stop wind time Day time Day Min Min Miles met? -->Break 30 minutes 21 Nuzvid 09:00AM 1 09:30AM 1 30 63.0 42 YES 36 Tadepallegudem 10:45AM 1 12:15PM 1 90 75.0 50 YES -->Break 30 minutes 39 Bhimavaram 01:39PM 1 03:09PM 1 90 54.0 36 YES Depot 05:51PM 1 ------- -- --162.0 108 Stop No description

Stop No description

Stop volume Weight Cube

Inc cost to serve stop In $ In $/unit

Capacity in use Weight Cube 90.0% .0% -63.00 -1.7 79.4 .0 -66.00 -.5 42.3 .0 93.00 .6 .0 .0 0 Cube: Del = 0 Pickups = 0

21 Nuzvid 37 36 Tadepallegudem 130 39 Bhimavaram 148 Totals Weight: Del = 315 Pickups

0 0 0 =

Route time: Driving Load/unload Break Total

5.9 hr 3.5 1.0 10.4 hr

Distance: To 1st stop From last stop On route Total

Max allowed

21.0 hr

Route costs: Driver (reg time) Driver (over time) Vehicle (mileage) Fixed Total

Max allowed

42 mi 108 86 236 mi 9999 mi

$.00 .00 708.00 .00 $708.00

70

*** DETAIL REPORT ON ROUTE NUMBER 18 *** A T310 leaves at 6:34AM on day 1 from the depot at Vijayawada

Stop No description 24 Jaggayyapeta 37 Eluru Depot

Stop Drive Distance Time Arrive Depart time to stop to stop wind time Day time Day Min Min Miles met? -->Break 30 minutes 09:00AM 1 09:30AM 1 30 115.5 77 YES -->Break 30 minutes 13:24AM 1 3:24PM 1 120 94.5 63 YES 5:00PM 1 ------- -- --115.5 77

Stop No description

Stop volume Weight Cube

Capacity in use Weight Cube 10.6% .0% 24 Jaggayyapeta 37 0 462.00 12.5 .0 .0 Totals Weight: Del = 37 Pickups = 0 Cube: Del = 0 Pickups = 0 Route time: Driving Load/unload Break Total

3.8 hr .5 .5 4.8 hr

Max allowed

21.0 hr

Route costs: Driver (reg time) Driver (over time) Vehicle (mileage) Fixed Total

Inc cost to serve stop In $ In $/unit

Distance: To 1st stop From last stop On route Total Max allowed

77 mi 63 136 276 mi 9999 mi

$.00 .00 828.00 .00 $828.00

*** DETAIL REPORT ON ROUTE NUMBER 19 *** A T310 leaves at 8:26AM on day 1 from the depot at Vijayawada

Stop No description 25 Hanuman 8 Chirala 27 Bapatia 16 Repalie Depot

Arrive time -->Break Junctio 09:00AM 11:11AM -->Break 02:01PM 04:08PM -->Break 07:49PM

Stop No description

Stop volume Weight Cube

25 Hanuman Junctio 8 Chirala 27 Bapatia 16 Repalie Totals Weight: Del = 280 Route time: Driving Load/unload

Stop Drive Distance Time Depart time to stop to stop wind Day time Day Min Min Miles met? 30 minutes 1 10:00AM 1 60 3.0 2 YES 1 01:11PM 1 120 72.0 48 YES 30 minutes 1 03:01PM 1 60 19.5 13 YES 1 05:08PM 1 60 67.5 45 YES 60 minutes 1 ------- -- --100.5 67

50 98 82 50 Pickups

4.4 hr 5.0

0 0 0 0 =

Inc cost to serve stop In $ In $/unit

Capacity in use Weight Cube 80.0% .0% -117.00 -2.3 65.7 .0 -57.00 -.6 37.7 .0 .00 .0 14.3 .0 108.00 2.2 .0 .0 0 Cube: Del = 0 Pickups = 0

Distance: To 1st stop From last stop

2 mi 67

71

Break Total

2.0 11.4 hr

On route Total

Max allowed

21.0 hr

Max allowed

Route costs: Driver (reg time) Driver (over time) Vehicle (mileage) Fixed Total

106 175 mi 9999 mi

$.00 .00 525.00 .00 $525.00

*** DETAIL REPORT ON ROUTE NUMBER 23 *** A T310 leaves at 4:23AM on day 1 from the depot at Vijayawada

Stop No description 31 Amaiapuram 32 Kakinada

Depot

Arrive time -->Break 09:00AM 11:22AM -->Break -->Break 07:52PM

Stop No description

Stop Drive Distance Time Depart time to stop to stop wind Day time Day Min Min Miles met? 30 minutes 1 10:00AM 1 60 247.5 165 YES 1 01:22PM 1 120 82.5 55 YES 30 minutes 60 minutes 1 ------- -- --300.0 200

Stop volume Weight Cube

Capacity in use Weight Cube 90.9% .0% 31 Amaiapuram 90 0 60.00 .7 65.1 .0 32 Kakinada 228 0 270.00 1.2 .0 .0 Totals Weight: Del = 318 Pickups = 0 Cube: Del = 0 Pickups = 0 Route time: Driving Load/unload Break Total

10.5 hr 3.0 2.0 15.5 hr

Max allowed

21.0 hr

Route costs: Driver (reg time) Driver (over time) Vehicle (mileage) Fixed Total

Inc cost to serve stop In $ In $/unit

Distance: To 1st stop From last stop On route Total Max allowed

165 mi 200 55 420 mi 9999 mi

$.00 .00 1260.00 .00 $1260.00

*** DETAIL REPORT ON ROUTE NUMBER 24 *** A T310 leaves at 5:12AM on day 1 from the depot at Vijayawada

Stop No description 34 Tanuku 35 Nidadvole 33 Kovvur Depot

Stop Drive Distance Time Arrive Depart time to stop to stop wind time Day time Day Min Min Miles met? -->Break 30 minutes 09:00AM 1 10:00AM 1 60 198.0 132 YES 10:33AM 1 11:33AM 1 60 33.0 22 YES 11:55AM 1 12:55PM 1 60 22.5 15 YES -->Break 30 minutes 05:04PM 1 ------- -- --219.0 146

72

Stop No description

Stop volume Weight Cube

Inc cost to serve stop In $ In $/unit

Capacity in use Weight Cube 65.4% .0% 60.00 .4 27.1 .0 -9.00 -.2 12.9 .0 81.00 1.8 .0 .0 0 Cube: Del = 0 Pickups = 0

34 Tanuku 134 35 Nidadvole 50 33 Kovvur 45 Totals Weight: Del = 229 Pickups

0 0 0 =

Route time: Driving Load/unload Break Total

7.9 hr 3.0 1.0 11.9 hr

Distance: To 1st stop From last stop On route Total

Max allowed

21.0 hr

Route costs: Driver (reg time) Driver (over time) Vehicle (mileage) Fixed Total

Max allowed

132 mi 146 37 315 mi 9999 mi

$.00 .00 945.00 .00 $945.00

*** DETAIL REPORT ON ROUTE NUMBER 29 *** A T310 leaves at 6:23AM on day 1 from the depot at Vijayawada

Stop No description 43 Piduguralia 10 Addanki 13 Vinukonda 12 Narasarapet Depot

Arrive time -->Break 09:00AM 11:57AM -->Break 01:52PM 03:56PM -->Break 07:49PM

Stop No description

Stop Drive Distance Time Depart time to stop to stop wind Day time Day Min Min Miles met? 30 minutes 1 10:00AM 1 60 127.5 85 YES 1 12:27PM 1 30 117.0 78 YES 30 minutes 1 02:52PM 1 60 55.5 37 YES 1 04:56PM 1 60 63.0 42 YES 60 minutes 1 ------- -- --114.0 76

Stop volume Weight Cube

Inc cost to serve stop In $ In $/unit

Capacity in use Weight Cube 87.1% .0% 147.00 1.8 64.3 .0 150.00 2.5 47.1 .0 54.00 .8 28.6 .0 .00 .0 .0 .0 0 Cube: Del = 0 Pickups = 0

43 Piduguralia 80 10 Addanki 60 13 Vinukonda 65 12 Narasarapet 100 Totals Weight: Del = 305 Pickups

0 0 0 0 =

Route time: Driving Load/unload Break Total

8.0 hr 3.5 2.0 13.4 hr

Distance: To 1st stop From last stop On route Total

Max allowed

21.0 hr

Route costs: Driver (reg time) Driver (over time) Vehicle (mileage) Fixed Total

Max allowed

85 mi 76 157 318 mi 9999 mi

$.00 .00 954.00 .00 $954.00

73

CHAPTER 8 FORECASTING SUPPLY CHAIN REQUIREMENTS 4 (a) The answer to this question is aided by using the FORECAST module in LOGWARE. A sample calculation is shown as carried out by FORECAST. The results are then summarized from FORECAST output. An example calculation for an  = 0.1 is shown. Other  values would be used, ranging 0.01 to 1.0. We first calculate a starting forecast by averaging the first four weekly requirements. That is,

[2,056 + 2,349 + 1,895 + 1,514]/4 = 1,953.50 Now we backcast this value and start the forecast at time 0. Thus, the forecasts and the associated errors would be:

F1 F2 F3 F4 F5 F6 F7 F8 F9 F10 F11

= = = = = = = = = = =

.1(2056) .1(2349) .1(1895) .1(1514) .1(1194) .1(2268) .1(2653) .1(2039) .1(2399) .1(2508)

+ + + + + + + + + +

.9(1953.50) .9(1963.75) .9(2002.20) .9(1991.48) .9(1943.73) .9(1868.76) .9(1908.00) .9(1982.50) .9(1988.15) .9(2029.24)

Forecast Error 1953.50 = 1963.75 = 2002.28 = 1991.48 = 1943.73 -749.73 = 1868.76 399.24 = 1908.00 745.00 = 1982.50 56.50 = 1988.15 410.85 = 2029.24 478.76 = 2077.12 Total squared error

Squared error

562,095.07 159,392.58 555,025.00 3,192.25 168,797.72 229,211.14 1,677,713.76

The standard error of the forecast is: SF 

Total squared error 1,677,713.76   528.79 6 N

Note: FORECAST does not use N-1 in the denominator. Repeating this type of analysis, the following table can be developed. The results from FORECAST are shown.

74

 .01 .05 .1 .2 .5 1.0

SF 528.72 528.42 528.46 528.89 535.55 566.07

The  that minimizes S F is 0.05. (b) The forecast errors are computed in part a. (c) If we assume that the errors are normally distributed around the forecast, we can then construct a 95% confidence band on the forecast. That is, if Y is the actual volume in period 11, then the range of the forecast (F11 = 2,017.81 for  = 0.05) will be: Y = F11 + z S F = 2,017.81 + 1.96528.42 Then, 982.11  Y  3,053.51 All values are in thousands. 5 (a) & (b) The solution to this problem was aided by the use of the exponential smoothing module in FORECAST. Using the first four week's data to initialize the level/trend version of the exponential smoothing model and setting  and  equal to 0.2, the forecast for next week is F11 = 2,024.47, with a standard error of the forecast of S F = 171.28.

(c) Assuming that the forecast errors are normally distributed around F11, a 95% statistical confidence band can be constructed. The confidence band is: Y = F11 + z S F = 2,024.47 + 1.96171.28 where z = 1.96 for 2.5% of the area under the two tails of a normal distribution. The range of the actual weekly volume is expected to be: 1,688.76  Y  2,360.18 6

75

(a) The data may be restated as shown below.

Sales, S 27,000 70,000 41,000 13,000 30,000 73,000 48,000 15,000 34,000 82,000 51,000 16,000 500,000

t 1 2 3 4 5 6 7 8 9 10 11 12 78

St 27,000 140,000 123,000 52,000 150,000 438,000 336,000 120,000 306,000 820,000 561,000 192,000 3,265,000

t2 1 4 9 16 25 36 49 64 81 100 121 144 650

Trend value,a St 41,087 41,192 41,298 41,403 41,508 41,613 41,719 41,824 41,929 42,035 42,140 42,245

Seasonal indexb 0.66 1.70 0.99 0.32 0.72 1.75 1.15 0.36 0.81 1.95 1.21 0.38

a

Computed from the linear trend line. For example, for period 1, S1 = 40,981.6 + 105.31 = 41,087. b The ratio of the actual sales S to the trend line value St. For example, for period 1, the seasonal index is 27,000/41,087 = 0.66.

Given the values from the above table and that t = 78/12 = 6.5, N = 12, and S = 500,000/12 = 41,666, the coefficients in the regression trend line would be: b

 S  t  N  S  t  3,265,000  12  41,666  6.5  105.3 650  12  6.5 t  N  t 2

2

2

and a  S  b  t  41,666  105.3  6.5  40,9816 . Therefore, the trend value St for any period t would be: St = 40,981.6 + 105.3t (b) The seasonal factors are determined by the ratio of the actual sales in a period to the trend value for that period. For example, the seasonal factor for period 12 (4th quarter of last year) would be 16,000/42,245 = 0.38. This and the seasonal factors for all past quarters are shown in the previous table. (c) The forecasts using the seasonal factors from the last 4 quarters are as follows. Seasonal 76

t 13 14 15 16

St 42,351 42,456 42,561 42,666

factors 0.81 1.95 1.21 0.38

Forecast 34,304 82,789 51,499 16,213

7 An exponential smoothing model is used to generate a forecast for period 13 (January of next year). The sales for January through April are used to initialize the model, and an  = 0.2 is used as the smoothing constant. The FORECAST module is used to generate the forecast. The results are summarized as follows.

Forecast, F13 Forecast error, SE

Region 1 219.73 26.89

Region 2 407.04 25.50

Region 3 303.30 17.54

Combined 938.26 61.41

Note that the sum of the forecasts by region nearly equals the forecast of the combined usage. However, whether a by-region forecast is better than an overall forecast that is disaggregated by region depends on the forecast error. The standard error of the forecast is the best indicator. A comparison of a bottoms-up forecast developed from regional forecasts to that of a forecast from combined data can be based on the law of variances. That is, if the usage rates within the regions are independent of each other, the estimate of the total error can be built from the individual regions and compared to that of the combined usage data. The total forecast error (variance) from the individual regions ST2 might be estimated as the weighted average of the variances as follows.

ST2 

F F1 2 F2 2 S E1  S E2  3 S E23 FC FC FC

where Fi = forecasts of each region FC = forecast based on combined data S E2 = variance of the forecast in each region i

S

2 T

= total variance of the forecast based on regional data

Therefore, 219.73 407.04 30330 . 26.89 2  25.50 2  17.54 2 930.07 930.07 930.07  0.236  723.07  0.438  650.25  0.326  307.65  555.74

S T2 

Then,

77

S T  555.74  2357 . Since ST < SC, it appears that a bottom-up, or regional, forecast will have a lower error than a top-down forecast. 9 (a) See the plot in Figure 8-1. It shows that there is a seasonal component with a very slight trend to the data as well as some random, or unexplained, variation. 300

Average monthly unit prices

250

200

150

100

50

Jly

Oct

Apr

Jan

Oct

Jly

Apr

Jan

Oct

Jly

Apr

Jan

Oct

Jly

Apr

Jan

Oct

Jly

Apr

Jan

0

Time, months

FIGURE 8-1 Plot of time series data for Problem 9

(b) A time series model typically will involve only two components: trend and seasonality. Using 2 years of data should be sufficient to establish an accurate trend line and the seasonal indices. We can develop the following table for computing a regression line and seasonal indices.

78

Prices, Pt 211 210 214 208 276 269 265 253 244 202 221 210 215 225 230 214 276 261 250 248 229 221 209 214 5,575

Time, t 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 300

Pt 211 420 642 832 1380 1614 1855 2024 2196 2020 2431 2520 2795 3150 3450 3424 4692 4698 4750 4960 4809 4862 4807 5136 69,678

t2 1 4 9 16 25 36 49 64 81 100 121 144 169 196 225 256 289 324 361 400 441 484 529 576 4,900

Trend,a Tt 232.4 232.4 232.4 232.4 232.2 232.3 232.3 232.3 232.3 232.3 232.3 232.2 232.3 232.2 232.3 232.3 232.2 232.2 232.2 232.2 232.2 232.2 232.2 232.2

Seasonal indexb 0.91 0.90 0.92 0.90 1.19 1.16 1.14 1.09 1.05 0.87 0.95 0.90 0.93 0.97 0.99 0.92 1.19 1.12 1.08 1.07 0.99 0.95 0.90 0.92

St

0.92 0.93 0.96 0.91 1.19 1.14 1.11 1.08 1.02 0.91 0.92 0.91

a

Computed from the trend regression line. For example, the period 1 trend is T1 = 232.39 - 0.0081 = 232.4. b The seasonal index is the ratio of the actual price to the trend for the same period. For example, the period 1 seasonal index is 211/232 = 0.91.

We also have N = 24, t = 300/24 = 12.5, and P = 5575/24 = 232.29. Now, b

 P  t  N  P  t  69,678  24( 232.29)(12.5)  0.008 4,900  24(12.5) t  N  t 2

2

2

and a  P t  b  t  232.29  ( 0.008)(12.5)  232.39 Therefore, the trend equation is:

79

Tt  232.29  0.008  t Note that the trend is negative for the last two years of data, even though the 5-year trend would appear to be positive. Now, computing the trend value Tt for each value of t gives the results as shown in the previous table. The seasonal index is a result of dividing Pt by Tt for each period t. The indices are averaged for corresponding periods that are one year apart. Forecasting into the 5th year shows the potential error in the method. That is, for January of the 5th year, the forecast is Ft = TtSt-12, or F25 = [232.39  0.00825][0.92] = 213.6. Repeating for each month, we have:

t 25 26 27 28 29 30 31 32 33 34 35 36 a

Actual price 210 223 204 244 274 246 237 267 212 211 188 188

Forecast Forecast price error 213.6 - 3.6 215.6 7.1 222.9 -18.9 211.3 32.7 276.3 - 2.3 264.6 -18.6 257.7 -20.7 250.7 16.3 236.8 -24.8 211.2 - 0.2 213.5 -25.5 211.2 -23.2 Total squared error

Squared error 13.0 50.4 357.2 1069.3 5.3 345.9 428.5 265.7 615.0 0.0 51.0 538.2 3,739.5

Revised seasonala 0.91

The seasonal index for period 25 is .90. The average of the seasonal index for period 25  12 = 13, and this period is (0.92 + 0.90)/2 = 0.91.

The standard error of the forecast is S F  3,739.5 / (12  2)  19.34 . Now, the forecast for period 37 would be: F37  ( 232.39  0.008  37)( 0.91)  21121 . (c) Using the exponential smoothing module in the FORECAST software, the forecast for the coming period is F = 201.26, with SF = 17.27. The smoothing constants given in the problem are the "best" that FORECAST could find. (d) Each model should be combined according to its ability to forecast accurately. We can give each a weight in proportion to its forecast error, or standard error of the forecast (SF). Hence, the following table can be developed.

80

(1) Model type Regression Exp. smooth. Total

Forecast error 19.34 17.27 36.61

(2) = (1)/36.61 Proportion of total error 0.528 0.472 1.000

(3)=1/(2) Inverse of error proportion 1.894 2.119 4.013

(4)=(3)/4.013 Model weights 0.472 0.528 1.000

Therefore, each of the model results is weighted according to the model weights. The weighted forecast for the January would be: (1) Model type Regression Exp. smooth.

(2)

Forecast Model weight 211.21 0.472 201.26 0.528 Weighted forecast

(3)=(1)(2) Weighted proportion 99.69 106.27 205.96

In a similar fashion, we can weight the forecast error variances to come up with a weighted forecast error standard deviation SFw. That is, S Fw  0.472  19.34 2  0.528  17.27 2  18.28 A 95% confidence band using the combined results might be constructed as: Y = 205.96  z18.28 where z is 1.96 for 95% of the area under the normal distribution. Y = 205.96  1.9618.28 Hence, we can be 95% sure that the actual price Y will be within the following range: 170.13  Y  241.79 10 The plot of the sales data is shown in Figure 8-2. The plot reveals a high degree of seasonality with a noticeable downward trend. A level-trend-seasonal model seems reasonable.

(b) Using the search capability within the FORECAST software, a Level-Trend-Seasonal form of the exponential smoothing model was found to give the lowest forecast error. A 14-period initialization and 6 periods to compute error statistics were used. The respective smoothing constants were  = 0.01,  = 0.08, and  = 0.60. This produced

81

a forecast for the period (January 2004) of F = 6,327.60 and a standard error of the forecast of SF = 1,120.81. 30000

Aggregate sales in 000s

25000

20000

15000

10000

Oct

Jly

Apr

Jan

Oct

Jly

Apr

Jan

Oct

Jly

Apr

Jan

Oct

Jly

Apr

Jan

Oct

Jly

Apr

0

Jan

5000

Time, months

FIGURE 8-2 Plot of Time Series Data for Hudson Paper Company

(c) Assuming that the forecast errors are normally distributed around the forecast, a 95% confidence band on the forecast is given by: Y = F + zSF Y = 6,327.60  1.961,120.81 where z = 1.96 for 95% of the area under the normal distribution curve. Therefore, we can be 95% sure that the actual sales Y should fall within the following limits: 4,130.8  Y  8,524.4 11 (a) For A569, the BIAS = 165,698 and the RMSE = 126,567 when using the 3-month moving average. However, if a level only exponential smoothing model with an  = 0.10, the BIAS drops to –9,556 and the RMSE is 118,689. The model fits the data better and there is a slight improvement in the forecasting accuracy. For A366, the BIAS = 18,231 and the RMSE = 144,973 when using the 3-month moving average. A level-trend-seasonal model offers the best fit, but it is suspect since the data show a high degree of random variability rather than seasonality. Overall, a simple level-only model is probably better in practice. The model has an  = 0.08, a BIAS = 3,227, and a RMSE = 136,256. This is an improvement over the 3-month moving average.

82

(b) Using the level-only models, the forecast for October for A569 = 193,230 and for A366 = 603,671. (c) The 3-sigma (99.7%) confidence band on the forecasts would be: For A569, Y = 193,230  3(118,689), or 0  Y  549,297. For A366, Y = 603,671  3(136,256), or 194,903  Y  1,012,439. The actual October usage falls within the 3-sigma confidence bands for each of these products. The difference of the actual from the forecast for each product is attributable to the substantial variability in the data, which is characteristic of purchasing in the steel processing industry.

83

WORLD OIL Teaching Note Strategy The purpose of this case study is to allow students to develop an appropriate forecasting model for some time series data. Discussion may begin with the nature of this productone with which most students should be very familiar. Based on the many available forecasting approaches, students should be encouraged to select several for consideration. In this note, both exponential smoothing and time series decomposition are evaluated. Both are appropriate here because (1) they can project from historical time series data, (2) they can handle seasonality, which appears to be present in the data, (3) there is enough data to construct and test the models, and (4) the forecast is for a short period into the future. Assistance with the computational aspects of this problem is available with the use of the FORECAST module in the LOGWARE software. Answers to Questions (1) Develop a forecasting procedure for this service station. Why did you select your method?

Both exponential smoothing and time series decomposition forecasting methods are tested using the FORECAST module in LOGWARE. For exponential smoothing, an initialization period of one seasonal cycle (52 weeks) plus two weeks are used for a total of 54 weeks, a minimum requirement in FORECAST. The last 30 weeks of data is used for computing the error statistics. This number of periods is arbitrary, but seems reasonably large so as to give stable statistical values. We wish to minimize the forecast error over time, and FORECAST computes both MAD and RMSE statistics that can be used to make comparisons among model types. Testing the various exponential smoothing model types and the time series gives the following statistics. Smoothing constants

Model type Level only.... Level-trend... Level-seasonal Level-trendseasonal...... TS decomp.....

 .4 .2 .3

 .5 

.01 .2  

  1.0

MAD 37.82 45.85 38.97

BIAS -5.27 7.13 11.30

RMSE 67.61 67.80 45.71

.4 

30.27 59.46

-6.05 37.18

44.17 71.85

Forecast week 6 of this year 817.35 860.26 648.75 770.74 731.33

The MAD and RMSE statistics show how well the forecast has been able to track historical fuel usage rates. They are an indication of the accuracy of the forecasting process in the future on the average. We favor forecasting methods that can minimize these statistics. In this case, the Level-Trend-Seasonal version of the exponential

84

smoothing model seems to do this best. Both MAD and RMSE are the lowest for this model type among the alternatives. Further evidence of the performance of a forecasting method is obtained from a plot of the forecast against the actual usage rates. This is shown in Figure 1. Note that the Level-Trend-Seasonal model tracks the usage rates quite well, especially in the more recent weeks. The modeling process has likely stabilized in the last 30 weeks of the data and is now tracking quite well.

FIGURE 1

Fit of Level-Trend-Seasonal Exponential Smoothing Model to Fuel Usage Data on Mondays of the Week

(2) How should the periods of promotions, holidays, or other periods where usage rates deviate from normal patterns, be handled in the forecast? If the deviations occur at the same time within the seasonal cycle and with the same relative intensity, no special procedures are required. The adaptive characteristic of the exponential smoothing process will automatically incorporate these deviations into the forecast. However, when the deviations are not regular, as promotions may be timed irregularly, they may best be handled as outliers in the time series and eliminated from the time series. The model may be fit without the outliers, and then the effect of them treated as modifications to the forecast. These modifications can be handled manually.

85

(3) Forecast next Monday's fuel usage and indicate the probable accuracy of the forecast. From the Level-Trend-Seasonal exponential smoothing model developed in question 1, where the smoothing constants are  = 0.01,  = 0.2, and  = 0.4, the forecast for Monday of week 6 would be 771 gallons. However, this forecast only represents the average fuel usage. Determining the accuracy of the forecast requires that the forecast track the mean of the actual usage, i.e., a bias of 0, and that the forecast errors be normally distributed. While the BIAS (sum of the forecast errors over the last 30 weeks) is not exactly 0, and will not likely ever be so, it is low (-6.05), such that we will assume good tracking by the forecast model. A histogram of the forecast errors can reveal whether they follow the familiar bell-shaped pattern. Such a histogram is given below. We can conclude that while the errors are not precisely normally distributed, we cannot reject the idea that they did not come from a normally distributed population. A goodness-of-fit test could be used to check this assumption. Although this test is not performed here, it is quite forgiving, such that the normal distribution of errors assumption is not likely to be rejected where the data show a reasonably normal distribution pattern. The distribution here qualifies. We can now proceed with developing a 95% confidence band around the forecast. The forecast of the actual fuel usage rate Y will be: F  z ( F )  Y  F  z ( F )

where  F is the standard error of the forecast. F is the forecast, and z is the number of standard deviations for 95% of the area under a normal distribution. FORECAST computes the root mean squared error (RMSE) as: N

(A

t 

RMSE =

Ft ) 2

t =1

N

86

HISTOGRAM FOR FORECAST ERROR OF LAST 30 WEEKS Class Width = 20.0000 Number of Classes = 10 MID CLASS < -80.0000 -70.0000 -50.0000 -30.0000 -10.0000 10.0000 30.0000 50.0000 70.0000 90.0000 110.0000 >= 120.0000

0% 50% 100% +----+----+----+----+----+----+----+----+----+----+ | | |****** | |*** | |******** | |******** | |***** | |******** | |****** | |* | |* | | | | | +----+----+----+----+----+----+----+----+----+----+

Since RMSE is uncorrected for degrees of freedom lost, we apply a correction factor (CF) as a multiplier to RMSE to get the unbiased estimate of the standard error of the forecast ( ˆ F ): CF =

N N-n

where n is the number of degrees of freedom lost in the model building process. We estimate n to be the number of smoothing constants in the model, or three in this case. Hence,

 F  RMSE  CF 30 30  3 .  1054 .  4417 .  4417

 46.56

Now, with z@95% = 1.96 from a normal distribution table, we can be 95% confident that the true 87-octane fuel usage Y on Monday of week 6 will be: 771  1.96(46.56) < Y < 771 + 1.96(46.56) 680 < Y < 862 gallons

87

METRO HOSPITAL

You are the materials manager at Metro Hospital. Approximately one year ago, the hospital began stocking a new drug (Ziloene) that helps the healing process for wounds and sutures. It is your responsibility to forecast and order the monthly supply of Ziloene. The goal is to minimize the combined cost of overstocking and understocking the drug. Orders are placed and received at the beginning of the month and demand occurs throughout the month. The following demand and cost data have been compiled. Costs. If more is ordered than is demanded, a monthly holding cost of $1.00 per case is incurred. If less is ordered than is demanded, a $2.00 per case lost sales cost is incurred. The drug has a short shelf life, and any overstocked product at the end of the month is worthless and no longer available to meet demand. Demand. The demand for the twelve months of last year was: Month Cases

1 43

2 36

Last year's demand 3 4 5 6 7 8 24 69 34 75 90 67

9 59

10 51

11 77

12 50

You believe this demand to be representative of Metro's normal usage pattern.

