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. A nine-activity precedence (activity-on-node) network. Durations are printed inside the boxes in working days and every link is finish-to-start with zero lag except the single labelled link D → E, which carries a six-day lag.

Find. The project duration and the critical path, the total float of every activity, and the schedule consequence of allowing activity E to slip by one day.

[Figure not reproduced: Figure 1.1 — the printed precedence network redrawn. Durations are in working days; the only lagged link is D → E with FS = 6. The critical chain found below is shown in red. See the official exam paper.]

Approach. Run a forward pass to obtain every earliest time, a backward pass from the project completion date to obtain every latest time, and read the total floats as the difference; the chain of zero-float activities is the critical path.

1. State the constraint that each link imposes. With i the predecessor, j the successor and L the lag, a finish-to-start link says that j may not start until L days after i finishes: $$ES_j \ge EF_i + L, \qquad EF = ES + d$$ Time is measured in working days from the start of the project, so the three opening activities A, D and G all have $ES = 0$. The lag on D → E is a delay, not an overlap: it holds E back six days after D is complete, which is how a cure period, a delivery lead time or a concrete-strength wait is modelled.
2. Forward pass — earliest start and finish of every activity. Taking the activities in logical order and using the maximum over all incoming links, $$\begin{aligned} A:&\quad ES = 0,\; EF = 0 + 10 = 10\\ D:&\quad ES = 0,\; EF = 0 + 5 = 5\\ G:&\quad ES = 0,\; EF = 0 + 3 = 3\\ B:&\quad ES = \max(EF_A,\, EF_D) = \max(10,\,5) = 10,\; EF = 22\\ C:&\quad ES = EF_B = 22,\; EF = 31\\ E:&\quad ES = \max(EF_D + 6,\, EF_G) = \max(5+6,\,3) = 11,\; EF = 18\\ F:&\quad ES = EF_E = 18,\; EF = 24\\ H:&\quad ES = EF_G = 3,\; EF = 7\\ I:&\quad ES = EF_H = 7,\; EF = 13 \end{aligned}$$ Three chains reach the End node, and the project cannot finish until the last of them does: $$T = \max(EF_C,\, EF_F,\, EF_I) = \max(31,\, 24,\, 13)$$ $$\boxed{T = 31 \text{ working days}}$$
3. Backward pass — latest finish and latest start. Every activity that runs to End is given $LF = T = 31$, and each predecessor is then pulled back to the earliest of its successors' latest starts, $LF_i = \min(LS_j - L)$ with $LS = LF - d$: $$\begin{aligned} C:&\quad LF = 31,\; LS = 31 - 9 = 22\\ B:&\quad LF = LS_C = 22,\; LS = 22 - 12 = 10\\ A:&\quad LF = LS_B = 10,\; LS = 10 - 10 = 0\\ F:&\quad LF = 31,\; LS = 25\\ E:&\quad LF = LS_F = 25,\; LS = 25 - 7 = 18\\ I:&\quad LF = 31,\; LS = 25\\ H:&\quad LF = LS_I = 25,\; LS = 21\\ D:&\quad LF = \min\!\big(LS_B,\; LS_E - 6\big) = \min(10,\, 12) = 10,\; LS = 5\\ G:&\quad LF = \min\!\big(LS_H,\; LS_E\big) = \min(21,\, 18) = 18,\; LS = 15 \end{aligned}$$ Activity A returns $LS = 0$, which is the standard self-check that the two passes are consistent: the backward pass must land exactly on zero at the project start.
4. Total floats. Total float is the slack an activity enjoys before it moves the completion date, $$TF = LS - ES = LF - EF$$ so, activity by activity, $TF_A = 0-0 = 0$, $TF_B = 10-10 = 0$, $TF_C = 22-22 = 0$, $TF_D = 5-0 = 5$, $TF_E = 18-11 = 7$, $TF_F = 25-18 = 7$, $TF_G = 15-0 = 15$, $TF_H = 21-3 = 18$ and $TF_I = 25-7 = 18$.
5. Identify the critical path. The activities carrying zero total float are A, B and C, and they form a continuous chain from Start to End: $$\boxed{\text{Critical path: Start} \to A \to B \to C \to \text{End},\; 10 + 12 + 9 = 31 \text{ days}}$$ Because every link on this chain is finish-to-start with no lag, the sum of the durations equals the project duration exactly. That identity is worth checking whenever it applies: on a network whose critical chain contains a lagged or overlapping link the traverse and the sum differ, and the difference is precisely the net lag carried on the path.
6. Effect of delaying activity E by one day. E owns seven days of total float, so a one-day slip is absorbed inside the float and none of it reaches the completion date: $$TF_E = 7 \text{ d} \;>\; 1 \text{ d} \quad\Longrightarrow\quad \Delta T = 0$$ $$\boxed{\text{No effect on the project: completion stays at day 31}}$$ The delay is not free, however. E’s own float falls from 7 d to 6 d, and because the free float of E is zero — its early finish on day 18 is exactly what releases F — the one-day slip pushes F’s early start from day 18 to day 19 and consumes one day of F’s float as well, leaving the D → E → F chain with six days of protection instead of seven. Only after seven such days of slippage would E become critical and begin to extend the project.