07-Str-B2 · May 2017

Question 1 of 6: Scheduling — CPM on an activity-on-arrow network and least-cost crashing

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

Notes on this paper

Paper format: National Exams, May 2017 — 07-Str-B2 Management of Construction. Three hours, closed book, one approved Casio or Sharp calculator permitted. Six questions of equal value (20 marks each); any five constitute a complete paper and only the first five answered in the answer book are marked. All six are worked below so the paper can be used for revision whichever five a candidate chooses.

Reference texts: Hegazy, T., Computer-Based Construction Project Management (Prentice Hall) — activity-on-arrow networks, event-time calculations, time–cost trade-off and least-cost crashing, project cash flow and overdraft financing, and labour productivity; these chapters cover Questions 1, 3 and 5. Hendrickson, C. & Au, T., Project Management for Construction (2nd ed., Carnegie Mellon) — Chapter 10 (fundamental scheduling procedures), Chapter 11 (advanced scheduling techniques) and Chapter 12 (cost control, monitoring and accounting), including the S-curve and the financing of construction operations. Halpin, D.W. & Senior, B.A., Construction Management (4th ed., Wiley) — construction financing and the interest cost of a negative cash position, labour productivity and motivation, and construction safety management. Sullivan, W.G., Wicks, E.M. & Koelling, C.P., Engineering Economy (17th ed., Pearson) — Chapters 5 and 6, present-worth analysis and the repeatability (common multiple of lives) assumption for alternatives with unequal lives, used in Question 4. Peurifoy, R.L. & Schexnayder, C.J., Construction Planning, Equipment and Methods (9th ed., McGraw-Hill) — site layout and the physical determinants of crew output. AACE International, Recommended Practice 29R-03, Forensic Schedule Analysis, and the Society of Construction Law, Delay and Disruption Protocol (2nd ed., 2017) — the delay-analysis taxonomy required by Question 2. Canadian Construction Documents Committee, CCDC 2 — Stipulated Price Contract (2020), CCDC 40 — Rules for Mediation and Arbitration and CCDC 220/221/222 bond forms — the Canadian contractual machinery for notice, claims and dispute resolution. Transportation Association of Canada, Manual of Uniform Traffic Control Devices for Canada (MUTCDC), and the BC Ministry of Transportation and Infrastructure Traffic Management Manual for Work on Roadways, together with WorkSafeBC's Occupational Health and Safety Regulation (Part 18 Traffic Control, Part 4 lighting and workplace conditions, Part 8 personal protective clothing) — the Canadian rule set behind Question 6.

Question 1 (network). The node numbers, the activity letters and the i–j pairs printed in the data table agree completely with the drawn arrows: solid arrows run 1→2, 1→3, 1→4, 2→3, 2→6, 3→5, 3→6, 4→5 and 5→6, and one dashed arrow runs 2→5. The dashed arrow carries no letter in the table, so it is the network's dummy activity with zero duration and zero cost. Every arrowhead was checked individually at 8× magnification.

Question 5(b) (cash-flow chart). The values below are read from the printed chart. The smooth curve (cash out) reads 4.0, 11.5, 15.5, 45.0, 54.0, 71.5 and 71.5 thousand at the ends of months 1 to 7; the staircase (payment received) rises at the end of months 2 to 7 to 4.0, 12.5, 16.5, 47.5, 57.5 and 80.0 thousand. Both series are read to the nearest $500, which is the resolution the printed chart supports and is consistent with the word “estimate” in the question.

Question 1: Scheduling — CPM on an activity-on-arrow network and least-cost crashing (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 nine-activity activity-on-arrow (AOA) network on six event nodes, with normal and crash durations and direct costs for each activity, and a project indirect cost of $500 per day. Costs in the exam table are printed in hundreds of dollars; the cost slope in the last column below is the extra direct cost of buying one day, computed as the crash-minus-normal cost divided by the days gained.

Activity	i–j	Normal duration (d)	Normal cost ($)	Crash duration (d)	Crash cost ($)	Days buyable	Cost slope ($/d)
A	1–2	8	2,000	5	3,300	3	433.33
B	2–3	4	1,600	2	3,200	2	800.00
C	2–6	5	1,000	5	1,000	0	—
D	1–3	4	4,000	4	4,000	0	—
E	3–6	8	1,600	3	3,600	5	400.00
F	3–5	2	200	1	500	1	300.00
G	1–4	4	1,500	4	1,500	0	—
H	4–5	3	1,600	3	2,100	0	—
I	5–6	4	1,650	2	3,050	2	700.00
Dummy	2–5	0	0	0	0	0	—

Check — activity H cannot be crashed. The table prints H with a normal duration of 3 days and a crash duration of 3 days but a higher crash cost ($2,100 against $1,600). Buying zero days for $500 is not a purchasable option, so H is treated as fixed at 3 days throughout, exactly as C, D and G are. The same conclusion follows from the float calculation below: H carries nine days of float and never becomes critical, so no reading of that row can change the answer.

