Lab Upgrade Schedule: Critical Path, Crashing Cost and an Impossible Deadline
The ICT department of a provincial school has approval to upgrade its computer laboratory before the new term, which starts 20 working days away. From the planning meeting the project manager recorded: A, gather requirements (3 days); B, order and receive hardware (2 days, after A); C, electrical wiring and air conditioning (5 days, after A); D, network cabling (4 days, after C); E, install and configure hardware (2 days, after B and D); F, install software, image machines and migrate user data (3 days, after E); G, train two teachers (2 days, after F); H, acceptance testing with the IT club (2 days, after F); I, handover to the principal (1 day, after G and H). Suppliers quoted: network cabling can be compressed from 4 days to 2, at Rs. 15,000 per day saved; software installation can be compressed from 3 days to 2, at Rs. 8,000 per day saved; wiring can be compressed from 5 days to 4, at Rs. 20,000 per day saved; hardware delivery can be cut by one day for Rs. 3,000. The principal wants handover completed on day 17 at minimum extra cost, and insists the teaching timetable must not slip.
1. Carry out the forward and backward pass, identify the critical path(s) and project duration, and state the total float of activity B and of activity C. Explain what the float values mean in practice for the project manager.
Read the worked answer
Forward pass (ES, EF), backward pass (LS, LF), float = LS − ES.
| Act | Dur | ES | EF | LS | LF | Float |
|---|---|---|---|---|---|---|
| A | 3 | 0 | 3 | 0 | 3 | 0 |
| B | 2 | 3 | 5 | 10 | 12 | 7 |
| C | 5 | 3 | 8 | 3 | 8 | 0 |
| D | 4 | 8 | 12 | 8 | 12 | 0 |
| E | 2 | 12 | 14 | 12 | 14 | 0 |
| F | 3 | 14 | 17 | 14 | 17 | 0 |
| G | 2 | 17 | 19 | 17 | 19 | 0 |
| H | 2 | 17 | 19 | 17 | 19 | 0 |
| I | 1 | 19 | 20 | 19 | 20 | 0 |
E's ES is max(EF_B = 5, EF_D = 12) = 12; I's ES is max(EF_G, EF_H) = 19. Project duration = 20 days.
Critical paths (two, sharing all but the G/H branch): A–C–D–E–F–G–I and A–C–D–E–F–H–I, both 20 days. Note the join at I: G and H run in parallel and both must finish, so neither has float.
Float of C = 0 (critical). Float of B = 7 days: B may start as late as day 10 and still finish by day 12 without delaying E. Float is not a licence to delay, though - it is a buffer. Spending B's float early removes the protection it gives against a hardware delivery delay, and because B is off the critical path the project manager must resist the temptation to crash it.
2. Determine the cheapest way to finish on day 17, showing the alternatives you rejected and why. State the extra cost and confirm the new schedule is feasible.
Read the worked answer
Three days must be removed from a critical path. Only critical activities can shorten the project; crashing B would cost money and buy nothing because B has 7 days of float.
Cost per day saved: F = Rs. 8,000 (max 1 day), D = Rs. 15,000 (max 2 days), C = Rs. 20,000 (max 1 day). Greedy order by cost: F first, then D, then C.
- Crash F by 1 day: duration 19, cost Rs. 8,000.
- Crash D by 2 days: duration 17, cost Rs. 30,000. Total: Rs. 38,000, duration 17 days.
Alternatives rejected:
- D × 2 + C × 1 = Rs. 30,000 + 20,000 = Rs. 50,000 for the same 3 days.
- D × 1 + F × 1 + C × 1 = 15,000 + 8,000 + 20,000 = Rs. 43,000.
- F × 1 + C × 1 + D × 1 is worse than taking the second day of D (Rs. 15,000) instead of C (Rs. 20,000).
Feasibility check: with D = 2 and F = 2, the critical chain A–C–D–E–F–G–I = 3+5+2+2+2+2+1 = 17. The parallel path through B: A–B–E–F = 3+2+2+2 = 9, plus 2 + 1 = 12, so B still has float and E's ES is now driven by D (day 10) rather than B. No new critical path has been created, so the 17-day target is genuinely achieved.
The project manager should also record the risk this buys: D is now 50% compressed, cabling is the activity most likely to overrun in an old building, and a 2-day cabling estimate with no float leaves nothing for hidden asbestos, conduit or wall-chasing problems. Contingency should be held in cost, not in the schedule.
3. The principal now demands handover on day 15. Using the logic of crashing, explain why the four quoted compressions cannot achieve this, and give two alternative managerial responses with their risks.
Read the worked answer
Maximum available compression on the critical path is F 1 + D 2 + C 1 = 4 days, and B's day is worthless to the schedule because B is not critical. Even if every option were bought at any price, the shortest achievable duration is 20 − 4 = 16 days. Fifteen days is therefore infeasible by crashing these activities, and no amount of money changes that - the constraint is the remaining work, not the budget.
It is also worth checking that the fully crashed schedule holds together: with C = 4, D = 2, F = 2, the critical chain is 3+4+2+2+2+2+1 = 16, and the B path is 3+2+2+2+2+1 = 12, so no alternative path becomes critical and swallows the gain. Compression beyond the point where a parallel path becomes critical yields zero schedule benefit - the general rule that makes crashing a diminishing-returns exercise.
Two alternative responses:
- Fast-track (overlap) instead of crash. Start F on half the machines while E is still finishing the rest, and run G and H in parallel on separate machine sets. Risk: rework and configuration drift between images, plus a much harder acceptance test, because the environment is a moving target during testing. Overlap also needs the principal to accept partial readiness.
- Reduce scope, not time. Hand over 20 working machines fully commissioned rather than 30 half-commissioned, and schedule the remainder for the third week of term. Risk: timetable pressure in the first weeks and possible double-booking of the lab, but it is honest and preserves quality of the delivered environment.
What should be resisted is silently accepting day 15 and hoping. The professional response is to present the 16-day floor with the evidence, quantify the risk of the compressed plan, and let the sponsor choose between scope, cost and date - that is a management decision, not a technical one.