Use your results from part B (i.e., Table 1.2, given PERT diagram, activities L-U) to do the following:
1. Identify the activities, considering all paths, to be crashed to meet the following criteria:
- completion of the project within 22 weeks
- achievement of the lowest possible cost
2. Determine the number of weeks each of the activities identified in part C1 should be crashed.
3. Determine the sum of the individual costs of crashing each activity identified in part C1. Computation of Action L
The maximum decrease in the period = Expected period of time - crash time= 2-1.5=0.5
Crash Cost per week =12500-9000= 3500/.5=7000
Maximum reduction in time = (Normal time) - (Crash time). Using Table 1.2 as reference. (Crash Cost) - (Normal Cost) / (Normal Time - Crash Time) = Crash Cost per unit
L, O, P, S, U
M, P, S, U
M, Q, S, U
M, Q, R, T, U
N, R, T, U
C1, C2, C3 : The critical path is M-P-S-U with a course equals 24. The expected project duration is 24 weeks. Therefore, to bring down the project duration to 22 weeks, we will crash activity P by 2 weeks. The extra cost of crashing = (2. *1,357.14) = The critical path is M-P-S-U. The project completion duration is 24 weeks, and we have only 22 weeks to complete the project. Therefore, 2 weeks have to be reduced. To shorten a project, only critical activities are crashed in order to reduce the project duration. Let's check the low cost options.
Activity ID Crash Cost Normal Cost Normal Time (days) Max Crash Time Crash cost per week
M 23500 10750 9 5 3187.50
P 18500 13750 8 4.5 1357.14
S 22750 11500 3 1.5 7500.00
U 21250 12000 4 1 3083.33
The lowest crash cost per week is activity P.
Cost to crash activity P by 2 weeks = 2 * 1357.14 = 2174.28
By crashing only activity P by 2 weeks, the project can be completed in 22 weeks.
Activity to be crashed = Activity P
Number of weeks = 2 weeks
Sum of Individual Cost = $2174.28