Question 1 of 6: Scheduling — PERT expected duration, critical path and total floats

Question 1: Scheduling — PERT expected duration, critical path and total floats (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.

(a) PERT expected completion time

Given. Six tasks whose expected durations and duration variances are already computed and tabulated, with the predecessor list defining the logic.

Find. The expected completion time of the whole project, that is the earliest finish of the terminal task F.

Approach. Draw the logic implied by the predecessor column, run a forward pass to obtain each task's earliest start and earliest finish, and read the project's expected duration off the terminal task; then confirm the answer by enumerating the three complete chains and use the variances on the governing chain to attach a confidence statement to the date.

1. Establish the logic from the predecessor column. A is the only task with no predecessor, so it starts the project. B and C both wait on A; D waits on B alone; E is a merge point waiting on both B and C; F is a second merge point waiting on both D and E. Nothing succeeds F, so F terminates the project. This is the network drawn in Figure 1.1.
2. Run the forward pass. For every task the earliest start is the latest earliest-finish among its predecessors, and the earliest finish follows by adding the expected duration:

   $$ES_j=\max_{i\,\in\,\text{pred}(j)}\left(EF_i\right),\qquad EF_j=ES_j+t_{e,j}$$

   Working left to right, A runs 0–30. B and C both start at day 30 and finish at days 70 and 90 respectively. D follows B alone and runs 70–95. E is held by the later of its two predecessors, and since C finishes at 90 against B's 70, E starts at day 90 and runs to day 135. F is held by the later of D (day 95) and E (day 135), so it starts at day 135 and finishes at day 155.

   Forward pass

   Task	$t_e$ (d)	ES (d)	EF (d)	Governing predecessor
   A	30	0	30	—
   B	40	30	70	A
   C	60	30	90	A
   D	25	70	95	B
   E	45	90	135	C (90 > 70)
   F	20	135	155	E (135 > 95)

3. Read the expected completion time. The project ends when its terminal task ends, so

   $$T_e=EF_F=\boxed{155\ \text{days}}$$

4. Confirm by enumerating the complete chains. Three chains run from A to F. Their lengths are $30+40+25+20=115$ days for A–B–D–F, $30+40+45+20=135$ days for A–B–E–F, and $30+60+45+20=155$ days for A–C–E–F. The longest is A–C–E–F at 155 days, which agrees with the forward pass and identifies the critical chain. Because every link here is a plain finish-to-start with no lag, this sum check is valid; it would not be on the lagged network of part (b).
5. Attach the variance the question supplies. The paper gives duration variances as well as expected durations, which is what makes this a PERT problem rather than a plain CPM one. Variances add along a chain of independent tasks, so on the critical chain

   $$\sigma_T^2=\sum_{\text{critical}}\sigma_i^2=25+81+36+9=151\ \text{days}^2,\qquad \sigma_T=\sqrt{151}=12.29\ \text{days}$$

   The expected duration is therefore a 50 % date, not a promise. Treating the sum of four independent durations as approximately normal by the central limit theorem, the probability of finishing by any date $T$ is $\Phi\!\left[(T-T_e)/\sigma_T\right]$, which gives 65.8 % by day 160, 88.8 % by day 170 and 97.9 % by day 180. Conversely a date the planner could commit to with 90 % confidence is

   $$T_{90}=T_e+z_{0.90}\,\sigma_T=155+1.2816\times12.29=170.7\approx 171\ \text{days}$$

(b) Critical path, total floats and the effect of delaying H

Given. The precedence network printed on page 2, comprising twelve activities with the durations shown in brackets and two lagged links.

Find. The critical path and the project duration, the total float of every activity, and the change in project duration caused by a six-day delay to activity H.

[Figure not reproduced: Figure 1.2 — the printed precedence network redrawn, with each activity's total float shown beneath its box and the critical path A–C–F–K–L in red. Note that only two links carry lags: SS = 2 from B to D and FS = 3 from F to K. See the official exam paper.]

Approach. Run a forward pass in which a finish-to-start link drives the successor from its predecessor's earliest finish while the start-to-start link drives it from the predecessor's earliest start; run the matching backward pass on latest starts; take total float as the difference; then re-run the forward pass with the delay imposed to measure its effect rather than assuming it.

