Question 2 of 6: Scheduling — critical path and late bar chart

Question 2: Scheduling — critical path and late bar chart (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 precedence (activity-on-node) network of ten activities whose durations are printed under the activity letters and whose relationships are labelled on the arrows; an unlabelled arrow is an ordinary finish-to-start link with zero lag.

Find. (a) the critical path through the network and its length in days, and (b) a bar chart drawn on every activity’s latest permissible dates.

[Figure not reproduced: Figure 2.1 — the precedence network as printed, redrawn. Each box carries the activity letter over its duration in days; labelled arrows carry the relationship type and its lag, and unlabelled arrows are finish-to-start with zero lag. See the official exam paper.]

Approach. Run a forward pass through the network using the precedence-diagram lag rules to obtain early dates and the project duration, run a backward pass from that duration for late dates, take total float as the difference, and read the critical path off the zero-float chain; then plot every activity between its late start and late finish.

1. State the three lag rules that govern the passes. In a precedence network the constraint depends on which ends of the two bars are tied together. Writing $L$ for the lag, the early start of a successor $j$ driven by a predecessor $i$ must satisfy$$ES_j \ge EF_i + L\ \text{(FS)},\qquad ES_j \ge ES_i + L\ \text{(SS)},\qquad EF_j \ge EF_i + L\ \text{(FF)}$$and the successor takes the largest value that any of its predecessors demands. The backward pass mirrors each rule: $LF_i \le LS_j - L$ for FS, $LS_i \le LS_j - L$ for SS, and $LF_i \le LF_j - L$ for FF.
2. Forward pass along the start chain. Taking day 0 as the project start, $ES_A = 0$ and $EF_A = 10$. The SS2 link lets B begin two days after A begins rather than after A ends, so $ES_B = ES_A + 2 = 2$ and $EF_B = 10$; the second SS2 link gives $ES_D = ES_B + 2 = 4$ and therefore $EF_D = 4 + 12 = 16$. Both overlaps are what stop the answer from being a simple sum of durations.
3. Forward pass through the middle band. E and C both follow D finish-to-start, so $ES_E = ES_C = 16$, giving $EF_E = 23$ and $EF_C = 24$. Activity F waits on all three of D, E and C, so its early start is the largest of $EF_D = 16$, $EF_E = 23$ and $EF_C = 24$:$$ES_F = \max(16,\;23,\;24) = 24 \quad\Rightarrow\quad EF_F = 24 + 9 = 33$$Activity H is driven by B with a six-day finish-to-start lag, $EF_B + 6 = 16$, and by C with no lag, $EF_C = 24$; the later of these governs, so $ES_H = 24$ and $EF_H = 34$.
4. Forward pass to the finish milestone. G starts eight days after F starts, $ES_G = 24 + 8 = 32$, so $EF_G = 44$. Activity I is driven twice: the SS8 link from F requires $ES_I \ge 32$, while the FF3 link from H requires $EF_I \ge EF_H + 3 = 37$, that is $ES_I \ge 37 - 8 = 29$. The SS link governs, so $ES_I = 32$ and $EF_I = 40$. The milestone J then takes the later of $EF_G = 44$ and $EF_I = 40$:$$\boxed{T = 44\ \text{days}}$$
5. Backward pass for late dates. Setting $LF_J = LS_J = 44$ and working backwards, G and I may both finish at day 44, so $LS_G = 32$ and $LS_I = 36$. Activity F feeds both by SS8, so its late start is limited by $LS_G - 8 = 24$ and by $LS_I - 8 = 28$; the tighter of the two governs, $LS_F = 24$ and $LF_F = 33$. The FF3 link fixes $LF_H = LF_I - 3 = 41$, hence $LS_H = 31$. Activity C must finish before F starts and before H starts, so $LF_C = \min(24,\,31) = 24$ and $LS_C = 16$, while E is limited only by F, giving $LF_E = 24$ and $LS_E = 17$. Continuing, $LF_D = \min(LS_E,\,LS_C,\,LS_F) = 16$ so $LS_D = 4$; the SS2 links then give $LS_B = 2$ and $LS_A = 0$.
6. Total float and the critical path. Total float is $TF = LS - ES$, and the activities with zero float form the critical chain:$$\boxed{\text{Critical path: } A \to B \to D \to C \to F \to G \to J,\ \text{length } 44\ \text{days}}$$E carries one day of float, I carries four and H carries seven, so none of them can extend the project unless it slips by more than that. The path runs through C and not through E, even though E is reached at the same time, because C is one day longer.
7. Check the length against the sum of the critical durations. Adding the durations of the critical activities gives $10+8+12+8+9+12+0 = 59$ days, which is fifteen days more than the answer. The difference is exactly the overlap bought by the three lead relationships on that chain: the SS2 into B saves $10-2 = 8$ days, the SS2 into D saves $8-2 = 6$ days and the SS8 into G saves $9-8 = 1$ day, and $59 - 8 - 6 - 1 = 44$. This is a useful independent check on the forward pass.
8. (b) Draw the late bar chart. A late bar chart plots each activity from its late start to its late finish, so every bar sits as far to the right as the network permits and the critical activities form an unbroken chain to day 44. It is the schedule that consumes all available float and is therefore the cash-flow-optimal, risk-maximal way of running the job.