Question

For the project outlined on the following table, (a) draw a network diagram (arrow or PDM,
as you prefer), and (b) identify the critical path and duration. (c) Identify the task(s) you
would crash and the incremental cost to reduce project duration by (i) one week; (ii) a second
week; (iii) a third week.

Duration (Weeks) Cost (Dollars) Task Follows Task(s) Normal Crash Normal Crash (Start) (Start) A, B B C D uun $500 400 600 $600 450 750 550 640 400 550

Answer 1

(a)

Network Diagram

w > Start End

(b)

Critical path is the path which takes longest time.

We will look at all possible paths and find the critical path with longest completion time. Project duration is same as the completion time for critical path activities

Possible paths are:

A-C-E : 3+5+4=12

B-C-E: 2+5+4 = 11

B-D-E: 2+5+4=11

Hence, the critical path is A-C-E with total project duration of 12 weeks

c)

Now to crash the project, we will first estimate the crash slope for all activities. Activities on critical path will be crashed to reduce over project duration, starting with activities with lowest cost slope and available weeks to crash for each critical activity.

Crash slope = (Crash cost - Normal cost)/ (Normal time - crash time)

We will calculate crash slope as below

	Duration (Weeks)		Cost (Dollars)		Crash slope
Task	Normal	Crash	Normal	Crash
A	3	2	500	600	(600-500)/ (3-2)=100
B	2	1	400	450	50
C	5	3	600	750	75
D	5	4	550	640	90
E	4	3	400	550	150

(i) Reduce project duration by one week

Among activities on critical path A-C-E, the lowest cost slope activity is C so first we will crash activity C by 1 week and project duration is reduced to 12-1 = 11 weeks. The cost of crashing is $75

Possible paths are: A-C-E (11), B-C-E (10), B-D-E (11). Two critical paths are A-C-E and B-D-E

(ii) Second week

Next we will crash B with lowest cost slope of 50 which will reduce time for B-D-E to 10 weeks. However to crash overall project duration, we need to crash another activity A-C-E by 1 week. C with lowest cost slope of 75 can be crashed again by one more week.

Hence, we will crash B by 1 week and C by 1 week. Project duration is reduced to 10 weeks. Total cost of crashing is $50+$75 = $125

Possible paths are: A-C-E (10), B-C-E (8), B-D-E (10). Two critical paths are A-C-E and B-D-E

(iii) Third week

Now, next looking at activities D, A and E which can crashed. We will crash E by 1 week which is a common activity among two critical paths. This will optimize the crash cost else we had to crash both D and A by 1 week which will have higher crash cost.

Crash activity E by 1 week to reduce the project duration to 9 weeks. Total crashing cost = $150

Possible paths are: A-C-E (9), B-C-E (7), B-D-E (9). Two critical paths are A-C-E and B-D-E.

Please give thumbs up/ likes if the answer is helpful. Thank you!