Scheduling: 16-Civ-B8 Management of Construction, December 2015, Q1

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. A five-activity precedence (activity-on-node) network with the durations printed inside each box and two lagged links, one start-to-start and one finish-to-finish.

Given data — durations and logical links
Activity	Duration (working days)	Predecessor link(s)
A	3	none — project start
B	3	A, start-to-start, lag 4 d
C	8	A, finish-to-start, lag 0
D	6	A, finish-to-start, lag 0
E	5	B finish-to-start; C finish-to-start; D finish-to-finish, lag 2 d

Find. The total float of every activity, the critical path and the project duration, and a bar chart drawn on the late schedule.

[Figure not reproduced: Figure 1.1 — the precedence network as printed, with the two lagged links shown in blue. Boxes carry the activity ID over its duration in days. See the official exam paper.]

Approach. Run a lag-aware forward pass to get early start and finish times, take the project duration as the largest early finish, run the backward pass for late times, and read total float as the difference between the late and early schedules.

Write the constraint that each link type imposes. In precedence diagramming a link carries a type and a lag L, and each type constrains a different pair of times. With i the predecessor and j the successor, $$\begin{aligned} \text{FS:}\quad ES_j &\ge EF_i + L\\ \text{SS:}\quad ES_j &\ge ES_i + L\\ \text{FF:}\quad EF_j &\ge EF_i + L \;\Longleftrightarrow\; ES_j \ge EF_i + L - d_j \end{aligned}$$ Times are measured in working days from the project start, so activity A begins at time 0 and EF = ES + d throughout.
Forward pass — earliest times. A starts the project, so $ES_A = 0$ and $EF_A = 3$. The start-to-start link releases B four days after A starts, not after A finishes, so $ES_B = 0 + 4 = 4$ and $EF_B = 7$. C and D both follow A finish-to-start with no lag, giving $ES_C = ES_D = 3$, $EF_C = 3 + 8 = 11$ and $EF_D = 3 + 6 = 9$.
Resolve the three links entering E. E cannot start until both B and C have finished, so $ES_E \ge \max(7,\,11) = 11$. The finish-to-finish link adds $EF_E \ge EF_D + 2 = 11$, which is equivalent to $ES_E \ge 11 - 5 = 6$ and is therefore not the governing constraint. Taking the largest, $ES_E = 11$ and $EF_E = 16$.
Read the project duration. No activity follows E, so the project finishes when E finishes: $$\boxed{T = EF_E = 16\ \text{working days}}$$
Backward pass — latest times. Set $LF_E = T = 16$, hence $LS_E = 11$. Working back through each link type, $LF_i \le LS_j - L$ for FS, $LS_i \le LS_j - L$ for SS, and $LF_i \le LF_j - L$ for FF: $$\begin{aligned} LF_C &= LS_E = 11 \Rightarrow LS_C = 3\\ LF_B &= LS_E = 11 \Rightarrow LS_B = 8\\ LF_D &= LF_E - 2 = 14 \Rightarrow LS_D = 8\\ LF_A &= \min\big(LS_C,\; LS_D,\; LS_B - 4 + d_A\big) = \min(3,\,8,\,7) = 3\\ LS_A &= LF_A - d_A = 3 - 3 = 0 \end{aligned}$$ The last line is worth reading slowly: A is bound by C through a finish-to-start link ($LF_A \le LS_C = 3$), by D through another ($LF_A \le LS_D = 8$), and by B through the start-to-start link ($LS_A \le LS_B - 4 = 4$, i.e. $LF_A \le 7$). The tightest of the three is C, so $LF_A = 3$ and $LS_A = 0$ — the backward pass returns exactly to zero at the start activity, which is the standard arithmetic check that both passes were run correctly.
Total float of each activity. Total float is the delay an activity can absorb without pushing the project completion date, $TF = LS - ES = LF - EF$, which collects the two passes into one table:
Schedule table — forward pass, backward pass and total float (days)
Activity d ES EF LS LF TF

