16-Civ-B8 Management of Construction · May 2014

Question 2 of 6: Scheduling — critical path, floats and the effect of a delay

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

Question 2: Scheduling — critical path, floats and the effect of a delay (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. An activity-on-node (precedence) network of eight activities whose durations and eleven relationships are listed below; B is the only activity with no predecessor, and G and I are the only activities with no successor.

Find. (a) The critical path, the project duration, and the total float of every activity; (b) the revised project duration if activity C is delayed by three days, and why.

Approach. Run a forward pass in which each relationship type contributes its own lower bound on a successor’s early start, take the project duration as the largest early finish, then run a backward pass with the mirror-image rules and report total float as late start minus early start; the zero-float chain is the critical path, and part (b) follows from whether C has any float to absorb the delay.

1. State the three lag rules that drive the forward pass. With $ES$ and $EF = ES + d$ measured in working days from a project start of day 0, a link from $i$ to $j$ with lag $L$ imposes $$\text{FS}: ES_j \ge EF_i + L, \qquad \text{SS}: ES_j \ge ES_i + L, \qquad \text{FF}: EF_j \ge EF_i + L \;\Leftrightarrow\; ES_j \ge EF_i + L - d_j.$$ Each activity takes the largest bound its predecessors impose.
2. Forward pass. B starts the project, $ES_B = 0$, $EF_B = 8$. The SS2 link lets D begin only two days into B, so $ES_D = 0 + 2 = 2$ and $EF_D = 14$ — note that D therefore finishes after B, which is exactly what an overlapped pair looks like. E and C both follow D conventionally, $ES = 14$, giving $EF_E = 21$ and $EF_C = 22$. F waits on the latest of its three predecessors, $\max(14,\,21,\,22) = 22$, so $ES_F = 22$ and $EF_F = 31$. H is bounded by B through the six-day FS lag, $EF_B + 6 = 8 + 6 = 14$, and by C at day 22; the second governs, so $ES_H = 22$, $EF_H = 32$.
3. Forward pass through the two SS8 links and the FF3 link. G may start eight days into F, $ES_G = 22 + 8 = 30$, hence $EF_G = 42$. For I the SS8 link gives $ES_I \ge 30$ while the FF3 link from H requires $EF_I \ge 32 + 3 = 35$, that is $ES_I \ge 35 - 8 = 27$; the first is larger, so $ES_I = 30$ and $EF_I = 38$. The project duration is the largest early finish over the two terminal activities: $$T = \max(EF_G,\,EF_I) = \max(42,\,38) = \boxed{42\ \text{days}}.$$
4. Backward pass. Set $LF = T = 42$ for the terminal activities G and I, giving $LS_G = 30$ and $LS_I = 34$, and apply the mirrored rules $LS_i \le LS_j - L$ for SS, $LF_i \le LS_j - L$ for FS and $LF_i \le LF_j - L$ for FF. F is bounded by G, $LS_F \le 30 - 8 = 22$, and by I, $LS_F \le 34 - 8 = 26$; the tighter governs, so $LS_F = 22$ and $LF_F = 31$. H is bounded only through the FF3 link, $LF_H \le 42 - 3 = 39$, so $LS_H = 29$. Then $LF_C \le \min(LS_F,\,LS_H) = \min(22,\,29) = 22$, $LF_E \le LS_F = 22$, $LF_D \le \min(LS_E,\,LS_C,\,LS_F) = \min(15,\,14,\,22) = 14$, and for B the SS2 link gives $LS_B \le LS_D - 2 = 0$ while the FS6 link gives $LS_B \le 29 - 6 - 8 = 15$, so $LS_B = 0$.
5. Assemble the schedule and the total floats. Total float is $TF = LS - ES = LF - EF$, and collecting the two passes gives the complete schedule:

   Forward and backward pass (working days from project start = 0)

   Activity	Duration	ES	EF	LS	LF	Total float
   B	8	0	8	0	8	0
   D	12	2	14	2	14	0
   E	7	14	21	15	22	1
   C	8	14	22	14	22	0
   F	9	22	31	22	31	0
   G	12	30	42	30	42	0
   H	10	22	32	29	39	7
   I	8	30	38	34	42	4

6. Identify the critical path. The zero-float activities are B, D, C, F and G, and they form a single connected chain, so $$\text{critical path} = \boxed{\text{B} \rightarrow \text{D} \rightarrow \text{C} \rightarrow \text{F} \rightarrow \text{G}, \quad T = 42\ \text{days}}.$$ Note that E carries one day of float even though it sits between the same two critical activities as C — C is one day longer, so it is C and not E that sets F’s start.
7. Check the length as a traverse, not as a sum. Adding the durations along the critical chain gives $d_B + d_D + d_C + d_F + d_G = 8 + 12 + 8 + 9 + 12 = 49$ days, seven more than the project duration. The difference is exactly the overlap bought by the two start-to-start links on the path: B→D with SS2 saves $d_B - 2 = 8 - 2 = 6$ days and F→G with SS8 saves $d_F - 8 = 9 - 8 = 1$ day, and $T = 49 - 7 = 42$. On a network with lags the critical path must always be traversed, never summed.
8. Part (b): delaying activity C by three days. C lies on the critical path with $TF_C = 0$, so it has no cushion and the delay transmits in full. Re-running the forward pass with C shifted to $ES_C = 17$, $EF_C = 25$ gives $ES_F = 25$, $EF_F = 34$, hence $ES_G = 33$ and $$T' = EF_G = 33 + 12 = \boxed{45\ \text{days}, \ \text{a slip of exactly } 3 \ \text{days}}.$$ The same 45 days results if the three days are read as an extension of C’s duration to 11 days rather than a postponement of its start, because either way C finishes on day 25. The non-critical branch absorbs the shift without complaint: H moves to 25–35 and I to 33–41, still inside the new 45-day envelope, so the delay costs three days of project time and nothing else.