FIGURE 1 Plot of last year's monthly demand in cases

88

Decision Worksheet Month 1 (13) 2 (14) 3 (15) 4 (16) 5 (17) 6 (18) 7 (19) 8 (20) 9 (21) 10 (22) 11 (23) 12 (24) Total

Month 1 (25) 2 (26) 3 (27) 4 (28) 5 (29) 6 (30) 7 (31) 8 (32) 9 (33) 10 (34) 11 (35) 12 (36) Total

Cases ordered

Actual demand

Over @ $1/case

Short @ $2/case

Cost, $

Cases ordered

Actual demand

Over @ $1/case

Short @ $2/case

Cost, $

89

METRO HOSPITAL Exercise Note Purpose Metro Hospital is an in-class exercise designed to illustrate the relationship between good forecasting and the control of inventory related costs. It shows that accurate forecasting is a primary factor in minimizing inventory costs. Participants in this exercise use a variety of methods, often intuition, to forecast demand and to come up with a purchase quantity. Their performance is measured as over- or understock costs. Using a simple exponential smoothing forecasting model and an understanding of the standard deviation of the forecast, an effective purchase plan can be constructed. This process results in costs that are significantly lower than the majority of the participants are able to achieve using intuitive methods. istration The descriptive material and the decision worksheet are to be distributed to the class at the time that the exercise is conducted. To hand out the material ahead of time may take away much of the drama from the exercise. About one half hour should be scheduled for running the exercise. The instructor asks the class to make a decision regarding the size of the order to be placed in the period and to record it on the worksheet. The participants are then informed of the demand for that period from Table 1 after the simulated time of one month has ed. Given that they now know the actual demand for the period, the participants are asked to record their costs and then to place an order for the next period. The pattern is repeated for at least twelve months, a full seasonal cycle. The participants are asked to sum their costs and to report them to the exercise leader. They are displayed in a public place, such as a chalkboard, for all to see. Then, the exercise leader announces his or her cost level that was achieved using a disciplined approach using a simple forecasting procedure and some basic statistics. TABLE 1 Actual Demand for Period 13 Through 36 Period 13 14 15 16 17 18 19 20 Demand 47 70 55 38 90 24 65 65

21 23

22 55

23 85

24 66

Period Demand

33 45

34 70

35 50

36 56

25 53

26 64

27 61

28 63

29 65

30 38

31 80

32 88

Quantitative Analysis The demand series was generated using a normal distribution with a reasonably high variance and a very slight upward trend. To illustrate the use of a quantitative approach to forecasting, an exponential smoothing model was selected, although other methods such as time series decomposition would also be appropriate. The twelve historical data points were submitted to the FORECAST module in LOGWARE. A 3-month initialization period and a 3-month time period for computing error statistics were chosen. The smoothing constants for the level, level-trend, and level-trend-seasonal models were examined. Based on the root mean squared error (RMSE), the best model

90

was the level-trend-seasonal (RMSE = 13.59), but the level model with  = 0.19 and RMSE = 13.89 performed very well and is used here. The model is: Ft 1  0.19 At  0.81Ft

where Ft 1  forecast for next period t  1 At  actual demand for current period t Ft  forecast for current period t LOGWARE gives a forecast value of 58.1 and this is used as the forecast value for period 13. Applying this simple, level only model to the second year demand as it is revealed in each period gives the following forecast values. TABLE 2 Simple Exponential Smoothing Forecast Values for the Next Year Actual Period demand Forecast 13 47 58.1 14 70 56.0 15 55 58.7 16 38 58.0 17 90 54.2 18 24 61.0 19 65 54.0 20 65 56.1 21 23 57.8 22 55 51.2 23 85 51.9 24 66 54.6

Now we must determine the order quantity. It can be calculated from QF  z(RMSE) t 1

Recall the RMSE was 13.89 for this model. To be precise, we calculate z by trial and error. The following order quantity and cost computations can be made for a z value of 0.8 (Table 3).

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TABLE 3

Period 13 14 15 16 17 18 19 20 21 22 23 24 *

Purchase Order Quantity and Associated Inventory Costs Actual Order Units Units demand Forecast quantity over short 47 58.1 69* 22 70 56.0 67 3 55 58.7 70 15 38 58.0 69 31 90 54.2 65 25 24 61.0 72 48 65 54.0 65 65 56.1 67 2 23 57.8 69 46 55 51.2 62 7 85 51.9 63 22 66 54.6 66

Cost, $ 22 6 15 31 50 48 0 2 46 7 44 0 271

Q = 58.1 + 0.8(13.89) = 69.21, or 69

We see from the following graph (Figure 1) that z = 0.8 is optimal. 310 305 300

Cost, $

295 290 285 280 275 270 265 0

0.2

0.4

0.6

0.8

1

1.2

1.4

Z

FIGURE 1 Plot of Total Annual Costs Against the Factor z

Figure 2 graphically shows the good purchase pattern of Table 3.

92

100 Order quantity

90 80

Cases

70 60 50 40

Demand Forecast

30 20 10 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 Time period FIGURE 2 Plot of Forecast and Purchase Order Quantity on Product Demand Summary The exercise leader should discuss that one of the problems with intuitively forecasting demand is overreacting to randomness in the demand pattern. This has the effect of causing extreme over and short costs in inventories. A model for short term forecasting that is integrated into the purchasing and inventory control process can help to avoid these extremes and give lower costs. Several forecasting models may perform well, such as exponential smoothing, a simple moving average, a regression model, or a times series decomposition model. One of the most practical for inventory control purposes is the exponential smoothing model. The results from a simple, level only model were illustrated above using the same information that was available to the participants. Recognizing that it is less costly to order too much than to order too little, the purchase quantity should exceed the forecast by some margin. The astute participant will likely approximate the standard deviation of demand from the range of the demand values, that is,  = (Max - Min)/6. Then, one or two  might be used to add a margin of safety to the forecast and size of the purchase order. This simple approximation procedure can lead to reasonable results.

93

CHAPTER 9 INVENTORY POLICY DECISIONS 1 The probability of finding all items in stock is the product of the individual probabilities. That is,

(0.95)(0.93) (0.87) (0.85) (0.94) (0.90) = 0.55 2 (a) The order fill rate is the weighted average of filling the item mix on an order. We can setup the following table.

(1)

(2) Frequency Order Item mix probabilities of order 0.20 1 .95.95.95.90.90 = .69 2 0.15 .95.95.95 = .86 3 0.05 .95.95.90.90 = .73 4 0.15 .95.95.95.95.95.90.90 = .62 5 0.30 .95.95.90.90.90.90 = .59 6 0.15 .95.95.95.95.95 = .77 Order fill rate

(3)=(1)(2) Marginal probability 0.139 0.129 0.037 0.094 0.178 0.116 0.693

Since 69.3% < 92%, the target order fill rate is not met. (b) The item service levels that will give an order fill rate of 92% must be found by trial and error. Although there are many combinations of item service levels that can achieve the desired service level, a service level of 99% for items A, B, C, D, E, and F, and 97% to 98% for the remaining items would be about right. The order fill rates can be found as follows. (1) Order 1 2 3 4 5 6

Item mix probabilities (.99)3(.975)2 = .922 (.99)3 = .970 (.99)2(.975)2 = .932 (.99)5(.975)2 = .904 (.99)2(.975)4 = .886 (.99)5 = .951

(2) Frequency of order 0.20 0.15 0.05 0.15 0.30 0.15 Order fill rate

(3)=(1) (2) Marginal probability 0.184 0.146 0.047 0.136 0.266 0.143 0.922

3

94

This is a problem of push inventory control. The question is one of finding how many of 120,000 sets to allocate to each warehouse. We begin by estimating the total requirements for each warehouse. That is, Total requirements = Forecast + zForecast error From Appendix A, we can find the values for z corresponding to the service level at each warehouse. Therefore, we have:

Warehouse 1 2 3 4 Total

(1) Demand forecast, sets 10,000 15,000 35,000 25,000 85,000

(2) Forecast error, sets 1,000 1,200 2,000 3,000

(3) Values for z 1.28 1.04 1.18 1.41

(4)=(1)+(2)(3) Total requirements, sets 11,280 16,248 37,360 29,230 94,118

We can find the net requirements for each warehouse as the difference between the total requirements and the quantity on hand. The following table can be constructed.

Warehouse 1 2 3 4

(1) Total requirements 11,280 16,248 37,360 29,230 94,118

(2)

(3)=(1)(2)

(4)

(5)=(3)(4)

On hand quantity 700 0 2,500 1,800

Net requirements 10,580 16,248 34,860 27,430 89,118

Proration of excess 3,633 5,450 12,716 9,083 30,882

Allocation 14,213 21,698 47,576 36,513 120,000

There is 120,000  89,118 = 30,882 sets to be prorated. This is done by assuming that the demand rate is best expressed by the forecast and proportioning the excess in relation to each warehouse's forecast to the total forecast quantity. That is, for warehouse 1, the proration is (10,000/85,000)30,882 = 3,633 sets. Prorations to the other warehouses are carried out in a similar manner. The allocation to each warehouse is the sum of its net requirements plus a proration of the excess, as shown in the above table. 4 (a) The reorder point system is defined by the order quantity and the reorder point quantity. Since the demand is known for sure, the optimum order quantity is: Q *  2 DS / IC  2(3,200 )(35) / ( 015 . )(55)  164.78, or 165 cases

The reorder point quantity is:

95

.  92 units ROP  d  LT  (3,200 / 52)  15 (b) The total annual relevant cost of this design is: TC  D  S / Q  I  C  Q * / 2 . )(55)(164.78 ) / 2  (3,200)(35) / 164.78  ( 015  679.69  679.97  $1,359.66 (c) The revised reorder point quantity would be:

ROP  (3,200 / 52)  3  185 units . The ROP is greater than Q*. It is possible under these circumstances the reorder quantity may not bring the stock level above the ROP quantity. In deciding whether the ROP has been reached, we add any quantities on order or in transit to the quantity on hand as the effective quantity in inventory. Of course, we start with an adequate in-stock quantity that is at least equal to the ROP quantity. 5 (a) The economic order quantity formula can be used here. That is, Q *  2 DS / IC  2(300)(8,500 ) / ( 010 . )(8,500 )  77.5, or 78 students

(b) The number of times that the course should be offered is: . or about 4 times per year N *  D / Q *  300 / 77.5  39, 6 This is a single-period inventory control problem. We have:

Revenue = $350/unit Profit = $350  $250 = $100/unit Loss = 0.2250 = $50/unit Therefore, n 

100  0.667 100  50

Developing a table of cumulative frequencies, we have:

96

Quantity 50 55 60 65 70 75

Frequency 0.10 0.20 0.20 0.30 0.15 0.05 1.00

Cumulative frequency 0.10 0.30 0.50 0.80 Q* 0.95 1.00

n lies between quantities of 60 and 65. We round up and select 65 as the optimal purchase order size. 7 This question can be treated as a single-order problem. We have:

Revenue = 1 + 0.01 = $1.01/$ Cost/Loss = 0.10(2/365) = $0.00055/$ which is the interest expense for 2 days Profit = 1.01  1.00055 = $0.00945/$ and n 

0.00945  0.945 0.00945  0.00055

For an area under the normal curve of 0.945 (see Appendix A), z = 1.60. The planned number of withdrawals is: Q* = D + z  D = 120 + 1.60(20) = 152.00 The amount of money to stock in the teller machine over 2 days would be: Money = Q*75 = 152.0075 = $11,400 8 This is a single-period inventory control problem.

(a) We have: Profit = 400  320 Loss = 320  300 Then,

97

n 

400  320  0.80 ( 400  320)  (320  300)

We now need to find the sales that correspond to a cumulative frequency of 0.80. In the following table:

Sales 500 750 1,000 1,250 1,500

Frequency 0.2 0.2 0.3 0.2 0.1 1.0

Cumulative frequency 0.2 0.4 0.7 0.9 Q* 1.0

Q* lies between 1,000 and 1,200 in the cumulative frequency table. We choose to roundup to Q* = 1,250 units. (b) Carrying the excess inventory to next year, n 

80  0.556 80  ( 0.2  320)

where the loss is the cost of holding a unit until the next year. The Q* now lies between 750 and 1,000 units. We choose 1,000 units. Holding the excess units means a potential loss of 0.2320 = $64/unit, whereas discounting the excess units represents a loss of only 320  300 = $20/unit. Therefore, Cabot will need fewer units if they are held over in inventory. 9 (a) The optimum order quantity is: Q *  2 DS / IC  2(1,250)(52)( 40) /(0.3)(56)  556 cases

and the reorder point quantity is: ROP  d  LT  z  sd' where sd'  sd LT  475 2.5  751 and z P0.80 = 0.84.

98

Now,

ROP  (1,250)( 2.5)  ( 0.84)( 751)  3,756 cases Policy: When the amount of inventory on hand plus any quantities on order or in transit falls below ROP, reorder an amount Q*. (b) For the periodic review system, we first estimate the order review time: T *  Q * / d  556 / 1,250  0.44 weeks The max level is: M *  d  (T *  LT )  z  sd' where sd' now is: sd'  sd T *  LT  475 0.44  2.5  814 cases Hence, M *  1,250( 0.44  2.5)  0.84(814 )  4,359 cases Policy: Find the amount of stock on hand every 0.44 weeks and place a reorder for the amount equal to the difference between the quantity on hand plus on order and the max level (M*) of 4,359 cases. (c) The total annual relevant cost for these policies is: TC  DS / Q  ICQ / 2  ICzsd'  kDs d' E ( z ) / Q

For the reorder point system: TCQ = 1250(52)(40)/556 + .3(56)(556)/2 + .3(56)(.84)(751) + 10(1250)(52)(751)(.1120)/556 = 4,676.26 + 4,670.40 + 10,598.11 + 98,332.37 = $118,277.14 For the periodic review system: T = 1250(52)(40)/556 + .3(56)(556)/2 99

+ .3(56)(.84)(814) + 10(1250)(52)(814)(.1120)/556 = 4,676.26 + 4,670.40 + 11,487.17 + 106,581.29 = $127,415.12 (d) The actual service level achieved is given by: SL  1 

sd' E( z ) Q

For the reorder point system: SLQ  1 

751( 01120 . )  1  015 . 556

or demand is met 85% of the time. For the periodic review system: SLP  1 

814( 01120 . )  1  016 . 556

or demand is met 84% of the time. (e) This requires an iterative approach as follows: Compute Q  2 DS / IC

Compute P  1  QIC / Dk , then z, then E (z)

Compute Q  2 D( S  ksd' E( z ) ) / IC Go back and stop when there is no change in either P or Q After the initial value of Q = 556.3, the process can be summarized in tabular form.

100

Step 1 2 3 4 5 6

Q 778.4 860.0 889.9 899.6 902.8 902.8

P 0.9856 0.9799 0.9778 0.9777 0.9767 0.9767

z 2.19 2.06 2.01 2.00 1.99 1.99

E(z) 0.0050 0.0072 0.0083 0.0085 0.0087 0.0087

Now, for P = 0.9767, z = 1.99 ROP = 1,250(2.5) + 1.99(751) = 4,620 cases and the total relevant cost is: TCQ  DS/Q  ICQ/2  ICzs'd  kDs'd E(z) /Q  65,000(40) / 902.8  0.3(56)(902.8) / 2  0.3(56)(1.99)(751)  10(65,000)(751)(0.0087) / 902.8  $40,275

This is considerably less than the $118,277.14 for the preset P at 0.80. If you solve this problem using INPOL, you will get a slightly different answer. That is, Q* = 858. This simply is because z is carried to two significant digits rather than the 4 significant digits used in the above calculations. 10 Refer to the solution of problem 10-9 for the general approach.

(a) Q* = 556.3 cases and 2 ROP  d  LT  z LT  sd2  d 2  sLT

 1,250( 2.5)  0.84 2.5  4752  1,250 2  0.52  3125  0.84(977.08 ) ,  3,946 cases

(b) An approximation for T* = Q*/d, or T* = 556/1,250 = 0.44 weeks and approximating sd' as

101

2 sd'  (T *  LT )sd2  d 2  sLT

 ( 0.44  2.5)( 4752 )  1,250 2 ( 0.5) 2  1,027 cases

So, Max  d (T *  LT )  z  sd'  1,250( 0.44  2.5)  0.84(1,027)  4,537 cases

(c) According to INPOL, TC = 4,686 + 4,686 + 128,195 + 13,862 = $151,429 Q

TC = 4,686 + 4,686 + 134,751 + 14,571 = $158,694 P

(d) According to INPOL, SL = 80.28% Q

SL = 79.27% P

(e) According to INPOL, Q* = 930 cases, ROP = 5,128 cases, TC = $49,532, SLQ = 99.22% Q

T* = 0.76 weeks, MAX = 6,257 cases TC = $52,894, SLP = 99.18% P

11 (a) The production run quantity is:

Q *p 

2 DS p   IC pd

2(100)( 250)( 250) 300   1,000 units 0.25(75) 300  100

(b) The production run cycle is: Q *p / p  1,000 / 300  3.33 days

102

(c) The number of production runs is: D / Q *p  100( 250 ) / 1,000  25 runs per year

12 (a) The order quantity is: Q *  2 DS / IC  2( 2,000 )( 250 )(100 . ) / ( 0.30)(35)  309 valves

and the reorder point quantity is:

ROP  d  LT  z  sd LT but sd  0 . Therefore,

ROP  d  LT  ( 2,000 / 8)(1)  250 valves (b) Boxes are set up that contain 309 valves—the optimum order quantity. When an order arrives from a supplier, 250 valves are set aside in a separate box and are treated as the backup stock. The residual 309  250 = 59 valves are used on the production line. When the 59 valves at the production line are used up, the backup box containing 250 valves is brought to the production line and the empty box is sent to the supplier for refilling. When the order arrives one hour later, there will be 0 valves remaining at the production line. Then, 250 valves are set aside and 59 are sent to the production line. The cycle is then repeated. This problem is similar to the KANBAN system. Lead times are very short so that lead times are virtually certain. Demand is certain, since it is fixed by the production schedule. Boxes or cards are used to assure movement of the most economic quantity. KANBAN is essentially classic economic reorder point inventory control under certainty. 13 (a) The economical quantity of cars to be called for at a time is found by the economic order quantity formula: Q *  2 DS / IC  2( 40 )(52 )(500 ) / ( 0.25)(90,000)(30) / 2,000  78.5, or 79 cars

(b) This is the reorder point quantity:

ROP  d  LT  z  sd LT where z = 1.28 from Appendix A for an area under the curve equal to 0.90. Therefore,

103

ROP  40(1)  128 . (333 . ) 1  44.3 cars, or 44.3(90,000 / 2,000) = 1,994 tons of soda ash 14 (a) This is a reorder point design under conditions of uncertainty for both demand and lead-time. We assume that the probability of an out of stock is given. Therefore, the order quantity is: Q *  2 DS / IC  2(50)(365)(50) / ( 0.30)( 45)  367.7 units

and

ROP  d  LT  z  sd' where

z = 1.04 (see Appendix A) for the area under the curve equal to 0.85 and 2 sd'  sd2 LT  d 2 sLT  152 ( 7)  (50 2 )( 2 2 )  107.6 units

Therefore,

ROP  50( 7)  104 . (107.6)  4619 . units (b) This is the periodic review system design under uncertainty. The complexity requires us to make some approximations here. The time interval for review of the stock level is:

T *  Q * / d  367.7 / 50  7.35 days The MAX level is:

MAX  d (T *  LT )  z  sd' where z = 1.04 and sd' is approximated as: 2 sd'  (T *  LT )( sd2 )  d 2 ( sLT )

 ( 7.35  7)(152 )  50 2 ( 2 2 )  115.0 units

104

Therefore,

MAX = 50(7.35 + 7) + 1.04(115.0) = 837.1 units (c) Since the service level is specified, the probability is not set at the optimum level. Knowing the out-of-stock cost allows us to find the most appropriate service level. Since this is an iterative process, we use INPOL to carry out the calculations. The optimized service level yields a reorder point design of

Q* = 410 units and ROP = 571 units and the total relevant cost drops from $12,642 in part a to $8,489. The demand in stock in part a was 97.74%, and it now increases to 99.81%. 15 (a) Find the common review time:

T *  2(O   s I ) / I  Ci Di  2(100  0) /[(0.3 / 52)( 2.25  2,000  1.90  500) ]  2.5 weeks Then, M A*  d A (T *  LT )  z A  sd A T *  LT

where z = 1.28 for P = 0.90 A

M A*  2,000( 2.5  1.5)  1.28(100) 2.5  1.5  8,256 units and

M B*  500( 2.5  15 . )  0.842( 70) 2.5  15 .  2,118 units where zB = 0.842 for P = 0.80. The control system works as follows: the stock levels of both items are reviewed every 2.5 weeks. The reorder size for A is the difference between the amount on hand (includes on-order) and 8,256 units. The reorder size for B is the difference between the amount on hand (includes on-order) and 2,118 units. (b) The average amount in inventory is expected to be:

105

AIL  d  T * / 2  z  sd T *  LT For A: AILA  2,000( 2.5) / 2  128 . (100) 2.5  15 .  2,756 units For B: AILB  500( 2.5) / 2  0.842( 70) 2.5  15 .  743 units (c) The service level is given by: SL  1  sd'  E( z ) / d  T *

For A: SL A  1  100 2.5  15 . ( 0.0475) / 2,000( 2.5)  0.998 For B: SLB  1  70 2.5  15 . ( 01120 . ) / 500( 2.5)  0.987 (d) We set T* = 4 and cycle through the previous calculations. Thus, we have: M A*  11,301 units

M B*  2,888 units

AILA = 4,301

AILB = 1,138

SLA = 0.999

SLB =0 .991

16 This problem is one of comparing the combined cost of transportation and in-transit inventory. In tabular form, we have the following annual costs:

106

Cost type Transportation

Formula RD

In-transit inventory

ICDT/365

Rail Truck 6(40,000)(1.25) 11(40,000)(1.25) = $300,000 = $550,000 0.25( 250)( 40,000)( 21) 0.25( 250)( 40,000)(7) 365 365 = $143,836 = $47,945 $597,945 $443,836

Total Select rail.

17 The two transport options from the consolidation point are diagrammed in Figure 9-1. Whether to choose one mode other the other depends more than transportation costs alone. Because the transport modes differ in the time in transit, the cost of the money tied up in the goods while in transit must be considered in the choice decision. This inICDt transit inventory cost is estimated from . The following design matrix can be 365 developed.

Cost type Transportation In-transit inventory *

Method RD ICDt/365 Total

Air $180,800 3,447* $184,247

Ocean $98,800 34,467 $133,267

ICDt/365 = 0.17(185)(20,000)(2)/365= 3,447

Ocean appears to be the lowest cost option even when a substantial in-transit inventory cost is included. The ocean option assumes that the trucking cost to move the goods from the consolidation point to the Port of Baltimore is included in the ocean carrier rate. FIGURE 9-1 The Consolidation Operation for a Hydraulic Equipment Manufacturer

Multiple sourcing points

Consolidation point

Baltimore

20 days 2 days Sao Paolo

107

18 The demand pattern is definitely lumpy, since s = 327 > d = 169. To develop the minmax system of inventory control, we first find Q*. That is, d

Q *  2 DS / IC  2(169 )(12 )(10) / 0.20( 0.96  0.048)  448.5 units

The ROP is ROP  d  LT  z  sd'  ED where z = 1.04 from Appendix A, ED = 8 unitsthe average daily demand rate, and 2 sd'  sd2 LT  d 2 sLT

 327 2 ( 4)  169 2 ( 0.8 2 )  667.8 units

So, ROP = 169(4) + 1.04(667.8) + 8 = 1,378.5 units The max level is: M* = ROP + Q*  ED = 1,378.5 + 448.5  8 = 1,819 units 19 (a) The basic relationship is:

IT  I i n

108

We know that I = $5,000,000. If there are 10 warehouses, the amount of inventory in a single one would be: T

I1  I T / 10  5,000,000 / 3162 .  1,581139 , The inventory in all 10 warehouses would be $1,581,13910 = $15,811,390. (b) The inventory in a single warehouse would be: I T  1,000,000 9  3,000,000 In each of 3 warehouses, we would have: I  3,000,000 / 3  $1,732,051 and in all 3 warehouses, we would have $1,732,0513 = $5,196,152. 20 (a) The turnover ratio is the annual demand (throughput) divided by the average inventory level. These ratios for each warehouse and for the total system are shown in the table below.

Warehouse 21 24 20 13 2 11 4 1 23 9 18 12 15 14 6 7 22 8 17 16

Annual warehouse thruput 2,586,217 4,230,491 6,403,349 6,812,207 16,174,988 16,483,970 17,102,486 21,136,032 22,617,380 24,745,328 25,832,337 26,368,290 28,356,369 28,368,270 40,884,400 43,105,917 44,503,623 47,136,632 47,412,142 48,697,015

Average inventory level 504,355 796,669 1,009,402 1,241,921 2,196,364 1,991,016 2,085,246 2,217,790 3,001,390 2,641,138 3,599,421 2,719,330 4,166,288 3,473,799 5,293,539 6,542,079 2,580,183 5,722,640 5,412,573 5,449,058

Turnover ratio 5.13 5.31 Avg. = 5.59 6.34 5.49 7.36 8.28 8.20 9.53 7.54 9.37 7.18 9.70 6.81 8.17 7.72 6.59 17.25 8.24 8.76 8.94 109

10 19 3 5

57,789,509 75,266,622 78,559,012 88,226,672 818,799,258

6,403,076 7,523,846 9,510,027 11,443,489 97,524,639

9.03 10.00 8.26 7.71 8.40

Avg. = 8.66

The overall turnover ratio is 8.40. Ranking the warehouses by throughput and averaging turnover ratios for the top 3 and the bottom 3 warehouses shows that the lowest volume warehouses have a lower turnover ratio (5.59) than the highest volume warehouses (8.66). There are several reasons why this may be so: 

The larger warehouses contain the higher-volume items such as the A items in the line. These may carry less safety stock compared with the sales volume. Conversely, the low-volume warehouses may have more dead stock in them.



There may be start-up (fixed) stock in the warehouses, needed to open them, that becomes less dominant with greater throughput.

(b) A plot of the inventory-throughput data is shown in Figure 10-1. A linear regression line is also shown fitted to the data. The equation for this line is: Inventory = 200,168 + 0.1132Throughput 12

Average inventory level, $ (Millions)

10

8 Estimating line

6

4

2

0 0

20

40

60

80

100

Annual warehouse thruput, $ (Millions)

FIGURE 10-1 Plot of Inventory and Warehouse Thruput for California Fruit Growers’ Association

110

(c) The total throughput for the three warehouses is: Warehouse 1 12 23 Total

Throughput $21,136,032 26,368,290 22,617,380 $70,121,702

Using this total volume and reading the inventory level from Fig. 10-1 or using the regression equation, we have: Inventory = 200,168 + .01132(70,121,702) = $8,137,945 (d) Warehouse 5 has a throughput of $88,226,672. Splitting this throughput by 30% and 70%, we have: 0.3088,226,672 = 26,468,002 0.7088,226,672 = 61,758,670 88,226,672 Estimating the inventory for each of the new warehouses using the regression equation, we have: Inventory = 200,168 + 0.113226,468,002 = $3,196,346 and Inventory = 200,168 + 0 .113261,758,670 = $7,191,249 for at total inventory in the two warehouses of $10,387,595 21 The order quantity for each item when there is no restriction on inventory investment is:

Q *  2 DS / IC We first find the unrestricted order quantities. Q A*  2(51,000)(10) / 0.25(17 . )  1,527 units QB*  2( 25,000 )(10) / 0.25(3.25)  784 units QC*  2(9,000)(10 ) / 0.25( 2.50)  537 units

111

The total inventory investment for these items is: IV  C A ( Q A / 2)  CB ( QB / 2)  CC ( QC / 2) . (1,527 / 2)  3.25( 784 / 2)  2.50(537 / 2)  175 .  $3,28138 Since the total investment limit is exceeded, we need to revise the order quantities. For each product: Q *  2 DS / [C( I   )]

For product A: Q A*  2(51,000)(10) / [175 . ( 0.25   )]

For product B: QB*  2( 25,000 )(10) / [3.25( 0.25   )]

For product C: QC*  2(9,000)(10 ) / [2.50( 0.25   )]

Now, the investment limit must be respected so that:

3,000  C A ( QA / 2)  CB ( QB / 2)  CC ( QC / 2) Expanding we have: 3,000  175 . 2(51,000 )(10) / [175 . ( 0.25   )] 3.25 2( 25,000 )(10) / [3.25( 0.25   )] 2.50 2(9,000 )(10 ) / [2.50( 0.25   )]

We now need to find an  value by trial and error that will satisfy this equation. We can set up a table of trial values.

112

Investment in



A 1,262.44 1,240.48 1,229.92 1,221.67 1,219.63 1,129.16

0.03 0.04 0.045 0.049 0.05 0.10

B 1,204.53 1,183.58 1,173.51 1,165.63 1,163.69 1,077.36

C 633.87 622.84 617.54 613.40 612.37 566.95

Total inventory value, $ 3,100.84 3,046.90 3,020.97 3,000.70 2,995.69 2,773.47

When the term I+ is the same for all products, as in this case,  may be found directly from Equation 10-30. We can substitute the value for  = 0.049 into the equation for Q* and solve. Hence, we have: Q A*  2(51000)(10) / [175 . ( 0.25  0.049)]  1,396 units QB*  2( 25,000 )(10) / [3.25( 0.25  0.049)]  717 units QC*  2(9,000)(10) / [2.50( 0.25  0.049 )]  491 units

Checking: 1.75(1,396)/2 + 3.25(717)/2 + 2.50(491)/2 = $3,000 22 We first check to see whether truck capacity will be exceeded. Since three items are to be placed on the truck at the same time, the items are tly ordered. The interval for ordering follows Equation 10-22, or: 2( O   S i )

T* 

I  Ci Di



2( 60  0 ) 0.25[50(100 )(52)  30(300)(52)  25( 200)(52)]

120  0.022 years, or 1.144 weeks 0.25(988,000 )



Now, from

DT i

*

wi  Truck capacity

i

[100(70) + 300(60) + 200(25)][1.144] = 34,320 lb.

