Scheduling — total floats, critical path and the AON equivalent: 16-Civ-B8 Management of Construction, December 2018, Q1

Question 1: Scheduling — total floats, critical path and the AON equivalent (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-arrow (AOA) network of eleven activities on nine events, read directly off the printed diagram. Every link is a plain finish-to-start connection with no lag, and no dummy arrows are drawn.

Given data — arrow list read from the page-2 network
Activity	Tail event i	Head event j	Duration (days)
A	1	2	4
B	1	4	6
C	1	7	2
D	2	3	8
E	3	6	4
F	4	5	10
G	4	8	16
H	5	6	8
I	6	9	6
J	7	8	6
K	8	9	10

Find. (a) the total float of every activity and the critical path; (b) the change in project duration if activity F is delayed by three days; (c) the precedence (activity-on-node) network that carries the same logic.

Figure 1.1 — the printed arrow network, annotated with the earliest event time E and latest event time L produced by the two passes below. The critical chain 1–4–8–9 (B–G–K) is drawn heavy.

Approach. Run a forward pass over the events to get earliest event times, a backward pass to get latest event times, then evaluate total float activity by activity as the slack between the latest time its head event may occur and the earliest its tail event can occur.

Forward pass — earliest event times. Setting the project start at zero, each event happens as soon as every arrow entering it has finished: $$E(j)=\max_{(i,j)}\left[E(i)+D_{ij}\right],\qquad E(1)=0$$ Working left to right, $E(2)=0+4=4$, $E(4)=0+6=6$ and $E(7)=0+2=2$; then $E(3)=4+8=12$ and $E(5)=6+10=16$. Event 8 is fed by G and J, so $E(8)=\max(6+16,\;2+6)=22$, and event 6 is fed by E and H, so $E(6)=\max(12+4,\;16+8)=24$. Finally I and K meet at event 9: $$E(9)=\max\left[E(6)+D_I,\;E(8)+D_K\right]=\max(24+6,\;22+10)=\boxed{T=32\ \text{days}}$$
Backward pass — latest event times. Fixing $L(9)=T=32$ and working right to left, $$L(i)=\min_{(i,j)}\left[L(j)-D_{ij}\right]$$ gives $L(8)=32-10=22$, $L(6)=32-6=26$, $L(5)=26-8=18$, $L(3)=26-4=22$ and $L(7)=22-6=16$. Event 4 releases both F and G, so $L(4)=\min(18-10,\;22-16)=6$; event 2 gives $L(2)=22-8=14$. The pass must close on zero at the start event, and it does: $L(1)=\min(14-4,\;6-6,\;16-2)=0$. That return to zero is the arithmetic proof that both passes are consistent.
Total and free float for every activity. With the two event vectors in hand, each arrow is assessed on its own: $$TF_{ij}=L(j)-E(i)-D_{ij},\qquad FF_{ij}=E(j)-E(i)-D_{ij}$$ Total float is the delay an activity can absorb without pushing the project completion out; free float is the delay it can absorb without disturbing the earliest start of anything that follows. Applying both expressions to the eleven arrows gives the schedule table below.

Complete activity schedule — earliest and latest times, total and free float
Activity	i–j	Duration	ES = E(i)	EF	LS	LF = L(j)	Total float	Free float
A	1–2	4	0	4	10	14	10	0
B	1–4	6	0	6	0	6	0	0
C	1–7	2	0	2	14	16	14	0
D	2–3	8	4	12	14	22	10	0
E	3–6	4	12	16	22	26	10	8
F	4–5	10	6	16	8	18	2	0
G	4–8	16	6	22	6	22	0	0
H	5–6	8	16	24	18	26	2	0
I	6–9	6	24	30	26	32	2	2
J	7–8	6	2	8	16	22	14	14
K	8–9	10	22	32	22	32	0	0