A 3 0 3 0 3 0

B 3 4 7 8 11 4

C 8 3 11 3 11 0

D 6 3 9 8 14 5

E 5 11 16 11 16 0
Identify the critical path. The zero-float activities form an unbroken chain from the start of the project to its completion: $$\boxed{\text{Critical path } A \rightarrow C \rightarrow E,\quad 3 + 8 + 5 = 16\ \text{days}}$$ Because every link on this chain is a plain finish-to-start with zero lag, the durations along the path add up to exactly the project duration. That identity is the second self-check on the pass, and it is the one that fails when a lagged link lies on the critical path: an overlap would make the traverse shorter than the sum.
Draw the late bar chart. Each bar is plotted from its late start to its late finish. B and D, which carry float, shift right of their earliest positions by four and five days respectively; the three critical activities cannot move at all.

Schedule table — forward pass, backward pass and total float (days)
Activity	d	ES	EF	LS	LF	TF
A	3	0	3	0	3	0
B	3	4	7	8	11	4
C	8	3	11	3	11	0
D	6	3	9	8	14	5
E	5	11	16	11	16	0

Figure 1.2 — late bar chart. Solid bars run from late start to late finish; the dashed outline behind B and D shows where the same activity would sit on the early schedule, so the horizontal offset is the total float.

The late bar chart is the schedule a planner would hand to a subcontractor who wants the latest possible mobilisation date, and it is also the honest picture of risk: every activity is critical on this chart, so any slippage at all delays completion. Note that B carries four days of free float as well (its successor E cannot start before day 11 in any case), while D’s five days of total float are also free because the finish-to-finish link lets D finish as late as day 14 without moving E.

Check: the paper states the start-to-start lag twice, with two different values. The prose says “Activities A and B have a Start-to-Start relationship with 2 lag days”, while the network itself is labelled SS = 4. The solution above uses the value printed on the link, SS = 4, because the drawing defines the network and its companion label FF = 2 agrees with the prose — the sentence appears to have quoted one lag for both links. The choice is not load-bearing: with SS = 2 activity B simply starts on day 2 instead of day 4, so its total float rises from 4 days to 6 days while the project duration (16 days), the critical path (A–C–E) and every other float are unchanged. In an examination, state the assumption in one line and solve, as the paper’s own Note 1 invites.

Final results — Question 1
Quantity	Value
Total float, A	0 days (critical)
Total float, B	4 days (6 days if SS = 2 is used)
Total float, C	0 days (critical)
Total float, D	5 days
Total float, E	0 days (critical)
Critical path	A → C → E
Project duration	16 working days
Late schedule (LS–LF)	A 0–3, B 8–11, C 3–11, D 8–14, E 11–16

Topic. Precedence diagramming (activity-on-node CPM) with start-to-start and finish-to-finish lags.

Key relations. $EF = ES + d$; $ES_j \ge EF_i + L$ (FS), $ES_j \ge ES_i + L$ (SS), $EF_j \ge EF_i + L$ (FF); $LS = LF - d$; $TF = LS - ES = LF - EF$; free float $= \min(ES_{\text{succ}}) - EF$.

Why this works. The forward pass is nothing more than the statement that an activity starts as soon as every constraint on it is satisfied, so each early start is the maximum over its incoming links; the backward pass is the mirror statement that an activity may finish as late as the tightest of its successors permits. Because both passes are monotone, the difference between them at any node measures the slack the network leaves there, and a chain on which that slack is zero must control the completion date. Lags do not change the logic, only the arithmetic of a single link: a start-to-start lag lets a successor overlap its predecessor, which is exactly how a schedule is compressed without changing any duration.

Common pitfall. Two errors dominate. The first is treating a start-to-start link as though it released the successor after the predecessor finishes, which inflates the project duration. The second is adding durations along the chain of zero-float activities and reporting that sum as the project duration — correct here, but wrong on any network whose critical chain contains a lag, where the traverse is shorter than the sum by the overlap the lags buy.

Source. Halpin & Senior, Construction Management, 4th ed., Ch. 12 (precedence networks and lags); Hendrickson, Project Management for Construction, 2nd ed., Ch. 10; PMBOK Guide 6th ed., §6.5 (develop schedule).

Question 1 of 6: Scheduling

Question 1: Scheduling (20 marks)