Scheduling — network, critical path and total floats: 16-Civ-B8 Management of Construction, December 2014, Q1

Question 1: Scheduling — network, critical path and total floats (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. Fourteen activities with durations in days and their immediate predecessors. All relationships are ordinary finish-to-start with zero lag — the table gives predecessors only, with no lag column, which is the standard reading for this question type.

Activity	Duration (days)	Predecessors	Activity	Duration (days)	Predecessors
B	5	—	Y	10	A
M	4	B	S	10	F, A
N	9	B	R	2	Q, F
X	15	B	T	5	C, Q
A	5	M, N	K	7	Y, S, R
F	6	N, X	U	3	K, T
Q	2	X	B is the only start activity; U is the only finish activity.
C	4	X

Find. The activity-on-node network diagram, the critical path and the project duration, and the total float of every one of the fourteen activities.

Figure 1.1 — Activity-on-node (precedence) network for the project, with the completed forward and backward passes. Each box carries the early start and early finish on the top row, the activity letter with its duration in the middle, and the late start and late finish on the bottom row. Boxes outlined in heavy red have zero total float and form the critical path B → X → F → S → K → U; the project duration is 46 days.

Approach. Build the activity-on-node network from the predecessor column, run a forward pass to obtain early start and early finish for every activity (early start is the largest early finish among its predecessors), take the project duration as the largest early finish, then run a backward pass from that duration for late finish and late start, and read total float as late start minus early start.

Lay out the network from the predecessor column. B has no predecessor, so it is the single start node; M, N and X all list B alone, so they burst from B. A follows M and N; F follows N and X; Q and C both follow X. Y follows A; S follows F and A; R follows Q and F; T follows C and Q. K merges Y, S and R, and U merges K and T, so U is the single finish node. Every activity appears exactly once as a successor except B, and exactly once as a predecessor except U, which confirms the network is complete and contains no dangling activity.
Forward pass — early start and early finish. For any activity i with predecessor set P(i), $$ES_i=\max_{j\in P(i)}EF_j ,\qquad EF_i=ES_i+d_i$$ Starting at day zero, $ES_B=0$ and $EF_B=0+5=5$. The three successors of B therefore all start on day 5: $EF_M=5+4=9$, $EF_N=5+9=14$ and $EF_X=5+15=20$.
Carry the forward pass through the merge points. Activity A merges M and N, so $ES_A=\max(9,14)=14$ and $EF_A=19$. Activity F merges N and X, so $ES_F=\max(14,20)=20$ and $EF_F=26$; the X branch governs. Q and C both depend on X alone, giving $ES_Q=ES_C=20$, $EF_Q=22$ and $EF_C=24$. Continuing, $ES_Y=19$ and $EF_Y=29$; S merges F and A so $ES_S=\max(26,19)=26$ and $EF_S=36$; R merges Q and F so $ES_R=\max(22,26)=26$ and $EF_R=28$; T merges C and Q so $ES_T=\max(24,22)=24$ and $EF_T=29$.
Close the forward pass and fix the project duration. K merges Y, S and R, so $ES_K=\max(29,36,28)=36$ and $EF_K=43$; the S branch governs. U merges K and T, so $ES_U=\max(43,29)=43$ and $EF_U=46$. U is the only terminal activity, so the project duration is $$T=\max_i EF_i=EF_U=\boxed{46\ \text{days}}$$
Backward pass — late finish and late start. Setting $LF_U=T=46$ and working back through the successor sets S(i), $$LF_i=\min_{j\in S(i)}LS_j ,\qquad LS_i=LF_i-d_i$$ so $LS_U=43$; then $LF_K=43$ and $LS_K=36$, while $LF_T=43$ and $LS_T=38$. Working left, Y, S and R all feed K, so each has $LF=36$, giving $LS_Y=26$, $LS_S=26$ and $LS_R=34$. C feeds T only, so $LF_C=38$ and $LS_C=34$. Q feeds R and T, so $LF_Q=\min(34,38)=34$ and $LS_Q=32$. F feeds S and R, so $LF_F=\min(26,34)=26$ and $LS_F=20$. A feeds Y and S, so $LF_A=\min(26,26)=26$ and $LS_A=21$.
Finish the backward pass at the start activity. X feeds F, Q and C, so $LF_X=\min(20,32,34)=20$ and $LS_X=5$. N feeds A and F, so $LF_N=\min(21,20)=20$ and $LS_N=11$. M feeds A only, so $LF_M=21$ and $LS_M=17$. Finally B feeds M, N and X, so $LF_B=\min(17,11,5)=5$ and $LS_B=0$. The backward pass returning exactly zero at the start node is the arithmetic check that both passes were run correctly.
Total float for every activity. Total float is the amount by which an activity may be delayed without pushing out the project finish, $$TF_i=LS_i-ES_i=LF_i-EF_i$$ Applying this to the completed table gives the values collected below. Six activities return zero and the remaining eight carry positive float.
Identify the critical path. The zero-float activities are B, X, F, S, K and U, and they form one continuous chain because each one starts on the day its predecessor in the chain finishes: B finishes day 5 and X starts day 5, X finishes day 20 and F starts day 20, F finishes day 26 and S starts day 26, S finishes day 36 and K starts day 36, K finishes day 43 and U starts day 43. The critical path is therefore $$\boxed{\text{B}\rightarrow \text{X}\rightarrow \text{F}\rightarrow \text{S}\rightarrow \text{K}\rightarrow \text{U}}$$ and its length, $5+15+6+10+7+3=46$ days, reproduces the project duration found in step 4. Because every relationship here is finish-to-start with zero lag, the path length is a simple sum of durations; that identity would fail if the network carried overlapping (start-to-start or finish-to-finish) links.

