07-Str-B2 · May 2017

Question 4 of 6: Engineering Economics — present-worth comparison of two projects with unequal lives

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 4: Engineering Economics — present-worth comparison of two projects with unequal lives (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. Two mutually exclusive projects with the cash flows below, at a discount rate of $i = 10\,\%$ per year. Operating cost and revenue are end-of-year annuities; major maintenance is a lump sum every fifth year of the life.

Item	Project A	Project B
Initial investment (year 0)	$70,000	$50,000
Yearly operating cost	$2,500	$1,000
Major maintenance, every 5 years	$5,000	$3,000
Yearly revenue	$13,500	$16,000
Net yearly cash flow (revenue − operating)	$11,000	$15,000
Life	15 years	10 years
Maintenance years	5 and 10	5

Check — two modelling assumptions, both stated as the question invites. First, major maintenance is taken to fall at years 5 and 10 for Project A and at year 5 for Project B, but not in the final year of each life: the asset is retired at the end of its life, so a major overhaul on the day of retirement would not be incurred. Second, no salvage value is stated for either project, so both are taken to be worth nothing at the end of their lives. Neither assumption changes the ranking: even if Project A's year-15 maintenance were charged, its present worth would fall further and Project B would still win by a wide margin.

Find. The present-value profit of each project and hence the more economical plan, compared on a basis that is fair despite the unequal lives.

Project A — cash-flow diagram over its 15-year life. Downward arrows are disbursements (the 70,000 investment at year 0 and 5,000 of major maintenance at years 5 and 10); upward arrows are the net annual receipt of 11,000. All amounts in Canadian dollars.

Project B — cash-flow diagram over its 10-year life: 50,000 invested at year 0, net annual receipt of 15,000, and 3,000 of major maintenance at year 5. All amounts in Canadian dollars.

Approach. Compute the present worth of each project over its own life using the uniform-series and single-payment present-worth factors, then — because the lives differ — place the two on a common 30-year study period (the lowest common multiple of 15 and 10) under the repeatability assumption, and confirm the ranking with the equivalent uniform annual worth.

