07-Str-B2 · May 2018

Question 1 of 6: Scheduling — CPM with a start-to-start lag, late bar chart and delay impact

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Question 1: Scheduling — CPM with a start-to-start lag, late bar chart and delay impact (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. The nine activities of the printed precedence diagram, their durations in working days, and the single lagged relation between A and D.

Find. Early and late start and finish times for every activity, total and free float, the project duration and critical path, a late bar chart, and the schedule consequence of a two-day delay to activity G.

Approach. Run a forward pass from day 0 that treats the A → D link as a constraint on starts rather than finishes, take the largest early finish as the project duration, run the backward pass with the same lag subtracted again, then read the floats and test the two-day delay against activity G's total float.

1. Set the forward-pass rule for each relation type. In a precedence network the early start of an activity is governed by the strongest constraint imposed by its predecessors, $$ES_j=\max_i\begin{cases}EF_i+\text{FS}_{ij}&\text{finish-to-start}\\ ES_i+\text{SS}_{ij}&\text{start-to-start}\end{cases}\qquad EF_j=ES_j+d_j$$ with day 0 as the instant the project opens, so an activity with $ES=0$ starts on the morning of working day 1.
2. Push the three chains forward. A, B and C have no predecessors, so $ES=0$ and $EF$ equals their durations: $EF_A=8$, $EF_B=4$, $EF_C=5$. The only lagged link then gives $$ES_D=ES_A+\text{SS}=0+3=3,\qquad EF_D=3+9=12$$ and the remaining finish-to-start links follow directly: $ES_H=EF_D=12$ so $EF_H=20$; $ES_E=ES_F=EF_B=4$ so $EF_E=18$ and $EF_F=10$; $ES_G=EF_C=5$ so $EF_G=12$.
3. Merge the two paths into activity I and read the project duration. Activity I waits for the later of F and G, so $ES_I=\max(10,12)=12$ and $EF_I=15$. The project cannot finish before its last activity does, and the three terminal activities finish at day 20 (H), day 18 (E) and day 15 (I): $$T=\max(EF_H,EF_E,EF_I)=\boxed{20\ \text{working days}}$$
4. Run the backward pass from day 20. Each terminal activity is given $LF=T=20$, and every other activity takes the tightest requirement of its successors, $LF_i=\min_j (LS_j-\text{FS}_{ij})$. Working right to left: $LS_H=20-8=12$, so $LF_D=12$ and $LS_D=3$; $LS_I=20-3=17$, so $LF_F=LF_G=17$, giving $LS_F=11$ and $LS_G=10$; $LF_C=LS_G=10$ so $LS_C=5$; $LF_E=20$ so $LS_E=6$; and $LF_B=\min(LS_E,LS_F)=\min(6,11)=6$ so $LS_B=2$.
5. Close the backward pass through the start-to-start link — the step most candidates drop. Activity A has no finish-to-start successor at all, so its late finish is not fixed by H; what the lag constrains is A's start. Subtracting the same lag on the way back, $$LS_A=LS_D-\text{SS}=3-3=0,\qquad LF_A=LS_A+d_A=0+8=8$$ Activity A therefore has zero float even though nothing directly follows its completion: if A starts a day late, D starts a day late, and the whole project slips.
6. Compute total and free float for every activity. Total float is the slack against the project end, free float the slack that can be used without disturbing any successor, $$TF_i=LS_i-ES_i=LF_i-EF_i,\qquad FF_i=\min_j\left(ES_j-EF_i\right)$$ Applying these to the passes above gives the table below. The activities with $TF=0$ are A, D and H.
7. State the critical path. The zero-float chain runs from A through the lagged link into D and on to H, and its length is checked directly against the lag rather than by adding all three durations: $$\text{SS}+d_D+d_H=3+9+8=\boxed{20\ \text{days along } A\rightarrow D\rightarrow H}$$ Note that $d_A+d_D+d_H=25\gt 20$; on a lagged network the critical path length is not the sum of the durations on it, which is why the pass arithmetic, not a path enumeration, must govern.
8. Draw the late bar chart. Every bar is plotted from its late start to its late finish, which is the schedule a contractor would follow to defer expenditure as long as possible without extending the project; the dashed lead on each non-critical bar is its total float.

The delay question is now answered by comparing the delay with the right float. Activity G carries five days of total float but zero free float, because its early finish on day 12 is exactly what sets activity I's early start. A two-day delay therefore does move work, but not the completion date:

Re-running the passes with $d_G$ effectively increased by two confirms the detail: $EF_G$ moves from day 12 to day 14, activity I is pushed from an early start of 12 to 14 and an early finish of 15 to 17, and the total float of C, G and I all fall from five days to three. The critical path is unchanged at A → D → H and no other activity is touched. In practical terms the delay is absorbed, but it consumes two-fifths of the cushion protecting the C → G → I chain, so a further three-day slip would make that chain critical as well.