Two secondary readings are worth stating because examiners often ask for them in the same breath. Total float belongs to a path, not to a single activity: M, A and Y are not independent, since A and Y share the seven days of float that A carries. Free float — the delay an activity can absorb without disturbing the early start of any successor — is the more useful number for a superintendent. For M it is $ES_A-EF_M=14-9=5$ days, for Y it is $ES_K-EF_Y=36-29=7$ days, and for C it is $ES_T-EF_C=24-24=0$ days even though C carries fourteen days of total float. In other words, C can slip only if T slips with it.

Forward and backward pass, and total float, for all fourteen activities (days)
Activity	Duration	ES	EF	LS	LF	Total float	Critical?
B	5	0	5	0	5	0	Yes
M	4	5	9	17	21	12	—
N	9	5	14	11	20	6	—
X	15	5	20	5	20	0	Yes
A	5	14	19	21	26	7	—
F	6	20	26	20	26	0	Yes
Q	2	20	22	32	34	12	—
C	4	20	24	34	38	14	—
Y	10	19	29	26	36	7	—
S	10	26	36	26	36	0	Yes
R	2	26	28	34	36	8	—
T	5	24	29	38	43	14	—
K	7	36	43	36	43	0	Yes
U	3	43	46	43	46	0	Yes
Project duration 46 days; critical path B – X – F – S – K – U.

Topic. Critical path method on an activity-on-node (precedence) network with finish-to-start relationships: forward pass, backward pass, total float and path criticality.

Key relations. Forward pass $ES_i=\max_{j\in P(i)}EF_j$ and $EF_i=ES_i+d_i$; project duration $T=\max_i EF_i$; backward pass $LF_i=\min_{j\in S(i)}LS_j$ (with $LF=T$ for terminal activities) and $LS_i=LF_i-d_i$; total float $TF_i=LS_i-ES_i=LF_i-EF_i$; free float $FF_i=\min_{j\in S(i)}ES_j-EF_i$. An activity is critical when $TF_i=0$, and for an all-finish-to-start network the critical path length equals the sum of the durations along it.

Why this works. The forward pass is nothing more than the statement that an activity cannot begin until every predecessor has finished, propagated left to right; taking the maximum at each merge enforces that constraint for all incoming arrows at once. The backward pass applies the mirror-image statement — an activity must finish early enough that no successor is pushed past the project deadline — propagated right to left, taking the minimum at each burst. The gap between the two answers is exactly the slack the schedule contains, so subtracting them gives float directly. Because the deadline was set equal to the earliest possible finish, at least one chain of activities must have no slack at all, and that chain is the critical path: it is the longest route through the network, and lengthening any activity on it lengthens the project day for day.

Common pitfall. Two errors dominate. The first is taking a minimum instead of a maximum at a merge in the forward pass (or the reverse in the backward pass), which shows up as a non-zero late start at the start activity — always check that $LS$ returns to zero at B. The second is treating total float as private property of one activity: M, A and Y lie on a common non-critical path, and if A consumes its seven days, Y has none left. Quoting total float to a superintendent without also quoting free float invites exactly that double-spend.

Source. Hendrickson & Au, Project Management for Construction, Ch. 10 (fundamental scheduling procedures); Halpin & Senior, Construction Management, 4th ed., Ch. 9–10; PMI PMBOK Guide §6.5.2 (critical path method, total and free float).

Question 1 of 6: Scheduling — network, critical path and total floats

Question 1: Scheduling — network, critical path and total floats (20 marks)