Question 1 of 6: Scheduling

Question 1: Scheduling (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. Eleven activities, A through K, with the predecessor logic, durations in working days, budgets in thousands of dollars, and the actual percent complete reported at the data date. Every link is finish-to-start with zero lag except C → E, which carries a three-day finish-to-start lag. Activities showing a dash under percent complete have not started.

Find. The activity-on-node network, the total float of every activity and the resulting critical path, the effect on project duration of a six-day delay to activity G, and the overall percent complete of the project at the data date.

Approach. Draw the precedence network from the predecessor column, run a forward pass to get early start and early finish, run a backward pass from the project duration to get late start and late finish, take total float as the difference, and then weight each activity's reported percent complete by its budget to obtain the project-level percent complete.

1. (a) Build the precedence network from the predecessor column. Activities A and C have no predecessors, so both start at time zero and the network has two entry points. The A branch runs A → B → D → G, and G then splits into the two terminal activities I and J. The C branch splits immediately into F and, through a three-day lag, E; E feeds H, and H merges with F at K. The network therefore has three terminal activities — I, J and K — and the project finishes when the last of them finishes. The drawing above uses the standard six-cell node, with early start, duration and early finish across the top and late start, total float and late finish across the bottom.
2. Run the forward pass for early start and early finish. For each activity, the early start is the largest early finish among its predecessors, increased by any lag on that link: $$\begin{aligned} ES_j &= \max_{i \to j}\left(EF_i + \text{lag}_{ij}\right) \\ EF_j &= ES_j + D_j \end{aligned}$$ Working down the list, A gives $EF_A = 0 + 2 = 2$ and C gives $EF_C = 2$. Activity E is the only one carrying a lag, so $ES_E = EF_C + 3 = 2 + 3 = 5$ and $EF_E = 9$, while its sibling F starts immediately at $ES_F = 2$ and runs to $EF_F = 12$. Continuing along the upper chain, $EF_B = 6$, $EF_D = 9$, $EF_G = 17$, and the two successors of G finish at $EF_I = 21$ and $EF_J = 22$. At the merge, $ES_K = \max(EF_H, EF_F) = \max(11, 12) = 12$, so $EF_K = 15$.
3. Read the project duration off the terminal activities. The three chains end at day 21 (I), day 22 (J) and day 15 (K), so the project cannot finish before the latest of them: $$T = \max\left(EF_I,\ EF_J,\ EF_K\right) = \max(21,\ 22,\ 15) = \boxed{22 \text{ working days}}$$
4. Run the backward pass for late finish and late start. Every terminal activity is given a late finish equal to the project duration, and each remaining activity takes the earliest late start among its successors, less any lag on that link: $$\begin{aligned} LF_i &= \min_{i \to j}\left(LS_j - \text{lag}_{ij}\right) \\ LS_i &= LF_i - D_i \end{aligned}$$ So $LF_I = LF_J = LF_K = 22$, giving $LS_I = 18$, $LS_J = 17$ and $LS_K = 19$. Activity G feeds both I and J, so $LF_G = \min(18, 17) = 17$ and $LS_G = 9$. Both H and F feed K, so each takes $LF = 19$. Activity C is the interesting one, because it feeds F directly and E through the three-day lag: $LF_C = \min(LS_F,\ LS_E - 3) = \min(9,\ 13 - 3) = \min(9, 10) = 9$. The pass closes correctly, because $LS_A$ returns to exactly zero.
5. (b) Take total float as the slack between the two passes. Total float is how long an activity can be delayed without pushing the project end date: $$TF_i = LS_i - ES_i = LF_i - EF_i$$ The completed schedule is tabulated below. Five activities come back with zero float, and they form a single unbroken chain from the project start to the project finish.

   Completed forward and backward pass, all values in working days

   Activity	Duration	ES	EF	LS	LF	Total float	Critical?
   A	2	0	2	0	2	0	yes
   B	4	2	6	2	6	0	yes
   C	2	0	2	7	9	7	no
   D	3	6	9	6	9	0	yes
   E	4	5	9	13	17	8	no
   F	10	2	12	9	19	7	no
   G	8	9	17	9	17	0	yes
   H	2	9	11	17	19	8	no
   I	4	17	21	18	22	1	no
   J	5	17	22	17	22	0	yes
   K	3	12	15	19	22	7	no

6. Name the critical path and prove it with the sum-of-durations identity. The zero-float activities are A, B, D, G and J, and they connect head to tail, so the critical path is $$\text{A} \to \text{B} \to \text{D} \to \text{G} \to \text{J}$$ Because every link on that chain is finish-to-start with zero lag, its length must equal the plain sum of its activity durations, and it does: $2 + 4 + 3 + 8 + 5 = 22$ days, matching the forward pass exactly. That identity is a genuine self-check here; it would fail if a lag or an overlapping relationship sat on the critical chain. Activity I is worth noting as near-critical with only one day of float, so the G → I branch is the first thing that would go critical if G slipped.
7. (c) Delay activity G by six days. Activity G is on the critical path with zero total float, so it has no slack to absorb a delay and the project end date moves day for day: $$T_{\text{delayed}} = T + \Delta = 22 + 6 = \boxed{28 \text{ working days}}$$ The two natural readings of “delaying an activity” agree on this network. If G's start slips six days it finishes on day 23 and J then runs from day 23 to day 28; if instead G's duration grows from 8 to 14 days it also finishes on day 23, with the same result. They agree because every successor of G hangs off G by a finish-to-start link, so nothing is released early; on a network with start-to-start overlaps the two readings would give different answers.
8. Check what the six-day delay does to the rest of the schedule. The critical path does not change: at 28 days the chains through K and I still finish inside the new deadline, since C → F → K needs only 15 days and A → B → D → G → I now needs 27. Activity I keeps its single day of float, while the floats on the C branch each grow by six days. Practically, the delay is entirely uncompensated unless G, I or J is crashed, and the cheapest recovery is normally sought on the longest critical activity, G itself.
9. (d) Weight each activity's progress by its budget. Overall percent complete is a value-weighted average, not an average of the reported percentages. Multiplying each activity's budget by its reported completion gives the earned value: $$\begin{aligned} EV &= \sum_i B_i \times p_i \\ BAC &= \sum_i B_i \end{aligned}$$ The budget at completion is $5 + 3 + 4 + 2 + 4 + 5 + 2 + 2 + 4 + 3 + 3 = 37$, or $37,000. Only five activities have reported progress, contributing $5.00 + 3.00 + 3.00 + 1.00 + 1.60 = 13.60$ thousand dollars of earned value.
10. Convert the earned value to a project percent complete. Dividing the earned value by the budget at completion: $$\text{Percent complete} = \frac{EV}{BAC} = \frac{13.60}{37.00} = 0.3676 = \boxed{36.8\%}$$ So roughly $13,600 of the $37,000 of budgeted work has been earned at the data date.