113

The truck capacity of 30,000 lb. has been exceeded, and the order quantity or the order interval must be reduced. Given the revised Equation 10-30, the increment to add to I can be found. That is,





2O  Truck capacity      Di wi 

I

2

C D i

i

2( 60 ) 2

  30,000   50(10)(52)  30(30)(52)  25( 20)(52)  [100(52 )( 70 )  300(52)( 60)  200(52)(10)] 120 2

 30,000    (988,000)  2,340,000 

 0.25

 0.73895  0.25  0.48895

Revise T*, the order interval by: T*  

2( O   S i )

( I   ) Ci Di



2( 60  0) ( 0.25  0.48895)[50(100)(52 )  30(300)(52 )  25( 200)(52)]

120  0.01282 years, or 0.6667 weeks 0.73895(988,000)

Once again, we check that the truck capacity has not been exceeded. [100(70) + 300(60) + 200(25)][0.66667] = 30,000 lb. Therefore, place an order every 4.7, or approximately 5 days. 23 The average inventory for each item is given by:

Q* AIL   z  sd' 2 2 DS . z@ 95% = 1.65 from the normal IC distribution in Appendix A. The results of these computations can be tabulated.

where sd'  sd LT and Q* is found by Q * 

114

 0.25

sd' Q*

AIL

A 7.75 188.38 106.98

B 15.49 238.28 144.70

C 19.36 421.23 242.56

D 11.62 361.98 200.16

E 27.11 565.14 327.30

Summing the AIL for each product gives a total inventory of 1,022 cases. 24 The peak quantity of an item to appear on a shelf can be approximated as the order quantity plus safety stock, or

Q  z  sd'  250 boxes where z@93% = 1.48 from Appendix A and sd'  sd LT  19 1  19 boxes. The economic order quantity is Q* 

2 DS  IC

2(123  52)(125 . )  255.42 boxes 019 . (129 . )

Checking to see if the shelf space limit will be exceeded by this order quantity 255.42 + 1.48(19) = 283.54 boxes The quantity is greater than the 250 allowed. Subtracting the safety stock from the limit gives 250  28 = 222 boxes. The order quantity should be limited to this amount. 25 The plot of average inventory to period facility throughput (shipments) gives an overall indication of how the company is managing collectively its inventory for all stocked items. We can see that the relationship is linear with a zero intercept. This suggests that the company is establishing its inventory levels directly to the level of demand (throughput). An inventory policy, such as stocking to a number of weeks of demand, may be in effect. Overall, the inventory policy seems to be well executed in that the regression line fits the point for each warehouse quite well. The terminal with an inventory level of $6,000 seems to be an outlier and it should be investigated. If its low turnover ratio were brought in line with the other terminals, an inventory reduction from $6,000 to $4,000 on the average could be achieved. The stock-to-demand inventory policy should be challenged. An appropriate inventory policy should show some economies of scale, i.e., the inventory turnover ratio should increase as terminal throughput increases. Whereas the current policy is of the form I  0.012 D , a better policy would be I  kD 0.7 , where D represents terminal throughput and I is the average inventory level. The coefficient 0.012 for the current policy is found as the ratio of 6,000/500,000 = 0.012 for the last data point in the plot.

115

The k value for the improved policy needs to be estimated. From the cluster of the lowest throughput facilities, the average inventory level is approximately $2,000 with an average throughput of about $180,000. Therefore, from I  kD 0.7 2,000  k (180,000) 0.7 2,000  k ( 4,771.894) 2,000 k 4,771.894 k  0.419 Reading values from the plot, the following table can be developed showing the inventory reduction that might be expected from revised inventory policy. (Note: If the inventory-throughput values cannot be adequately read from the plot, the values in the following table may be provided to the students.)

Terminal 1 2 3 4 5 6 7 8 9 Totals

Actual Inventory, $ 2,000 1,950 2,000 2,050 3,900 6,000 4,500 4,300 5,500 32,200

Shipments, $ 150,000 195,000 200,000 200,000 320,000 330,000 390,000 410,000 500,000 2,695,000

Estimated inventory, $ I  0.012 D 1,800 2,340 2,400 2,400 3,840 3,960 4,680 4,920 6,000 32,340

Revised inventory, $ I  0.419 D 0.7 1,760 2,115 2,152 2,152 2,991 3,056 3,435 3,558 4,088 25,307

Revising the inventory control policy has the potential of reducing inventory from the 32,340  25,307 x100  21.7% . linear policy by 32,340 26 We can use the decision curves of Figure 9-23 in the text answer this question since it applies to a fill rate of 95% and an  = 0.7. First, determine K for an inventory throughput curve for the item, which is

K

D1 (117 x12) 0.3   1.466 6 TO

Next,

116

X 

tD10.7 12(117 x12) 0.3   0.90 ICK 0.20( 400)(1.466)

and with z 1.96 from Appendix A Y 

zs LT 1.96(15) 2   0.18 a KD (1.466)(117 x12) 0.7

The demand ratio r is 42/177 = 0.36. The intersection of r and X lies below the curve Y (use curve Y = 0.25), so do not cross fill. 27 Regular stock For two warehouses, estimate the regular stock for the three products.

Product A

Product B

2dS Q RS   IC 2 2 2(3,000)( 25) 0.02(15) RS A1   354 units 2 2(5,000)( 25) 0.02(15) RS A2   457 units 2

RS B1  RS B 2 

Product C

RS C1  RS C 2 

2(8,000)( 25) 0.02(30)  408 units 2 2(9,500)( 25) 0.02(30)  445 units 2

2(12,500)( 25) 0.02( 25)  559 units 2 2(15,000)( 25) 0.02( 25)  612 units 2

Regular system inventory for two warehouses is RS2W = 354 + 457 + 408 + 445 + 559 + 612 = 2,835.

117

Regular stock for a central warehouse

RS A  RS B  RS C 

2(8,000)( 25) 0.02(15)  577 units 2 2(17,500)( 25) 0.02(30)  604 units 2 2( 27,500)( 25) 0.02( 25)  829 units 2

Total central warehouse regular stock is RS1W =577 + 604 + 828 = 2,009 units. Safety Stock Product A SS  zsd LT SS A1  1.65(500) 0.75  714 units SS A2  1.65(700) 0.75  1,000 units

where [email protected] = 1.65 from Appendix A Product B SS B1  1.65( 250) 0.75  357 units SS B 2  1.65(335) 0.75  479 units Product C SS  zsd LT SS C1  1.65(3,500) 0.75  5,001 units SS C 2  1.65( 2,500) 0.75  3,572 units

System safety stock is SS2W = 714 + 1,000 + 357 + 479 + 5,001 + 3,572 = 11,123 units For each product, the estimated standard deviation of demand on the central warehouse is s A  s12  s22  500 2  700 2  860 units s B  250 2  3352  418 units s B  3,500 2  2,500 2  4,301 units The safety stock is

118

SS  zs LT SS A  1.65(860) .75  1,229 units SS B  1.65( 418) .75  597 units SS C  1.65( 4,301) .75  6,146 units Total safety stock in the central warehouse SS1W = 1,229 + 597 + 6,146 = 7,972 units. Total inventory with two warehouses RS2W + SS2W = 2,835 + 11,123 = 13,958 units and for a central warehouse RS1W + SS1W = 2,009 + 7,972 = 9,981 units. Centralizing inventories reduces them by 13,958 – 9,981 = 3,977 units. 28 The solution to this multi-echelon inventory control problem is approached by using the base-stock control system method. The idea is that inventory at any echelon is to plan its inventory position plus the inventory from all downstream echelons. First, compute the average inventory levels for each customer. This requires finding Q and the safety stock. Q is found from the EOQ formula.

For customer 1

Q1 

2( 425 x12)(50)  270 units 0.2(35)

AIL1 

Q1 270  zsd1 LT1   1.65(65) 0.5  211 units 2 2

where [email protected] =1.65 from Appendix A For customer 2 Q2 

2(333x12)(50)  239 units 0.2(35)

AIL2 

Q2 239  zsd 2 LT2   1.65(52) 0.5  180 units 2 2

For customer 3 Q3 

2( 276 x12)(50)  218 units 0.2(35)

AIL3 

Q3 218  zs d 3 LT3   1.65( 43) 0.5  159 units 2 2

119

Total customer echelon inventory is AILC = 211 + 180 + 159 = 550 units For the distributors echelon QD  2,000 units as given AILD 

QD 2,000  zsd D LTD   1.28(94) 1.0  1,120 units 2 2

where [email protected] =1.28 from Appendix A The expected inventory that the distributor will hold is the distributor echelon inventory less the combined inventory for the customers, or 1,120  550 = 570 units.

120

COMPLETE HARDWARE SUPPLY, INC. Teaching Note Strategy Complete Hardware Supply is an exercise involving the control of inventoried items collectively. Data for a random sample of 30 items from the company's total of 500 items held in inventory are given. The objective is to manage the total dollar value allowed to be held as inventory. Several alternatives can be considered for changing inventory levels, some of which require an investment other than in inventory. The number of items that must be analyzed and the multiple scenarios that are to be examined can be computationally time consuming. It is strongly suggested that students use the INPOL module within LOGWARE to aid analysis. The current database has been prepared and is available in the LOGWARE software. The Base Case We begin with the current data optimized as a reorder point design. The optimum order quantities and associated inventory levels are found. The base case costs are shown as follows:

Fixed order quantity policy Purchase cost Transport costa Carrying cost Order processing cost Out-of-stock cost Safety stock cost Total cost Total investment

$556,912 0 4,425 4,425 0 2,529 $568,291 $27,801

aIncluded in the purchase cost

We note that optimizing the current design shows that investment of $27,801 exceeds the allowed investment level of $18,000. Ways need to be explored to reduce this. Transmit Orders More Rapidly Instead of mailing orders to vendors, Tim O'Hare can buy a facsimile machine and transmit orders electronically. This scenario can be tested by reducing the lead times in the base case by 2 days, or (2/5) = 0.40 weeks and increasing order processing costs by $2, and then optimizing again. INPOL shows that there will be a slight increase in operating costs from $568,291 to $568,640, an incremental increase of $349. Projecting this to all 500 items, we have 349(500/30) = $5,817. Since both operating cost and inventory investment level increase, there is no economic incentive to implement this change. Faster Transportation Suggesting that vendors who are located some distance (>600 miles) from the warehouse use transportation is a possible way of reducing lead times, and therefore safety

121

stock levels. Of course, the increase in transportation cost for those affected vendors is likely to lead to a price increase to cover these costs. This scenario is tested by reducing the lead-time in weeks to 2.2 for those vendors over 600 miles from the warehouse. For these same vendors, a 5% price increase is made. Compared with the base case, there is little change in the inventory investment ($27,801 vs. $27,746); however, operating costs increase. The total costs now are $585,490 compared with the base case of $568,291, an increase of $27,199. The major portion ($17,159) of this comes from the increase in price. We conclude that this is not a good option for Tim. Reduce Forecast Error Reducing the forecast error involves reducing the standard deviation of the forecast error. Testing this option requires taking 70% of the base-case forecast error standard deviations and optimizing the design once again. These changes have a positive impact on operating costs and inventory investment. Operating cost now is $567,529 and inventory investment is $24,739. This is a saving in operating costs of $762 per year. For all 500, we can project the savings to be 762(500/30) = $12,700. Based on a simple return on investment, we have:

ROI 

12,700 0.25, or 25% / year 50,000

This would appear to be attractive since carrying costs are 25% per year and the company's return on investment probably makes up about 80% of this value. Reduce Customer Service At this point, we have only accepted the idea of reducing the forecast error. However, inventory investment remains too high. We can now try to reduce it by reducing the service levels. This is tested by dropping the service index from its current 0.98 level to a level where inventory investment approximates $18,000. This is done, assuming the forecast software will be purchased and the forecast error reduced by 30%. By trial and error, the service index is found to be 0.54, which gives an investment level of $18,028. The revised service level compared with the base case is summarized below for the 30 items.

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Item 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Base case 99.88% 99.92 99.96 99.98 99.98 99.96 99.97 99.96 99.92 99.98 99.99 99.99 99.92 99.98 99.96

Revised 96.26% 98.02 98.54 99.15 99.45 98.60 98.84 98.61 97.29 99.26 99.70 99.43 97.30 99.14 98.84

Item 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

Base case 99.98% 99.90 99.95 99.89 99.97 99.69 99.97 99.97 99.96 99.92 99.97 99.93 99.89 99.97 99.91

Revised 99.56% 97.57 97.81 95.96 98.15 89.53 98.96 98.96 97.58 99.33 96.68 97.45 98.78 96.92 96.78

Notice how little the service level changes, even with a substantial reduction in the service index. Conclusions Tim can make a good economic argument for purchasing software that will reduce the forecast error. The only questions here are whether the software can truly produce at least the error reduction noted and whether a 25% return on investment is adequate for the risks involved. Arguing to accept a service reduction in order to lower the investment level is a little less obvious since we do not know the effect that service levels have on sales. However, Tim may point out that the service levels need to be changed so little that it is unlikely that customers will detect the change. He might also raise the question as to whether customer service levels were too high initially, and suggest that customers be surveyed as to the service levels that they do need.

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AMERICAN LIGHTING PRODUCTS Teaching Note Strategy American Lighting Products is a manufacturer of fluorescent lamps in various sizes for industrial and consumer use. As frequently happens in business, top management has requested that inventories be reduced across the board, but it does not want to sacrifice customer service. Sue Smith and Bryan White have been asked to eliminate 20 percent of the finished goods inventory. Their plan is to reduce the number of stocking locations and, thereby, eliminate the amount of inventory needed. Of course, they must recognize that with fewer stocking points, transportation costs are likely to increase and customer delivery times may increase as well. On the other hand, facility fixed cost may be reduced. The purpose of this case is to allow students to examine inventory policy and planning through aggregate inventory management procedures. They also can see the connection between location and inventory levels. Answer to Questions (1) Evaluate the company’s current inventory management procedures.

The company’s procedures for controlling inventory levels are at the heart of whether inventory reductions are likely to be achieved through inventory consolidation. The company appears to be using some form of reorder point control for the entire system inventory, but it is modified by the need to produce in production lot sizes. It is not clear how the reorder point is established. If it is based on economic order quantity principles, then the effect of the principles becomes distorted by the need to produce to a lot size that is different from the economic order quantity. Therefore, average inventory levels in a warehouse will not be related to the square root of the warehouse’s throughput (demand), i.e., throughput raised to the 0.5 power.5 Rather, the throughput will be raised to a higher exponent between 0.5 and 1.0. The above ideas can be verified by plotting the data given in Table 1 of the case and then fitting a curve of the form I  TP  . Note: The curve can be found from standard linear regression techniques when the equation is converted to a linear form through a logarithmic transformation, i.e., lnI = ln + lnTP. The results are shown in Figure 1. The inventory curve is I  2.99TP 0.816 with r = 0.86, where I and TP are in lamps. The projected inventory reduction can be calculated by using this formula. From the plot of the inventory data, we can see that there is substantial variation about the fitted inventory curve. There is not a consistent turnover ratio between the warehouses. This probably results from the centralized control policy. On the other hand, improved control may be achieved by using a pull procedure at each MDC. The data available in the case do not let us explore this issue.

5

Based on the economic order quantity formula, the average inventory level (AIL) for an item held in inventory can be estimated as AIL  Q / 2  2 DS / IC / 2 . Collecting all constants into K, we have AIL=K(D)0.5, where D is demand, or throughput.

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FIGURE 1 Plot of MDC average inventory vs. annual throughput.

(2) Should establishing the LOC be pursued? One of the ideas proposed in the case is to consolidate all Consumer product line items into one large order center (LOC). Evaluating the impact of the LOC on inventory reduction requires that an assumption be made as to how much demand and associated inventory of the total belongs to Consumer products. Table 2 of the case gives the order and back order breakdown by sales channel. Using these data, total consumer demand is 312,211 line items, or 33.4% of the total line items. The assumption is that the same percentage applies to total demand. Hence, Consumer demand is 33.4%169,023,000 = 56,453,682 lamps. From the inventory-throughput curve, we can estimate the amount of inventory needed at the single LOC. That is, I = 2.997(56,453,682)0.816 = 6,339,684 lamps. If Consumer products for 33.4% of total inventory, then there are 33.4%23,093,500 = 7,713,229 lamps in Consumer inventory. The reduction that can be projected is 7,713,229  6,339,684 = 1,373,545 lamps for a reduction of Reduction 

1,373,545  100  17.8% 7,713,229

in Consumer inventory levels, but only a 6% reduction in overall inventory levels. The 20% reduction goal is not achieved. Other alternatives need to be explored. (3) Does reducing the number of stocking locations have the potential for reducing system inventories by 20%? Is there enough information available to make a good inventory reduction decision? The second alternative proposed in the case is to reduce the number of MDCs from eight to a smaller number. In order to evaluate this proposal, it needs to be determined which MDCs will be consolidated and the associated total demand flowing through the consolidated facilities. The inventory-throughput relationship can then be used to estimate the resulting inventory levels. For example, if the Seattle and Los Angeles MDCs are combined, the consolidated demand would be 4,922,000 + 21,470,000 = 26,392,000 lamps. The combined inventory is projected to be I = 2.997(26,392,000)0.816 =

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3,408,852 lamps, compared with the inventory for the two locations of 4,626,333, as shown in Table 1. This yields a 26.3% reduction from current levels. Table 1 shows other possible MDC consolidations and the resulting inventory reductions that can be projected. TABLE 1 Inventory Reduction for Selected MDC Combinations, in Lamps Combined Combined Inventory MDC combination demand inventory reduction Seattle/Los Angeles 26,392,000 3,408,852 1,217,481 Kansas City/Dallas 29,194,000 3,701,403 50,181 Chicago/Ravenna 49,174,000 5,664,257 -557,590 Atlanta/Dallas 39,314,000 4,718,862 1,224,721 Kansas City/Chicago 39,271,000 4,714,650 -933,900 Ravenna/Hagerstown 64,046,000 7,027,231 1,715,607 K City/Dallas/Chicago 52,515,000 5,976,377 -36,377 Ravenna/H’town/Chicago 87,367,000 7,508,054 3,423,196 Atlanta/Dallas/K City 55,264,000 5,242,351 2,293,566

From the MDC combinations in Table 1, proximity to each other is a primary consideration in order to not increase transportation costs or jeopardize delivery service any more than necessary. Several options can be identified that yield a 20% inventory reduction. These are:

Option 1

2

3

MDC combinations LA/Seattle Ravenna/H’town/Chicago Total reduction

Inventory reduction, lamps 1,217,481 3,423,196 4,640,677

Total inventory reduction

LA/Seattle Kansas City/Hagerstown Ravenna/Hagerstown Total reduction

1,217,481 1,224,721 1,715,602 4,157,804

18.0%

LA/Seattle Ravenna/Hagerstown Atlanta/Dallas/K City Total reduction

1,217,481 1,715,602 2,293,566 5,226,649

22.6%

20.1%

Options 1 and 3 achieve the 20% reduction goal, although other MDC combinations not evaluated may also do so. The maximum reduction would be achieved with one MDC. The total inventory would be I = 2.997(169,023,000)0.816 = 15,512,812 lamps, for a system reduction of 32.8%. However, we must recognize that as the number of warehouses is decreased, outbound transportation costs will increase. Inbound transportation costs to the combined MDC will remain about the same, since replenishment shipments are 126

already in truckload quantities. Some difference in cost will result from differences in the length of the hauls to the warehouses. On the other hand, outbound costs may substantially increase, since the combined MDC locations are likely to be more removed from customers then they are at present. Outbound transportation rates will be higher, as they are likely to be for shipments of less-than-truckload quantities. If the sum of the inbound and outbound transportation cost increases is greater than the inventory carrying cost reduction, then the decision to reduce inventories must be questioned. Calculating all transportation cost changes is not possible, since the case study does not provide sufficient data on outbound transportation rates. However, they should be determined before and after consolidation to assess the tradeoff between inventory reduction and transportation costs increases. On the other hand, inbound transportation costs can be found, as shown below for option 1, where the consolidation points are Los Angeles and Hagerstown. Annual demand, Location lamps Seattle 4,922,000 Los Angeles 21,470,000 Ravenna 25,853,000 Hagerstown 38,193,000 Chicago 23,321,000 Total 113,759,000 a (4,922,000/35,000)1800 = 253,131 TL rate, $/TL 1800 1800 250 475 350

Transport cost, $ 253,131a 1,104,171 184,664 518,334 233,210 2,293,510

Combined annual demand, lamps

Transport cost, $

26,392,000

1,357,302

87,367,000

1,185,695

113,759,000

2,542,997

There will be a net increase in inbound transportation costs of $2,542,997  2,293,510 = $249,487 for option 1. In addition, the annual fixed costs for the MDCs will be less, since the total space needed in the consolidated facilities should be less than that for the existing facilities. Again, the case study does not estimate the fixed costs for existing or potential locations. We do know that taking them into would favor consolidation. In summary, the costs associated with option 1, that just meets the 20% inventory reduction goal, would be: Cost type Inventory carrying cost reduction Warehouse cost Warehouse fixed cost Outbound transportation cost Inbound transportation cost

Cost savings 0.200.8824,640,677 = $818,615 0.104,640,677 = $464,068 Unknown, but may be included in warehouse cost Unknowndata not given ($249,487)

Although Sue and Bryan could report a substantial savings in inventory related costs, they should be encouraged to include fixed costs and transportation costs so as to report the true benefits of the inventory reduction plan. (4) How might customer service be affected by the proposed inventory reduction?

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The general effect of inventory consolidation is to reduce the number of stocking points and make them more remote from customers. That is, the delivery distance will be increased if inventory consolidation is implemented. Therefore, delivery customer service may be jeopardized and must be considered before deciding to consolidate inventories. From Table 3 of the case, it can be seen that customer lead times remain constant for a variety of locations with the exception of Kansas City. Since consolidation points will be selected among the existing locations, outbound lead times will remain unaffected. Customer service due to location should be constant, at least for a moderate degree of consolidation. Customer service due to stock availability will be affected if safety stock levels are reduced after consolidation. Although the inventory-throughput relationship projects adequate safety stock to maintain the current first-time delivery levels, it does not for any increase in lead times that may occur between the current system of MDCs and the consolidated ones. By comparing the weighted inbound lead times for the existing distribution system and option 1, as shown in Table 2, the average inbound leadtime is slightly reduced through consolidation. Lead-time variability is usually related to average lead-time. This should have a favorable affect on inventory levels since uncertainty is reduced. First-time deliveries should not be adversely affected by consolidation, according to option 1. TABLE 2

A Comparison of Inbound Lead Times for the Existing Distribution System and a Consolidated Distribution System (Option 1) (a) Current Distribution System Inbound Weighted lead time, lead time, Master Distribution Center Shipments days days Atlanta 26,070,000 2 0.308 Chicago 23,321,000 1 0.138 Dallas 13,244,000 3 0.235 Hagerstown 38,193,000 1 0.226 Kansas City 15,950,000 2 0.094 Los Angeles 21,470,000 5 0.635 Ravenna 25,853,000 1 0.153 6 0.175 Seattle 4,922,000 Total 169,023,000 1.964

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(b) Consolidation Option 1

Master Distribution Centera Atlanta Dallas H’town/Ravenna/Chicago Kansas City Los Angeles/Seattle Total a

Shipments 26,070,000 13,244,000 87,367,000 15,950,000 26,392,000 169,023,000

Inbound lead time, days 2 3 1 2 5

Weighted lead time, days 0.308 0.235 0.517 0.094 0.781 1.935

Consolidation is assumed to take place at the MDC with the largest number of current shipments.

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AMERICAN RED CROSS: BLOOD SERVICES Teaching Note Strategy The American Red Cross Blood Services has a mission to provide the highest quality blood components at the lowest possible cost. High quality blood products are provided to regional hospitals, but managing the inventory to meet demand as it occurs is a difficult problem. Blood is considered a precious product, especially by those who give it voluntarily. So, managing this perishable product carefully is a foremost concern. Blood is a vital product to those in need of it for emergencies and a precious product to those requiring it for elective surgery and other treatments. The goal is to always have what is needed but never so much that this perishable product has to outdated. Managing the blood inventory is quite difficult because (1) forecasting demand is not particularly accurate, (2) the planning horizon for collections can be up to a year long with uncertain yields, (3) the life of blood products ranges from 42 days to as short as 5 days, (4) once scheduled, blood donors are never turned away except for medical reasons, and (5) there is a limited opportunity to sell blood outside of the local region if too much is on hand. Overall, this situation has many characteristics of a “supply driven” inventory management problem, which requires inventory management techniques different from those for typical consumer products. The intended purpose of this case study is for students to examine an inventory situation where there is limited control over the amount of the product flowing into inventory. This supply-driven inventory situation is likely to be quite different from that discussed on the introductory level. Students are encouraged to consider the various elements that affect inventory levels of individual products and how they interact. These elements are (1) demand forecasting, (2) collections, (3) decision rules for creating blood derivatives, (4) product prices, and (5) inventory policy. It is expected that students will be able to make general suggestions for improvement. Questions (1) Describe the inventory management problem facing blood services at the American Red Cross.

One of the major problems facing the American Red Cross (ARC) is that the availability of blood is supply-driven, meaning that quantities of blood received for processing to meet demand in the short term are unknown, yet they must be placed in inventory if demand is less than the collected quantities. Blood availability is a function of number of factors that cannot be well-controlled by the regional blood center in the short run, causing wide variability in supply. The usage of blood at hospital blood banks, which creates the demand on ARC’s blood inventories, is also uncertain and varies from day to day and between hospital facilities. The yield of blood at the point of collection is random and does not necessarily give the product mix needed to meet demand. Different blood types can only be known by a probability distribution as to the percentage of the blood types that exist in the general population. In the short term, the demand for blood types may differ from the collected

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quantities, resulting in a potential for under- and over-stocking, since blood is drawn from all qualified donors as they arrive at collection sites. Forecasting demand for blood products will likely be reasonably accurate for a base load. Surgery loads on hospitals are scheduled in advance so that blood needs will be known with a fair degree of certainty, although each operation will not typically use the full amount of blood allocated to it. However, emergency blood needs are not well predicted, and they can cause spikes in demand and unplanned draws on inventory. A problem is establishing how much accuracy is needed for good inventory management. Inventory policy for managing inventory levels is a mixed strategy of product pricing, derivative product selection for processing at the time of collection, conversion to other products later in the product life cycle, product sell off, emergency supply (call for blood), discount pricing, and stocking rules for hospitals. Although there are many avenues to controlling inventory levels, shortages and outdating cannot always be avoided. It is not clear that these procedures lead to an optimal control of inventory levels. Competition from local independent blood banks that sell selected blood products at low prices makes it difficult for ARC to cover costs. ARC provides a wider range of products, but it has difficulty-differentiating price among derivative products so that it might compete effectively. Given pressures for hospitals to increase efficiency, they will shop around for the lowest-priced blood products. ARC is having difficulty maintaining its position as the dominant supplier of blood products in the region, which results in the greater uncertainty in managing inventory levels. In summary, blood is a precious product given by volunteers for the benefit of others. Donors have the right to expect that their contribution will be handled responsibly. To ARC, this means managing the blood supply so that recipients receive a high-quality product at the lowest possible price. To achieve this goal, ARC manages the blood supply through four inter-connected elements: (1) estimating the blood product needs over time, (2) planning the collection of whole blood, (3) deciding which derivative products and their amounts should be created from whole blood, and (4) controlling the inventory levels to avoid outdating. The volunteer nature of the blood giving and donor attitudes surrounding it, long planning lead times and the associated uncertainties, rising competition among some products from local blood banks, and the uncertainties of blood needs all make blood supply management a unique inventory management problem. (2) Evaluate the current inventory management practices in light of ARC’s mission. Performance of blood management can be evaluated on two levels: customer service and cost. Tables 8 and 9 of the case show that in March standards were not quite met overall. Within specific product types, there was up to an 8 percent deficit. Both order fill rate and item fill rate were less than 100 percent for most products. There would seem to be some room for improvement, especially in managing the variation among product types. From a cost standpoint, it is not known how efficiently the blood supply is managed since no costs are reported. In addition, the revenue that the blood products generate is not known. We would like to know how prices of the various products are set so that revenues might be maximized, considering competition among some of the product line. We do expect that demand is price elastic, since hospitals do shop around for blood

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products that are available from local, commercial, and community blood banks. On the other hand, ARC is the sole regional supplier of certain products such as platelets. Setting product fill-rate standards at various levels can influence costs. We do not know this effect. Setting inventory levels by a “number of days of inventory” rule of thumb is simple but not as effective as planning inventory levels based on the uncertainties that occur in demand forecasts and supply lead times. The number-of-days-of-inventory rule does tend to lead to too much inventory or to too many out-of-stock situations. The plan for evaluation, if enough data were available, would be to establish a base case of cost and service. This, then, would provide a basis for evaluating the effect of change in the supply procedures. (3) Can you suggest any changes in ARC’s inventory planning and control practices that might lead to cost reduction or service improvement? Suggestions for improvement in blood supply management stem from a basic understanding of the nature of the demand-supply relationship. When supply is uncertain and all supply must be taken that is available, there is the possibility that significant excess inventory will occur. The goal is to “manage” the demand in the short run to reduce inventory levels when overstocking occurs, rather than focusing on managing supply. Several approaches for doing this are:  Aggressively price selected products that are in excess supply and are nearing their expiration dates, e.g. run a sale or offer price discounts.  Sell off excess supply to secondary demand sources or other regions of the ARC.  Temporarily adjust return rules for hospitals.  Bring demand more in line with supply by converting products into derivative ones that have excess demand, e.g., reprocess whole blood into plasma.  Encourage hospitals to buy certain products in excess supply for a more favorable status in buying other products that are in short supply, such as phersis platelets and rare whole blood types.  Try to create excess demand for all products, especially those items that are available from local blood banks, through promotion of ARC’s distinct advantages, such as quality, high service levels, and a wide range of blood derivative products.  Offer “two-for-one” sales, such that if a hospital buys one blood product, it may receive another at a favorable price.  Pool the risk of uncertain demand by maintaining a central inventory for all hospitals, or managing the inventories at all hospitals, as well at ARC, collectively. Provide quick deliveries or transfers among inventory locations. ARC should attempt to be the premier provider of blood products and leverage the advantage. This will allow it to maintain a degree of control over the demand for blood. Effectively controlling demand in turn allows it to control its costs and avoid product outdating.

