In: Operations Management
Table below contains information about an environmental clean-up project in the township of Hiles. Shorten the project three weeks by finding the minimum-cost schedule. Assume that project indirect costs and penalty costs are negligible. Identify activities to crash while minimizing the additional crash costs.
Table 1 - Environmental Project Data
Activity |
Normal Time (weeks) |
Crash Time (weeks) |
Cost to Crash ($ per week) |
Immediate Predecessor(s) |
A |
7 |
6 |
200 |
None |
B |
12 |
9 |
250 |
None |
C |
7 |
6 |
250 |
A |
D |
6 |
5 |
300 |
A |
E |
1 |
1 |
- |
B |
F |
1 |
1 |
- |
C, D |
G |
3 |
1 |
200 |
D, E |
H |
3 |
2 |
350 |
F |
I |
2 |
2 |
- |
G |
Network Diagram:
Activity | Normal Time | Crash time | Time can be decreased | Cost to crash per week |
A | 7 | 6 | 1 | 200 |
B | 12 | 9 | 3 | 250 |
C | 7 | 6 | 1 | 250 |
D | 6 | 5 | 1 | 300 |
E | 1 | 1 | 0 | -- |
F | 1 | 1 | 0 | -- |
G | 3 | 1 | 2 | 200 |
H | 3 | 2 | 1 | 350 |
I | 2 | 2 | 0 | -- |
In this project, there are two critical paths. So to shorten the project by three weeks we need to crash the activities on both paths simultaneously.
Step 1:
On Path A-C-F-H, min crashing cost is for A, while on path B-E-G-I min crashing cost is for G.
So we can crash the both activity A and G by 1 week.
Crashing Cost = 200 +200 = 400
Step 2:
Now, D is also become a critical activity.
Further, we can crash Activity G by 1 week, and simultaneously we can crash activity H by 1 wee.
Crashing Cost = 200 +350 = 550
Step 3:
Further, we can crash Activity B by 1 week, and simultaneously we need to crash Activity C and D (because both are on the critical path) by 1 week.
Crashing Cost = 250+ 250 +300 =800
Total Crashing Cost = 400 + 550 +800 = 1750