Question 1 of 6: Scheduling

Question 1: Scheduling (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Nine activities on a six-event arrow network, each with a normal and a crash duration and the matching direct cost (costs are tabulated in hundreds of dollars and are converted to dollars below), plus a dashed dummy running from event 2 to event 5 on the drawing. Project indirect cost is $500 per day.

Find. (a) the earliest and latest event times, every total float, the critical path and the normal project duration; (b) the duration at which the sum of direct and indirect cost is a minimum, and the crashing programme that reaches it.

[Figure not reproduced: Figure 1 — the arrow network as drawn on the paper, redrawn with the event times obtained below. Each circle carries the event number above the divider and the earliest | latest event times below it. See the official exam paper.]

Approach. Run a forward pass for earliest event times and a backward pass for latest event times over the arrow list, take total float as \(L_j - E_i - D_{ij}\) to identify the critical path, then buy duration down one day at a time by crashing the cheapest critical activity (or cheapest cut across all critical paths) for as long as the cost slope stays below the $500 per day of indirect cost that is saved.

1. Part (a) — write the network down as an arrow list. Each table entry already carries its event pair, so the drawing only has to supply the dummy. Tracing each polyline back to the circle it leaves confirms the nine solid arrows 1–2 (A), 2–3 (B), 2–6 (C), 1–3 (D), 3–6 (E), 3–5 (F), 1–4 (G), 4–5 (H) and 5–6 (I), plus the dashed 2–5 dummy.
2. Forward pass — earliest event times. Taking \(E_1 = 0\) and \(E_j = \max_i (E_i + D_{ij})\) over every arrow entering event \(j\): $$E_2 = 0 + 8 = 8, \qquad E_4 = 0 + 4 = 4$$ Event 3 is fed by D from event 1 and B from event 2, and event 5 by H, F and the dummy, so each is the larger of the competing arrivals: $$E_3 = \max(0 + 4,\; 8 + 4) = 12, \qquad E_5 = \max(4 + 3,\; 12 + 2,\; 8 + 0) = 14$$ Event 6 closes the network from C, E and I: $$E_6 = \max(8 + 5,\; 12 + 8,\; 14 + 4) = \boxed{20 \text{ days}}$$
3. Backward pass — latest event times. Setting \(L_6 = E_6 = 20\) and working back with \(L_i = \min_j (L_j - D_{ij})\): $$L_5 = 20 - 4 = 16, \qquad L_4 = 16 - 3 = 13$$ $$L_3 = \min(20 - 8,\; 16 - 2) = 12, \qquad L_2 = \min(12 - 4,\; 20 - 5,\; 16 - 0) = 8$$ and \(L_1 = \min(8 - 8,\; 12 - 4,\; 13 - 4) = 0\). Returning to exactly zero at the start event is the check that the backward pass has been run correctly; any other value means an arrow has been missed.
4. Total floats and the critical path. With \(TF_{ij} = L_j - E_i - D_{ij}\) the floats are as tabulated below. Three activities have zero float and they form a continuous chain of events, which is what makes them a path rather than three isolated tight arrows.

The chain of zero-float arrows runs 1 → 2 → 3 → 6, so the critical path is A – B – E. Because every link on an arrow network is a plain finish-to-start relation with no lag, the durations along that path must sum to the project duration, and they do: \(8 + 4 + 8 = 20\) days. That identity is a genuine self-check here and is exactly what fails on the precedence networks with start-to-start lags.

5. Part (b) — cost slopes and the starting position. The slope is the premium paid per day saved, \(S = (C_c - C_n)/(D_n - D_c)\); four activities (C, D, G, H) have equal normal and crash durations and so cannot be shortened at all. Note that H does carry a higher crash cost with no time saving, which is simply money for nothing. The normal-duration position is $$C_{\text{direct}} = \$15{,}150, \qquad C_{\text{indirect}} = 20 \times \$500 = \$10{,}000$$ $$C_{\text{total}} = \$15{,}150 + \$10{,}000 = \$25{,}150$$
6. First crash cycle — the cheapest critical activity. Only A, B and E are critical and can be crashed, and E is the cheapest at $400 per day against an indirect saving of $500 per day, a net gain of $100 per day. Shortening E from 8 days to 6 days brings the project to 18 days, at which point the parallel chain A–B–F–I (\(8 + 4 + 2 + 4 = 18\) days) becomes critical as well and no further gain can be had from E alone.
7. Second crash cycle — a cut across both critical paths. With two critical paths, a day is only saved if every path loses a day. A and B lie on both paths, so either alone is a valid cut; the alternatives are the pairs E+F at $700 per day and E+I at $1,100 per day. The cheapest cut is A at $$S_A = \frac{\$3{,}300 - \$2{,}000}{8 - 5} = \$433.33 \text{ per day}$$ which still undercuts the $500 indirect saving. Crashing A over its full three-day range takes the project to 15 days.
8. Stopping test. At 15 days A is at its crash limit and B, at $800 per day, is dearer than the indirect saving; the cheapest remaining cut is E+F at \(\$400 + \$300 = \$700\) per day. Every further day therefore costs $200 more than it saves, so 15 days is the least-cost duration: $$C_{\text{total}} = \$17{,}250 + 15 \times \$500 = \boxed{\$24{,}750 \text{ at } 15 \text{ days}}$$

Tabulating the whole trade-off confirms that the total-cost curve really does turn at 15 days rather than merely flattening, and shows how shallow the optimum is — anywhere between 14 and 17 days the total sits within $200 of the minimum, which is worth saying to a client who has a reason to prefer a slightly different date.

The crashing programme is therefore: shorten E from 8 to 6 days and A from 8 to 5 days, leaving every other activity at its normal duration. The 15-day network has A, B, E, F and I all critical, so the compressed programme has almost no float left in it and needs closer day-to-day control than the 20-day one.