07-Str-B2 · December 2017

Question 1 of 6: Scheduling — precedence network, floats, critical path, delay impact and percent complete

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

Question 1: Scheduling — precedence network, floats, critical path, delay impact and percent complete (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 with the predecessor, duration, budget and progress data printed above. Every relation is finish-to-start with zero lag except C → E, which carries a finish-to-start lag of three days. Durations are in working days, budgets in thousands of dollars, and the progress column is the physical percent complete reported at the data date; the five activities showing "----" have not started.

Find. (a) the precedence network; (b) the total float of every activity and the critical path; (c) the effect on project completion of delaying activity G by six days; and (d) the overall percent complete of the project at the data date.

Approach. Draw the activity-on-node network from the predecessor column, run a forward pass to get early start and early finish times (adding the three-day lag on the one arrow that carries it), take the project duration as the largest early finish, run a backward pass from that duration to get late start and late finish, compute total float as the difference, read the critical path off the zero-float chain, and finally measure progress by earned value — the budget-weighted sum of the reported percentages.

1. Build the network from the predecessor column. Two activities, A and C, have no predecessor and therefore both start at the project origin. Following the column downwards gives the chains $$\text{A} \to \text{B} \to \text{D} \to \text{G} \to \{\text{I},\ \text{J}\}$$ and $$\text{C} \to \text{F} \to \text{K}, \qquad \text{C} \xrightarrow{\ FS=3\ } \text{E} \to \text{H} \to \text{K}.$$ Activity K is the only merge node (it waits for both H and F), and G is the only burst node on the upper chain (it feeds both I and J). Nothing on the upper chain touches the lower chain, so the two are independent until the project end. This is the network asked for in part (a) and is drawn in Figure 1.1.
2. Run the forward pass. With day 0 as the project start, the early start of an activity is the largest early finish among its predecessors, increased by any lag on that link: $$ES_j = \max_{i \in P(j)}\left(EF_i + Lag_{ij}\right), \qquad EF_j = ES_j + D_j .$$ Working down the list, A runs 0–2 and C runs 0–2. B follows A, so 2–6; D follows B, so 6–9; G follows D, so 9–17. F follows C with no lag, so 2–12. The one link that needs care is C → E: the lag pushes E three days beyond the finish of C, giving $ES_E = 2 + 3 = 5$ and $EF_E = 9$. H then runs 9–11, and K, which merges H and F, cannot start before $\max(11,\,12) = 12$, so it runs 12–15. On the upper chain I runs 17–21 and J runs 17–22.
3. Read the project duration. The project finishes when its last activity finishes, so $$T = \max\left(EF_I,\ EF_J,\ EF_K\right) = \max(21,\ 22,\ 15)$$ $$\boxed{T = 22\ \text{working days}}$$ The controlling activity is J, not the longer-looking F–K chain: F is a ten-day activity but it starts on day 2 and its successor K adds only three days, so that branch is spent by day 15.
4. Run the backward pass. Setting the late finish of every terminal activity to the project duration and working backwards, $$LF_i = \min_{j \in S(i)}\left(LS_j - Lag_{ij}\right), \qquad LS_i = LF_i - D_i ,$$ so J has $LF = 22,\ LS = 17$; I has $LF = 22,\ LS = 18$; K has $LF = 22,\ LS = 19$. G feeds both I and J, so $LF_G = \min(18,\,17) = 17$ and $LS_G = 9$. H and F both feed K, so both take $LF = 19$, giving $LS_H = 17$ and $LS_F = 9$. Continuing, $LS_E = 13$, $LS_D = 6$, $LS_B = 2$ and $LS_A = 0$. Activity C needs the lag subtracted a second time: it feeds F (late start 9) and E (late start 13, less the three-day lag = 10), so $LF_C = \min(9,\,10) = 9$ and $LS_C = 7$.
5. Tabulate total float and free float. Total float is the slack an activity has before it delays the project, and free float is the slack it has before it delays any successor: $$TF_i = LS_i - ES_i = LF_i - EF_i , \qquad FF_i = \min_{j \in S(i)}\left(ES_j - Lag_{ij}\right) - EF_i .$$ Applying both to every activity gives the schedule table below.