Assemble the interest factors at 10 %. The two factors needed are the uniform-series present worth and the single-payment present worth: $$(P/A,\,i,\,n) = \frac{1-(1+i)^{-n}}{i}, \qquad (P/F,\,i,\,n) = (1+i)^{-n}$$ $$(P/A,10\%,15) = 7.6061, \quad (P/A,10\%,10) = 6.1446, \quad (P/A,10\%,30) = 9.4269$$ $$(P/F,10\%,5) = 0.6209, \quad (P/F,10\%,10) = 0.3855, \quad (P/F,10\%,15) = 0.2394, \quad (P/F,10\%,20) = 0.1486$$
Present worth of Project A over 15 years. The net annual cash flow is $13{,}500 - 2{,}500 = 11{,}000$ per year, against which the investment and the two maintenance events are set: $$PW_A = -70{,}000 + 11{,}000\,(P/A,10\%,15) - 5{,}000\left[(P/F,10\%,5) + (P/F,10\%,10)\right]$$ $$PW_A = -70{,}000 + 11{,}000(7.6061) - 5{,}000(0.6209 + 0.3855)$$ $$PW_A = -70{,}000 + 83{,}666.88 - 5{,}032.32 = \boxed{\$8{,}634.55}$$ The project is profitable at 10 %, but only just: the annuity barely covers the investment.
Present worth of Project B over 10 years. Here the net annual cash flow is $16{,}000 - 1{,}000 = 15{,}000$ and there is a single maintenance event, at year 5: $$PW_B = -50{,}000 + 15{,}000\,(P/A,10\%,10) - 3{,}000\,(P/F,10\%,5)$$ $$PW_B = -50{,}000 + 15{,}000(6.1446) - 3{,}000(0.6209) = -50{,}000 + 92{,}168.51 - 1{,}862.76$$ $$PW_B = \boxed{\$40{,}305.74}$$ Project B costs less to build, earns more each year and needs less maintenance, so the direction of the result is unsurprising; what remains is to show that the comparison is legitimate.
Put the two on a common study period. Present worths computed over 15 years and over 10 years are not directly comparable, because the longer project is being credited with five extra years of earning. The standard remedy is the repeatability assumption: each alternative is assumed to be replaced at the end of its life by an identical one, and both are evaluated over the lowest common multiple of the lives, $$n_{\text{study}} = \operatorname{lcm}(15,\,10) = 30 \ \text{years}$$ so Project A runs twice (cycles starting at years 0 and 15) and Project B three times (cycles starting at years 0, 10 and 20). Discounting each repetition to the present: $$PW_A^{30} = PW_A\left[1 + (P/F,10\%,15)\right] = 8{,}634.55\,(1 + 0.2394) = \$10{,}701.59$$ $$PW_B^{30} = PW_B\left[1 + (P/F,10\%,10) + (P/F,10\%,20)\right] = 40{,}305.74\,(1 + 0.3855 + 0.1486) = \$61{,}836.54$$
Cross-check with equivalent uniform annual worth. Converting each single-cycle present worth to an annuity over its own life gives an equal-footing comparison without invoking the 30-year horizon explicitly: $$AW_A = PW_A\,(A/P,10\%,15) = 8{,}634.55 \times 0.131474 = \$1{,}135.22 \ \text{per year}$$ $$AW_B = PW_B\,(A/P,10\%,10) = 40{,}305.74 \times 0.162745 = \$6{,}559.57 \ \text{per year}$$ Multiplying each by $(P/A,10\%,30) = 9.4269$ reproduces the 30-year present worths above exactly, which confirms that the two routes are the same calculation seen from different ends.
Decision. On the common 30-year basis, $$\boxed{PW_B^{30} = \$61{,}836.54 \;>\; PW_A^{30} = \$10{,}701.59}$$ so Project B is the more economical plan, by $51,134.95 of present-value profit, equivalently by $5,424 per year of annual worth. The margin is so large that it is insensitive to the modelling assumptions: it survives charging maintenance in the terminal year of each life, and it survives any plausible salvage value for Project A.

Quantity	Project A	Project B
Net annual cash flow	$11,000/yr	$15,000/yr
Present worth over one life	$8,634.55 (15 yr)	$40,305.74 (10 yr)
Equivalent uniform annual worth	$1,135.22/yr	$6,559.57/yr
Present worth over the common 30-year period	$10,701.59	$61,836.54
Decision	Select Project B — higher present-value profit by $51,134.95 over 30 years

Topic. Present-worth comparison of mutually exclusive alternatives with unequal lives.

Key relations. $(P/A,i,n) = \left[1-(1+i)^{-n}\right]/i$; $(P/F,i,n) = (1+i)^{-n}$; $(A/P,i,n) = i/\left[1-(1+i)^{-n}\right]$; repeatability over $\operatorname{lcm}$ of the lives; $AW = PW \times (A/P,i,n)$, and ranking by $AW$ is identical to ranking by $PW$ over the common multiple.

Why this works. Present worth is only a valid ranking criterion when the alternatives are compared over the same span of time, because a project evaluated over a longer life is being given credit for cash flows the shorter one has not been allowed to earn. The repeatability assumption fixes this by supposing each alternative is renewed indefinitely on identical terms, which makes the lowest common multiple of the lives a fair horizon. Annual worth reaches the same ranking in one step, because converting a single cycle to an annuity implicitly assumes exactly the same repetition; that is why the two answers agree to the cent here and why annual worth is usually the faster route in an examination.

Common pitfall. Comparing $8,634 against $40,306 directly and declaring B the winner without justifying the comparison — the conclusion is right but the reasoning earns no marks, and on a differently balanced pair of projects it would be wrong. The second frequent slip is charging major maintenance in the terminal year of the life, and the third is discounting the annuity over the study period rather than over the project life, which double-counts the repetitions.

Source. Sullivan, Wicks & Koelling, Engineering Economy, 17th ed., §5.3 (present worth), §5.5 (annual worth) and §6.4 (comparing alternatives with unequal lives — repeatability and the study-period assumption).

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