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(4) Is pricing policy an appropriate mechanism to control inventory levels? If so, how should price be determined? From the previous discussion, it can be seen that price plays a role in controlling demand. Since there appears a relationship between demand and price for some products, especially among those products offered by local blood banks that compete with ARC blood products, price may be an effective weapon to meet competition. Rather than setting price based on the cost of production, ARC might consider raising the price on products for which it is the sole provider, such as platelets, and then meeting the price of competitors on whole blood. Although ARC strives to be a nonprofit organization, the increased volume that an effective pricing strategy promotes would allow more of the fixed costs to be covered. This may lead to lower overall average prices for ARC’s products. Blood could also be priced as a function of its freshness at two or more levels. Although blood that has been donated within 42 days legally can be utilized, the quality of blood does not remain the same for the entire 42-day period. A chemical compound found in blood, called 2,3-DPG, decreases with the age of the stored blood, and is believed to be important in oxygen delivery. For this reason, certain procedures such as heart transplants and neonatal procedures require that blood be fresh, usually donated within 10 days or less. Thus, a simple pricing policy could be to charge a higher price for blood that is less than 10 days old, and a lower price for blood that is between 10 and 42 days old. Price differences here are based on product quality.

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CHAPTER 10 PURCHASING AND SUPPLY SCHEDULING DECISIONS 1 (a) The following requirements schedules will lead to the proper timing and quantities for the purchase orders.

Desk style A Sales forecast Receipts Qty on hand 0 Releases to prod.

1 150 200 50 300

2 150 300 200

3 200

1 60

2 60 100 60

3 60

2 120 100 80 100

3 100 100 80

0 300

Week 4 200 300 100 300

5 150 300 250

6 200

7 200 300 150

8 150

6 100 100 40 100

7 80 100 60

8 60

6 60 100 60

7 60

8 80 100 60

50 300

0

Desk style B Sales forecast Receipts Qty on hand 80 Releases to prod.

20 100

0 100

Week 4 5 80 80 100 100 20 40 100 100

0

Desk style C Sales forecast Receipts Qty on hand 200 Releases to prod.

1 100 100 100

Week 4 5 80 80 100 0 20 100 100

0 100

Summing the releases for these three desk release schedules gives a production requirements schedule for desks in general and sheets of plywood in particular. That is, Week 1 2 3 4 5 6 7 8 Desk requirements 500 100 400 500 200 400 100 1500 300 1200 1500 600 1200 300 0 Plywood sheetsa a Desk requirements times 3

0

Now, find the purchase order releases for the plywood sheets.

Sales forecast Receipts Qty on hand 2400 Releases to prod.

1 2 1500 300 600 900 1200 1000 1000

3 1200 1000 1000 1000

Week 4 5 6 7 1500 600 1200 300 1000 1000 1000 500 900 700 400 1000

8 0 400

Therefore, purchase orders should be placed in weeks 1, 2, 3, and 4 for 1000 sheets each.

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(b) Using Equation 10-2 in the text, the probability of not having the plywood sheets at the time needed would be Pr  1 

Pc 5  1  0.02 01 . 5 Cc  Pc

From Appendix A, [email protected] = 2.05. Therefore, the lead-time should be . days T *  LT  z  sLT  14  2.05( 2)  181 Another ½ week should be added to the current lead-time of 2 weeks. 2 (a) Using Equation 10-2, the probability of not having the item when needed for production is: Pr 

Pc 150   0.9999 C c  Pc (0.2  35 / 365)  150

The time to place an order ahead of need is: T *  LT  z  s LT  14  3.6( 4)  28 days where [email protected] = 3.6 from Appendix A. (b) Use part period cost balancing. The unit carrying cost is (0.2/52)35 = 0.134. Then, (Q=250) Week 4 0.134[500 + 200]/2 = 46.9 (Q=1350) Weeks 4 + 5 0.134[(1350 + 1050)/2 + (1050 + 200)/2] = 244.6 The carrying cost closest to the order cost of $50 is Q = 250. Order this amount. 3 Using the requirements planning procedure, we can develop a schedule of material flows through the network over the next 10 weeks. Whse 1 Requirements Schd receipts On-hand qty 1700 Releases

1 2 3 4 5 6 7 1200 1200 1200 1200 1200 1200 1200 7500 500 6800 5600 4400 3200 2000 800 7500 7500

8 9 10 1200 1200 1200 7500 7100 5900 4700

135

Whse 2 Requirements Schd receipts On-hand qty 3300 Releases

1 2 3 2300 2300 2300 7500 1000 6200 3900 7500

Whse 3 Requirements Schd receipts On-hand qty 3400 Releases

1 2 3 4 2700 2700 2700 2700 7500 700 5500 2800 100 7500 7500

Regnl whse A Requirements Schd receipts On-hand qty 52300 Releases to plant

1 2 22500

3 0

4 5 6 2300 2300 2300 7500 1600 6800 4500 7500 5 6 2700 2700 7500 4900 2200 7500

4 5 0 15000

7 8 9 10 2300 2300 2300 2300 7500 2200 7400 5100 2800 7500 7 8 9 2700 2700 2700 7500 7000 4300 1600 7500

6 7 8 0 7500 15000 0 15000

29800 29800 29800 14800 14800 7300 15000 4 5 4100 4100 7500 4300 200 7500

10 2700 7500 6400

9 7500 15000

7300 7300

10 0

1300

1300

15000

Whse 4 Requirements Schd receipts On-hand qty 5700 Releases

1 2 3 4100 4100 4100 7500 1600 5000 900 7500 7500

Whse 5 Requirements Schd receipts On-hand qty 2300 Releases

1 2 3 4 1700 1700 1700 1700 7500 600 6400 4700 3000 7500

Whse 6 Requirements Schd receipts On-hand qty 1200 Releases

1 2 3 4 5 6 7 8 9 10 900 900 900 900 900 900 900 900 900 900 7500 7500 300 6900 6000 5100 4200 3300 2400 1500 600 7200 7500 7500

Regnl whse B Requirements Schd receipts On-hand qty 31700 Releases to plant

1 2 3 4 22500 0 7500 15000

Plant Requirements Schd receipts On-hand qty 0 Releases-matls

1

5 6 7 0 15000 7500 15000

15000 2

3 0 0

0 0 0 20000 20000

7 8 4100 4100 7500 7000 2900 7500

9 10 4100 4100 7500 6300 2200

5 6 7 8 9 10 1700 1700 1700 1700 1700 1700 7500 1300 7100 5400 3700 2000 300 7500

9200 24200 16700 16700

0

6 4100 7500 3600 7500

4 15000 20000 5000 20000

1700

8 9 10 0 7500 7500 0 15000

9200 1700 9200

9200 9200

15000

5 6 7 15000 0 30000 20000 20000 10000 10000 0

8 0

9 0

10 0

0

0

0

Summing the releases to the plant shows that the plant should place into production 15,000 cases in weeks 4 and 5 and 30,000 cases in week 7. Orders for materials should be placed in weeks 1, 2, and 4 in an order size to make 20,000 units. Because demand is shown to be constant, the average inventory must be one-half the order quantity. For the six field warehouses and a shipping quantity of 7500, the average long run inventory would be (7500/2)6 = 22,500 cases. For the regional warehouses, 136

the average inventory would be (15,000/2)2 = 15,000 cases. For the plant, the average inventory would be 20,000/2 = 10,000 cases. The total system average inventory would be 22,500 + 15,000 + 10,000 = 47,500 cases. 4 (a) The leverage principle shows the relative change that must be made in cost, price, or sales volume to affect a given change in the profit level. Usually it is used in reference to the cost of goods sold to show the impact that small changes in the cost of goods will have on profits and the important role that purchasing plays in the profitability of the firm. The following simple profit and loss statements will show how much change is needed in various activities to increase profits to $5,000,000. Sales Price L&S OH COG Current (+4%) (1%) (-3%) (-6%) (-2%) Sales $55.0 $57.2a $55.5 $55.0 $55.0 $55.0 Cost of goods 27.5 28.6 27.5 27.5 27.5 27.0 Labor & salaries 15.0 15.6 15.0 14.5 15.0 15.0 Overhead 8.0 8.0 8.0 8.0 7.5 8.0 Profit $ 4.5 $ 5.0 $ 5.0 $ 5.0 $ 5.0 $ 5.0 a Sales - .7727xSales -8 = 5, where L&S is 0.2727 of Sales and COG is 0.5 of Sales. So, Sales = (5 + 8)/(1 – 0.7727) = 57.2

Due to the magnitude of cost of goods sold, it requires less than a 2 percent change in COG to increase profits to $5,000,000. (b) The current ROA as: Profit margin = (4.5/55)100 = 8.2% Investment turnover = 55/20 = 2.75 ROA = 2.758.2 = 22.6%

Or, ROA = Profit/Assets

Reducing cost of goods by 7% will increase profits to 55  27.50.93  15  8 = $6.43 and the profit margin now is 6.43100/55 = 11.7%. Inventory at 20% of total assets is $4 million. If the cost of goods is reduced by 7%, inventory value will decline to $40.93 = $3.72. Total assets will be 3.72 + 16 = $19.72 million. The investment turnover is 55/19.72 = 2.789. The ROA now will be 11.72.789 = 32.63% 5 (a) A mixed purchasing strategy will generally be beneficial when prices show a definite seasonality, they are predictable, and inventory costs associated with forward buying are not excessive. In the problem, we should consider forward buying in the first half of the year and hand-to-mouth buying in the last half. To test the various strategies, compare (1) hand-to-mouth buying, (2) forward buying every 2 months, (3) forward buying every 3 months, and (4) forward buying for the first 6 months. The results are summarized in Table 10-1.

137

The inventory for the hand-to-mouth buying strategy can be approximated as 50,000/2 = 25,000. The carrying cost would be 0.304.9825,000 = $37,350 per year. The carrying cost for the 2-mouth forward buying strategy is: 0.304.88[(0.5100,000/2) + (0.550,000/2)] = $54,900 For the 3-month forward buying strategy:

For 2nd half of year

0.34.56[(0.5300,000/2) + (0.550,000/2)] = $119,700 From the total costs in Table 10-1, the best strategy is to forward buy the first sixmonth's requirements in January and hand-to-mouth buy for the last six months. (b) Some possible disadvantages are:  Prices may fall rather than rise in the first six months  There may not be adequate storage space to accommodate such a large purchase.  The materials may be perishable and not easily stored.  Uncertainties in the requirements and carrying costs may void the strategy.

138

TABLE 10-1 A Comparison of Various Forward Buying Strategies with Hand-to-Mouth Buying Hand-to-mouth buy

Jan Feb Mar Apr May Jun Jly Aug Sep Oct Nov Dec

Price, Quantity, $/unit units 4.00 50,000 4.30 50,000 4.70 50,000 5.00 50,000 5.25 50,000 5.75 50,000 6.00 50,000 5.60 50,000 5.40 50,000 5.00 50,000 4.50 50,000 4.25 50,000 Subtotals Inventory costs Totals Average price/unit

Total $200,000 215,000 235,000 250,000 262,000 287,500 300,000 280,000 270,000 250,000 225,000 212,000 $2,987,500 37,350 $3,024,850 $4.98

2-month forward buy

Price, $/unit 4.00

Quantity, units 100,000

Total $400,000

4.70

100,000

470,000

5.25

100,000

525,000

6.00 5.60 5.40 5.00 4.50 4.25

50,000 50,000 50,000 50,000 50,000 50,000

300,000 280,000 270,000 250,000 225,000 212,000 $2,932,500 54,900 $2,987,400 $4.88

3-month forward buy

6-month forward buy

Price, $/unit 4.00

Quantity, units 150,000

Total $600,000

5.00

150,000

750,000

6.00 5.60 5.40 5.00 4.50 4.25

50,000 50,000 50,000 50,000 50,000 50,000

300,000 280,000 270,000 250,000 225,000 212,500 $2,887,500 72,150 $2,959,650 $4.81

Price, $/unit 4.00

Quantity, units 300,000

6.00 5.60 5.40 5.00 4.50 4.25

50,000 50,000 50,000 50,000 50,000 50,000

Total $1,200,000

300,000 280,000 270,000 250,000 225,000 212,500 $2,737,500 119,700 $2,857,200 $4.56

139

6 (a) On the average, a total expenditure of 1.1025,000 = $27,500 should be made for copper each month.

(b) For the next 4 months, the dollar averaging purchases would be: (1) (2) Price, No. of Month $/lb. lb. 1 1.32 20,833 2 1.05 26,190 3 1.10 25,000 4 0.95 28,947 100,970 a 50,486/4 = 12,622

(3)=(1)(2) Total cost,$ 27,500 27,500 27,500 27,500 $110,000

(4)=(2)/2 Average inventory, lb. 10,417 13,095 12,500 14,474 12,622a

The average per-lb. cost would be $110,000/100,970 = $1.089. The inventory carrying cost over 4 months would be 0.201.089(4/12) 12,622 = $916. If hand-to-mouth were used, we would have: (1) (2) Price, No. of Month $/lb. lb. 1 1.32 25,000 2 1.05 25,000 3 1.10 25,000 4 0.95 25,000 100,000 a 50,000/4 = 12,500

(3)=(1)(2) Total cost,$ 33,000 26,250 27,500 23,750 $110,500

(4)=(2)/2 Average inventory, lb. 12,500 12,500 12,500 12,500 12,500a

The average per-lb. cost would be $110,500/100,000 = $1.105. The inventory carrying cost over 4 months would be 0.201.105(4/12) 12,500 = $921. If 100,000 lb. of copper is purchased, the two strategies can be compared as follows. Purchase Inventory Total Strategy cost cost cost Dollar averaging $108,900 + 916 = $109,816 Hand-to-mouth 110,500 + 921 = 111,421 Dollar averaging buying would be preferred.

7 For an inclusive quantity discount price incentive plan, we first compute the economic order quantities for each range of price. Using

Q *  2 DS / IC we compute

140

Q1*  2(500 )(15) / ( 0.20 )( 49.95)  38.75 cases Q2*  2(500 )(15) / ( 0.20 )( 44.95)  40.85 cases

Since Q2* is outside of the second price bracket, Q1* is the only relevant quantity. Now we check the total cost at Q1* and at the minimum quantities within the price break. We solve:

TCi  Pi D  DS / Qi  ICi Qi / 2 At Q = 38.75 TC = 49.95500 + 50015/38.75 + 0.249.9538.75/2 = $25,362 At Q = 50 TC = 44.95500 + 50015/50 + 0.244.9550/2 = $22,850 At Q = 80 TC = 39.95500 + 50015/80 + 0.239.9580/2 = $20,388 Floor polish should be purchased in quantities of 80 cases. 8 This noninclusive price discount problem requires solving the following relevant total cost equation for various order quantities until the minimum cost is found.

TCi  Pi D  DS / Qi  ICi Qi / 2 The computations can be shown in the table below given that D = 1,400, S = 75, and I = 0.25.

141

Q 20 50 100 200 300 400 500

550

600

Price 795 795 795 795 200795+100750 300 200795+200750 400 200795+200750 +100725 500 200795+200750 +150725 550 200795+200750 +200725 600

P D +DS/Q +ICQ/2 = Total cost 1,113,000.00 5,250.00 1,987.50 $1,120,237.50 1,113,000.00 2,100.00 4,968.75 1,120,068.75 1,113,000.00 1,050.00 9,937.50 1,123,987.50 1,113,000.00 525.00 19,875.00 1,133,400.00 1,092,000.00 350.00 29,250.00 1,121,600.00 1,081,500.00

262.50

38,625.00

1,120,387.50

1,068,200.00

210.00

47,687.50

1,116,097.50

1,063,363.64

190.91

52,218.75

1,115,773.27

1,059,333.33

175.00

56,750.00

1,116,258.33

The optimal purchase quantity is 550 motors. 9 (a) This problem is a good application of the transportation method of linear programming. We begin by determining the costs for the current sourcing arrangement. Source Dayton Dayton Kansas City Minneapolis

Destination Cincinnati Baltimore Dallas Los Angeles

Price 3.40 3.40 3.45 3.25

Transport 0.05 0.15 0.08 0.24

Volume Cost 5,000 $17,250 1,000 3,550 2,500 8,825 1,200 4,188 Total $33,813

To optimize, we establish the following transportation cost matrix and solve it using any appropriate method, such as the TRANLP module in LOGWARE. Cincinnati 3.40

Dallas 3.44

Minneapolis 3.55

3.53

Los Angeles 3.49 1200 3.65

Baltimore 3.46

1200 3.63

Kansas City

4800 3.52

3.45 Dayton Requirements

Capacity

5000 5000

3.67 2500 2500

3.55 1200

1000 1000

9999

The total cost for this solution is $33,788 or a savings of $25 over the current sourcing.

142

(b) Because Minneapolis is at capacity, this supplier should be examined further. If unlimited capacity were available at Minneapolis, all requirements would be met by this supplier for a total cost of $33,248, or a savings of $565 for this material. (c) The above analysis does indicate that too many suppliers are being used. Only two are needed if Minneapolis continues to supply at the current level. If Minneapolis can be expanded, it becomes the only supplier. Of course, whether the company would risk a single supplier for this material must be left unanswered. 10 (a) The deal-buying equation (Equation 10-5) can be applied to this problem. First, find the optimal order quantity before the discount.

Q* 

2 DS  IC

2(120,000)( 40)  566 units 0.30(100)

Next, find the adjusted order quantity after the discount has been applied. Qˆ 

dD pQ * 5(120,000) 100(566)     21,648 units (p  d)I p  d (100  5)(0.30) (100  5)

A large order size of 21,648 units should placed. (b) The time that an order of this size will be held before it is depleted is given by Qˆ 21,648   0.18 years, or 9.4 weeks D 120,000

143

INDUSTRIAL DISTRIBUTORS, INC. Teaching Note Strategy The purpose of the Industrial Distributors case study is to illustrate the computation of purchase quantities under inclusive and noninclusive price discounts and transport rateweight breaks. The INPOL module of LOGWARE is helpful in conducting the analysis. As a teaching strategy, it may be worthwhile to begin any class discussion with the cost tradeoffs that are present in such a problem as this. This will help to establish the nature of the total cost equation that needs to be solved in this problem. Answers to Questions (1) What size of replenishment orders, to the nearest 50 units, should Walter place, given the manufacturer's noninclusive price policy?

When price discounts are offered, purchase quantities are not simply determined by a single formula. Due to discontinuities in the total cost curve as a function of order quantity, the optimal order quantity is found by computing total costs for different quantity values. In this case of both price and transport rate breaks plus warehousing costs that can be affected by the order size, the following annual total cost formula is to be solved. TC = PD + RD +

SD ICQ + + W ( Q - 300) Q 2

where TC = total cost for quantity Q, $ PD = purchase cost for price P, $ RD = transport costs at rate R, $ SD/Q = ordering cost at quantity Q, $ ICQ/2 = carrying cost at quantity Q, $ W(Q-300) = public warehousing cost if Q is greater than 300 units, $ W = public warehousing rate, $ per unit per year D = annual demand, units P = price for orders of size Q, $ per unit R = transport per unit for shipments of size Q, $ per unit S = order processing cost, $ per order I = annual carrying cost, % C = product value, $ per unit Q = size of purchase order, units Under noninclusive price discounts, price is an average, determined by the number of units in each break. For example, if 250 units are to be ordered, the average price per unit would be computed as:

144

P250 =

(100  $700) + (100  $680) + (50  $670)  $686.00 250

A table of annual costs can now be developed, as shown in Table 1. To the nearest 50 units, the optimal purchase quantity should be 250 units. (2) If the manufacturer's pricing policy were one where the prices in each quantity break included all units purchased, should Walter change his replenishment order size? The average price per unit is more easily determined in this case than the previous one. Since all units are included in the price break back to the first unit, the average price is simply the price associated with a given purchase quantity. Finding the optimal purchase quantity is simply a matter of determining the total cost for the quantities, found by the economic order quantity formula, assuming these quantities are feasible, and for the quantities at the transport rate-weight break. The comparison is made among the total costs of these alternatives. These costs are shown in Table 2. The order quantities, as determined by the economic order quantity formula for the base price of $700, would be Q* =

2 DS 2(1500)( 25) = = 18.3, or 18 units 0.3(700 + 45) IC

where C is the $700 price per unit at Baltimore plus the $45 transport cost from Baltimore, as determined by an LTL shipment (18 units  250 lb. = 4,500 lb.) at $18 2.5 cwt. = $45 per unit. The Q values for the other prices in the schedule lie outside the feasible range of the price used to compute Q. The optimal strategy is to purchase 201 units per order, which is one unit into the last price break. Yes, Walter should alter his buying strategy. TABLE 1 Annual Costs by Quantity Purchased for Noninclusive Price Discounts Average Purchase Transport Quantity price cost cost 18a $700.00 $1,050,000 $67,500 50 700.00 1,050,000 67,500 100 700.00 1,050,000 67,500 693.33 1,039,995 67,500 150 692.50 1,038,750 45,000 160 200 690.00 1,035,000 45,000 686.00 1,029,000 45,000 250 300 683.33 1,024,995 45,000 400 680.00 1,020,000 45,000 a EOQ at a price of ($700 + 45+ 0.625) per unit. b First price break. c Transport rate break. d Second price break.

Ordering cost $2,083 750 375 250 234 188 150 125 94

Carrying cost $2,013 5,592 11,184 16,619 17,355 21,619 26,873 32,128 42,638

Warehouse cost $0 0 0 0 0 0 0 0 1,000

Total cost $1,121,596 1,123,842 1,129,059 1,124,364 1,101,339 1,101,807 1,101,023Opt. 1,102,248 1,108,732

145

TABLE 2 Annual Costs by Quantity Purchased for Inclusive Price Discounts Average Purchase Transport Ordering Quantity price cost cost cost $700.00 $1,050,000 $67,500 $2,083 18 680.00 19 670.00 19 680.00 1,020,000 67,500 371 101 680.00 1,020,000 45,000 234 160 670.00 1,005,000 45,000 187 201 a Feasible EOQ at a price of ($700 + 45 + 0.625) per unit. b Infeasible EOQ at a price of ($680 + 45 + 0.625) per unit. c Infeasible EOQ at a price of ($670 + 30 + 0.625) per unit. d First price break. e Transport rate break. f Second price break.

Carrying cost $2,013

10,993 17,175 21,124

Warehouse cost $0

0 0 0

Total cost $1,121596 Infeasible Infeasible 1,098,864 1,082,409 1,071,311Opt.

146

CHAPTER 11 THE STORAGE AND HANDLING SYSTEM

All questions in this chapter require individual judgment and response. No answers are offered.

147

CHAPTER 12 STORAGE AND HANDLING DECISIONS 2 Various alternatives are evaluated in Tables 12-1 to 12-4. The annual costs of each alternative are plotted in Figure 12-1. The best economic choice is to use all public warehousing.

148

or a Pure Public Warehouse Strategy

Privately-operated Rented Space equirePrivate Monthly Monthly Rented Monthly Monthly ments, allofixed cost variable allostorage handling Monthly sq. ft.b cation cost cation cost cost total cost 62,500 0% $0 $0 100% $30,000c $50,000d $80,000 50,000 0 0 0 100 24,000 40,000 64,000 37,500 0 0 0 100 18,000 30,000 48,000 25,000 0 0 0 100 12,000 20,000 32,000 12,500 0 0 0 100 6,000 10,000 16,000 3,125 0 0 0 100 1,500 2,500 4,000 15,625 0 0 0 100 7,500 12,500 20,000 28,125 0 0 0 100 13,500 22,500 36,000 37,500 0 0 0 100 18,000 30,000 48,000 43,750 0 0 0 100 21,000 35,000 56,000 50,000 0 0 0 100 24,000 40,000 64,000 56,250 0 0 0 100 27,000 45,000 72,000 21,875 $0 $0 $202,500 $337,500 $540,000 ($5/lb.) t.) = Thruput (lb.)  (1/ 2 turns)  (1/0.40 storage space ratio)  (0.1 cu. ft./$)  (1/10 ft.)  (5 $/lb.)

2 and 100% of the demand through the rented warehouse, then 1,000,000  (1.00/2)  0.06 = $30,000 = $50,000

TABLE 12-2 Costs for a Mixed Warehouse Strategy Using a 10,000 Square Foot PrivatelyOperated Warehouse Privately-operated Rented Space WarerequirePrivate Monthly house Monthly Rented Monthly Monthly ments, allofixed cost variable thruput, allostorage handling Monthly sq. ft.b cation Month lb.a cost cation cost cost total cost 84% $25,200f $42,000g $80,192 Jan. 1,000,000 62,500 16% $9,792d $3,200e Feb. 800,000 50,000 20 9,792 3,200 80 19,200 32,000 64,192 Mar. 600,000 37,500 27 9,792 3,200 73 13,140 21,900 48,032 Apr. 400,000 25,000 40 9,792 3,200 60 7,200 12,000 32,192 May 200,000 12,500 80 9,792 3,200 20 1,200 2,000 16,192 June 50,000 3,125 100 9,792 1,000 0 0 0 10,792 July 250,000 15,625 64 9,792 3,200 36 2,700 4,500 20,192 Aug. 450,000 28,125 36 9,792 3,200 64 8,640 14,400 36,032 Sept. 600,000 37,500 27 9,792 3,200 73 13,140 21,900 48,032 Oct. 700,000 43,750 23 9,792 3,200 77 16,170 26,950 56,112 Nov. 800,000 50,000 20 9,792 3,200 80 19,200 32,000 64,192 56,250 18 9,792 3,200 82 22,140 36,900 72,032 Dec. 900,000 Totals 6,750,000 421,875 $117,504 $36,200 $147,930 $246,550 $548,184 a Thruput (lb.) = Sales ($)/($5/lb.) b Space requirements (sq. ft.) = Thruput (lb.)  ½  (1/0.40)  (0.1/10)  5 = Thruput (lb.)  0.0625 c 10,000/62,500 = 0.16 d (3510,000/20) + 1010,000/12 = $9,792 per month e 1,000,0000.160.02 = $3,200 f Given a turnover ratio of 2 and 84% of the demand through the rented warehouse, then 1,000,000  (0.84/2)  0.06 = $25,200 d 1,000,000  0.84  0.05 = $42,000

150

TABLE 12-3 Costs for a Mixed Warehouse Size Strategy Using a 30,000 Square Foot PrivatelyOperated Warehouse Privately-operated Rented Space WarerequirePrivate Monthly house Monthly Rented Monthly Monthly ments, allofixed cost variable thruput, allostorage handling Monthly sq. ft.b cation Month lb.a cost cation cost cost total cost $29,375d $9,600e 52% $15,600f $26,900g $80,575 Jan. 1,000,000 62,500 48%c Feb. 800,000 50,000 60 29,375 9,600 40 9,600 16,000 64,575 Mar. 600,000 37,500 80 29,375 9,600 20 3,600 6,000 48,575 Apr. 400,000 25,000 100 29,375 8,000 0 0 0 37,375 May 200,000 12,500 100 29,375 4,000 0 0 0 33,375 June 50,000 3,125 100 29,375 1,000 0 0 0 30,375 July 250,000 15,625 100 29,375 5,000 0 0 0 34,375 Aug. 450,000 28,125 100 29,375 9,000 0 0 0 38,375 Sept. 600,000 37,500 80 29,375 9,600 20 3,600 6,000 48,575 Oct. 700,000 43,750 69 29,375 9,600 31 6,510 13,020 58,505 Nov. 800,000 50,000 60 29,375 9,600 40 9,600 16,000 64,575 56,250 53 29,375 9,600 47 12,690 21,150 72,815 Dec. 900,000 Totals 6,750,000 421,875 $352,500 $94,200 $61,200 $104,170 $612,070 a Thruput (lb.) = Sales ($)/($5/lb.) b Space requirements (sq. ft.) = Thruput (lb.)  ½  1/0.40  (0.1/10)  5 = Thruput (lb.)  0.0625 c 30,000/62,500 = 0.48 d (3530,000/20) + 1030,000/12 = $29,375 per month e 1,000,0000.480.02 = $9,600 f Given a turnover ratio of 2 and 52% of the demand through the rented warehouse, then 1,000,000  (0.52/2)  0.06 = $15,600 d 1,000,000  0.52  0.05 = $26,000

151

TABLE 12-3 Costs for a Mixed Warehouse Size Strategy Using a 40,000 Square Foot PrivatelyOperated Warehouse Privately-operated Rented Space WarerequirePrivate Monthly house Monthly Rented Monthly Monthly ments, allofixed cost variable thruput, allostorage handling Monthly sq. ft.b cation Month lb.a cost cation cost cost total cost $39,167 $12,800e 36% $10,800f $18,000g $80,767 Jan. 1,000,000 62,500 64%c Feb. 800,000 50,000 80 39,167 12,800 20 4,800 8,000 64,767 Mar. 600,000 37,500 100 39,167 12,800 0 0 0 51,167 Apr. 400,000 25,000 100 39,167 8,000 0 0 0 47,167 May 200,000 12,500 100 39,167 4,000 0 0 0 43,167 June 50,000 3,125 100 39,167 1,000 0 0 0 40,167 July 250,000 15,625 100 39,167 5,000 0 0 0 44,167 Aug. 450,000 28,125 100 39,167 9,000 0 0 0 48,167 Sept. 600,000 37,500 100 39,167 12,000 0 0 0 51,167 Oct. 700,000 43,750 91 39,167 12,800 0 1,890 3,150 57,007 Nov. 800,000 50,000 80 39,167 12,800 20 4,800 8,000 64,767 56,250 71 39,167 12,800 29 7,830 13,050 72,847 Dec. 900,000 Totals 6,750,000 421,875 $470,004 $115,000 $30,120 $50,200 $665,324 a Thruput (lb.) = Sales ($)/($5/lb.) b Space requirements (sq. ft.) = Thruput (lb.)  ½  1/0.40 ( 0.1/10)  5 = Thruput (lb.)  0.0625 c 40,000/62,500 = 0.64 d (3540,000/20) + 1040,000/12 = $39,167 per month e 1,000,0000.640.02 = $12,800 f Given a turnover ratio of 2 and 52% of the demand through the rented warehouse, then 1,000,000  (0.36/2)  0.06 = $10,800 d 1,000,000  0.36  0.05 = $18,000

152

FIGURE 12-1 Total Annual Costs for a Combined Warehouse Size Using Private and Public Warehouse Space

670

Total cost, $000s

650 630 610 590 570 550 530 0

10,000

30,000

40,000

Private warehouse space, sq. ft.

