Chapter 14 — Quiz

B. The critical path is a calculated result. It is the longest path through the network, and its activities have zero total float (or the minimum float in the network, if a constraint has driven float negative). Nobody designates it; it emerges from durations and logic, and it moves when either changes.

C — the largest. A merging activity must wait for its slowest predecessor. Taking anything other than the maximum produces a schedule that is optimistic in a way that is very hard to spot later.

B — the smallest. A bursting activity must finish in time for its most urgent successor. You cannot satisfy one successor and abandon another. Max going forward, min coming back.

B — 7. TF = LS − ES = 37 − 30 = 7, and it must equal LF − EF = 49 − 42 = 7. Computing it both ways is a free error check; if the two disagree, you have made a mistake in one of the passes.

B. Total float protects the project; free float protects your successor. Total float without free float is interfering float — spending it consumes a downstream party's room without moving the completion date. This is exactly how the Northgate steel procurement chain got emptied out.

C. Only C is physics. A and D are preferences (soft logic) — you could reverse them without changing the building. B is resource logic — you could reverse it by hiring a second crew.

C. Mandatory constraints override logic. The activity holds its date regardless of its predecessors, which means delay stops propagating at that point. The schedule can then report that a five-week delay had no impact on completion, and nothing in the output looks wrong.

B. With no successor, an activity's late finish defaults to the end of the project, giving it enormous float. Those activities then disappear from every float-filtered report until one of them stops the job.

B. A FS−10 is a badly written SS relationship. It behaves unpredictably when the predecessor's duration changes and is very hard to explain to a third party. If two activities overlap, model the overlap with SS and a lag — that is what SS exists for.

C. $5,150/CD of extended general conditions (from a $2,900,000 general-conditions budget over 565 calendar days) plus $5,500/CD of liquidated damages. Both accrue on calendar days, which is why every duration in a schedule must be labeled WD or CD.

B. The curve is essentially the accumulation of the manpower curve weighted by the value each trade installs. A cash-flow curve that is a straight line means somebody divided the contract sum by the number of months instead of loading the schedule.

True. Free float is limited by the earliest successor's early start; total float is limited by the project finish, which is never earlier. FF can equal TF, but it can never exceed it.

False. It gets shorter only until a parallel path takes over. The Cottonwood Creek drill in §14.6.10 shows the knee exactly: shortening girder fabrication buys four days and then buys nothing, because the west abutment chain governs the merge. Re-run the network before you buy acceleration.

False. Standard industry contract language generally reserves means, methods, techniques, sequences, and procedures to the contractor. Acceptance is a review for conformance with the specified requirements; it does not warrant that the sequence works or relieve you of the obligation to finish on time. It is still enormously valuable to you, because it fixes the agreed starting point for every future delay analysis.

False. Weather belongs in the calendar, where it automatically makes a January exterior activity longer than the same activity in July. A "weather" bar in the middle of a network is the first thing consumed and the last thing anybody admits was consumed.

True. Shorten one and the other still governs the completion date. This is the first question to ask of any acceleration proposal: how many critical paths are there, and does this proposal shorten all of them?

The two numbers are quantity and production rate. Duration = quantity ÷ (production rate × number of crews); labor cost = (quantity ÷ production rate) × crew size × hours × wage rate. Change the production rate and both the schedule and the estimate move together, which is why an estimator and a scheduler who disagree about a production rate are having the same argument twice — and why the schedule and the budget are one conversation, not two.

(1) A constraint on the finish milestone set later than the contract date; (2) open ends — activities with no successor; (3) soft logic modeling work as parallel that must actually interleave; (4) durations padded at the activity level, so the same contingency is counted twice. The constraint is the silent one — it changes every float number in the schedule and produces no flag, no red bar, and no exception report. That is why the constraint count is the first check, not the last.

Take a copy of the file, add a large duration (say 100 days) to one activity on the computed critical path, and recalculate. The project finish should move by exactly 100 days. If it moves less, something downstream — a constraint, a calendar, a mandatory date — is absorbing delay, and the schedule cannot compute a true critical path. Repeat on an activity with, say, 12 days of float: the finish should move exactly 88 days. If it does not, the float numbers are wrong.

Forward pass: A 0→4. B 4→10. C 4→13. D 10→15. E 13→16. F starts at max(D 15, E 16) = 16 → 23. Project duration = 23 work days.

Backward pass (LF of F = 23): F 16→23. D LF = 16, LS = 11. E LF = 16, LS = 13. B LF = LS(D) = 11, LS = 5. C LF = LS(E) = 13, LS = 4. A LF = min(LS(B) 5, LS(C) 4) = 4, LS = 0. ✓ (A's LS returns to 0.)

Critical path: A → C → E → F = 4 + 9 + 3 + 7 = 23 ✓

Note B and D: both carry one day of total float, but only D has free float. If B slips one day, the project is fine — but D's early start moves and D's free float disappears. One day of float on that chain, shared between two activities.

(a) Four work days. Once you have compressed the critical path by 4 days, the parallel path — which had 4 days of float — becomes critical too and governs the completion date. The remaining 3 days of compression buy nothing on their own.

(b) 4 WD × 7/5 = 5.6 calendar days. 5.6 × $4,300 = $24,080.

(c) Marginally yes — $24,080 of benefit against $22,000 of cost is a thin $2,080 margin, and it assumes the acceleration performs exactly as priced. What you do next is more important than the yes/no: find out what it would cost to compress the parallel path as well. If the two together can be bought for a reasonable number, you may be able to convert a 4-day gain into a 7-day gain and change the economics entirely. If the parallel path cannot be compressed at all, decline or renegotiate, because you are being asked to pay for seven days and receive four.