   Forward and backward pass results (working days from project start)

   Activity	Duration	Predecessor	ES	EF	LS	LF	Total float	Free float
   A	2	—	0	2	0	2	0	0
   B	4	A	2	6	2	6	0	0
   C	2	—	0	2	7	9	7	0
   D	3	B	6	9	6	9	0	0
   E	4	C (FS = 3)	5	9	13	17	8	0
   F	10	C	2	12	9	19	7	0
   G	8	D	9	17	9	17	0	0
   H	2	E	9	11	17	19	8	1
   I	4	G	17	21	18	22	1	1
   J	5	G	17	22	17	22	0	0
   K	3	H, F	12	15	19	22	7	7

6. Identify the critical path. The chain of zero-float activities runs continuously from the project start to the project finish: $$\boxed{\text{Critical path } = \text{A} \to \text{B} \to \text{D} \to \text{G} \to \text{J}}$$ and its length checks the forward pass exactly, since $2 + 4 + 3 + 8 + 5 = 22$ days. Notice how little float protects the near-critical work: activity I is only one day off the critical path, so the burst at G controls two paths that differ by a single day, whereas the whole lower chain carries seven or eight days of float and can absorb ordinary disruption without touching the completion date. That answers part (b).
7. Assess a six-day delay to activity G (part c). Activity G lies on the critical path and therefore has zero total float, so it has no slack to absorb the delay; every day lost pushes the finish of the project by the same day: $$T' = T + \left(\text{delay} - TF_G\right) = 22 + (6 - 0)$$ $$\boxed{T' = 28\ \text{working days, a slip of 6 days}}$$ The result is the same whether G is started six days late or takes six days longer, because in both readings the finish of G moves from day 17 to day 23. The critical path does not change: it is still A–B–D–G–J, and the activities downstream of G simply move with it, so I and J start on day 23 and I keeps exactly the one day of float it had. Only the independent lower chain gains slack: C, F and K rise from seven to thirteen days of float, and E and H from eight to fourteen. Because the whole six days reach the completion date, who carries them depends entirely on the cause — an owner-caused delay is excusable and compensable (time and money), a neutral event such as abnormal weather is excusable but non-compensable (time only), and a contractor-caused delay is non-excusable and exposes the contractor to liquidated damages — which is exactly why the contract's notice provisions matter here.
8. Measure progress by earned value (part d). Percent complete for a project is not the average of the activity percentages; each activity must be weighted by the share of the budget it represents. With $BAC$ the budget at completion and $EV$ the earned value, $$EV = \sum_i \left(\text{Budget}_i \times \text{\% complete}_i\right), \qquad \text{\% complete} = \frac{EV}{BAC} \times 100 .$$ Summing the cost column gives $BAC = 5+3+4+2+4+5+2+2+4+3+3 = 37$, that is $37,000. Only five activities have earned anything:

   Earned value at the data date (costs in $×1,000)

   Activity	Budget	Percent complete	Earned value
   A	5	100 %	5.0
   B	3	100 %	3.0
   C	4	75 %	3.0
   D	2	50 %	1.0
   E	4	40 %	1.6
   F, G, H, I, J, K	19	0 %	0.0
   Total	37	—	13.6

   Substituting, $$\text{\% complete} = \frac{13.6}{37.0} \times 100 = 36.76\ \%$$ $$\boxed{EV = \$13{,}600 \quad\text{and}\quad \text{overall percent complete} = 36.8\ \%}$$
9. Sanity-check the progress figure against duration weighting. Weighting by planned duration instead of by cost gives $\left(2 + 4 + 1.5 + 1.5 + 1.6\right)/47 = 10.6/47 = 22.6\ \%$. The two measures differ because the completed work is comparatively expensive per day (A alone is $2,500 per day) while the untouched work includes the cheap ten-day activity F. Reporting must therefore state which weighting is used; the cost-weighted figure of 36.8 % is the standard earned-value answer and the one intended here, since the question supplies a cost column precisely so that it can be used as the weight.