Question 1 of 6: Scheduling — cheapest-option plan, and the fastest plan under a labour cap

Question 1: Scheduling — cheapest-option plan, and the fastest plan under a labour cap (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. Five activities, each with a “slow machine” estimate and (except for site preparation) a “fast machine” alternative. The Depends on column gives the predecessor by row number, so the logic is a simple fork: site preparation releases both trench excavations, and each trench releases its own pipe-laying and backfill. All links are finish-to-start with no lag.

Find. (a) the duration, total cost and peak daily labour demand of the all-cheapest plan, together with the network and its critical path; and (b) the least-cost mix of options that finishes in eight days without ever exceeding four workers on site in a day, its cost, and its bar chart.

Approach. Choose the cheapest option for each activity, build the activity-on-node network from the Depends on column, run a forward and backward pass for the duration and the floats, add the option costs, and read the peak labour demand off a day-by-day staffing histogram; then, for part (b), use the eight-day deadline to eliminate infeasible option pairs chain by chain before pricing what survives.

1. Select the cheapest option for every activity. Estimate 1 costs $5,000 for each of the five activities, while the Estimate 2 alternatives cost $10,000, $7,000, $7,000 and $9,000. The cheapest plan is therefore Estimate 1 throughout, and site preparation has no alternative in any case. Every activity then takes four days.
2. Lay out the network logic. The Depends on column reads 1 → {2, 3}, 2 → 4 and 3 → 5, so site preparation forks into two independent chains that never rejoin. Writing $A$ for site preparation, the two paths through the network are $A\to B\to D$ and $A\to C\to E$.
3. Forward pass for the project duration. With $ES$ the earliest start and $EF=ES+d$ the earliest finish, working left to right from time zero gives $$\begin{aligned} EF_A &= 0+4 = 4 \\ EF_B &= EF_C = 4+4 = 8 \\ EF_D &= EF_E = 8+4 = 12 \end{aligned}$$ so the project finishes at the end of day 12: $\boxed{T = 12\ \text{days}}$.
4. Backward pass and total floats. Setting $LF = 12$ at both terminal activities and working right to left, $LS_D = LS_E = 8$, hence $LF_B = LF_C = 8$, $LS_B = LS_C = 4$, and $LF_A = 4$ with $LS_A = 0$. Every activity returns $TF = LF - EF = 0$, so both chains are critical. The all-finish-to-start check confirms it: the durations along either path sum to $4+4+4 = 12$ days, exactly the project duration, which is the identity that must hold when a network carries no lags.
5. Total direct cost of the cheapest plan. Five activities at the Estimate 1 price give $$C_a = 5 \times 5{,}000 = \boxed{25{,}000\ \text{CAD}}$$ which is the least the work can be done for at any duration, since Estimate 1 is the cheaper option everywhere.
6. Staffing histogram and the required number of workers. Adding the crews of the activities that overlap in each period: days 1–4 carry site preparation alone at 3 workers/day; days 5–8 carry both trench excavations at $3+2 = 5$ workers/day; and days 9–12 carry both pipe-laying activities at $3+3 = 6$ workers/day. The crew size the plan must be able to field is therefore the peak, $\boxed{6\ \text{workers per day}}$, and the labour content of the job is $4(3)+4(3)+4(2)+4(3)+4(3) = 56$ worker-days.

Part (b) changes the question from “what does the cheapest plan cost?” to “what is the cheapest plan that satisfies two hard constraints?” — eight days and four workers a day. Because site preparation has only one estimate, four of the eight days are already spent before either chain can start, and that single observation does almost all of the work.

1. Establish the time available to each chain. Site preparation is fixed at four days and both chains hang off it, so each chain has $8 - 4 = 4$ days in which to finish. The two chains are independent, so they can be screened separately.
2. Screen the Trench 1 chain. The four possible duration pairs for (Trench 1, Lay Pipe 1) are $4+4 = 8$, $4+1 = 5$, $2+4 = 6$ and $2+1 = 3$ days. Only the last is within the four days available, so both activities must take their Estimate 2 option; the chain then finishes in three days and carries one day of float.
3. Screen the Trench 2 chain. The pairs are $4+4 = 8$, $4+2 = 6$, $2+4 = 6$ and $2+2 = 4$ days, and again only the all-fast pair fits. This chain uses the whole four days and is the controlling chain, so the second path is critical while the first is not.
4. Confirm the labour cap is met. With every activity on its Estimate 2 crew of two, the histogram reads 3 workers/day on days 1–4 (site preparation), $2+2 = 4$ on days 5–6 (the two excavations), $2+2 = 4$ on day 7 (both pipe-laying activities) and 2 on day 8. The peak is $\boxed{4\ \text{workers per day}}$, exactly the cap, so the schedule is feasible without levelling. The one day of float on Lay Pipe 1 could be used to shift it to day 8 if a lower peak were wanted, but the cap is already satisfied.
5. Price the resulting plan and the premium it carries. Adding the selected options, $$\begin{aligned} C_b &= 5{,}000 + 10{,}000 + 7{,}000 + 7{,}000 + 9{,}000 \\ &= \boxed{38{,}000\ \text{CAD}} \end{aligned}$$ so the additional cost of buying the schedule down from twelve days to eight is $38{,}000 - 25{,}000 = \boxed{13{,}000\ \text{CAD}}$, a premium of about 52 percent for a one-third reduction in duration.