(a) Identify the critical path. Exactly three activities carry zero total float — B, G and K — and they form an unbroken chain of events: $$1\;\xrightarrow{\;B(6)\;}\;4\;\xrightarrow{\;G(16)\;}\;8\;\xrightarrow{\;K(10)\;}\;9$$ Because every link in this network is finish-to-start with no lag, the durations along the critical chain must add up to the project duration, and they do: $6+16+10=32$ days. That identity is a free self-check and it is the reason F and H, at only two days of float each, are described as near-critical rather than critical. Note also that E is the only activity whose free float (8 days) differs materially from its total float (10 days): E may finish eight days late without disturbing the earliest start of I, but a ninth day of delay would begin eating float that belongs to the D–E chain as a whole.
(b) Effect of delaying activity F by three days. F carries two days of total float, so the first two days are absorbed without any effect and only the third day reaches the completion date: $$\Delta T = \max\left(0,\;\text{delay}-TF_F\right)=\max(0,\;3-2)=1\ \text{day}$$ Re-running the forward pass with F stretched from ten to thirteen days confirms it: $E(5)=6+13=19$, $E(6)=\max(12+4,\;19+8)=27$, and $E(9)=\max(27+6,\;22+10)=33$. The project therefore finishes one day late: $$\boxed{T'=33\ \text{days, a one-day extension}}$$ The more important consequence is structural. At 33 days the chain B–F–H–I now sums to $6+13+8+6=33$ and becomes the new critical path, while B–G–K falls back to one day of float. A superintendent who keeps pressing G after F slips is accelerating work that no longer governs. Note that the two readings of "delaying F" coincide on this network: whether F's start slips three days or F's duration grows by three days, its earliest finish moves to day 19 either way, because every link leaving F is a plain finish-to-start with no lag.
(c) Equivalent activity-on-node network. The translation rule is mechanical: each arrow becomes a node, and activity Y is a successor of activity X whenever the head event of X is the tail event of Y. Reading the arrow list that way gives A, B and C as start activities; D follows A; E follows D; F and G both follow B; H follows F; I follows E and H; J follows C; and K follows G and J. I and K are the terminal activities. No dummy arrows appear in the printed diagram, so no logical restrictions are lost in the translation and the precedence network reproduces the same 32-day duration and the same critical path.

Figure 1.2 — the equivalent AON (precedence) network. Every link is finish-to-start with zero lag; the critical chain B–G–K is drawn heavy.

Final results — Question 1
Quantity	Result
Project duration	32 days
Critical path	1–4–8–9, i.e. B → G → K (6 + 16 + 10 = 32 days)
Activities with zero total float	B, G, K
Total floats (days)	A 10, B 0, C 14, D 10, E 10, F 2, G 0, H 2, I 2, J 14, K 0
Free floats (days)	E 8, I 2, J 14; all others zero
(b) Effect of a 3-day delay to F	Project extends by 1 day, to 33 days
(b) New critical path	B → F → H → I (6 + 13 + 8 + 6 = 33 days)
(c) AON precedences	A, B, C start; D←A; E←D; F←B; G←B; H←F; I←E,H; J←C; K←G,J

Topic. Critical path method on an activity-on-arrow network: event passes, total and free float, and translation between the arrow and precedence notations.

Key relations. Forward pass $E(j)=\max\left[E(i)+D_{ij}\right]$; backward pass $L(i)=\min\left[L(j)-D_{ij}\right]$; total float $TF=L(j)-E(i)-D_{ij}$; free float $FF=E(j)-E(i)-D_{ij}$; delay effect $\Delta T=\max(0,\,\text{delay}-TF)$.

Why this works. The forward pass propagates the earliest possible occurrence of each event forward through the logic; the backward pass propagates the latest permissible occurrence backward from a fixed completion date. The difference between the two, measured across a single arrow, is the slack that arrow enjoys. Total float belongs to a path, not to an activity: the ten days shared by A, D and E is one pool of ten days, and if A consumes it, D and E have none. Free float is the portion an activity may consume entirely on its own, and it is what a foreman can spend without telephoning anyone. Because floats are computed from event times rather than from a path enumeration, the method scales to networks far too large to inspect by eye, and it gives the delay-effect rule directly: a delay shorter than the total float changes nothing, and a longer delay pushes the completion out by the excess.

Common pitfall. Two recur in this subject. The first is adding an activity's total float to a project duration as though every day of delay counted — here three days of delay to F produce only one day of extension. The second is trusting the "critical durations must sum to the project duration" check on networks that carry start-to-start or finish-to-finish lags, where overlaps buy time back and the sum exceeds the duration; the check is valid here only because every link is a plain finish-to-start with zero lag.

Source. Hendrickson, Project Management for Construction, 2nd ed., Ch. 10 (Fundamental Scheduling Procedures, activity-on-branch and activity-on-node formulations); Halpin & Senior, Construction Management, 4th ed., Ch. 8–9.

Question 1 of 6: Scheduling — total floats, critical path and the AON equivalent

Question 1: Scheduling — total floats, critical path and the AON equivalent (20 marks)