Find. (a) the earliest and latest event times, every activity's total float, the critical path and the normal project duration; (b) the project duration that minimises the sum of direct and indirect cost, the activities to crash to reach it, and the resulting total cost.

[Figure not reproduced: The activity-on-arrow network as printed, redrawn with each arrow labelled by its activity letter and normal duration in days. The dummy 2→5 (zero duration) is the tenth arrow. Nodes are events, not activities: node 6 is project completion. See the official exam paper.]

Approach. Run a forward pass for earliest event times and a backward pass for latest event times on the event nodes, take total float as $TF_{ij} = L_j - E_i - d_{ij}$ to identify the critical path, then crash one day at a time by buying the cheapest set of activities that shortens every currently critical path, continuing only while that set's combined cost slope is below the $500 per day of indirect cost it saves.

Forward pass — earliest event times. Each event cannot occur until every arrow entering it has finished, so $E_j = \max_i \left( E_i + d_{ij} \right)$ with $E_1 = 0$. $$E_2 = E_1 + d_A = 0 + 8 = 8$$ $$E_3 = \max\left( E_1 + d_D,\; E_2 + d_B \right) = \max(0+4,\; 8+4) = 12$$ $$E_4 = E_1 + d_G = 0 + 4 = 4$$ Node 5 is entered by H, by F and by the dummy, so all three must be tested: $$E_5 = \max\left( E_4 + d_H,\; E_3 + d_F,\; E_2 + 0 \right) = \max(4+3,\; 12+2,\; 8) = 14$$ $$E_6 = \max\left( E_2 + d_C,\; E_3 + d_E,\; E_5 + d_I \right) = \max(8+5,\; 12+8,\; 14+4) = 20$$ The project therefore finishes in $\boxed{20 \ \text{days}}$ on normal durations.
Backward pass — latest event times. Setting $L_6 = E_6 = 20$ and working backwards with $L_i = \min_j \left( L_j - d_{ij} \right)$: $$L_5 = L_6 - d_I = 20 - 4 = 16, \qquad L_4 = L_5 - d_H = 16 - 3 = 13$$ $$L_3 = \min\left( L_6 - d_E,\; L_5 - d_F \right) = \min(20-8,\; 16-2) = 12$$ $$L_2 = \min\left( L_3 - d_B,\; L_6 - d_C,\; L_5 - 0 \right) = \min(8,\; 15,\; 16) = 8$$ $$L_1 = \min\left( L_2 - d_A,\; L_3 - d_D,\; L_4 - d_G \right) = \min(0,\; 8,\; 9) = 0$$ The return of $L_1 = 0$ is the arithmetic check on both passes: any other value would mean a pass had been mis-taken.

Total floats and the critical path. Applying $TF_{ij} = L_j - E_i - d_{ij}$ to each arrow gives the table below. The chain of arrows carrying zero float runs 1 → 2 → 3 → 6, that is $$\boxed{\text{critical path } = \text{A} \to \text{B} \to \text{E}, \quad T = 20 \ \text{days}}$$

Activity	i–j	$E_i$	$L_j$	$d_{ij}$	$TF = L_j - E_i - d_{ij}$	Critical?
A	1–2	0	8	8	0	yes
B	2–3	8	12	4	0	yes
C	2–6	8	20	5	7	no
D	1–3	0	12	4	8	no
E	3–6	12	20	8	0	yes
F	3–5	12	16	2	2	no
G	1–4	0	13	4	9	no
H	4–5	4	16	3	9	no
I	5–6	14	20	4	2	no
Dummy	2–5	8	16	0	8	no

The dummy is worth a comment because it is the only feature of the drawing that a candidate can misread. It forces $E_5 \ge E_2$, but node 5 is already pushed to day 14 by F, so the constraint is slack by eight days and does not affect either part of the answer.