3 The annual cost of public warehousing is:

Handling Storage Total

$ 600,000 300,000 $ 900,000

The costs of private warehousing are: Annual operating $ 250,000 Annual lease payment 3150,000 = 450,000 Other fixed (one time) 400,000 The savings in operating costs of lease vs. public warehousing is: Savings = $900,000  250,000 = $650,000/yr.

153

TABLE 12-5 Ten-Year Cash Flow Stream for Public vs. Leased Warehouse Comparison Savings DepreSavings less Savings: Pre-tax ciation less depreAfter-tax Lease vs. net cash schedepreTaxes ciation Savings net cash Year public flow dule ciation (35%) & tax less tax flow 0 $0 (3,050)a $0 0 0 0 0 ($3,050) 593 208 385 442c 442 1 650 650 57b 2 650 650 57 593 208 385 442 442 3 650 650 57 593 208 385 442 442 4 650 650 57 593 208 385 442 442 5 650 650 57 593 208 385 442 442 6 650 650 57 593 208 385 442 442 7 650 650 58 592 207 385 443 443 8 650 650 0 650 228 422 442 442 9 650 650 0 650 228 422 442 442 650 0 650 228 422 442 442 10 650 $6,500 $3,450 $400 $6,100 $2,139 $3,961 $4,361 $1,311 a Capitalization lease plus initial cash outlay, i.e., $2,650,154 + 400,000 = $3,050,154 b Depreciation charge for each of seven years is 1/7 = 0.1429 such that 400,0000.1429 = $57,143 c Add back depreciation, i.e., 385 + 57 = $442

Discount factor 1/(1+i)j 0.9009 0.8116 0.7312 0.6587 0.5935 0.5346 0.4817 0.4339 0.3909 0.3522 NPV =

Discounted cash flow ($3,050) 398 359 323 291 262 236 213 183 165 149 ($471)

154

Capitalizing the lease over ten years, we have: (1  011 . )10  1 PV  450,000  $2,650,154 011 . (1  011 . )10 The initial investment in $000s then is: Initial investment = $2,650 + 400 = $3,050 The ten-year cash flow stream is shown in Table 12-5. Since the savings are expressed to favor leasing and the net present value is negative, choose public warehousing. 4 Given:

k = $210/sq ft. S = 100,000 sq. ft. C = $0.01/ft.10,000 = $100/ft. The width is: C  8k S 2C  8k

W* 

100  8( 210) 100,000 2(100)  8( 210)



 308 ft. The length is: L*  S / W *  100,000 / 308  325 ft.

5 Space layout according to text Fig. 12-4(a) can be determined by the application of equations 12-8 and 12-9. These equations specify the best number of shelf spaces and the best number of double racks, respectively. Equations 12-10 and 12-11 give the length and width of the building. The optimal number of shelf spaces would be:

m1* 

1 L

 dCh  2aCs  2C p   K ( w  a ) L     2h  2( dCh  C p )  

1  400,000( 0.001)  2(10)( 0.50)  2(3.00)   50,000(8  10)( 4)    4  2( 400,000)( 0.001)  3.00 2( 4)    120.48, or 121 

155

The optimal number of double racks would be: n1*  

1 wa

 2( dCh  C p )   K ( w  a ) L     2h  dCh  2aCs  2C p  

   50,000(8  10 )( 4)  1 2[( 400,000)( 0.001)  3.00]   8  10  400,000( 0.001)  2(10)( 0.50)  2(3.00)   2( 4 ) 

 52 The warehouse length would be: u1  n1* ( w  a )  52(8  10)  936 ft. and the width would be: v1  2a  m1* L  2(10)  121( 4)  504 ft. 6 According to Equation 12-17, the number of truck doors can be estimated by:

N

DH CS

Therefore, N = (7512,000) 3/(312,000) 8 = 9.37, or 10 doors 7 Summarizing the given information as follows:

Initial investment Useful life Salvage value @15% of initial cost Annual operating expenses Return on investment before tax

Three Five Seven type 1 type 2 type 3 units units units $60,000 $50,000 $35,000 10 yr. 10 yr. 10 yr. $ 9,000

$ 7,500

$ 5,250

$ 6,000

$12,500

$21,000

20%

20%

20%

An initial solution to this problem can be found through a discounted cash flow analysis. Three alternatives are to be evaluated.

156

 (1  0.2)10  1   1  9,000 PV1  60,000  6,000 10  10   (1  0.2)   0.2(1  0.2)  . )  60,000  6,000( 4.2)  9,000( 016  $83,760 PV2  50,000  12,500( 4.2 )  7,500( 016 . )  50,000  52,500  1,200  $101,300 . ) PV3  35,000  21,000( 4.2)  5,250( 016  35,000  88,200  840  $122,360 The low present value of the Type 1 truck indicates that from among these three alternatives, this would be the best buy. 8 Given:

Initial cost of equipment = $4,000 Operating costs 500 + 40(t - 1)2 - 30(t - 1) Salvage value Sn = I(1 – t/7) Rate of return on investment = 20% Replacement is expected to be with equipment of like kind The best replacement year can be found by comparing the equivalent annual cost of a sequence of similar equipment replaced every n years. The equivalent annual cost is:







AC n  I   j1 ( C j /(1  i ) j )  ( S n /(1  i ) n ) i (1  i ) n /(1  i ) n  1 n

Solving this equation for different years is facilitated if the equation is set up in tabular form, as shown in Table 12-6. The equipment should be replaced at the end of the third year of service although a 5year replacement cycle is also attractive.

157

TABLE 12-6 Equivalent Annual Cost Computations for Problem 8 (4) (5)=(1+2-3)(4) (3) (1) (2) Factor Salvage Operating Equivalent annual Year, value, Initial costs, i(1  i ) n n j n S n / (1  i) cost, ACn n investment, I  C j / (1 i) (1  i )  1 1 $4,000 $416 $2,857 1.20 $1,871 a 1,984 0.65 1,821 2 4,000 770 1,323 0.47 3 4,000 1,117b 1,783  4 4,000 1,488 827 0.39 1,818 5 4,000 1,898 459 0.33 1,795 6 4,000 2,350 191 0.30 1,848 7 4,000 2,841 0 0.28 1,915 a b

$416 + [500 + 40(2-1)2 – 30(2-1)]/(1+0.20)2 = $770 $770 + [500 + 40(3-1)2 – 30 (3-1)]/1+0.20)3 = $1,117

9 (a1) Layout by popularity involves locating the more frequently ordered items closest to the outbound dock. Based on the average number of daily orders on which the item appears, the items closest to the outbound dock would be ranked as follows: B,I,E,A,F,H,J,C,G,D The storage space might then be used as follows.

Inbound

D

H, F, A

A, E

B

J, C, G

H, J

E

B, I, E

Outbound (a2) Layout by cube places the smallest items nearest the outbound dock. Using the individual item size, the ranking would be as follows: A,E,I,C,J,H,G,B,F The layout of items in the storage bays would be:

158

Inbound

F, D

B

H, D

H, G, B

D, J, C

E, A

E, I, C

E

Outbound (a3) The cube-per-order index is created by ratioing the average required cubic footage of a product to the average number of daily orders on which the item is requested. Hence, this index is found as follows: (1) (2) Space required Daily Product cu. ft. orders A 5,000a 56 B 30,000 103 C 15,000 27 D 17,000 15 E 55,000 84 F 11,000 55 G 7,000 26 H 28,000 45 I 13,000 94 J 9,000 35 a 500 sq. ft. stacked 10 ft. high

(3)=(1)/(2) O index 89 291 556 1,133 655 200 269 622 138 257

Locating the products with the lowest index values nearest to the outbound dock results in the following ranking and layout: A,I,F,J,G,B,C,H,E,D

159

Inbound

D

E

E, F

E

H, C

B

F, I, A

F, J, G, B

Outbound (b) All of the above methods assume (1) that the product is moved to the storage locations in large unit loads but retrieved from the storage locations in relatively small quantities and (2) that only one product is retrieved during an out-and-back trip. Therefore, these methods do not truly apply to the situation of multiple picks on the same trip. However, they may be used with some degree of approximation if the products can be aggregated as one and grouped together or zoned in the same section of the warehouse. 10 This is an extra challenging problem that requires some knowledge of linear programming. It may be formulated as follows.

Let Xij represent the amount per 1,000 units of product j stored in location i. Let Cij be the handling time associated with storage bay i and product j. Gj is the capacity of a bay for product j and Rj is the number of units of product j required to be stored. The linear programming statement is: Objective function Zmin = .90X11 + .75X12 + .90X13 + .80X21 + .65X22 + .95X23 + .60X31 + .70X32 + .65X33 + .70X41 + .55X42 + .45X43 + .50X51 + .50X52 + .45X53 + .40X61 + .45X62 + .35X63

Subject to: Capacity restrictions on bays

160

20X11 + 33.3X12 + 16.7X13  100 20X21 + 33.3X22 + 16.7X23  100 20X31 + 33.3X32 + 16.7X33  100 20X41 + 33.3X42 + 16.7X43  100 20X51 + 33.3X52 + 16.7X53  100 20X61 + 33.3X62 + 16.7X63  100

and storage requirements restrictions on products X11 + X21 + X31 + X41 + X51 + X61  11 X12 + X22 + X32 + X42 + X52 + X62 

4

X13 + X23 + X33 + X43 + X53 + X63  12

Solving the linear programming problem by means of any standard transportation code of linear programming, such as LNPROG in LOGWARE, yields: X12 X21 X22 X31 X43 X51 X53 X63

= = = = = = = =

The total minimum handling time is 138.68 hours

1.610 1.020 2.390 5.000 5.988 4.980 0.024 5.988

where Xs are in thousands of units. That is, product 1 should be stored in bays 3, 4, and 5 in quantities of 1,020, 5,000, and 4,980, respectively. Product 2 should be stored in bays 1 and 2 in quantities of 1,610 and 2,390, respectively. Product 3 should be stored in bays 4, 5, and 6 in quantities of 5,988, 24, and 5,988, respectively. Graphically, this is:

Bay Product 1 2 3 % of bay capacity

1

2

3 5,000

1,610

1,020 2,390

53.7

100.0

100.0

4

5

6

4,980 5,988

24

5,988

100.0

100.0

100.0

Require -ments 11,000 4,000 12,000

161

CHAPTER 13 FACILITY LOCATION DECISIONS 1 (a) The center-of-gravity method involves finding the X,Y coordinates according to the formulas:

V R X X V R i

i

i

i

i

i

i

and

V R Y Y  V R i

i i

i

i

i

i

These formulas can be solved for in tabular form as follows. Point P1 P2 M1 M2 M3

X Y 3 8 8 2 2 5 6 4 8 8 Totals

Vi 5,000 7,000 3,500 3,000 5,500

Ri .040 .040 .095 .095 .095

ViRi 200.0 280.0 332.5 285.0 522.5 1620.0

ViRiXi 600.0 2240.0 665.0 1710.0 4180.0 9395.0

ViRiYi 1600.0 560.0 1662.5 1140.0 4180.0 9142.5

Now, X

9,395.0  5.8 1,620.0

and Y 

9,142.5  5.64 1,620.0

This solution has a total cost for transportation of $53,614.91. This problem may also be solved using the COG module in LOGWARE. (b) Solving for the exact center-of-gravity method requires numerous computations. We now use the COG module of LOGWARE to assist us. A table of partial results is shown below.

162

Iteration number 0 1 2 3 . . . 50

X coord 5.799383 5.901199 5.933341 5.941554 . . . 5.939314

Y coord 5.643518 5.518863 5.446919 5.402429 . . . 5.317043

Total cost 53,614.91  COG 53,510.85 53,483.97 53,474.60 . . . 53,467.71

After 50 iterations, there is no further change in total cost. The revised coordinates are X = 5.94 and Y = 5.32 for a total cost of $53,467.71. (c) The center-of-gravity solution can be one that is close to optimum when there are many points in the problem and no one point has a dominant volume, that is, has a larger volume relative to the others. Otherwise, the best single location can be at a dominant location. The exact center-of-gravity approach has the capability to find the minimal cost location. Although the COG model only considers transportation costs that are constant per mile, the transportation cost can be the major consideration in single facility location. However, other costs such as labor, real estate, and taxes can also be important in selecting one location over another. These are not directly considered by the model. Although the COG model may seem of limited capability, it is a useful tool for locating facilities where transportation costs are dominant. Location of oil wells in the Gulf, truck terminals, and single warehouses are examples of application. It also can be quite useful to provide a starting solution to more complex location models. (d) Finding multiple locations by means of the center-of-gravity approach requires assigned supply and demand volumes to specific facilities and then solving for the center of gravity for each. In this problem, there are 3 market combinations that need to be considered. This creates 3 scenarios that need to be evaluated. They can be summarized as follows. Point volumes appear in the body of the table. Scenario

Whse 1

P1 P2 1458a 2042

2

3542

4958

1

2708

3792

2

2292

3208

1

3750

5250

M1 3500

M2

M3

I 3000 3500

5500

3000

II 5500 3500

5500

III 2 1250 1750 3000 Allocated as a proportion of the volume to be served through the warehouse. That is, 50003500/(3500 + 3000 + 5500) = 1458. The volumes associated with other supply points are computed similarly. a

163

The COG module in LOGWARE was used to find the exact centers of gravity for each warehouse in each scenario. The computational results are: Scenario I II III

Warehouse 1 X Y 2.00 5.00 5.84 4.04 7.06 7.28

Warehouse 2 X Y 7.88 7.80 8.00 8.00 6.00 4.00

Total cost $39,050 35,699  46,568

Scenario II appears to be the best. 2 (a) The center-of-gravity formulas (Eqs. 13-5 and 13-6) can be solved using the COG module of LOGWARE, or they can be solved in tabular form as shown below. Point A B C D E F G H I J

X Y 50 0 10 10 30 15 40 20 10 25 40 30 0 35 5 45 40 45 20 50 Totals

Vi 9,000 1,600 3,000 700 2,000 400 500 8,000 1,500 4,000

Ri .75 .75 .75 .75 .75 .75 .75 .75 .75 .75

ViRi 6,750 1,200 2,250 525 1,500 300 375 6,000 1,125 3,000 23,025

ViRiXi 337,500 12,000 67,500 21,000 15,000 12,000 0 30,000 45,000 60,000 600,000

ViRiYi 0 12,000 33,750 10,500 37,500 9,000 13,125 270,000 50,625 150,000 586,500

Now, X

600,000 .  261 23,025

and Y 

586,500  25.5 23,025

The total cost of this location is $609,765. The exact center-of-gravity coordinates are: X  2351 . , Y  26.98 with a total cost of $608,478. (b) The number of points, even in this small problem, requires us to apply some heuristics to find which patient clusters should be assigned to which warehouses. We will use a clustering technique whereby patient clusters are grouped by proximity until two clusters are found. The procedure works as follows.

164

 There are as many clusters as there are points, which is 10 in this case.  The closest points are found and replaced with a single point with the combined volume located at the center of gravity point. There is now one less cluster.  The next closest two points/clusters are found, and they are further combined and located at their center of gravity.  The process continues until only two clusters remain. The centers of gravity for these two clusters will be the desired clinic locations. Applying the clustering technique, we start by combining points D and F into cluster DF. X

40( 700)( 0.75)  40( 400)( 0.75)  40.00 700( 0.75)  400( 0.75)

and Y 

20( 700 )( 0.75)  30( 400)( 0.75)  23.64 700( 0.75)  400( 0.75)

Continuing this process, we would form two clusters containing A, C, D, and F, and B, E, G, H, I, and J. The centers of gravity would be: Cluster 1 - ACDF X  50.00, X  0.00 Cluster 2 - BEGHIJ X  5.00, Y  45.00 for a total cost of $241,828. These are the same results obtained from the MULTICOG module in LOGWARE. (c) The second clinic can save $608,278  241,828 = $366,458 in direct costs annually. This savings does not exceed the annual fixed costs of $500,000 required to maintain a second clinic. On economic grounds, it should not be built. 3 (a) The center-of-gravity location can be determined by forming the following table or by using the COG module in LOGWARE. The coordinates for each location must be approximated.

165

Point A B C D E F G H I

X Y 1.5 6.6 4.7 7.3 8.0 7.1 1.5 4.0 5.0 4.9 8.5 5.1 1.5 1.3 4.4 1.8 7.8 1.8 Totals

Vi 10,000 5,000 70,000 30,000 40,000 12,000 90,000 7,000 10,000

Ri .10 .10 .10 .10 .10 .10 .10 .10 .10

ViRi 1,000 500 7,000 3,000 4,000 1,200 9,000 700 1,000 27,400

ViRiXi 1,500 2,350 56,000 4,500 20,000 10,200 13,500 3,080 7,800 118,930

ViRiYi 6,600 3,650 49,700 12,000 19,600 6,120 11,700 1,260 1,800 112,430

The center-of-gravity coordinates are: X

118,930  4.34 27,400

and Y 

112,430 .  410 27,400

with a total cost of $195,966. (b) After 100 iterations in the COG module, the exact center of gravity was found to be: X  4.75, Y  4.62 with a total cost of $195,367. In this case, using the exact center of gravity coordinates as compared with the approximate ones reduced costs by only: 195,966  195,367  100  0.3% 195,966 (c) Additional costs can be included in the analysis, although not necessarily in the model. The COG model can evaluate the variable costs of location. Other costs are compared with these. 4 We begin by developing a 3-dimensional transportation problem. The cost matrix is developed in the same manner as that in text Fig. 13-11. The initial throughput of W1 and W2 is found by assuming that an equal amount of the customer demand flows through each warehouse. The cell cost for W1-C1 would be:

[100(200,000/2)0.7]/(200,000/2 )+ 2 + 4 + 0 = 3.2 + 2 + 4 = 9.2

166

In fact, the cell costs are identical to those in text Fig. 13-11, except that there is no fixed cost element. Using the transportation method of linear programming (e.g., the TRANLP module in LOGWARE), the cell cost and solution matrix for iteration 1 is shown in Fig. 13-1. The solution shows that only W2 remains, and the solution process can be terminated. A summary of the costs is shown in Table 13-1. The total cost is $2,213,714, and the product is produced in plant P2 and stocked in warehouse W2. No further iterations are needed since only one warehouse is used and no further dropping of warehouses is possible. TABLE 13-1 Summary Information for Solution to Problem 4 Whse 1 Whse throughput Costs: Transportation Inbound Outbound Inventory Warehousing Fixed Production Total

Whse 2

0

200,000

$0 0 0 0 0 0

$400,000 300,000 513,714 200,000 0 800,000 $2,213,714

Iteration 2 A repeat of the iteration 1 solution.

Plants

P1 P2

Whses

W1

W2 Capacity/ Req’mts a

Warehouses W1 W2 9 4 0 0 6 8 0 200,000 99a 0 60,000 99 0 799,999 60,000

999,999

Stop iterating.

Customers C2 C3 99 99

C1 a

99

Capacity 60,000

99

99

99 999,999

9.2

8.2

10.2

0

0

0

60,000

6.2 50,000

5.2 100,000

6.2 50,000

999,999

50,000

100,000

50,000

High rate of $99/unit for an inissible cell.

FIGURE 13-1 Cell Cost and Solution Matrix for Iteration 1 of Problem 4 5 We begin by forming the cell cost matrix of a 3-dimensional transportation problem, as shown in Figure 13-2. It is similar to the text Figure 13-1 except that the capacity for warehouse 1 is set at 75,000. Solving the problem by means of the transportation method shows the solution given in Figure 13-2.

167

Plants

P1 P2

Whses

W1

Warehouses W1 W2 9 4 60,000 6 8 40,000 100,000 100 0 899,999 100 0

100

Customers C2 C3 100 100

100

100

100

9.7

8.7 100,000 7.2

10.7

C1

Capacity 60,000 999,999

8.2 50,000

999,999 8.2 50,000

W2 100,000 Capacity/ Req’mts 999,999 100,000 50,000 100,000 50,000 FIGURE 13-2 Cell Cost and Solution Matrix for Iteration 1 of Problem 5 Given the solution from iteration 1, the per-unit inventory and fixed costs are revised. Inventory W1

$100(100,000) 0.7  $3.16 / unit 100,000 units

W2

$100(100,000) 0.7  $3.16 / unit 100,000 units

Fixed W1 W2

$100,000/100,000 = $1.00/unit $400,000/100,000 = $4.00/unit

Adding outbound transportation and warehouse handling to per-unit inventory and fixed costs gives the following cell costs.

W1 W2

C1 10.2 10.2

C2 9.2 9.2

C3 11.2 10.2

Revising the warehouse-customer cell costs and solving gives the same warehouse throughputs, so cell costs will no longer change. A stopping points is reached. The solution is the same as that in Figure 13-2. A summary of the costs is:

168

Cost type Production Inbound transportation Outbound transportation Fixed Inventory carrying Handling Subtotal Total

Warehouse 1 100,000 cwt. 60,0004 = $240,000 40,0004 = 160,000 60,0000 = 0 40,0004 = 160,000 100,0003 = 300,000

Warehouse 2 100,000 cwt. 100,0004 = $400,000 100,0002 =

200,000

50,0002 = 50,0002 =

100,000 100,000 100,000 400,000 100(100,000)0.7 = 316,228 100(100,000)0.7 = 316,228 100,000 100,0002 = 200,000 100,0001 = $1,476,228 $1,616,228 $3,092,456

Compared with the costs from the text example, the cost difference is $3,092,456  2,613,714 = $478,742. This is the penalty for restricting a warehouse with economic benefit to the network. 6 Prepare a matrix for a 3-dimensional transportation problem like that in text Figure 1311, except that the per-unit cell costs for warehouse 2 to customer are reduced by $1/unit to reflect the reduction in that warehouse’s fixed costs. That is, $200,000/200,000 = $1/unit instead of $400,000/200,000 = $2/unit. The matrix setup and first iteration solution are shown in Figure 13-3. Customers

Warehouses

W1 4a Plants

C1 99b

C2 99

C3 99

P1

60,000 8

P2 Warehouses

W2 9

Plant & warehouse capacities

W1

W2 Warehouse capacity & customer demand

0 60,000 99b

60,000

6 200,000 99

99 9.7

0 799,999

999,999c

99

99 999,999c

d

8.7

10.7

7.2e 50,000

6.2 100,000

7.2 50,000

50,000

100,000

50,000

60,000 999,999c

a

Production plus inbound transport rates, that is, 4 + 0 = 4. Used to represent an infinitely high cost. c Used to represent unlimited capacity. d Inventory carrying, warehousing, outbound transportation, and fixed rates, that is, 3.2 + 2 + 4 + 0.5 = 9.7. e 3.2 + 1 + 2 + 1 = 7.2. b

FIGURE 13-2

Cell Cost and Solution Matrix for Iteration 1 of Problem 6

169

The results show that one warehouse is to be used. Further computations are not needed, as further warehouse consolidation is not possible. The total network costs are the same as those in the text example minus the $200,000 reduction in fixed costs for a total coat of $2,413,714. 8 This problem requires us to rework the dynamic programming solution to the example problem given in the text. The only change is that the cost of moving from one location to another is now $300,000 instead of $100,000. We begin with the last year and determine the best action based on the highest net profits. The action will be to move (M) or to stay (S). For example, given the discounted moving cost of 300,000/(1 + 0.20)(4) = $144,676, we evaluate each course of action, assuming that we are in location alternative A at the end of year 4. From the location profits of text Table 13-6, we generate the following table for location A. Alternative (x) A B P5(A) = max C D E

Location profit $1,336,000 1,398,200 1,457,600 1,486,600 1,526,000

Moving cost -

0 144,676 144,676 144,676 144,676

Net profit = $1,336,000 = 1,253,524 = 1,312,924 = 1,341,924 = 1,381,324

The best action in the beginning of the 5th year, if we are already in location A, is to move to location E. This is an entry in Table 13-2. Once each of the five alternatives is evaluated for the 5th year, then the 4th year alternatives are evaluated. The moving cost is 300,000/(1 + 0.2)(3) = $173,611. We now include the profits for the subsequent years in our calculations. After Table 13-2 is completed, we search the first column for the highest cumulative profit. This is initially to locate in location D and remain there throughout the subsequent years. 9 (a) Using PMED software in LOGWARE and the PMED02.DAT database, solve for the number of locations from 1 to 9. The best locations for each number of sites are given in the table below.

170

TABLE 13-2 Location-Relocation Strategies Over a Five-Year Planning Horizon with Cumulative Profits Shown from Year j to Year 5 for Problem 8 Year from present date j Warehouse 5th 1st 4th 3rd 2nd location StraStraStraStraaltertegya P2(x) tegya P3(x) tegya P4(x) tegya P5(x) P1(x) natives (x) A $3,557,767 SA $3,363,767 SA $3,007,667 MD $2,268,289 MD $1,381,324 3,379,667 SB 3,007,667 MD 2,268,289 MD 1,398,000 B 3,556,167 SB 3,500,900 SC 3,156,200 SC 2,319,800 SC 1,457,600 C 3,673,200 SC 3,553,600 SD 3,216,000 SD 2,459,900 SD 1,486,600 D b 3,720,300 SD 3,374,700 SE 3,071,300 SE 2,355,800 SE 1,526,000 E 3,534,100 SE a Strategy symbol refers to “staying” (S) in the designated location or “moving” (M) to a new location as indicated. b Arrows indicate maximum profit location plan when warehouse is initially located at D.

Strategya ME SB SC SD SE

171

Number of sites 1 2 3 4 5 6 7 8 9

Total cost $11,694,821 7,634,242 7,596,604 8,381,775 9,455,339 11,536,669 13,909,997 16,581,348 20,000,338

The optimal number of sites is 3 and they are to be located at Cincinnati, Phoenix, and Denver.

(b) The optimal cost for four sites is $8,381,775, as found in part a. The company operates the same sites as found in the optimal solution for four sites. Therefore, the cost savings comes from a reassignment of customers to the sites. The savings is $35,000,000  8,381,775 = $26,618,225 without any major investment. All of this may not be recovered since there may be other operating costs included that are not directly associated with location. The savings between the optimal 4 sites and the optimal 3 sites is $8,381,775  7,596,604 = $785,171. There is also the reclamation of the salvage value of two incinerators (close Chicago and Atlanta) and the construction of a new one (Cincinnati). If the real estate recovery cost exceeds the new construction cost, that would add to the savings, otherwise a ROI estimation is needed to see there is an adequate return from the savings on the net investment. Reallocation of customers among the existing incinerators is certainly attractive and the reduction of the number of incinerators from 4 to 3 is also attractive as long as there is no net investment required.

(c) Increase the annual volume for Los Angeles and Seattle markets by a factor of 10 and re-solve the problem as in part a. Selected results are as follows: Number of sites 2 3 4 5

Total cost $10,506,286 9,831,846 9,810,216 10,595,387

The optimal number of sites is four. An additional site at Seattle is needed, compared with the three locations found in part a.

172

10 (a) We can apply Huff's model of retail gravitation to this problem. The solution table (Table 13-3) can be developed. Summarizing, branch A can be expected to attract 11,735/(11,735 + 11,765) = 49.9% of the customers, and branch B should attract the remaining 50.1%. TABLE 13-3 Estimate of the Number of Customers Attracted to Each Branch Bank Pij  S j / Tij2

Time to ja Customer i 1 2 3 4 5 6 7 8 9 a

A 0.28 0.10 0.28 0.40 0.20 0.50 0.45 0.57 0.67

B 0.72 0.50 0.45 0.20 0.40 0.50 0.28 0.20 0.54

Tij2 A 0.08 0.01 0.08 0.16 0.04 0.25 0.20 0.32 0.45

S j / Tij2 B 0.52 0.25 0.20 0.04 0.16 0.25 0.08 0.04 0.29

A 12.5 100.0 12.5 6.3 25.0 4.0 5.0 3.1 2.2

B 1.4 2.8 3.5 17.5 4.4 2.8 8.8 17.5 2.4

S j

A 0.90 0.97 0.78 0.26 0.85 0.59 0.36 0.15 0.48

j

E ij  Pij Ci

/ Tij2 B 0.10 0.03 0.22 0.74 0.15 0.41 0.64 0.85 0.52

A 900 1,940 3,120 1,820 850 885 1,440 300 480 11,735

B 100 60 880 5,180 150 625 2,560 1,700 520 11,765

Time = ( X i  X )2  (Yi  Y )2 / 50

(b) The economic analysis of site A would be: Revenue (100No. of customers) $1,173,500 Operating expenses 300,000 Profit $ 873,000

Return on investment 873,000/750,000 = 116.4% The ROI seems sufficiently high so that the branch should be constructed. (c) The size of a branch and its proximity to customers may be too simple to explain the market share of each. The nature of the services offered, the accessibility of the site, and the reputation of the bank may be just as important in estimating patronage. What will the countermoves be of the competing branch? Adding another branch and locating near branch A could substantially reduce its market share. How likely is this to happen? Are the customer numbers stable? Will a third or fourth bank be locating branches in the region? Can we expect that customers will drive such long distances to seek banking services?

173

11 This problem can be solved as an integer linear programming problem similar to the Ohio Trust Company example in the text. First, we create a table showing the counties that are adjacent to each county. That is, Counties under consideration 1. Williams 2. Fulton 3. Lucas 4. Ottawa 5. Defiance 6. Henry 7. Wood 8. Sandusky 9. Paulding 10. Putnam 11. Hancock 12. Seneca 13. Van Wert 14. Allen 15. Hardin 16. Wyandot 17. Mercer 18. Auglaize 19. Marion 20. Shelby 21. Logan

Adjacent counties by number 2,5,6 1,3,6 2,4,6,7 3,7,8 1,6,9,10 1,2,3,5,7,10,11 3,4,6,8,10,11,12 4,7,12 5,10,13 5,6,7,9,11,13,14 6,7,10,12,14,15,16 7,8,11,16 9,10,14,17,18 10,11,13,15,18 11,14,16,18,19,21 11,12,15,19 13,18 13,14,15,17,20,21 15,16 18,21 15,18,20

Next, according to the problem formulation given in the Ohio Trust Company example, we can build the matrix as given in the prepared database called ILP03.DAT. The problem formulation is shown in Table 13-4. This matrix can be solved by the integerprogramming module (MIPROG) in LOGWARE for solution. Note that all coefficients are 1s or 0s. A coefficient of 1 is given to each county and its adjacent counties. The sum of all constraints must be 1 or greater. The Xs take on the values of 0 or 1. An X of 1 means that the branch is located in the county. Solving this problem using the integer-programming module in LOGWARE shows that a minimum of 5 principal places of business are needed. They should be located in Henry, Wood, Putnam, Hardin, and Auglaize counties.