1. Set the driving rule for each link type. With $L$ the lag, a finish-to-start and a start-to-start link impose respectively

   $$ES_j\ \ge\ EF_i+L_{ij}\quad\text{(FS)},\qquad ES_j\ \ge\ ES_i+L_{ij}\quad\text{(SS)}$$

   and each activity takes the largest of the demands made on it, $ES_j=\max(\cdot)$, with $EF_j=ES_j+D_j$. The single SS link is what makes this network more than an exercise in addition: D does not wait for B to finish, only for B to have been under way two days.

2. Forward pass. A starts at day 0 and finishes at 2. B and C both follow A finish-to-start and so start at day 2, finishing at days 6 and 5. D is released by the start-to-start link at $ES_B+2=4$ and runs 4–9 — two days before B has finished, which is the whole point of the lag. E and F both follow C and start at day 5, finishing at 9 and 13. G follows D and runs 9–11; H follows E and runs 9–13; I and J both follow G, running 11–15 and 11–18. K is the interesting merge: H offers $EF_H=13$ while F offers $EF_F+3=16$, so the lagged link governs and K runs 16–20. L is held by the latest of I (15), J (18) and K (20), so it runs 20–22.

   $$T=EF_L=\boxed{22\ \text{days}}$$

3. Backward pass. Fixing $LF_L=T=22$ and working right to left, each activity's latest start is the smallest demand its successors make. A finish-to-start successor requires $LS_i\le LS_j-L_{ij}-D_i$, whereas a start-to-start successor constrains only the predecessor's start, $LS_i\le LS_j-L_{ij}$. Applying these gives the schedule below, and the pass closes correctly with $LS_A=0$ — the standard self-check that the two passes are consistent.
4. Compute total and free float. Total float is the slack an activity has against the project completion date, $TF=LS-ES=LF-EF$; free float is the slack it can use without disturbing any successor's earliest start, $FF=\min_j\left(ES_j-L_{ij}\right)-EF_i$ over its finish-to-start successors.

   Complete schedule — forward pass, backward pass and floats (days)

   Activity	Duration	ES	EF	LS	LF	Total float	Free float
   A	2	0	2	0	2	0	0
   B	4	2	6	4	8	2	0
   C	3	2	5	2	5	0	0
   D	5	4	9	6	11	2	0
   E	4	5	9	8	12	3	0
   F	8	5	13	5	13	0	0
   G	2	9	11	11	13	2	0
   H	4	9	13	12	16	3	3
   I	4	11	15	16	20	5	5
   J	7	11	18	13	20	2	2
   K	4	16	20	16	20	0	0
   L	2	20	22	20	22	0	0

5. Identify the critical path. The activities with zero total float are A, C, F, K and L, and they form a connected chain from the start to the finish, so

   $$\text{critical path}=\boxed{\text{A}\rightarrow\text{C}\rightarrow\text{F}\rightarrow\text{K}\rightarrow\text{L},\ \ T=22\ \text{days}}$$

   Walk it link by link rather than adding durations: the five critical durations sum to $2+3+8+4+2=19$ days, three days short of the answer, and the missing three days are exactly the FS = 3 lag on F → K. On any network carrying lags the familiar “critical durations must sum to the project duration” check is invalid, and a candidate who applies it here will conclude, wrongly, that the pass contains an error.

6. Measure the effect of delaying H by six days. H carries three days of total float, so the general rule

   $$\Delta T=\max\left(0,\ \text{delay}-TF\right)=\max(0,\ 6-3)=3\ \text{days}$$

   predicts a three-day extension. Re-running the forward pass confirms it: pushing H out to 15–19 makes H, not the lagged F link, the governing predecessor of K at day 19, so K runs 19–23 and L runs 23–25.

   $$T'=\boxed{25\ \text{days}\quad(\text{a 3-day extension})}$$

   The critical path swings away from F and now runs A → C → E → H → K → L, whose durations under the delay sum to 25 days with no lag left on the path.

7. Show that the usual ambiguity collapses here. “Delaying activity H by 6 days” can mean that H starts six days late or that H takes six days longer. Because H's only successor link is a plain finish-to-start to K, both readings push $EF_H$ from day 13 to day 19 and both give 25 days. The two readings would diverge only if H fed its successor through a start-to-start link, in which case a duration extension would be absorbed entirely. It is worth saying so explicitly rather than quietly picking one meaning.