Question 1 of 6: Scheduling — CPM, late bar chart and the effect of delaying G

Question 1: Scheduling — CPM, late bar chart and the effect of delaying G (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 with one lagged link. Durations are working days.

Find. The forward and backward passes (ES, EF, LS, LF), the total float of every activity, the project duration and critical path, a late bar chart drawn to the late schedule, and the schedule consequence of a two-day delay to activity G.

[Figure not reproduced: Figure 1.1 — the precedence network as printed, redrawn with the critical activities highlighted. The bracket over A and D is the start-to-start link with a three-day lag; every other link is a plain finish-to-start. See the official exam paper.]

Approach. Convert the drawing into a link list, run a forward pass in which D is released three days after A starts rather than after A finishes, run the backward pass from the project finish, then read the critical path off the zero-float activities and draw each bar at its late start.

1. Write the network logic down before calculating. Three activities (A, B, C) have no predecessor and three (E, H, I) have no successor, so the network has three parallel chains that never rejoin. The only non-standard link is A → D. For a start-to-start link with lag \(SS_{ij}\) the constraint is on start times, \(ES_j \ge ES_i + SS_{ij}\), whereas the ordinary finish-to-start link gives \(ES_j \ge EF_i\). Activity A therefore governs when D may begin but its own finish releases nothing.
2. Forward pass — earliest times. Working left to right with \(EF = ES + D\) and day 0 as the project start:$$ES_A = ES_B = ES_C = 0 \quad\Rightarrow\quad EF_A = 8,\; EF_B = 4,\; EF_C = 5$$The lagged link then releases D at \(ES_D = ES_A + 3 = 3\), not at \(EF_A = 8\); this single distinction is worth five days of project duration. Continuing:$$EF_D = 3 + 9 = 12,\qquad EF_E = 4 + 14 = 18,\qquad EF_F = 4 + 6 = 10$$$$EF_G = 5 + 7 = 12,\qquad EF_H = 12 + 8 = 20$$Activity I merges two chains and takes the later of them, \(ES_I = \max(EF_F, EF_G) = \max(10, 12) = 12\), so \(EF_I = 15\).
3. Project duration. The project cannot finish until the last terminal activity does, so$$T = \max(EF_E,\, EF_H,\, EF_I) = \max(18,\, 20,\, 15) = \boxed{20 \text{ working days}}$$
4. Backward pass — latest times. Set \(LF = T = 20\) for the three terminal activities E, H and I and work right to left with \(LS = LF - D\). For an ordinary link \(LF_i = \min(LS_j)\); for the start-to-start link the constraint is again on starts, \(LS_A \le LS_D - 3\). This gives$$LS_H = 20 - 8 = 12 \;\Rightarrow\; LF_D = 12 \;\Rightarrow\; LS_D = 3 \;\Rightarrow\; LS_A = 3 - 3 = 0$$and, down the other two chains, \(LS_E = 6\), \(LS_I = 17\), \(LF_F = LF_G = 17\) so \(LS_F = 11\) and \(LS_G = 10\); B is limited by the earlier of its two successors, \(LF_B = \min(LS_E, LS_F) = \min(6, 11) = 6\), and \(LF_C = LS_G = 10\). Every start activity returns a latest start of zero or more, which is the arithmetic check that the pass closes.
5. Total float and the critical path. With \(TF = LS - ES = LF - EF\) the zero-float activities are A, D and H, so$$\text{critical path} = \boxed{A \;\xrightarrow{SS=3}\; D \;\rightarrow\; H \quad (20 \text{ days})}$$Walk the path link by link to confirm the length: A starts on day 0, the lag releases D on day 3, D runs nine days to day 12, and H runs eight days to day 20, i.e. \(3 + 9 + 8 = 20\). Note that A is critical because of the lag on its start: it is A's start date, not A's finish, that releases D, so A could run well past eight days without affecting the project, while a one-day slip in its start would push the completion date out day for day.
6. Draw the late bar chart. A late bar chart places every bar at its latest permissible position, from \(LS\) to \(LF\); the gap between the early and late positions is the total float. The critical bars occupy the same position in both charts.

The full set of times is collected below; it is the standard CPM tabulation a marker looks for.

Effect of delaying activity G by two days. The phrase is genuinely ambiguous on an exam paper, so both readings are answered. Activity G carries five days of total float but zero free float, because its successor I already starts on day 12, exactly when G finishes.

1. Reading (i) — G's start slips two days. G then runs from day 7 to day 14 and pushes I to \(ES_I = 14\), \(EF_I = 17\). Since \(LF_I = 20\), the chain still finishes three days early:$$T_{\text{new}} = \max(18,\, 20,\, 17) = \boxed{20 \text{ days (unchanged)}}$$
2. Reading (ii) — G's duration grows by two days. G then runs day 5 to day 14 and I again finishes on day 17, so the project duration is again 20 days. On this network the two readings happen to agree, because the delay is absorbed entirely within the float of the C–G–I chain.
3. What actually changes. The float on that chain is consumed, not destroyed: \(TF_G\) falls from 5 to 3 days, \(TF_I\) falls from 5 to 3, and G's free float is unchanged at zero, so any further delay is felt immediately by I. The critical path, the 20-day duration and every float on the A–D–H and B–E chains are untouched. A third day of delay would still be absorbed; a sixth would push the project out day for day.