Question 2 of 6: Scheduling — Critical Path and Late Bar Chart in a Precedence Network

Question 2: Scheduling — Critical Path and Late Bar Chart in a Precedence Network (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 inside the boxes and whose logic is written on the arrows; an unlabelled arrow is an ordinary finish-to-start link with zero lag, and activity J is a zero-duration completion milestone.

Find. (a) the chain of zero-float activities that fixes the project duration, and that duration in days; (b) a bar chart drawn on late start and late finish dates.

[Figure not reproduced: Figure 2.1 — the examination network redrawn. Boxes carry the activity letter over its duration in days; the red boxes and arrows are the zero-float chain established below. Unlabelled arrows are finish-to-start links with zero lag. See the official exam paper.]

Approach. Run a forward pass with the three precedence rules to get early dates and the project duration, run a backward pass for late dates, take total float as the difference, and read the critical path off the zero-float activities; the late bar chart is then simply every activity drawn from its late start to its late finish.

1. Part (a) — state the three precedence rules before touching a number. In a precedence network with lags, each link type constrains a different pair of dates: $$\text{FS}_L:\ ES_j \ge EF_i + L,\qquad \text{SS}_L:\ ES_j \ge ES_i + L,\qquad \text{FF}_L:\ EF_j \ge EF_i + L$$ An activity may be governed by several links at once, so its early start is the largest value any incoming link demands. Dates are counted in elapsed days from a project start at day 0, so an activity that starts on day 0 and lasts 10 days finishes at day 10.
2. Forward pass through the opening chain. Activity A starts the project, so $ES_A = 0$ and $EF_A = 10$. The A–B link is start-to-start with a two-day lag, which releases B two days after A begins rather than after it ends: $$ES_B = ES_A + 2 = 2,\qquad EF_B = 2+8 = 10$$ The same reasoning carries through B–D, $$ES_D = ES_B + 2 = 4,\qquad EF_D = 4+12 = 16$$ Overlapping in this way is the whole point of a start-to-start lag: three activities totalling 30 days of duration have consumed only 16 days of calendar.
3. Forward pass through the middle band. Both E and C follow D finish-to-start with no lag, so both are released at day 16: $$EF_E = 16+7 = 23,\qquad EF_C = 16+8 = 24$$ Activity F waits on three predecessors and takes the latest of them, $$ES_F = \max(EF_D,\ EF_E,\ EF_C) = \max(16,\ 23,\ 24) = 24,\qquad EF_F = 33$$ so C, not the longer E, is what actually holds F back. Activity H is governed by B–H with a six-day finish-to-start lag ($10+6=16$) and by C–H with none ($24$), so $$ES_H = \max(16,\ 24) = 24,\qquad EF_H = 34$$
4. Forward pass through the closing activities. Both G and I hang off F by an eight-day start-to-start lag, so each may begin on day $24+8 = 32$. Activity I carries a second, finish-to-finish constraint from H: $$EF_I \ge EF_H + 3 = 37 \;\Longrightarrow\; ES_I \ge 37-8 = 29$$ which is weaker than the 32 already imposed, so it does not govern and $ES_I = 32$, $EF_I = 40$. Meanwhile $$ES_G = 32,\qquad EF_G = 32+12 = 44$$ The milestone J follows both, so $$\boxed{T = \max(EF_G,\ EF_I) = \max(44,\ 40) = 44\ \text{days}}$$
5. Backward pass. Set $LF_J = 44$ and work the same three rules in reverse, each predecessor taking the smallest late finish its successors will tolerate: $$\text{FS}_L:\ LF_i \le LS_j - L,\qquad \text{SS}_L:\ LS_i \le LS_j - L,\qquad \text{FF}_L:\ LF_i \le LF_j - L$$ so $LF_G = 44$ and $LS_G = 32$; $LF_I = 44$ and $LS_I = 36$; and F is squeezed by its two start-to-start successors to $$LS_F = \min(LS_G-8,\ LS_I-8) = \min(24,\ 28) = 24,\qquad LF_F = 33$$ Continuing back, $LF_H = LF_I-3 = 41$, $LF_C = \min(LS_F,\ LS_H) = \min(24,31) = 24$, $LF_E = 24$, $LF_D = \min(LS_E,\ LS_C,\ LS_F) = \min(17,16,24) = 16$, and the lags carry B and A back to $LS_B = 2$ and $LS_A = 0$.
6. Total float and the critical path. Total float is the slack between the early and late schedules, $$TF_i = LS_i - ES_i = LF_i - EF_i$$ which is zero for A, B, D, C, F, G and J, one day for E, four for I and seven for H. The critical path is therefore $$\boxed{A \to B \to D \to C \to F \to G \to J,\qquad T = 44\ \text{days}}$$ driven by the links SS2, SS2, FS0, FS0, SS8 and FS0 in that order. It is worth checking the arithmetic along the chain rather than trusting the table: starting at day 0, the two start-to-start lags put D at day 4, its 12 days end at 16, C runs 16–24, F runs 24–33, the eight-day lag starts G at 32 and its 12 days end at 44.
7. Note what the lags do to the usual arithmetic. The durations along the critical path sum to $10+8+12+8+9+12+0 = 59$ days, yet the project takes 44. The difference of 15 days is exactly the overlap bought by the three start-to-start links: B starts on day 2 rather than day 10, saving 8 days; D starts on day 4 rather than day 10, saving 6; and G starts on day 32 rather than day 33, saving 1. In a precedence network the length of the critical path is a path traverse, not a sum of durations, and it is the single most common source of a wrong answer on this question.
8. Part (b) — draw the late bar chart. A late bar chart plots each activity from its late start to its late finish, so critical activities sit exactly where they do on the early chart and every floating activity slides right by its total float: E by one day (17–24 instead of 16–23), I by four (36–44) and H by seven (31–41). Scheduling the whole job this way is the latest-possible programme that still finishes on day 44 — it maximises the deferral of cash outflow but leaves no protection whatever, since every activity has become critical the moment it is drawn there.