174

TABLE 13-4 Coefficient Matrix Setup for Problem 11 Obj Fun Constraints Williams Fulton Lucas Ottawa Defiance Henry Wood Sandusky Paulding Putnam Hancock Seneca Van Wert Allen Hardin Wyandot Mercer Auglaize Marion Shelby Logan

X1 1

X2 1

1 1

1 1 1

1 1

1

X3 1

1 1 1 1 1

X4 1

X5 1

X6 1

1

1 1 1

1 1 1 1 1 1 1 1

1 1

1 1 1

1 1 1 1

X7 1

1 1

X8 1

X9 1

1 1

1 1 1

X10 X11 X12 X13 X14 X15 X16 X17 X18 X19 X20 X21 1 1 1 1 1 1 1 1 1 1 1 1

1 1

1 1 1 1 1 1

1 1

1 1 1 1 1 1 1 1

1 1

1 1 1 1 1 1

1 1 1 1 1 1 1

1 1

1 1

1 1 1

1

1 1

1 1 1

1

1

1 1 1 1

1 1 1 1

1 1

1 1

1

1

1

1

1 1

1 1

1 1 1

RHS            

1 1 1 1 1 1 1 1 1 1 1 1

       

1 1 1 1 1 1 1 1

175

12 This problem can be solved with the aid of the PMED program in LOGWARE. A database has been prepared for it called PMED04.DAT. The database shows the fixed costs for a single site. When other numbers are to be evaluated, the FOC must be recalculated and entered into the database. The scaling factor is set at 1 for this problem. The fixed operating cost must be calculated for each possible number of locations. Using PMED in LOGWARE gives the following results based on recalculated FOC values, an estimation of vendor to laboratories transportation cost, and an enumerative search. No. of locations 1 2

Volume, Sites lb. Chicago 680,000 Cleveland 515,000 L. Angeles 165,000 Total 680,000 3 New York 235,000 Chicago 280,000 L. Angeles 165,000 Total 680,000 4 New York 200,000 Atlanta 145,000 Chicago 170,000 L. Angeles 165,000 Total 680,000 5 New York 200,000 Atlanta 100,000 Chicago 170,000 Dallas 60,000 L. Angeles 150,000 Total 680,000 6 New York 200,000 Atlanta 65,000 Miami 35,000 Chicago 170,000 Dallas 60,000 L. Angeles 150,000 Total 680,000 1 515,000x310x0.02=3,193,000 2

Outbound cost, $ 28,350,000

Inbound distance, mi. 0 310 1,750

15,083,500 713 0 1,750 10,641,999 713 585 0 1,750 7,515,250 713 585 0 790 1,750 6,079,249 713 585 1180 0 790 1,750 5,029,250

Inbound cost, $ 0 3,193,0001 5,775,000 8,968,000 3,351,100 0 5,775,000 9,126,100 2,852,000 1,696,500 0 5,775,000 10,313,500 2,852,000 1,170,000 0 948,000 5,250,000 10,220,000 2,852,000 760,500 826,000 0 948,000 5,250,000 10,636,500

FOC, $ 5,000,000 3,535,5342 3,535,534 7,071,068 2,886,751 2,886,751 2,886,751 8,660,253 2,500,000 2,500,000 2,500,000 2,500,000 10,000,000 2,236,067 2,236,067 2,236,067 2,236,067 2,236,067 11,180,335 2,041,241 2,041,241 2,041,241 2,041,241 2,041,241 2,041,241 12,247,446

Total cost, $ 33,350,000

31,122,568

28,428,352

27,828,750

27,479,584

27,913,196

5,000,000 2 / 2  3,535,534

The PMED program is used to find the best combination of sites for a particular number of sites to be found. The fixed cost must be adjusted for the number of sites being evaluated. It should be recognized that the model handles only the outbound leg of the network (sites to serve laboratories). The vendor to site transportation cost is included externally, as shown in the previous table. Calculating the distances between vendor and the selected sites easily can be done by using the MILES module in LOGWARE with a scaling factor of 1. Then, inbound transport costs are a product of site volume, distance, and the inbound rate.

176

Searching from 1 to N sites shows that outbound transportation costs decrease while inbound and fixed costs increase with increasing numbers of sites. Initially, total cost declines until 5 sites are reached after which total cost increases. We select 5 sites as economically the best number. Their customer assignments are Location number 1 2 3 4 5

Assignments New York Atlanta Chicago Dallas Los Angeles

Volume 200,000 100,000 170,000 60,000 150,000

Customers 1, 2, 3, and 12 4 and 5 6, 7, 8, 9, 10, and 11 13, 14, and 16 15, 17, 18, 19, and 20

A map of the solution is as follows.

The total cost for 5 sites is $27,479,584. 13 After changing the fixed costs in the problem setup matrix, the following solution is found.

177

OPTIMAL SOLUTION Variable X(1) = X(2) = X(3) = X(4) = X(5) = X(6) = X(7) = X(8) = X(9) = X(10) = X(11) = X(12) = X(13) = X(14) = X(15) = X(16) = X(17) = X(18) = X(19) = X(20) = X(21) = X(22) = X(23) = X(24) = X(25) = X(26) = X(27) = X(28) = X(29) = X(30) = X(31) = X(32) =

Value .0000 10000.0000 50000.0000 .0000 .0000 .0000 50000.0000 90000.0000 .0000 .0000 .0000 .0000 .0000 30000.0000 20000.0000 .0000 .0000 .0000 20000.0000 .0000 40000.0000 .0000 .0000 .0000 1.0000 .0000 1.0000 1.0000 1.0000 .0000 .0000 .0000

Rate 8.0000 7.0000 9.0000 11.0000 10.0000 11.0000 12.0000 11.0000 13.0000 8.0000 7.0000 8.0000 6.0000 5.0000 7.0000 11.0000 10.0000 11.0000 9.0000 8.0000 10.0000 7.0000 6.0000 7.0000 100000.0000 500000.0000 140000.0000 260000.0000 220000.0000 70000.0000 130000.0000 110000.0000

Objective function value =

Cost .0000 70000.0000 450000.0000 .0000 .0000 .0000 600000.0000 990000.0000 .0000 .0000 .0000 .0000 .0000 150000.0000 140000.0000 .0000 .0000 .0000 180000.0000 .0000 400000.0000 .0000 .0000 .0000 100000.0000 .0000 140000.0000 260000.0000 220000.0000 .0000 .0000 .0000

Variable label P1S1W1C1 P1S1W1C2 P1S1W1C3 P1S1W2C1 P1S1W2C2 P1S1W2C3 P1S2W1C1 P1S2W1C2 P1S2W1C3 P1S2W2C1 P1S2W2C2 P1S2W2C3 P2S1W1C1 P2S1W1C2 P2S1W1C3 P2S1W2C1 P2S1W2C2 P2S1W2C3 P2S2W1C1 P2S2W1C2 P2S2W1C3 P2S2W2C1 P2S2W2C2 P2S2W2C3 zW1 zW2 yW1C1 yW1C2 yW1C3 yW2C1 yW2C2 yW2C3

3700000.00

Note that warehouse 2 is no longer used in favor of all products flowing through warehouse 1. 14 (a) The demand of customer 1 for product 1 increases to 100,000 cwt. In the problem matrix of ILP02.DAT the following cell values are changed.

Cell Dem P1W1C1, yW1C1 Dem P1W2C1, yW2C1 Cap-W1, yW1C1 Cap-W2, yW2C1 Obj. coef., yW2C1 Obj. coef., yW2C1

From -50000 -50000 70000 70000 140000 70000

To -100000 -100000 120000 120000 240000 120000

The result shows that warehouse 2 is still the only warehouse used, and the products are sourced from plant 2. However, the costs have increased to $3,500,000.

178

(b) Using the ILP02.DAT file in MIPROG of LOGWARE, the following changes are made to the following cells. Cell

Obj. Coef., P2S2W1C1 Obj. Coef., P2S2W1C2 Obj. Coef., P2S2W1C3 Obj. Coef., P2S2W2C1 Obj. Coef., P2S2W2C2 Obj. Coef., P2S2W2C1

From 9 8 10 7 6 7

To 12 11 13 10 9 10

The result shows that warehouse 2 is still the only warehouse used, and the products are sourced from plant 2. However, the costs have increased to $3,380,000. (c) Using the ILP02.DAT file in MIPROG of LOGWARE, the changes are made to the following cells. Cell

Obj. Coef., yW2C1 Obj. Coef., yW2C2 Obj. Coef., yW2C3

From 70,000a 130,000 110,000

To 280,000 520,000 440,000

a

The sum of demand for the same customer for all products multiplied by the handling rate, i.e., (50,000 + 20,000)  $4/cwt. = 280,000.

The solution for product 1 shows that 50,000 cwt. flows from plant 1 through warehouse 1 and on to customer 1. The remainder flows from plant 2 through warehouse 2 and on to customers 2 and 3. For product 2, plant 1 supplies warehouse 1 and customer 1 with 20,000 cwt. The remaining 90,000 cwt. flows from plant 2 through warehouse 2 to customers 2 and 3. The total cost is $3,920,000. (d) Making some slight revisions in file ILP02.DAT can adjust the capacities on plant 1. The cell changes to make are: Cell Cap-P1S1, RHS Cap-P1S2, RHS

From 60,000 999,999

To 150,000 90,000

Both warehouses are now used for both products. summarized as:

The solution can be

179

Product 1 1 1 1 2 2 2

Plant 1 – 50,000 cwt. 1 – 60,000 2 – 40,000 2 – 50,000 1 – 20,000 2 – 30,000 2 – 60,000

Warehouse 1 – 50,000 2 – 60,000 2 – 40,000 2 – 50,000 1 – 20,000 2 – 30,000 2 – 60,000

Customer 1 – 50,000 cwt. 2 – 60,000 2 – 40,000 3 – 50,000 1 – 20,000 2 – 30,000 3 – 60,000

The total cost is $3,270,000 (e) Again revising the ILP02.DAT file by changing cells Obj. coef., P2S1W2C3 and Obj. Coef., P2S2W2C3 to have a very high cost (999), these cells are locked out of consideration. The solution is the same as the text example except that the customers both products are serve from warehouse 1. The total cost is $3,340,000. 15 This problem follows the form of the Ohio Trust Company example in the text. First, identify the zones that are within 30 minutes of any particular zone. That is, Zone no. 1 2 3 4 5 6 7 8 9 10

Zones within 30 minutes 1,2,4,7,8,9,10 1,2,3,4,7,9,10 2,3,4,5,8,9,10 1,2,3,4,5,6,10 3,4,5,6,7,8 4,5,6,7,8 1,2,5,6,7,9,10 1,3,5,6,8,9,10 1,2,3,7,8,9,10 1,2,3,4,7,8,9,10

Using the MIPROG module in LOGWARE, the following matrix can be defined.

180

The solution from MIPROG is: OPTIMAL SOLUTION Variable Value Rate X(1) = 1.0000 1.0000 X(2) = .0000 1.0000 X(3) = .0000 1.0000 X(4) = 1.0000 1.0000 X(5) = .0000 1.0000 X(6) = .0000 1.0000 X(7) = .0000 1.0000 X(8) = .0000 1.0000 X(9) = .0000 1.0000 X(10) = .0000 1.0000 Objective function value =

Cost 1.0000 .0000 .0000 1.0000 .0000 .0000 .0000 .0000 .0000 .0000

Variable label 1 2 3 4 5 6 7 8 9 10

2.00

The optimal solution is to place claims adjuster stations in zones 1 and 4. 16 This is a location problem where the dominant location factor is transportation cost is and this cost is determined from optimizing the multi-stop routes originating at the material yard. The ROUTER module in LOGWARE can be used to generate these routes for each yard location. A database file for this problem (RTR13.DAT) has been prepared. Solving this problem requires balancing the cost of transporting the merchandise to the customers with the operating cost of the material yards at various locations. Optimizing the routing from the current material yard gives the route design shown in Figure 13-3. A minimum of 9 trucks are required to meet all constraints on the problem. The total daily cost for this location is the route cost + vehicle costs + yard operating cost, or P4,500.87 + 9 x P200 + P350 = P6,650.87.

FIGURE 13-3 Optimized Routing from Current Material Yard Location *** SUMMARY REPORT ***

181

TIME/DISTANCE/COST INFORMATION

Route no 1 2 3 4 5 6 7 8 9 Total

Route Run Stop Brk Stem time, time, time, time, time, Start Return No of Route hr hr hr hr hr time time stops dist,Mi 9.9 6.4 2.5 1.0 5.0 08:00AM 05:56PM 4 205 3.4 1.6 1.8 .0 .7 08:00AM 11:25AM 3 52 10.0 6.5 2.5 1.0 3.1 08:00AM 05:58PM 4 207 7.4 4.1 2.3 1.0 2.2 08:00AM 03:24PM 3 130 9.6 6.4 2.3 1.0 5.3 08:00AM 05:37PM 3 204 9.4 5.8 2.6 1.0 4.3 08:00AM 05:22PM 4 185 9.6 6.7 1.9 1.0 6.2 08:00AM 05:34PM 2 214 9.6 5.9 2.7 1.0 2.8 08:00AM 05:35PM 4 189 5.9 2.8 2.2 1.0 1.5 08:00AM 01:56PM 3 89 74.8 46.1 20.7 8.0 31.1 30 1476

Route cost,$ 602.45 220.25 608.59 416.05 599.64 552.96 625.79 563.48 311.65 4500.87

VEHICLE INFORMATION Route Veh Weight Delvry Pickup Weight Cube Delvry Pickup no typ capcty weight weight util capcty cube cube 1 1 1000 925 0 92.5% 9999 0 0 2 1 1000 625 0 62.5% 9999 0 0 3 1 1000 900 0 90.0% 9999 0 0 4 1 1000 950 0 95.0% 9999 0 0 5 1 1000 900 0 90.0% 9999 0 0 6 1 1000 950 0 95.0% 9999 0 0 7 1 1000 825 0 82.5% 9999 0 0 8 1 1000 1000 0 100.0% 9999 0 0 9 1 1000 850 0 85.0% 9999 0 0 Total 9000 7925 0 88.1% 89991 0 0

Cube util .0% .0% .0% .0% .0% .0% .0% .0% .0% .0%

Vehicle description Truck #1 Truck #1 Truck #1 Truck #1 Truck #1 Truck #1 Truck #1 Truck #1 Truck #1

Substituting yard location A for the current yard location and solving for the route design in ROUTER yields Figure 13-4. No routes can be put end-to-end so that one truck can be used instead of two, so the minimum number of trucks remains at nine. The total daily cost for this location is 3872.02 +9 x 200 + 480 = P6,152.092. Yard Location A

FIGURE 13-4 Route Design for Yard Location A

182

*** SUMMARY REPORT *** TIME/DISTANCE/COST INFORMATION

Route no 1 2 3 4 5 6 7 8 9 Total

Route Run Stop Brk Stem time, time, time, time, time, Start Return No of Route hr hr hr hr hr time time stops dist,Mi 9.8 7.0 1.8 1.0 5.4 08:00AM 05:47PM 4 224 8.4 5.1 2.3 1.0 3.8 08:00AM 04:26PM 3 163 8.4 5.0 2.4 1.0 3.9 08:00AM 04:26PM 3 161 8.6 4.8 2.9 1.0 2.2 08:00AM 04:37PM 5 152 7.2 4.0 2.2 1.0 2.6 08:00AM 03:12PM 3 128 6.3 2.9 2.4 1.0 1.6 08:00AM 02:18PM 3 93 5.3 2.1 2.2 1.0 1.6 08:00AM 01:15PM 3 66 8.9 5.7 2.1 1.0 3.7 08:00AM 04:51PM 3 183 5.1 1.7 2.3 1.0 .6 08:00AM 01:03PM 3 55 68.0 38.3 20.7 9.0 25.5 30 1225

Route cost,$ 649.88 498.40 491.57 470.04 410.15 321.27 254.22 548.68 227.82 3872.02

VEHICLE INFORMATION Route Veh Weight Delvry Pickup Weight Cube Delvry Pickup no typ capcty weight weight util capcty cube cube 1 1 1000 475 0 47.5% 9999 0 0 2 1 1000 950 0 95.0% 9999 0 0 3 1 1000 1000 0 100.0% 9999 0 0 4 1 1000 975 0 97.5% 9999 0 0 5 1 1000 875 0 87.5% 9999 0 0 6 1 1000 1000 0 100.0% 9999 0 0 7 1 1000 875 0 87.5% 9999 0 0 8 1 1000 825 0 82.5% 9999 0 0 9 1 1000 950 0 95.0% 9999 0 0 Total 9000 7925 0 88.1% 89991 0 0

Cube util .0% .0% .0% .0% .0% .0% .0% .0% .0% .0%

Vehicle description Truck #1 Truck #1 Truck #1 Truck #1 Truck #1 Truck #1 Truck #1 Truck #1 Truck #1

Continuing with location B, nine trucks are required and the total daily cost for the route design in Figure 3 is 3370.42 + 9 x 200 + 450= P5,620.42. Yard Location B

FIGURE 3 Design for Yard Location B

*** SUMMARY REPORT ***

183

TIME/DISTANCE/COST INFORMATION

Route no 1 2 3 4 5 6 7 8 9 Total

Route Run Stop Brk Stem time, time, time, time, time, Start Return No of Route hr hr hr hr hr time time stops dist,Mi 7.1 3.7 2.4 1.0 2.3 08:00AM 03:06PM 4 120 6.5 3.1 2.3 1.0 1.8 08:00AM 02:27PM 3 100 7.9 4.6 2.4 1.0 3.0 08:00AM 03:56PM 4 146 5.9 2.8 2.2 1.0 1.4 08:00AM 01:56PM 3 89 5.9 2.6 2.3 1.0 1.1 08:00AM 01:55PM 3 84 6.7 3.6 2.1 1.0 2.3 08:00AM 02:41PM 3 114 5.8 2.5 2.3 1.0 2.1 08:00AM 01:50PM 3 80 6.1 3.3 1.9 1.0 2.8 08:00AM 02:07PM 2 104 9.7 5.8 2.8 1.0 3.0 08:00AM 05:40PM 5 187 61.7 32.0 20.7 9.0 19.8 30 1024

Route cost,$ 389.30 340.53 454.97 311.84 300.41 375.39 290.22 350.46 557.31 3370.42

VEHICLE INFORMATION Route Veh Weight Delvry Pickup Weight Cube Delvry Pickup no typ capcty weight weight util capcty cube cube 1 1 1000 825 0 82.5% 9999 0 0 2 1 1000 950 0 95.0% 9999 0 0 3 1 1000 825 0 82.5% 9999 0 0 4 1 1000 850 0 85.0% 9999 0 0 5 1 1000 925 0 92.5% 9999 0 0 6 1 1000 825 0 82.5% 9999 0 0 7 1 1000 950 0 95.0% 9999 0 0 8 1 1000 825 0 82.5% 9999 0 0 9 1 1000 950 0 95.0% 9999 0 0 Total 9000 7925 0 88.1% 89991 0 0

Cube util .0% .0% .0% .0% .0% .0% .0% .0% .0% .0%

Vehicle description Truck #1 Truck #1 Truck #1 Truck #1 Truck #1 Truck #1 Truck #1 Truck #1 Truck #1

Finally, the route design from location C is shown in Figure 4. Although 10 routes are in the design, two of these can be dovetailed so that only nine trucks are needed. The total daily cost is 4479.99 + 9 x 200 + 420 = P6,699.99. Yard Location C

FIGURE 4 Design for Yard Location C *** SUMMARY REPORT ***

184

TIME/DISTANCE/COST INFORMATION

Route no 1 2 3 4 5 6 7 8 9 10 Total

Route Run Stop Brk Stem time, time, time, time, time, Start Return No of Route hr hr hr hr hr time time stops dist,Mi 1.5 .6 .9 .0 .6 08:00AM 09:30AM 1 20 6.7 3.7 2.0 1.0 1.9 08:00AM 02:41PM 2 120 8.0 4.1 2.9 1.0 1.5 08:00AM 04:00PM 5 132 9.4 5.6 2.8 1.0 3.4 08:00AM 05:25PM 5 180 7.5 4.2 2.3 1.0 2.7 08:00AM 03:29PM 3 134 7.7 4.2 2.4 1.0 1.9 08:00AM 03:39PM 3 136 7.0 3.7 2.3 1.0 2.8 08:00AM 02:59PM 3 117 7.3 4.0 2.3 1.0 3.5 08:00AM 03:18PM 3 127 9.6 7.2 1.4 1.0 5.6 08:00AM 05:35PM 2 229 9.8 7.4 1.4 1.0 6.3 08:00AM 05:47PM 3 236 74.5 44.7 20.7 9.0 30.3 30 1432

Route cost,$ 140.91 389.21 420.20 540.48 425.69 429.10 383.04 408.53 663.40 679.42 4479.99

VEHICLE INFORMATION Route Veh Weight Delvry Pickup Weight Cube Delvry Pickup no typ capcty weight weight util capcty cube cube 1 1 1000 375 0 37.5% 9999 0 0 2 1 1000 875 0 87.5% 9999 0 0 3 1 1000 975 0 97.5% 9999 0 0 4 1 1000 925 0 92.5% 9999 0 0 5 1 1000 925 0 92.5% 9999 0 0 6 1 1000 1000 0 100.0% 9999 0 0 7 1 1000 950 0 95.0% 9999 0 0 8 1 1000 950 0 95.0% 9999 0 0 9 1 1000 550 0 55.0% 9999 0 0 10 1 1000 400 0 40.0% 9999 0 0 Total 10000 7925 0 79.3% 99990 0 0

Cube util .0% .0% .0% .0% .0% .0% .0% .0% .0% .0% .0%

Vehicle description Truck #1 Truck #1 Truck #1 Truck #1 Truck #1 Truck #1 Truck #1 Truck #1 Truck #1 Truck #1

From an economic analysis, it appears that yard location B is the best choice.

185

SUPERIOR MEDICAL EQUIPMENT COMPANY Teaching Note Strategy The purpose of the Superior Medical Equipment Company case study is to encourage students to apply the center-of-gravity methodology to a problem involving a single warehouse location. Although the methodology is somewhat elementary, it can be useful in providing some first approximations to good warehouse locations. It allows students to evaluate the financial implications of alternative network designs, and it is intended to be solved with the aid of the COG module provided in the LOGWARE software. Answers to Questions (1) Based on information for the current year, is Kansas City the best location for a warehouse? If not, what are the coordinates for a better location? What cost improvement can be expected from the new location?

When a single warehouse is to be located, the primary location costs are transportation, both inbound to the warehouse and outbound from it, and the warehouse lease, which varies with the location. The current location serves as a benchmark against which the costs for other locations can be compared. That is, based on information given in the case, the total relevant cost for the Kansas City location is: Inbound transportation Outbound transportation Lease $2.75/sq. ft. × 200,000 sq. ft. Total relevant cost

$

$

2,162,535 4,819,569 550,000 7,532,104

Using the COG module with inbound transport rates from Phoenix set at $16.73/1163 = $0.014/cwt./mile and from Monterrey set at $9.40/1188 = $0.008/cwt./mile, and the outbound transport rate from the unknown warehouse location set at $0.0235/cwt./mile, the coordinates for the best location are X = 7.61 and Y = 4.51, or approximately Oklahoma City. Total transportation cost for this location would be $6,754,082. The total relevant cost would be: Transportation Lease $3.25/sq. ft. × 200,000 sq. ft. Total

$ 6,754,082 650,000 $ 7,404,082

The annual cost savings can be projected as $7,532,104  7,404,082 = $128,022. (2) In five years, management expects that Seattle, Los Angeles, and Denver markets to grow by 5%, but the remaining markets to decline by 10%. Transport costs are expected to be unchanged. Phoenix output will increase by 5%, and Monterrey's output will decrease by 10%. Would your decision about the location of the warehouse change? If so, how?

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A new benchmark for the 5th year can be computed from the data given in Tables 1 and 2. After adjusting the plant and market volumes according to the changes indicated, the 5th-year benchmark costs can be computed as follows. Volume,

Rate,

Transport

Point cwt. 1 64,575 2 108,540 3 17,850 4 33,600 5 13,125 6 8,550 7 26,550 8 18,900 9 37,170 10 7,740 11 9,630 Totals 346,230

$/cwt. $16.73 9.40 33.69 30.43 25.75 18.32 25.24 19.66 26.52 26.17 27.98

cost, $ $1,080,340 1,020,276 601,367 1,022,448 337,969 156,636 670,122 371,574 985,748 202,556 269,447 $6,718,483

The total 5th-year benchmark relevant cost would be Transportation Lease $2.75/sq. ft. × 200,000 sq. ft. Total

$ 6,718,483 550,000 $ 7,268,483

Optimizing the location with the 5th-year data gives a location at X = 7.05 and Y = 4.52. The relevant costs for this location are Transportation Lease $3.25/sq. ft. × 200,000 sq. ft. Total

$ 6,464,206 650,000 $ 7,114,206

The annualized cost savings would be $7,268,483  7,114,206 = $154,277. The average annual cost savings are (128,022 + 154,277)/2 = $141,150. The simple annual return on investment (moving cost) can be computed as: ROI 

$141,150  100  471% . $300,000

Management must now judge whether 47.1% annual return is worth the risk of changing warehouse locations. (3) If by year 5 increases are expected of 25% in warehouse outbound transport rates and 15% in warehouse inbound rates, would your decision change about the warehouse location?

187

It is assumed here that the 5th-year demand level applies. A revised 5th-year benchmark can be recomputed by applying the cost growth factors to the overall 5th-year transport costs. That is, Inbound 1.15 × 2,100,616 =

$2,415,708 Outbound 1.25×4,617,867 =

Subtotal

$8,188,042

Total

$8,738,042

Lease

5,772,334 550,000

Running COG shows that the minimum transport cost location would be at coordinates X = 7.20 and Y = 4.62, which is near the previous location in question 2. The shift in location is minimal. The cost for this location is: Transportation $ 7,939,545 Lease $3.25/sq. ft. × 200,000 sq. ft. 650,000 Total $ 8,589,545 The annualized cost savings would be $8,738,042  8,589,545 = $148,497. It can be concluded that: 1. Location is similar to the optimized 5th-year location. 2. The increase in possible cost savings further encourages relocation from Kansas City and toward a site near Oklahoma City, OK. (4) If the center-of-gravity method is used to analyze the data, what are its benefits and limitations for locating a warehouse? The center-of-gravity method locates a facility based on transportation costs alone. This is reasonable when only one facility is being located and the general location for it is being sought. Such costs as inventory carrying, production, and warehouse fixed are not included, but they are not particularly relevant to the problem. However, costs such as warehouse storage and handling, and other costs that vary by the particular site are not included but may be relevant in a given situation. Transportation costs are assumed linear with distance. This may not be strictly true, although distance may be nonlinear. The obvious benefits of the method are (1) it is a fast solution methodology; (2) it considers all possible locations (continuous); (3) it is simple to use; (4) its data are readily available; and (5) it gives precise locations through a coordinate system. Some potential limitations are (1) coordinates need to be linear; (2) transportation rates on a per-mile basis are constant; (3) volumes are known and constant for given demand and source points; and (4) locations may be suggested that are not feasible such as in lakes, central cities, or restricted lands. Concluding Comments

188

The analysis in the case seems to suggest a move from Kansas City to a region around Oklahoma City would be advantageous. A return on investment of 47 percent or higher is possible, however management must now seek a particular site in the area whose choice may add or detract from this savings potential. In any case, the COG method has assisted in the selection of good potential locations and testing their sensitivity to changes in costs and volumes. APPENDIX 1

LOGWARE COG Module Input Data for the Current Year

Title: SUPERIOR MEDICAL EQUIPMENT COMPANY Power factor: .5 Scale factor: 230 Point X-coordinate Y-coordinate Volume 1 3.60 3.90 61,500 2 6.90 1.00 120,600 3 0.90 9.10 17,000 4 1.95 4.20 32,000 5 5.60 6.10 12,500 6 7.80 3.60 9,500 7 10.20 6.90 29,500 8 11.30 3.95 21,000 9 14.00 6.55 41,300 10 12.70 7.80 8,600 11 14.30 8.25 10,700

Rate 0.0140 0.0080 0.0235 0.0235 0.0235 0.0235 0.0235 0.0235 0.0235 0.0235 0.0235

189

OHIO AUTO & DRIVER'S LICENSE BUREAUS Teaching Note Strategy The purpose of this case study is to introduce students to a logistics problem in a service area. It also provides an application for the multiple center of gravity location methodology. The MULTICOG module in the LOGWARE software can effectively be applied. The module allows students to quickly evaluate alternatives as to the number of bureaus to use, the bureau locations, and the size of the territory that each bureau should serve. The case may be assigned as a homework problem, a short case study project, or as a case for class discussion. The later would be appropriate especially if adequate attention is given to the issues of how Dan should go about solving a problem such as this, what data he needs and where to obtain it, and what concerns he should have about changing the existing network design. This would encourage students to think beyond the computational aspects of the problem. Answers to Questions (1) Do you think there is any benefit to changing the network of license bureaus in the Cleveland area? If so, how should the network be configured?

