Kinta Rail Systems: Crashing a Signalling Project with Two Critical Paths
Kinta Rail Systems must upgrade a signalling system. The activity list, with normal durations in days and crash limits and costs, is:
| Act | Predecessor | Normal | Crash limit | Crash cost (RM/day) |
|---|---|---|---|---|
| A | none | 4 | 0 | - |
| B | none | 6 | 0 | - |
| C | A | 5 | 2 | 700 |
| D | B | 3 | 1 | 1,100 |
| E | A | 7 | 0 | - |
| F | C,D | 4 | 0 | - |
| G | E | 2 | 0 | - |
| H | C,D | 6 | 2 | 800 |
| I | F,G | 3 | 0 | - |
| J | H | 5 | 2 | 1,200 |
The client wants completion in 18 days instead of the normal 20. The contract carries a penalty of RM2,000 for each day beyond 18. Crashing is paid per day and can be done in whole days only. The project manager has already suggested crashing C, because C lies on a path through A. The client asks whether that is sensible and what the least-cost plan is.
1. Carry out a forward and backward pass. State the project duration, the critical path or paths, and the float on each non-critical activity.
Read the worked answer
Forward pass: E0 = 0; E1 = 4; E2 = 6; E3 = max(4+5, 6+3) = 9; E4 = 4+7 = 11; E5 = max(9+4, 11+2) = 13; E6 = 9+6 = 15; E7 = max(13+3, 15+5) = 20.
Backward pass: L7 = 20; L6 = 15; L5 = 17; L4 = 15; L3 = min(15-6, 17-4) = 9; L1 = min(9-5, 15-7) = 4; L2 = 9-3 = 6; L0 = 0.
Zero-float activities: A, B, C, D, H and J. There are two critical paths, A-C-H-J and B-D-H-J, each 4+5+6+5 = 20 days or 6+3+6+5 = 20 days.
Non-critical float: E has 4 days, F has 4 days, G has 4 days, I has 4 days. The existence of two critical paths matters: shortening only one of them does not shorten the project.
2. Evaluate the project manager's suggestion to crash C. Then determine the least-cost way to meet the 18-day target and calculate the net benefit.
Read the worked answer
Crashing C by 2 days costs 2 x RM700 = RM1,400. But C lies only on A-C-H-J. The other path, B-D-H-J, remains 20 days, so the project duration is still 20. The money is wasted.
The cheapest shared activity is H, which lies on both critical paths. Crash H by 2 days at 2 x RM800 = RM1,600. Both paths become 18 days, so the project reaches 18 days. Penalty avoided = 2 x RM2,000 = RM4,000. Net benefit = RM4,000 - RM1,600 = RM2,400.
Crashing J by 2 days would also work, at RM2,400, but it costs RM800 more than crashing H and gives the same time reduction. So the least-cost plan is H by 2 days.
3. The client then asks for 17 days. Find the least-cost additional crashing and decide whether it is worth doing.
Read the worked answer
After H has been crashed by 2 days, both critical paths are 18 days. To reach 17, both paths must be shortened by one day, or one activity common to both must be shortened.
Option 1: crash J by one more day. J is common to both paths, so duration becomes 17. Cost = RM1,200.
Option 2: crash C by one day and D by one day. C shortens A-C-H-J and D shortens B-D-H-J. Cost = RM700 + RM1,100 = RM1,800.
Option 1 is cheaper. The penalty avoided for one day is RM2,000, so the net saving is RM2,000 - RM1,200 = RM800. It is worth doing if the penalty is real and there are no other constraints. Note that the decision depends on both paths being shortened together; crashing C alone at this stage would again have no effect on project duration.