Cost at the normal schedule. Summing the normal direct costs, $2{,}000 + 1{,}600 + 1{,}000 + 4{,}000 + 1{,}600 + 200 + 1{,}500 + 1{,}600 + 1{,}650$, and adding twenty days of overhead: $$C_{\text{direct}} = \$15{,}150, \qquad C_{\text{indirect}} = 20 \times \$500 = \$10{,}000$$ $$\boxed{C_{\text{total,normal}} = \$25{,}150 \ \text{at } 20 \ \text{days}}$$ This is the datum every crashing step is measured against.
Crashing step 1 — buy days from E. Only activities on the critical path can shorten the project, and of A ($433.33/d), B ($800/d) and E ($400/d) the cheapest is E. Each day bought from E costs $400 of direct cost and saves $500 of overhead, a net saving of $100 per day, so it is worth taking. E has five days available, but the near-critical path A–B–F–I is only two days shorter than the critical path: $$L_{\text{A-B-E}} = 8+4+8 = 20, \qquad L_{\text{A-B-F-I}} = 8+4+2+4 = 18$$ Crashing E by two days therefore brings the project to 18 days and makes A–B–F–I critical as well; going further on E alone would achieve nothing. $$T = 18 \ \text{d}, \quad C_{\text{total}} = 15{,}150 + 2(400) + 18(500) = \$24{,}950$$
Crashing step 2 — two parallel critical paths. With A–B–E and A–B–F–I both at 18 days, a further day must be removed from both simultaneously. The candidate purchases are the activities common to both paths, A at $433.33/d and B at $800/d, or a matched pair with one activity on each path, E + F at 400 + 300 = $700 per day, or E + I at 400 + 700 = $1,100 per day. Only A is below the $500/d overhead saving, so A is bought to its limit of three days: $$d_A: 8 \to 5, \qquad T: 18 \to 15 \ \text{days}$$ $$C_{\text{total}} = 15{,}150 + 2(400) + 3(433.33) + 15(500) = \$24{,}750$$ Both critical paths are now 15 days long: $5+4+6 = 15$ and $5+4+2+4 = 15$.

Stopping test. With A fully crashed, the cheapest way to remove a sixteenth day is the matched pair E + F at $700 per day, or B alone at $800 per day. Both exceed the $500 per day of overhead they would save, so the fourteenth-day option raises the total to $24,950 and the search stops: $$\boxed{T_{\text{opt}} = 15 \ \text{days}, \quad C_{\text{total,opt}} = \$24{,}750}$$ The saving against the normal schedule is $25,150 − $24,750 = $400. The full time–cost trade-off is tabulated below; total cost falls monotonically to 15 days and rises thereafter, confirming a single interior optimum.

Duration (d)	Crashing applied	Direct cost ($)	Indirect cost ($)	Total cost ($)
20	none (normal)	15,150	10,000	25,150
19	E −1 d	15,550	9,500	25,050
18	E −2 d	15,950	9,000	24,950
17	E −2 d, A −1 d	16,383	8,500	24,883
16	E −2 d, A −2 d	16,817	8,000	24,817
15	E −2 d, A −3 d	17,250	7,500	24,750
14	E −3 d, A −3 d, F −1 d	17,950	7,000	24,950

Quantity	Value
Normal project duration	20 days
Critical path (normal durations)	1–2–3–6, i.e. A → B → E
Total floats	A 0, B 0, C 7, D 8, E 0, F 2, G 9, H 9, I 2, dummy 8 (days)
Total cost at normal duration	$25,150 (direct $15,150 + indirect $10,000)
Optimum crashing strategy	Crash E from 8 to 6 days and A from 8 to 5 days; leave all other activities at normal duration
Optimum project duration	15 days (both A–B–E and A–B–F–I critical)
Least total cost	$24,750 (direct $17,250 + indirect $7,500)
Saving against the normal schedule	$400

Topic. Event-time CPM on an activity-on-arrow network, and the time–cost trade-off (least-cost crashing).

Key relations. Forward pass $E_j = \max_i (E_i + d_{ij})$; backward pass $L_i = \min_j (L_j - d_{ij})$; total float $TF_{ij} = L_j - E_i - d_{ij}$; free float $FF_{ij} = E_j - E_i - d_{ij}$; cost slope $s = (C_c - C_n)/(d_n - d_c)$; crash while $\sum s < \text{indirect cost per day}$.

Why this works. In an AOA network the numbers belong to the events, not the activities, which is why the passes are written over nodes and why the dummy — an arrow with no work in it — still carries a real logical constraint. Total float measures how far an activity can slip before it pushes the project end date, so the zero-float chain is by definition the longest path through the network. Crashing then works because direct cost rises as duration falls while indirect cost falls linearly with duration: the total-cost curve is convex, so buying the cheapest available day repeatedly, and stopping the moment the marginal direct cost exceeds the daily overhead, lands on the minimum. The subtlety that carries most of the marks is that once a second path becomes critical, a day must be removed from every critical path at once, so the relevant price is the sum of the slopes in the cheapest cutting set, not the cheapest single activity.

Common pitfall. Crashing an activity past the point where another path becomes critical — spending on E for all five available days here buys nothing after the second day, because A–B–F–I holds the project at 18 days. The second frequent error is crashing an activity with float (F or I look cheap, but shortening either alone leaves the 20-day critical path untouched), and the third is forgetting that indirect cost must be added before the optimum can be identified at all: on direct cost alone the answer would always be the normal schedule.

Source. Hegazy, Computer-Based Construction Project Management, chapters on CPM calculations and time–cost trade-off; Hendrickson & Au, Project Management for Construction, §10.3–10.5 and §11.5.

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