The nature of the costs and the number of possible alternative network designs make it impractical to seek an optimal solution. Therefore, a possible approach to the analysis is outlined as follows. First, establish a benchmark against which changes to the network can be compared. Much of the data for this is given in the case write up. The costs can simply be applied to the size of each bureau and its associated staff. The cost for residents traveling to the bureaus is not known because the bureau territories are not known. However, an estimate can be made of travel costs by solving the problem in MULTICOG for eight bureaus. Since MULTICOG attempts to optimize bureau location, this travel cost is probably understated. The benchmark costs are summarized in Table 1. The location costs for the current operation are estimated to be $1,355,706. Second, what improvements can be made on the existing eight locations? Besides moving the locations of the bureaus, which results in resizing the facilities and adjusting the staff numbers, there are no obvious improvements to be made. Therefore, the benchmark remains the base for comparison. Third, it is now necessary to estimate the approximate number of bureaus that are needed to serve the area. Since the costs for a particular network design depend on the size of each bureau, which cannot be known until the problem is solved, an initial assumption must be made. It will be assumed that all bureaus are of the same size. Hence, for 5 bureaus, the average number of residents in each bureau’s territory would be the total number of residents divided by the number of bureaus, or 691,700/5 = 138,340. Rent, staff salaries, and utility expenses can be derived from this estimate. Table 2 is developed to show the bureau size and the number of staff for 1 to 10 bureaus. Table 3 extends the average costs from these estimates. A reduced number of bureaus, in the range of two, is about right.

190

TABLE 1

Bureau 1 2 3 4 5 6 7 8 Totals

Benchmark Costs for the Current Network of License Bureaus Customer Size, Rent, Travel, Salaries, Utilities, sq. ft. Staff $ $ $ $ 1,700 4 37,400 84,000 6,800 1,200 4 26,400 84,000 4,800 2,000 5 44,000 105,000 8,000 1,800 4 39,600 84,000 7,200 1,500 4 33,000 84,000 6,000 2,200 5 48,400 105,000 8,800 2,700 5 59,400 105,000 10,800 5 33,000 105,000 6,000 1,500 14,600 36 321,200 756,000 58,400 220,106*

*Approximated by running MULTICOG at 8 bureaus.

TABLE 2 Average Size and Staff for Various Numbers of Bureaus (1) 691,700/(1) Avg. Avg. no. of bureau No. of residents per size, Staff per bureaus bureau sq. ft. bureau 1 691,700 4,500 10 2 345,850 3,000 7 3 230,567 2,500 6 4 172,925 2,000 5 5 138,340 2,000 5 6 115,283 2,000 5 7 98,814 1,500 4 8 86,463 1,500 4 9 78,855 1,500 4 10 69,170 1,500 4

191

TABLE 3 Average Costs by Number of Bureaus

No. of bureaus 1 2 3 4 5 6 7 8 9 10

Rent, $ 99,000 132,000 165,000 176,000 220,000 264,000 231,000 264,000 297,000 330,000

Staff salaries, $ 210,000 294,000 378,000 420,000 525,000 630,000 588,000 672,000 756,000 840,000

Utilities, $ 18,000 24,000 30,000 32,000 40,000 48,000 42,000 48,000 54,000 60,000

Resident travel, $ 662,319 430,922 354,239 298,000 278,181 249,287 237,635 220,106 206,496 198,600

Annual total cost, $ 989,319 880,922 927,239 926,965 1,063,181 1,191,287 1,098,636 1,204,106 1,313,496 1,428,600

The cost estimates can now be refined around two bureaus. A comparison with the benchmark costs and a return of the initial investment (costs related to changing the network design) are sought. A sample analysis for two bureaus, based on a design provided by MULTICOG, is shown below.

Bureau 1 2 Totals

Residents 290,200 401,500 691,700

Size, sq. ft. 2,500 3,500 6,000

Staff 6 8 14

Rent, $ 55,000 77,000 132,000

Staff salaries, Utilities, $ $ 126,000 10,000 168,000 14,000 294,000 24,000

Resident travel cost, $ 166,332 264,590 430,922

The total annual variable cost is $132,000 + 294,000 + 24,000 + 430,922 = $880,922. There are one-time costs due to staff separation and equipment moves. Compared with the benchmark, 36  14 = 22 staff will be separated for a cost of 22  $8,000 = $176,000. Equipment movement costs to two bureaus would be 2  10,000 = $20,000. Total movement costs would be $176,000 + 20,000 = $196,000. Annual variable cost savings compared with the benchmark would be $1,355,706  880,922 = $474,784. A simple return on investment would be: ROI 

$474,784  100  242% $196,000

Similar calculations are carried out for various numbers of bureaus. These results are tabulated in Table 4.

192

TABLE 4 Cost Savings and Return of Investment for Alternate Network Designs as Produced by MULTICOG Total Annual Return on No. of size, sq. Total variable cost, Moving investSavings, bureaus ft. staff $ cost, $ ment, % $ 1 4,500 10 989,319 218,000 366,387 168 2 6,000 14 880,922 196,000 242 474,784 3 7,500 18 927,239 176,000 428,467 243 4 8,500 21 960,965 160,000 394,741 246 5 9,500 24 1,029,181 146,000 326,525 223 6 11,500 29 1,157,287 106,000 198,419 187 7 12,000 31 1,200,635 110,000 155,071 141 8 13,000 34 1,272,106 96,000 83,600 87 Benchmark 14,600 36 1,355,706 ---------

The maximal annual savings occurs with a network containing two bureaus. However, the maximal return on investment occurs with four bureaus. ROI is selected as the appropriate measure on which to base this economic decision. The details for a design with four bureaus are given in Table 5. The design is shown pictorially in Figure 1 of this note. TABLE 5 Design Details for a Network with Four Bureaus Column grid Row grid coordicoordiBureau nate nate Residents Grid box number assignment 1 3.36 2.74 218,200 1,2,3,4,5,6,7,8,9,10,11,12,13, 14,15,16,17,18,19,20,21,22,23, 24,25,26,27,28,29,30,31,32,33 2 7.00 3.00 168,700 34,35,36,37,38,39,40,41,43,44, 45,46,47,50,51,52,53 3 9.74 5.66 195,000 42,48,49,54,55,56,60,61,62,63, 67,68,69,70,74,75,76,77,81,82, 83,84, 4 10.00 2.00 109,800 57,58,59,64,65,66,71,72,73,78, 79,80

(2) Do you think Dan Roger's study approach is sound? Overall Dan can be praised for the simplicity of the methodology that he has chosen. The center-of-gravity approach is appropriate in this problem since there are no capacity limitations on the facilities, locations across a continuous space are desired, and transportation cost (except for some fixed costs) is the primary location variable. The data requirements are relatively straightforward, although they are not always easy to fulfill. This would likely be a problem with any other solution approach as well. The results of this methodology can only be used as a first approximation at best. Several criticisms of this particular approach can be offered as follows:

193

 The effect of bureau location on the resident's perception of service is not as well known as portrayed in the case. In addition, service may need to be represented by more than location.  Travel to the bureaus is assumed straight line. However, location in the area is likely to be influenced by a road network. Time may be more important than distance to residents.  The fixed costs associated with location are not handled directly by the center-ofgravity approach.  Residents are assumed to travel to the locations within their assigned territories. They may not strictly do this.  Good facilities may not be available at the indicated location coordinates. The analysis is particularly weak around the estimate of the resident travel cost. While an exact cost is not likely to be known, Dan should conduct a sensitivity analysis around this cost. He may find that the design does not change a great deal over a wide range of assumed values. If this is the case, he can feel comfortable that his recommendation is fundamentally sound. If not, he should seek to find a more precise value. (3) What concerns besides economic ones should Dan have before suggesting that any changes be made to the network? A quantitative approach to location will rarely give the precise locations to be implemented. Rather, it provides a starting point for further analysis. There are a number of other factors to be considered before the revised network design can be implemented. First, there are site selection factors to be taken into such as the availability of adequate space near the location coordinates, proximity to good highway linkages, and reasonable neighborhood reactions to this type of operation. Second, there are political concerns. Reducing the number of locations will result in a releasing some of the staff. Dan may experience some political resistance to this. Since staff is a large expense in the operation, currently about 2/3 of the costs, retention of a larger number of bureaus may be required. Of course, transferring staff to other governmental operations may be a way of dealing with this issue. This assumes their willingness to be relocated, although this is not likely to be a strong issue if relocation were to occur in the same area. Third, there may be difficulty in demonstrating the economics of network redesign. Although others may appreciate the costs of rent, salaries, and utilities, the cost of resident travel is subject to much interpretation. Those favoring many bureaus may argue the high cost while those wanting to reduce the number of bureaus may perceive it as not very significant.

194

Grid row number

Current bureau locations

Grid column number

Revised bureau locations

FIGURE 1 Four Bureau Locations and Their Territories

SUPPLEMENT Sample Input Data File for MULTICOG in LOGWARE for the Ohio Auto & Driver's License Bureau Case Study Title: LICENSE BUREAU Number of sources: 4 Number of demand points: 84 Scaling factor: 2.5 POINT X-COORDINATE Y-COORDINATE 1 1 1 2 1 2 3 1 3 4 1 4 5 1 5 6 1 6 7 1 7 8 2 1 9 2 2 10 2 3 11 2 4 12 2 5 13 2 6 14 2 7 15 3 1 16 3 2 17 3 3 18 3 4 19 3 5 20 3 6

VOLUME 4100 6200 7200 10300 200 0 0 7800 8700 9400 11800 100 0 0 8100 10500 15600 10500 200 0

RATE .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12

195

21

3

7

0

.12

SUPPLEMENT (Continued) POINT 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78

X-COORDINATE 4 4 4 4 4 4 4 5 5 5 5 5 5 5 6 6 6 6 6 6 6 7 7 7 7 7 7 7 8 8 8 8 8 8 8 9 9 9 9 9 9 9 10 10 10 10 10 10 10 11 11 11 11 11 11 11 12

Y-COORDINATE 1 2 3 4 5 6 7 1 2 3 4 5 6 7 1 2 3 4 5 6 7 1 2 3 4 5 6 7 1 2 3 4 5 6 7 1 2 3 4 5 6 7 1 2 3 4 5 6 7 1 2 3 4 5 6 7 1

VOLUME 10700 12800 13800 15600 400 0 0 11500 13900 14500 13700 600 0 0 9300 14900 13700 10200 1200 0 0 10100 12600 16700 15800 12400 2600 0 8800 13700 15200 14100 10800 17200 500 5300 16700 13800 11900 13500 18600 12000 5100 17400 10300 9800 10300 15500 11700 7700 9200 7500 8500 7800 9900 8700 4300

RATE .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12

196

79 80

12 12

2 3

6700 5800

.12 .12

Y-COORDINATE 4 5 6 7

VOLUME 6800 5400 7100 6400

RATE .12 .12 .12 .12

SUPPLEMENT (Continued) POINT 81 82 83 84

X-COORDINATE 12 12 12 12

197

SOUTHERN BREWERY Teaching Note Strategy The purpose of this case study is to provide students with the opportunity to design a distribution network where plant location is at issue. They first should identify the major costs and alternatives that are important to such a design problem. Second, they should be encouraged to apply the transportation method of linear programming to assist in the analysis of alternatives using the TRANLP module in LOGWARE. Finally, they should consider factors other than those in the analysis that might alter the course of their recommendation and be sensitive to the limitations and benefits of linear programming as a solution methodology. Answers to Questions (1) If you were Carolyn Carter, would you agree with the proposal to build the new brewery? If you do, what plan for distribution would you suggest?

If growth is uniform over the next five years, Southern can expect that demand for its products will exceed the currently available plant capacity. That is, demand is increasing at the rate of (0 + (595,000  403,000)) ÷ 5 = 38,400 barrels per year. Thus, the current annual capacity of 500,000 barrels will be used up in (500,000  403,000) ÷ 38,400 = 2.5 years. A major concern is whether it would be profitable to construct the new plant. Rough estimates of its profitability can be made, projecting profits with and without the new plant. We know that 595,000  500,000 = 95,000 barrels of beer would not be sold annually in the 5th year if additional capacity is not constructed. This represents a potential average lost revenue of $280/barrel  0.20  [(0 + 95,000) ÷ 2.5] = $2,128,000. (The figure of 2.5 years assumes that the new brewery can be brought on stream at approximately the time when capacity will be used up in the existing plants.) From the benchmark1 costs for the current system, as shown in Table 1, Southern is currently producing and distributing 403,000 barrels at a total cost of $60,015,000. This is an average cost per barrel of $60,015,000 ÷ 403,000 = $148.92. The overhead and sales expense is 27 percent, or $280  0.27 = $75.60 per barrel. Total costs per barrel are $148.92 + 75.60 = $224.52, which is about 80 percent of the sales dollar. The 20 percent profit margin seems valid. Therefore, the benefit of serving the potentially lost demand with a new brewery can be estimated using on a simple return on investment: ROI 

$2,128,000  0.21, or 21% $10,000,000

If management feels that this is an adequate return for such a project, Carolyn should proceed with her analysis. Let's assume that she has this approval. Next, she may wish to explore the opportunities available by improving upon the existing distribution system without the presence of the new plant. This is an improved 1

A benchmark refers to the costs of producing and distributing demand as currently allocated throughout the network.

198

benchmark2 and it can be found by solving a transportation-type linear programming problem of the type shown in Table 2. The results given in the TRANLP module of LOGWARE can be summarized in Table 3. This shows that with some slight reallocation of demand among the plants, production and distribution costs can be reduced by $60,015,000  59,804,000 = $211,000 annually. A comparison of the benchmark results in Table 1 and the improved benchmark results in Table 3 shows that Knoxville's demand should be shifted from Columbia to Montgomery and 28,000 barrels of Columbia's market should be shifted from the Columbia plant to the Montgomery plant. The Columbia plant is a high-cost producer and cost savings are achieved by trading production costs for transportation costs. That is, production costs will be reduced by $400,000 while transportation costs will be increased by $189,000. Note that this $211,000 savings can be realized without any capital investment. TABLE 1 Benchmark of Production and Transportation Costs ($000s) for Current Demand Demand TransBrewery of in 000s Produort Market area origin barrels tion costs costs Total costs 1 Richmond Richmond 56 $7,840 $475 $8,315 2 Raleigh Richmond 31 4,340 332 4,672 3 Knoxville Columbia 22 3,190 282 3,472 4 Columbia Columbia 44 6,380 306 6,686 5 Atlanta Montgomery 94 12,878 959 13,837 6 Savannah Montgomery 13 1,781 179 1,960 7 Montgomery Montgomery 79 10,823 550 11,373 8 Tallahassee Montgomery 26 3,562 355 3,917 5,206 577 5,783 9 Jacksonville Montgomery 38 Total 403 $56,000 $4,015 $60,015 TABLE 2 TRANLP Problem Setup RICH-

RA-

KNOX-

COL-

AT-

SAVAN-

MONT-

TALL

JACK-

From\To

MOND

LEIGH

VILLE

UMBIA

LANTA

NAH

GOMRY

AHAS

SONVL

RICHMD

148.49

150.70

156.38

152.54

155.48

154.64

159.98

164.30

158.84

100

COLMBA

157.54

154.78

157.81

151.96

156.85

154.54

157.93

160.18

157.27

100

MONTGM

156.98

153.35

150.80

149.93

147.20

150.80

143.69

150.65

152.18

300

JACKVL

152.13

149.25

150.48

146.16

148.80

144.54

148.80

144.72

142.68

0

Demand

56

31

22

44

94

13

79

26

38

Supply

2

An improved benchmark refers to a reallocation of current demand in an optimal way, respecting plant capacity restrictions.

199

TABLE 3 Improved Benchmark of Production and Transportation Costs ($000s) for Current Demand Demand TransBrewery of in 000s Produort Market area origin barrels tion costs costs Total costs 1 Richmond Richmond 56 $7,840 $475 $8,315 2 Raleigh Richmond 31 4,340 332 4,672 3 Knoxville Montgomery 22 3,014 304 3,318 4 Columbia Columbia 16 2,320 111 2,431 4 Columbia Montgomery 28 3,836 362 4,198 5 Atlanta Montgomery 94 12,878 959 13,837 6 Savannah Montgomery 13 1,781 179 1,960 7 Montgomery Montgomery 79 10,823 550 11,373 8 Tallahassee Montgomery 26 3,562 355 3,917 5,206 577 5,783 9 Jacksonville Montgomery 38 Total 403 $55,600 $4,204 $59,804

Adding a plant at Jacksonville with a capacity of 100,000 barrels per year provides enough capacity to satisfy demand out to the 5th year. If the new plant were constructed and producing immediately, total costs could be reduced from the improved benchmark by $59,804,000  59,090,000 = $714,000 per year. (Compare the total costs in Tables 3 and 4.) The Columbia plant would not be needed if the lower-cost Jacksonville plant were on line. We do not know the savings for the 5th-year demand level since not all demand can be served without the presence of the new plant. Therefore, a future-year benchmark cannot be determined. However, we do know how the new plant should be utilized within the system (see Table 5) and how demand allocation should be adjusted to accommodate it. Also, note that the Columbia plant is needed once again although its capacity is not required until the last one-half year of the 5-year planning horizon. This suggests that the Columbia plant should not be sold, but perhaps some alternate use could be made of the facility in the interim, such as subcontracting beer production to a noncompeting company. The new plant is not likely to be brought on stream immediately nor is it needed for 2.5 years, so Carolyn might suggest a distribution plan similar to that in Table 5. A careful inspection of this plan shows that only 1 barrel of demand in the Knoxville region is assigned to Richmond. Splitting demand to this extent is probably not practical and can be assigned to Columbia where there is excess capacity. Costs will rise only slightly. An interesting question is whether the Columbia plant, through modernization, could be made as efficient as the new brewery, and what the implications for distribution might be. We know that this could potentially save $145  135 = $10 per barrel in production costs. At a 100,000-barrel capacity, this is $1,000,000 in cost savings. If the modernization were to cost no more than $5,000,000, this option might be attractive. Of course, we would need to resolve the linear programming problem with Columbia's per barrel costs at $135 plus transportation costs. This would tell us how and to what extent demand would be allocated to Columbia and give a more accurate basis for determining

200

the cost savings. Similarly, it would be interesting to explore what it means to expand the capacity of an existing brewery at a lower investment cost per barrel than the construction of a new facility. TABLE 4 Production and Transportation Costs ($000s) for Current Demand with the Jacksonville Plant Demand TransBrewery of in 000s Produort Market area origin barrels tion costs costs Total costs 1 Richmond Richmond 56 $7,840 $475 $8,315 2 Raleigh Richmond 31 4,340 332 4,672 3 Knoxville Montgomery 22 3,014 304 3,318 4 Columbia Montgomery 21 2,877 272 3,149 4 Columbia Jacksonville 23 3,105 257 3,362 5 Atlanta Montgomery 94 12,878 959 13,837 6 Savannah Jacksonville 13 1,755 124 1,879 7 Montgomery Montgomery 79 10,823 550 11,373 8 Tallahassee Jacksonville 26 3,510 253 3,763 5,130 292 5,422 9 Jacksonville Jacksonville 38 Total 403 $55,272 $3,818 $59,090

(2) If the new brewery is not to be constructed, what distribution plan would you propose to top management? Table 6 shows a linear programming solution where the new plant is not brought on stream and the demand in the markets is set at the 5-year level. An interesting solution occurs when demand exceeds capacity. The most costly demand region to serve is not assigned to any plant. As can be seen in Table 6, portions of the demand in Tallahassee and Jacksonville should not be served, and essentially the entire Knoxville market should not be served at all. Top management may wish to adjust this plan for reasons other than economic ones.

201

TABLE 5 Production and Transportation Costs ($000s) for Projected 5th-Year Demand With the Jacksonville Plant Demand TransBrewery of in 000s Produort Market area origin barrels tion costs costs Total costs 1 Richmond Richmond 64 $8,960 $543 $9,503 2 Raleigh Richmond 35 4,900 375 5,275 3 Knoxville Richmond 1 140 16 156 3 Knoxville Columbia 20 2,900 256 3,156 3 Knoxville Montgomery 12 1,644 166 1,810 4 Columbia Columbia 55 7,975 383 8,358 5 Atlanta Montgomery 141 19,317 1,438 20,755 6 Savannah Columbia 20 2,900 191 3,091 7 Montgomery Montgomery 119 16,303 828 17,131 8 Tallahassee Montgomery 28 3,836 382 4,218 8 Tallahassee Jacksonville 24 3,240 233 3,473 10,260 584 10,844 9 Jacksonville Jacksonville 76 Total 595 $82,375 $5,395 $87,770

TABLE 6 Production and Transportation Costs ($000s) for Projected 5thYear Demand Without the Jacksonville Plant Expected Served demand in demand in Brewery of 000s 000s Market area origin barrels barrels Total costs 1 Richmond Richmond 64 64 $9,503 2 Raleigh Richmond 35 35 5,275 3 Knoxville Richmond 33 1* 156 4 Columbia Columbia 55 55 8,358 5 Atlanta Montgomery 141 141 20,755 6 Savannah Columbia 20 20 3,091 7 Montgomery Montgomery 119 119 17,131 8 Tallahassee Montgomery 52 40* 6,026 25* 3,931 9 Jacksonville Columbia 76 Total 595 500 $74,226 *Indicates market demand is not fully served due to inadequate plant capacity.

(3) What additional considerations should be taken into before reaching a final decision? A number of assumptions have been implied in the analysis shown above. For example,  Demand has been assumed to grow at a constant rate in the markets.  Production is assumed limited to exactly the values given without the possibility for expansion through overtime, additional shifts, or subcontracting. 202

 Per-unit production and transportation costs are assumed to remain unchanged with the reallocation of demand throughout the network.  Customer service effects are not considered in reallocation of demand.  There is no change in per-unit costs throughout the 5-year planning horizon. This case might end with a discussion of the appropriateness of using linear programming as a vehicle for analysis in a problem such as this. Mentioning that linear programming does not consider such factors as fixed costs, return on investment, or the many subjective factors (top management's intuition about location, vested interests, etc.) that are typically a part of such problems means that linear programming, at best, is a facilitating vehicle for analysis. It does not provide the final answer.

203

CHAPTER 14 THE LOGISTICS PLANNING PROCESS 3 The MILES module within the LOGWARE software is used to solve this problem. It computes distance based on the great circle distance formula using longitude and latitude.

(a) The estimated road distance is 1,380 miles. (b) The estimated road distance is 830 miles. (c) Since both latitudes are in the same hemisphere, no adjustments need to be made. The estimated distance is 244 miles, or 2441.61 = 393 km. (d) In this case, one point is east and the other west of the Greenwich line. Therefore, we need to set a sign convention. Let's set west longitudes as + and east longitudes as . Thus, 2.20o E longitude is entered into MILES as 2.20 o. The estimated distance is 250 miles, or 2501.61 = 402.5 km. 4 Suppose that a certain linear grid coordinate system has been overlaid on a map of the United States. The grid numbers are calibrated in miles, and there is a road circuity factor of 1.21. Find the expected road distances between the following pairs of points: Equation 14-1 in the text is used to approximate distances from linear coordinates. The K factor in the equation is set at 1.21.

(a) Lansing, MI to Lubbock, TX Location a. From To b. From To c. From To d. From To

X Coordinate Lansing, MI Lubbock, TX El Paso, TX Atlanta, GA Boston, MA Los Angeles, CA Seattle, WA Portland, OR

Y Coordinate 924.3 1488.6 1696.3 624.9 374.7 2365.4 2668.8 2674.2

1675.2 2579.4 2769.3 2318.7 1326.6 2763.9 1900.8 2039.7

D  121 . (924.3  1,488.6) 2  (1,675.2  2,579.4) 2  1,290 miles (b) El Paso, TX to Atlanta, GA D  121 . (1,696.3  624.9) 2  ( 2,769.3  2,318.7) 2  1,406 miles (c) Boston, MA to Los Angeles, CA

204

D  121 . (374.7  2,365.4) 2 (1,326.6  2,7639 . )2  2,971 miles (d) Seattle, WA to Portland, OR D  121 . ( 2,668.8  2,674.2) 2  (1,900.8  2,039.7) 2  168 miles 5 The plot of the truck class rates is shown in Figure 14-1. The rates show a high degree of linearity. A linear regression was found with aid of the MULREG module in LOGWARE. The rate equation was determined to be:

R = 5.1745 + 0.0041D The standard error of the estimate SE is 0.9766 The coefficient of determination r2 is 0.928 The best single estimate of the rate at 500 miles is R = 5.1745 + 0.0041500 = $7.23/cwt. Assuming the error around the regression line is normally distributed, a 95% confidence band would give a range for the actual rate. That is, Y = R  1.96SE = 7.23  1.914 where 1.96 is the normal deviate for the normal distribution representing 95% of the area in a two-tailed distribution. The range of the estimate is: $5.32/cwt.  Y  $9.14/cwt. The r2 value of 0.928 indicates that a linear rate equation explains about 93% of the variation in the data with distance. Such a simple relationship seems to represent the rates quite well.

205

20 18

Class rate, $/cwt.

16 14 12 10

Estimating line

8 6 4 2 0 0

500

1000

1500

2000

2500

3000

3500

Distance, miles

FIGURE 14-1 Plot of Truck Class Rates 6 A plot of the average inventory level versus warehouse throughput is shown in Figure 142. The multiple regression software in LOGWARE was used to test two equation forms. The first was of the form

I  aTP b and the other was of the form

I  a  bTP Both forms showed high r2 values, with the exponential form being slightly higher at 0.9406. It was selected as the equation form to use. This equation was:

I  0.704  TP 0.83 where TP and I are both expressed in thousands of dollars. We can now estimate that for an annual warehouse throughput of $50,000,000, the average inventory would be: I  0.704  50,000 0.83  5,593939 . , or $5,593,939

Warehouse 22 has a much higher inventory turnover ratio than the average of the other warehouses. This would suggest that the inventory control procedures might be different from the others. One reason might be that person in control of the inventory in this warehouse attempts to keep inventories at low level, demand may be high such that the inventory level has been restored to a normal level, or lead times have been extended to the point where replenishment has been delayed. The reason should be investigated. 206

Average inventory level, $(Millions)

This type of inventory-throughput relationship is very useful in network planning, especially warehouse location, to estimate how inventory levels will change when sales are reallocated to a varying number of warehouses.

12 10 8 6 4

Estimating line

2 0 0

20

40

60

80

100

Annual warehouse thruput, $(Millions)

FIGURE 14-2 Plot of Inventory Levels and Warehouse Throughput for California Fruit Growers’ Association

207

USEMORE SOAP COMPANY Teaching Note

The purpose of this case study is to provide students with the opportunity to evaluate and design a large-scale production-distribution network using real data and cost relationships. To assist in the substantial amount of computational effort in this problem, an interactive computer program (WARELOCA) is available in the LOGWARE collection of software modules. Major Issues The text of the case suggests a number of questions that are critical to productiondistribution network design. These reduce to three major issues, namely:

(1) Should plant capacity be added and, if so, when and where? (2) How many warehouses are optimal and where should they be located? (3) Should the current customer service level be retained? Although no change can be made in the network without potentially affecting other variables, the attempt here will be to treat these questions sequentially to converge on a good network design. Numerous computer runs were made to provide the basic information needed in the analysis. The more meaningful runs are summarized in Appendix A to this note. Tables 1 and 2 compare selected runs for both the current-year and the future-year time periods. This information is used throughout the analysis of the major issues. The Plant Expansion Issue An attempt to meet 5-year growth goals using current plant capacity will cause the system having a total capacity of 1,630,000 cwt. to be out of capacity in 1.7 years. That is,

5th-year demand Current demand Net increase

1,908,606 cwt. 1,477,026 431,580 cwt.

Therefore, the average annual growth rate is 431,580/5 = 86,316 cwt. So, in (1,630,000  1,477,026)/86,316 = 1.7 years all available capacity will be depleted. If no expansion of plant capacity occurs, then 1,908,606  1,630,000 = 278,606 cwt. will potentially be lost by the 5th year. Sales are $100 million on 1.477 million cwt. in volume for a product value of $67.7/cwt. With a profit margin of 20%, the profit per cwt. would be 20%$67.7/cwt, or $20/1.477, = $13. Thus, 278,60613 = $3.16 million in lost sales. The weighted profit loss over the five-year period would be: 2/5  (0) + (3/5)  ([0 + 3.6])/2) = $1.08m/yr.

208

TABLE 1 Current-Year Comparison of Network Alternatives ($000s) Cost type

Benchmark

Production $30,762 Warehouse operations 1,578 Order processing 369 Inventory carrying 457 Transportation Inbound 2,050 Outbound 6,896 Total costs $42,112

Improved benchmark

Optimum number of whses

Optimum number of whses

Relaxed service (1)

Relaxed service (2)

Maximum opportunity

$30,678 1,468 354 431

$30,673 1,608 370 508

$30,675 1,572 358 490

$30,678 1,296 349 390

$30,673 1,420 354 445

$30,386 1,529 358 500

1,802 6,991 $41,725

1,976 6,310 $41,447

1,860 6,365 $41,321

1,249 7,238 $41,201

1,178 6,698 $41,043

1,178 6,458 $40,409

Customer service:  300 mi.  600 mi.

93% 98%

93% 98%

98% 100%

92% 100%

75% 98%

88% 100%

81% 94%

No. of stocking points

22

21

31

30

19

26

40

No. of plants

4

4

4

4

4

4

6

Savings vs. benchmark

$0

$387

$665

$791

$911

$1,069

$1,703

$0

$278

$404

$524

$

$1,316

Service to match benchmark

600 mi constraint on current warehouses

600 mi constraint on opt no. of warehouses

Savings vs. improved benchmark $0 Comments:

682

Unlimited service, whses, and plant cap.

