Resource Allocation I 6z6i1p

Project Management A Managerial Approach

Resource Allocation

Critical Path Method - Crashing a Project M includes a way of relating the project schedule to the level of physical resources allocated to the project  This allows the project manager to trade time for cost, or vice versa  In M, two activity times and two costs are specified, if appropriate for each activity 

Critical Path Method - Crashing a Project The first time/cost combination is called normal, and the second set is referred to as crash  Normal times are “normal” in the same sense as the „m‟ time estimate of the three times used in PERT  Crash times result from an attempt to expedite the activity by the application of additional resources 

Critical Path Method - Crashing a Project Careful planning is critical when attempting to expedite (crash) a project  Expediting tends to create problems; and the solution to one problem often creates several more problems that require solutions  Some organizations have more than one level of crashing 

Slope =( crash cost - normal cost)/ (crash time – normal time)  That is, the cost per day of crashing a project. Negative slope indicates that as the time required for a project task is decreased, the cost is increased. 

Exercise 1 Given the following network, determine the first activity to be crashed by the following priority rules: a. Shortest task first b. Minimum slack first c. Most critical followers d. Most successors A4

D5 2

1

C7 B3 3

4

Shortest task is B  B, C, or D have zero slack.  Critical path is B-C-D.  A has slack of 6 days.  B has the most critical followers.  B has the most successors, 2.  Critical Time is 15 days. 

Exercise 2 Using the previous network and the additional information below, find:  The crash cost per day.  Which activities should be crashed to meet a project deadline of 13 days at minimum cost. Assume partial crashing is allowed.

Activity

Crash time (days)

Crashed cost (total)

Normal Time (days)

Normal cost

A

3

$500

4

$300

B

1

325

3

250

C

4

550

7

400

D

3

250

5

150

Crash cost per day Activity A

Crash days=normal time-crash time 1

Crash cost 200

Crash cost/day 200

B

2

75

37.5

C

3

150

50

D

2

100

50

Task B is on the critical path and has the lowest crash cost per day.  Hence crashing task B will imply additional crash cost of $75 only.  Crashing B by 2 days reduces the duration to 13 days without making any other new tasks critical. 

Exercise 3: Given the following network(time in days).

Find the lowest cost to complete the project in 10 days. 2 d a 1

c

b

4

e

3

Activity

Crash time,cost $

Normal time,cost

Partial crashing

a

3, 60

3, $60

No

b

6, 80

7, 30

Yes

c

2, 90

5, 50

No

d

5, 50

6, 30

No

e

2, 100

4, 40

Yes

Current time & cost: 12 days & $ 210.

0

a3 1

3 2

d 6

Hence best solution: Cut e by 2 days at a cost of $60

4 12 5 c e 4 b 7 3 8 Since the critical path is a-c-e, we initially need consider only these 3 activities:

a: cannot be crashed c: can cut 3days at an extra cost of $ 40 but only results in project completion by day 11, due to b. to reach 10 days, cut b by 1 day, total extra cost $90. e: can cut e by 2 days for an extra cost of $60 & results in project completion by day 10.