209

TABLE 2 Future-Year Comparison of Alternatives ($000s) No plant expansion

Add plant @ Memphis

Add plant @ Memphis & Chicago

Memphis and opt no. of whses

Memphis and opt no. of whses

$33,965 1,496 393 431

$39,517 1,842 462 505

$39,548 1,847 454 497

$39,524 2,028 470 591

$39,522 1,976 460 573

1,647 7,230 $45,164

2,350 9,030 $53,705

2,000 9,036 $53,382

2,614 8,117 $53,342

2,426 8,222 $53,179

Customer service:  300 mi  600 mi

98% 99%

94% 98%

95% 98%

98% 100%

92% 100%

No. of stocking points

20

21

20

31

30

No. of plants

4

5

6

5

5

Comments:

Not all demand met

Cost type Production Warehouse operations Order processing Inventory carrying Transportation Inbound Outbound Total costs

Service at benchmark

210

Based on a simple rate of return on investment, capturing this profit potential would yield 1.08/4 = 27% annually on a $4,000,000 investment for expansion. The return would increase to 90% per year with the full loss in the 5th year. The potential seems great enough to justify one unit of expansion (1,000,000 cwt.). Two units of expansion probably cannot be justified, since adequate capacity would be available from the first capacity unit to meet demand requirements. The only benefit would be from the network design improvement. The savings would be about $323,000 per year in the 5th year (see Table 2) comparing one additional plant with two additional plants and keeping the current number of warehouses. The simple return on investment using 5th-year savings would only amount to about 8% (323,000100/4,000,000 = 8.1%). The next question is: Where should the expansion take place  at an existing plant or at one of the two proposed locations? From a test of expanding any of the four existing plants or the two proposed plant locations (runs 10 through 16 in Appendix A of this note), it would appear that Memphis would be the lowest cost site in the 5th year with Chicago next at only an additional cost of $76,000 per year (compare runs 14 and 15 in Appendix A). Adding a plant at a new location rather than expanding an existing plant site saves a minimum of $281,000 annually (compare runs 11 and 14 in Appendix A of this note), which results from placing plant capacity closer to warehouses. Selecting Warehouses A simple test on the number of warehouses in the network shows that transportation costs are dropping more rapidly than inventory related costs are increasing (see Figure 1). This means that 40 active warehouses will have the lowest total cost. However, some of these warehouses will have low throughput. In order to maintain a minimum replenishment frequency and shipment size, a minimum throughput needs to be met. Approximately a truckload every two weeks, or 10,400 cwt. of throughput per year, is the minimum activity needed to open a warehouse. Therefore, any warehouse showing less than this throughput will be eliminated from consideration. Under various assumptions about plants and their capacities, demand growth, and service levels, 30 to 31 warehouses seem most economical with no deterioration on service over the benchmark network. The following table shows selected results.

211

Type of run Benchmark Improved benchmark Improved benchmark Current yr. whses 5th yr. whses

Percent of demand  300 mi.

Year

Plant capacities

Total cost

No. of whses

Current

Current

93

$42,112

22

Current

93

41,725

21

5th yr.

Current Current + Memphis

94

53,705

31

Current Current + Memphis

92

41,321

30

5th year

92

53,179

30

Note that this conclusion about the number of warehouses depends on the previous conclusion that a Memphis plant should be added by the 5th year. The number of warehouses should be increased from the present 22 in both the current year and the 5th year. 42

100 99

41.8

98 97

41.6

% of demand < 300 mi.

Total cost, $(000,000s)

Service (right scale)

96 95

41.4

94 Cost (left scale)

93 41.2

92

Practical design

91 41

90 22

26

30

31

36

40

Number of warehouses

FIGURE 1 Cost and Customer Service Profiles for Alternative Network Designs

More detailed economic analysis shows that if the plants are held at current throughput levels, a savings realized from 30 warehouses would be $41,725,000  41,321,000 = $404,000 (see previous table). If current plant capacities are used and the Memphis plant is on-stream in year 5, the savings of the added warehouse would be: $53,705,000  53,179,000 = $526,000

212

On the average, there can be savings of approximately ($404,000 + 526,000)/2 = $465,000 per year by increasing the number of warehouses to 30 from the current 22. Since these are public warehouses, little or no investment would be required to implement the change. Although the number of warehouses remains relatively unchanged from the current year to the 5th year, there is some shifting among the particular warehouses in the mix. The 30 warehouses in the current year should be numbers: 1,2,3,4,5,7,8,11,13,14,15,16,17,18,19,20,21,25,28,31,32,33,34,35,36,37,38,40,44,45 providing that the loading on the current plants is allowed up to the limits of their current capacity. When the Memphis plant is brought on-stream by the end of the second year, the warehouse mix should begin to evolve to numbers: 1,2,3,4,5,7,8,11,13,14,15,17,18,19,20,21,25,28,29,31,32,34,35,36,37,38,40,44,45,47 As the Memphis plant is bought on stream, the Memphis public warehouse is closed and the volume is shifted to the Memphis plant as a warehouse. In addition, the Richmond, VA warehouse is closed and the Las Vegas, NV warehouse is opened. The number of warehouses remains at 30. Both in the base year and in the future year, the throughputs in the plants serving as warehouses are within acceptable limits as the following summary shows. Plant as a warehouse

Thruput limits

Current year solution

Future year solution

Covington New York Arlington Long Beach

450,000 cwt. 380,000 140,000 180,000

254,471 cwt. 302,043 66,592 95,943

306,478 cwt. 380,523 66,161 117,288

Customer Service Currently, a high proportion of demand (93%) is located within 300 miles of a stocking point. Since the service distance may be up to 600 miles and still meet the company's service policy, should the service level be reduced somewhat to effect a cost saving? For example, using the improved benchmark as the base case (run 2), 93% of the demand is within 300 miles and 97.5% is within 600 miles. If a 600-mile constraint is applied to the current network configuration (run 23), 75% of the demand is within 300 miles and 98% is still within 600 miles. The total costs are reduced from $41,725,000 to $41,201,000, or a savings of $524,000 per year. In addition, if the number of warehouses in the network is optimized, the costs can be reduced by another $158,000 per year (run 23 vs. run 22). However, $278,000 of the total $524,000 + 158,000 = $628,000 can be realized without a service change. This leaves approximately $404,000 that can be saved by a relaxed service restriction. The question now becomes one of whether the higher costs associated with the more restrictive service level are justified. Since there is no sales-service relationship for this

213

problem, we can only estimate the worth of the service. That is, can enough sales be generated to cover the higher service level? If physical distribution costs for the company are 15 percent of sales, which is probably a conservative estimate, then 1/0.15 = $6.70 in sales must be generated for each dollar that is added to distribution costs. Therefore, to cover $404,000 in cost would require $404,000  $6.70  38,124 cwt. $0.71 / lb.100lb./cwt. increase in sales. In of overall demand, this would be 38,124100/1,477,026 = 2.5%. But not all customers would experience a higher service level. Comparing the demand centers for 299,818 cwt. of demand shows a reduction in warehouse to customer miles. Thus, moving from a minimum cost network to one with a high service level, where the percent of demand less than 300 miles increases from 75 percent to 93 percent, requires that the 38,124 cwt. increase in demand occur in the 299,818 cwt. of demand affected by the change. This would be a 13 percent increase. The products are not highly differentiated from others in the marketplace so that service plays an important role in selling these products. Whether a 93  75 = 18 percentage points increase in service can result in a 2.5 percent increase in overall sales cannot be judged by the distribution department alone. The sales department must play an important part in indicating whether the additional sales are possible. If they are not likely to be realized, there is no incentive for a network other than the minimum cost one. If this information is not available from sales, the conclusion is likely to be to maintain the status quo as represented by the benchmark. That is, one-day service is most likely to guide the design. Overall Analysis and Summary The recommended design would involve an immediate increase in the number of warehouses from 22 to 30. In addition, there should be an immediate reallocation of demand among the existing plants. No reduction in the customer service level seems justified at this time. Therefore, a total cost reduction of $42,112,000  41,321,000 = $791,000 per year seems immediately achievable (run 1 vs. run 18). By the end of the 2nd year, the Memphis plant should be brought on stream and the network should begin to evolve from the current design (run 24) to that for the 5th year (run 25). The addition of a plant is justified from the high rate of return realized from the profit potential of being able to continue meeting the growth in demand. For the current year, a breakdown of the service and the cost changes show the following.

214

Cost type Production Whse operations Order processing Inventory carrying Transportation Inbound Outbound Total costs ($000s)

Benchmark

Currentyear design

Change from benchmark

$30,762 1,578 369 457

$30,675 1,572 358 490

$ -87 - 6 -11 +33

-0.3% -0.4 -3.0 +7.2

2,050 6,896 $42,112

1,860 6,365 $41,320

-190 -531 $-792

-9.3 -7.7 -1.9%

By the 5th year, total distribution costs should be $53,179,000, or $53,179,000/1,908,606 = $27.86, compared with the current-year cost of 42,112,463/1,477,026 = $28.51 per cwt. If current year costs are projected to the 5thyear demand level, the 5th-year production/distribution costs might be 28.511,908,606 = $54,414,357, or a savings of $54,414,357  53,179,000 = $1,235,357 per year. Of course, these savings can only be realized through the addition of capacity at Memphis for $4,000,000. If this capacity is useful for at least 15 years, the amortization of $4,000,000/15 = $267,000 per year would yield a net savings of $532,000 per year. Overall, the design change appears to be justified.

215

APPENDIX A Listing of Selected Computer Runs Service constraint

Run Run no. description

No of plants

Plant capacity

Demand level

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25

4 4 4 6 4 4 4 4 4 4 4 4 4 5 5 6 4 4 4 5 4 4 4 4 5

Current Current Current Crnt+1m Current Current Current Current Current Current Current Current Current Current Current Current Current See cmt Current Current Current Current Current See cmt

Current - mi Current 300 Current 9000 Current 9000 5th yr. 300 Current 300 Current 300 Current 300 Current 300 5th yr. 300 5th yr. 300 5th yr. 300 5th yr. 300 5th yr. 300 5th yr. 300 5th yr. 300 5th yr. 300 Current 300 Current 300 5th yr. 300 Current 600 Current 600 Current 600 Current 375 5th yr. 375

Benchmark Improved benchmark No serv constraint Max opportunity Future yr-imp bmk Test 27 whses Test 32 whses Test 37 whses Test 42 whses Exp Covington Exp New York Exp Arlington Exp Long Beach Add Memphis Add Chicago Add Mem & Chi No plant expansion Optimum whses Optimum whses Optimum whses Test cust service Test cust service Test cust service Optimum whses Optimum whses

No of whses

Total costs

Percent of demand within  300  600 Comments

22 21 18 40 21 26 31 36 40 21 21 21 21 21 20 20 20 31 30 31 31 26 19 30 30

$42,112 41,725 40,896 40,409 53,777 41,744 41,615 41,501 41,486 54,145 53,986 54,709 55,251 53,705 53,781 53,382 45,164 41,447 41,563 53,342 40,996 41,043 41,201 41,321 53,179

93% 93 71 81 93 95 98 99 99 94 93 94 94 94 94 95 98 98 97 98 80 88 75 92 92

98% 98 89 94 98 100 100 100 100 98 98 98 98 98 98 98 100 100 100 100 100 100 98 100 100

Current network design No investment required Added plants at 1m cwt Plant cap + 1m cwt

Covington cap + 1m cwt New York + 1m cwt Arlington cap + 1m cwt Long Beach cap + 1m cwt Add Memphis at 1m cwt Add Chicago at 1m cwt Add Chi & Mem at 1m cwt ea Only 85.4% of demnd served Plants at current capacity Plants at current thruput Memphis at 1m cwt Whses at opt no = 31 Whses from opt no = 31 Whses from current 22 Service level at bmk Serv at bmk/Mem @ 1m cwt

216

ESSEN USA Teaching Note Strategy

Essen USA is concerned with entire supply channel performance. The supply channel consists of four echelons ranging from factory to customers. The purpose of this case study is for the student to manipulate the supply channel variables through the use of a channel simulator in order to improve individual member and system-wide performance. The channel variables include forecasting methods, inventory policies, transportation services, production lot sizes, order processing costs, and stock availability levels. Students should seek to optimize channel performance, although it is not expected that the optimum actually can be found or verified. However, improving performance over existing levels is achievable. The SCSIM module of LOGWARE is used to simulate the demand and product flows throughout the multi-echelon supply chain. SCSIM is an ordinary Monte Carlo day-today type of simulator. Using a simulator for performance improvement requires thinking of it in of as an experimental methodology. That is, a single run of the simulator is a particular event sequence generated from random numbers. Changing the seed number in the simulator causes a different set of random numbers to be generated and possibly another outcome from the same input data. A simulation run with a specified seed number should be viewed as a single statistical observation and multiple outcomes from various seed numbers should be treated as a statistical sample and analyzed accordingly, i.e., comparing means and standard deviations. Each simulation is run for a period of 11 years with results taken from years 2 through 11. The first year is not used since it can show unstable results due to startup conditions. The results appear to reach steady state by the second year, and the results for the 10 years thereafter are averaged to give a reasonable representation of channel performance for a given run. The database used to represent the current performance of the channel, as derived from the case study, is summarized in the Appendix A of this note and a typical run report is shown in Appendix B. This case provides students with the opportunity to observe the operation of a multiechelon supply channel and to assess the impact of changing key operating variables on individual as well as on channel-wide performance. The effect on cost and customer service as well as sales, inventory, and back order levels of demand patterns, demand forecasting methods, inventory control methods, transportation performance, production lot sizing, order processing procedures, and item fill rates can be observed in both graphical and report forms. Most importantly, students can see the effects of supply chain decisions rather than project the results statistically. Questions

1. What can you say about the logistics performance throughout the supply channel for Essen and its customers?

217

General observations

It is recognized that Essen must deal with demand that has significant seasonal peaks at gift giving times of the year as shown in Figure 1. Compared with a smooth demand pattern, this can cause increasing demand variability upstream from the customers, as illustrated in Figure 2. This “bull whip” effect is partly a result of the demand for an upstream member being derived from the order size and pattern of its immediate downstream channel member. Forecast accuracy, lead-time uncertainty, and inventory control method also affect demand variability and the resulting cost of that variability.

Figure 1 Typical Demand Pattern for Essen Over the Period of One Year

Retailer Distri warehouse

Essen Warewarehouse house

Factory Factory

-butor

Retailer Retailer

Figure 2 Increasing Demand Variability of Upstream Channel for a Four-Year Period

218

Benchmark

Running the simulator (SCSIM) with a seed number of 123456 and simulated period of 11 years with results taken from the last 10 years, the channel generates average annual sales of $109.5 million for a net average annual system profit contribution of $24.4 million, as shown in Table 1. The question arises as to whether channel performance can be improved and profits increased. At least two observations can be made that suggest there is room for improvement. First, the inventory levels for both the retailer’s warehouse and Essen’s warehouse are quite high compared with the Retailer (see Figure 3). It is possible that Retailer inventories are too low. However, the inventory turnover ratio is about 7 for the Essen’s warehouse (see Table 1). This is not particularly high for a food product that might have a turnover at least in the range of 10 to 12. The turnover for the retailer’s warehouse appears more in line with industry norms of about 13 (see Table 1).

Retailer warehouse

Essen warehouse

Retailer

Figure 3 Inventory Levels for Four Years Using Benchmark Data

Second, the backorders at the Retailer level do not seem to recover well from the seasonal spike in demand. Correspondingly, the Retailer inventory turnover is 81 (see Table 1), which is quite high. The low percentage of demand filled on request (<50%) suggests that inadequate inventory is being maintained to meet reasonable fill rates. Third, customer service levels are also low for the retailer’s warehouse and Essen’s warehouse. Backorder occurrences are high for both channel . Although inventory levels are adequate most of the time, seasonal demand rippling through the supply chain causes a significant number of back orders before inventory can be replenished. The observation is that there is an opportunity to improve channel performance, especially in of customer service. A major concern is how to mange the seasonal demand pattern that is causing the cyclical behavior throughout the echelons of the channel. Current performance of the channel is summarized in Table 1 for 4 simulation runs using different seed numbers.

219

Table 1

Average Annual Performance of Channel and the System at Benchmark

Channel member

Run 1

Run 2

Run 3

Run 4

Average

$73,105,904 37,918 $1,928

$72,967,088 37,846 $1,928

$72,616,192 37,664 $1,928

$72,477,376 37,592 $1,928

$72,791,640 37,755 $1,928

6.52 <50% $5,578,291 $147.02

6.59 <50% $5,549,202 $146.50

6.61 <50% $5,521,236 $146.32

6.58 <50% $5,540,447 $146.65

6.58 <50% $5,547,294 $146.62

12.95 <50% $3,873,236 $101.94

13.01 <50% $3,895,406 $102.15

12.96 <50% $3,853,890 $102.14

12.89 <50% $3,912,699 $103.07

12.95 <50% $3,883,808 $102.33

81.02 <50% 38,017 $2,884,527 $76.19

80.38 53.02% 37,983 $2,768,697 $72.89

80.45 54.09% 37,774 $2,939,464 $77.82

80.60 <50% 37,804 $2,981,607 $78.87

80.61 <50% 37,895 $2,893,574 $76.44

$24,426,593 22.23% 123456

$24,591,053 22.40% 444444

$24,235,498 22.20% 555555

$24,340,563 22.28% 666666

$24,398,426 22.28%

Essen’s factory Total cost

Units produced Cost per unit Essen’s warehouse TO ratio Fill rate Cost

Cost per unit Retailer’s warehouse TO ratio Fill rate Cost

Cost per unit Retailer TO ratio Fill rate Units sold Cost

Cost per unit System Profit Profit as % of sales Seed Number

2. What steps would you suggest taking to improve logistics performance throughout the channel? Do any of the changes involve Essen? If so, does the company directly realize any cost and/or operating performance improvements? There are a number of actions that generally can be taken to lower costs and improve customer service. Improving the forecast, shortening the lead times, changing the inventory control policy, and changing production lot sizes are all variables that can be altered for possible performance improvement. The interactions among these variables and the large number of variable combinations preclude finding the optimal set. However, they can be explored in a systematic way to find improvement. The primary focus of this analysis will be to increase the fill rates at the risk of increasing costs. Ultimately, revenues, through improved customer service, may be preserved or increased to more than compensate for reduced profits. Retailer Level

Start with the retailer because of the proximity to the customer. Fill rates need to be improved, probably in the 95-99% range as specified in the database. Inventory turns can

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be guided by the industry average of 12 turns per year. Where the two cannot be tly met, service will prevail. Clearly, putting additional inventory at the retail point will improve customer service. Using the company’s current inventory policy of stocking to demand, the target level can be raised without changing the review time. Exploring different target levels shows 14 days to offer about 35 turns and a 99+% fill rate. Because of the high cost of a back order, total costs at the retail level drop significantly. Altering the forecasting method and the settings associated with the method yield little opportunity for improvement. Using an exponential smoothing model with a high smoothing constant to better follow the seasonal changes in demand results in increased costs. Lowering the smoothing constant to 0.1 did not offer improvement either. Altering the number of periods in the moving average model did not improve costs and only degraded performance. Shortening the review time in the stock-to-demand reorder policy did have a positive effect on fill rate, but resulted in high costs and lower inventory turns. The tradeoff did not seem beneficial, given the fill rate and turnover targets. Retail Warehouse Level

Determining an improved policy at this level is difficult because a 95% fill rate and 10 to 12 inventory turns is an illusive goal. Using service as the primary target, a stock-todemand control policy is used with a review period of 7 days and a target of 25 days of inventory. The forecasting method is moving average with a period of 7 days. The performance achieved at this channel level is about 9 inventory turns per year and a 97% fill rate. Essen’s Warehouse Level

The performance at Essen’s warehouse level seems to mimic that at the retail warehouse level except that there is more demand variability. Again, an inventory turnover ratio in the target range cannot be achieved while maintaining a high fill rate level. Trying to achieve high service levels with high levels of inventory is difficult, probably due to the extensive demand variability that filters back to this member of the channel. Multiple simulation runs show that a high fill rate cannot consistently be achieved even when on the average inventory levels are high. However, average performance shows an 82% fill rate and 1.5 inventory turns per year based on a 7-day moving average forecasting model and a stock-to-demand inventory control policy with a review time of 7 days and an inventory target of 25 days. Essen’s Factory

The concern with the factory level in the channel is whether product should be manufactured in a larger lot size, but with slightly higher production time variability. The reduced costs seem to out weigh the negative effects of increased variability. Producing in the larger lot size is favored. Overall

Using the objective of improving customer service, it is not surprising that supply channel costs increase as shown from the reduced profit in Table 2 compared with Table 1. The average fill rate has increased for all of the channel, but the cost effects 221

are spread disproportionately among the . Even with a higher fill rate, the retailer benefits from a substantial reduction in the cost per unit sold. On the other hand, the cost for handling a unit of the product at Essen’s warehouse is substantially increased. Essen should take advantage of the cost reduction from producing in the larger batch size, but this does not offset the higher cost at the company’s warehouse. As an upstream member of the supply channel, Essen undoubtedly suffers from the variability in demand, which cannot entirely be controlled. The retailer benefits from the action to increase fill rates across the channel. However, Essen is put at a disadvantage and may take a counter action to improve its cost position. Essen may simply lower its inventory level by reducing the reorder target quantity from 25 to 10 days. This reduces Essen’s per-unit warehouse cost, but it also increases the costs for the retailer. The reduced inventory level at the Essen warehouse causes lower fill rates for the downstream retailer. Unless the retailer can find an incentive to reward Essen for its good service, it will be difficult for Essen to provide the level of service that the retailer would like and that is economically beneficial to Essen. Table 2

Average Annual Performance of Channel and the System as Revised

Channel member

Run 1

Run 2

Run 3

Run 4

Average

$68,638,738 35,950 $1,909

$74,761,258 39,286 $1,903

$78,418,188 41,268 $1,900

$75,400,666 39,622 $1,903

$74,304,713 39,032 $1,904

1.47 90.65% $12,060,444 $317.51

1.42 100% $12,449,170 $326.28

1.51 63.16% $11,895,743 $311.87

1.46 72.92% $12,178,581 $319.61

1.47 81.68% $12,145,984 $318.82

9.24 96.64% $4,018,143 $105.70

9.02 96.29% $4,080,160 $107.09

9.14 97.60% $4,043,973 $106.55

9.28 98.09% $3,992,791 $105.51

9.17 97.16% $4,033,767 $106.21

35.44 99.52% 38,017 $727,378 $19.13

35.36 99.53% 37,936 $717,363 $18.91

34.99 99.70% 37,979 $709,772 $18.69

35.14 99.40% 37,730 $748,198 $19.75

35.23 99.54% 37,916 $725,677 $19.12

$24,423,849 22.23% 123456

$17,627,666 16.08% 111111

$14,690,765 13.38% 222222

$17,184,464 15.69% 333333

$18,481,686 16.85%

Essen’s factory Total cost

Units produced Cost per unit Essen’s warehouse TO ratio Fill rate Cost

Cost per unit Retailer’s warehouse TO ratio Fill rate Cost

Cost per unit Retailer TO ratio Fill rate Units sold Cost

Cost per unit System Profit Profit as % of sales Seed Number

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3. Would shipping by airfreight from be a benefit to channel performance? To Essen? No. Selling candies to end customers at $2,890 per thousand lb. using airfreight shipping results in obvious loss to Essen. The cost of shipping by air is $1,833 per thousand lb., plus $1,000 material cost and $850 production cost results in a total cost much higher than selling price. There is no point to using airfreight. Running the simulation with the higher freight rate but lower variability confirms that channel profits would be negative. 4. Is there a benefit to producing in the larger 20,000-pound batch size? Yes. This was tested in question 2. From Tables 1 and 2, it can be seen that production costs drop from $1,928 per unit to $1,904 per unit. The overall channel cost reduction overshadows the negative effects of greater length and variability in production time.

Appendix A Simulation Database for Essen USA Under Current Conditions Title: ESSEN USA Initialization 123456 11 2890

Seed value Length of simulation, years Annual price, $/unit

Customer demand pattern Generate daily demand 100 Average daily demand, units 15 Standard deviation of daily demand, units 1 Annual demand growth increment, % Monthly seasonal indices Month Index Month Index Month Index Month 1 0.25 4 0.75 7 0.75 10 2 1.25 5 0.75 8 0.75 11 3 1.25 6 0.75 9 0.75 12 Retailer/Level 1 Product item data 2220 1 35 25 1 0 98 670

Index 0.75 1.50 2.50

Item value in inventory, $/unit Customer order filling cost, $/unit Purchase order processing cost, $/order Inventory carrying cost, %/year Average customer order fill time, days Customer order fill time standard deviation, days In-stock probability, % Back order cost, $/unit

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Forecasting method Moving average 7 Number of periods Reorder policy Stock-to-demand control method 10 Target days of inventory 7 Review time in days Distributor/Level 2 Product item data 2220 Item value in inventory, $/unit 20 Retailer order filling cost, $/unit 75 Purchase order processing cost, $/order 25 Inventory carrying cost, %/year 2 Average retailer order fill time, days 0.2 Retailer order fill time standard deviation, days 95 In-stock probability, % 100 Back order cost, $/unit Forecasting method Moving average 30 Number of periods Reorder policy Stock-to-demand control method 45 Target days of inventory 30 Review time in days Warehouse/Level 3 Product item data 1710 Item value in inventory, $/unit 15 Distributor order filling cost, $/unit 75 Purchase order processing cost, $/order 20 Inventory carrying cost, %/year 3 Average distributor order filling time, days 0.3 Distributor order fill time, days 95 In-stock probability, % 25 Back order cost, $/unit Forecasting method Moving average 360 Number of periods Reorder policy Stock-to-demand control method Target days of inventory 90 Review time in days 30

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Factory/Source Product item data

850 10 10 8 2 1000

Production cost, $/unit Minimum production lot size, units Warehouse order filling cost, $/unit Average production time, days Production time standard deviation, days Purchase cost, $/unit Transportation

Transport between Distributor and Retailer

25 1 0

Transport cost, $/unit Average time in-transit, days Transit time standard deviation, days

Transport between Warehouse and Distributor

70 5 1

Transport cost, $/unit Average time in-transit, days Transit time standard deviation, days

Transport between Factory and Warehouse

78 9 3

Transport cost, $/unit Average time in-transit, days Transit time standard deviation, days

Appendix B

Benchmark Simulation Results with Seed Number 123456 and Simulation Length of 11 Years

SUPPLY CHANNEL REPORT FOR SIMULATED YEARS 2 TO 11 Yearly Simulated average period FINANCIAL PERFORMANCE $109,868,552 $1,098,685,520 Revenue 37,918,000 379,180,000 Cost of purchased goods 71,950,552 719,505,520 Gross margin 32,230,300

322,303,000

949,025 2,656,010 2,957,604

9,490,250 26,560,100 29,576,040

38,017 759,890 569,145

380,168 7,598,900 5,691,450

1,638 683 675

16,380 6,825 6,750

Production cost Transportation costs: Distributor to retailer Warehouse to distributor Factory to warehouse Sales order handling cost for: Customer orders Retailer orders Distributor orders Order processing cost for: Orders to distributor Orders to warehouses Orders to factory Inventory costs

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260,414 1,627,909 1,988,984

2,604,145 16,279,092 19,889,837

Retailer Distributor Warehouse

2,584,458 535,730 363,478

25,844,580 5,357,300 3,634,775

Back order costs Retailer Distributor Warehouse

$24,426,593

$244,265,928

Net profit contribution

Appendix C Simulation Database for Essen USA as Revised for Serivce Improvement Title: ESSEN USA Initialization 123456 11 2890

Seed value Length of simulation, years Annual price, $/unit

Customer demand pattern Generate daily demand 100 Average daily demand, units 15 Standard deviation of daily demand, units 1 Annual demand growth increment, % Monthly seasonal indices Month Index Month Index Month Index Month 1 0.25 4 0.75 7 0.75 10 2 1.25 5 0.75 8 0.75 11 3 1.25 6 0.75 9 0.75 12

Index 0.75 1.50 2.50

Retailer/Level 1 Product item data 2220 Item value in inventory, $/unit 1 Customer order filling cost, $/unit 35 Purchase order processing cost, $/order 25 Inventory carrying cost, %/year 1 Average customer order fill time, days 0 Customer order fill time standard deviation, days 98 In-stock probability, % 670 Back order cost, $/unit Forecasting method Moving average 7 Number of periods Reorder policy Stock-to-demand control method 14 Target days of inventory

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7

Review time in days

Distributor/Level 2 Product item data 2220 Item value in inventory, $/unit 20 Retailer order filling cost, $/unit 75 Purchase order processing cost, $/order 25 Inventory carrying cost, %/year 2 Average retailer order fill time, days 0.2 Retailer order fill time standard deviation, days 95 In-stock probability, % 100 Back order cost, $/unit Forecasting method Moving average 7 Number of periods Reorder policy Stock-to-demand control method 35 Target days of inventory 7 Review time in days Warehouse/Level 3 Product item data 1710 Item value in inventory, $/unit 15 Distributor order filling cost, $/unit 75 Purchase order processing cost, $/order 20 Inventory carrying cost, %/year 3 Average distributor order filling time, days 0.3 Distributor order fill time standard deviation, days 95 In-stock probability, % 25 Back order cost, $/unit Forecasting method Moving average 7 Number of periods Reorder policy Stock-to-demand control method 25 Target days of inventory 7 Review time in days Factory/Source Product item data

825 20 10 10 2.1 1000

Production cost, $/unit Minimum production lot size, units Warehouse order filling cost, $/unit Average production time, days Production time standard deviation, days Purchase cost, $/unit

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Transportation Transport between Distributor and Retailer

25 1 0

Transport cost, $/unit Average time in-transit, days Transit time standard deviation, days

Transport between Warehouse and Distributor

70 5 1

Transport cost, $/unit Average time in-transit, days Transit time standard deviation, days

Transport between Factory and Warehouse

78 9 3

Transport cost, $/unit Average time in-transit, days Transit time standard deviation, days

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CHAPTER 15 LOGISTICS/SUPPLY CHAIN ORGANIZATION

All questions in this chapter require individual judgment and response. No answers are offered.

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CHAPTER 16 LOGISTICS/SUPPLY CHAIN AND CONTROL

All questions in this chapter require individual judgment and response. No answers are offered.

230