07-Str-B2 · May 2013

Question 4 of 6: Engineering Economics — breakeven revenue for alternatives of unequal life

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

Notes on this paper

National Examinations — May 2013 — 07-Str-B2 Management of Construction. Three hours, closed book; candidates may use one of the two approved calculators (Casio or Sharp). The paper prints six questions of equal value (20 marks each) and states that any five questions constitute a complete paper, only the first five appearing in the answer book being marked. Candidates are urged to record any interpretive assumptions with their answers. All six questions are worked below, because the set is intended as a study resource rather than as a single exam sitting.

Reference texts: Hegazy, T., Computer-Based Construction Project Management (Prentice Hall) — bar charts, earned-value control and precedence networks with SS/FS/FF lags, which is the notation this paper uses; Hendrickson, C. & Au, T., Project Management for Construction (2nd ed., Carnegie Mellon) — cost control and earned value; Halpin, D.W. & Senior, B.A., Construction Management (4th ed., Wiley) — scheduling, cash flow and bonding; Sullivan, W.G., Wicks, E.M. & Koelling, C.P., Engineering Economy (17th ed., Pearson) — annual-worth comparison of alternatives with unequal lives; Canadian Construction Documents Committee, CCDC 2 — Stipulated Price Contract (2020) with CCDC 220/221/222 bond forms — bid, performance and labour-and-material payment bonds and the holdback provisions; Goldsmith, I. & Heintzman, T.G., Goldsmith on Canadian Building Contracts (5th ed., Thomson Reuters) — delay, notice and surety law in Canada; AACE International, Recommended Practice 29R-03: Forensic Schedule Analysis — but-for and windows methods; WorkSafeBC, Occupational Health and Safety Regulation (Parts 4, 11, 14 and 20) and the BC Workers Compensation Act — construction health and safety duties.

Check — values scaled from the printed figures. Questions 1 and 2 carry hand-drawn figures with no written numbers on the time axis. The activity durations and lag labels in Question 2 are printed inside the network boxes and are read directly. The interpretation adopted here is stated in the Given of each question; it reproduces the drawing and yields round results (a project cost performance index of exactly 0.80 and a 44-day critical path), which is the usual signature of a correct reading.

Question 4: Engineering Economics — breakeven revenue for alternatives of unequal life (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 whose service lives differ, appraised at a discount rate of $i = 10\,\%$ per year, with Project A’s annual revenue $R$ left as the unknown.

Given data — cash flows (CAD)
Item	Project A	Project B
Initial investment (year 0)	90,000	80,000
Yearly operating cost	1,500	1,000
Major maintenance (every 3 years)	5,000	3,000
Yearly revenue	R (unknown)	20,000
Service life	4 years	3 years
Discount rate	10 % per year

Find. The annual revenue $R$ at which Project A is exactly as attractive as Project B.

Figure 4.1 — Project A. The single major maintenance charge inside the four-year life falls at the end of year 3.

Figure 4.2 — Project B. Its three-year life carries one major maintenance charge, at the end of year 3.

Approach. Because the two lives differ, compare them on equivalent uniform annual worth — which is the correct treatment whenever the alternatives are assumed repeatable — forming the annual worth of each project and solving the equation $AW_A = AW_B$ for $R$.

Choose the comparison basis. Present worth may only be compared over a common study period, and here the lives are 4 and 3 years. Either repeat both over the least common multiple of 12 years or, equivalently and far more quickly, convert each project to an equivalent uniform annual worth over its own life. Annual worth already embeds the repeatability assumption, so no explicit replication is needed.
Evaluate the two interest factors that are needed. With $i = 0.10$,$$(A/P,10\%,4) = \frac{i(1+i)^{4}}{(1+i)^{4}-1} = \frac{0.10(1.4641)}{0.4641} = 0.315471,\qquad (A/P,10\%,3) = \frac{0.10(1.331)}{0.331} = 0.402115$$and the single-payment factor $(P/F,10\%,3) = 1.10^{-3} = 0.751315$ is used to bring the year-3 maintenance charge back to time zero.
Annualise Project A’s capital and maintenance. The capital recovery charge on the initial investment is$$CR_A = \$90{,}000\,(A/P,10\%,4) = \$90{,}000 \times 0.315471 = \$28{,}392.37\ \text{per year}$$The major maintenance occurs once inside the four-year life, at the end of year 3, so it is first discounted to time zero and then spread over the four years: $\$5{,}000 \times 0.751315 = \$3{,}756.57$, and $\$3{,}756.57 \times 0.315471 = \$1{,}185.09$ per year.
Write the annual worth of Project A. Collecting the revenue, the operating cost, the capital recovery and the annualised maintenance,$$AW_A = R - \$1{,}500 - \$28{,}392.37 - \$1{,}185.09 = R - \$31{,}077.46$$so Project A must generate $\$31{,}077.46$ per year merely to break even against its own costs.
Evaluate Project B in the same way. Its capital recovery is $CR_B = \$80{,}000 \times 0.402115 = \$32{,}169.18$ per year, and its year-3 maintenance annualises to $\$3{,}000 \times 0.751315 \times 0.402115 = \$906.34$ per year. With a known revenue of $\$20{,}000$ and an operating cost of $\$1{,}000$,$$AW_B = \$20{,}000 - \$1{,}000 - \$32{,}169.18 - \$906.34 = -\$14{,}075.53\ \text{per year}$$
Set the two annual worths equal and solve. Indifference requires $AW_A = AW_B$, that is $R - \$31{,}077.46 = -\$14{,}075.53$, giving$$\boxed{R = \$31{,}077.46 - \$14{,}075.53 = \$17{,}002\ \text{per year}}$$At an annual revenue of about $\$17{,}000$ the two projects are economically identical; above it Project A is preferred, below it Project B is.
Check the answer over the least common multiple. Repeating A three times and B four times over a common twelve-year horizon and discounting every cash flow at 10 % gives a present worth of $-\$95{,}906$ for each alternative at $R = \$17{,}001.93$, confirming the annual-worth result by an independent route. Both figures are negative, which is worth saying plainly in an appraisal report: at 10 % neither project earns its cost of capital, and the question asks only where they are equally unattractive.

Check — treatment of the “every 3 years” maintenance. The charge is taken to occur once inside each service life, at the end of year 3, for both projects. For Project B that instant coincides with the end of its three-year life, and some appraisers would argue the overhaul is never performed because the asset is retired at that moment. Under that alternative reading $AW_B = -\$13{,}169.18$ and the indifference revenue rises to $\$17{,}908.28$ per year, about 5 % higher, because removing a cost from Project B makes B the harder benchmark for Project A to match. The reading used above is the conventional one; in an examination, state whichever assumption you adopt, as the rubric on the cover page invites.

Final results — annual-worth comparison at 10 % per year (CAD per year)
Quantity	Project A	Project B
Capital recovery of the initial investment	28,392.37	32,169.18
Annualised major maintenance	1,185.09	906.34
Yearly operating cost	1,500.00	1,000.00
Total annual cost	31,077.46	34,075.53
Yearly revenue	R	20,000.00
Annual worth	R − 31,077.46	−14,075.53
Breakeven revenue R for Project A		17,001.93 ≈ 17,000 per year
Cross-check: present worth of each over 12 years at that R		−95,906 (equal)

Topic. Comparison of mutually exclusive alternatives with unequal service lives by equivalent uniform annual worth, and solution for a breakeven cash-flow parameter.

Key relations. $(A/P,i,n) = \dfrac{i(1+i)^{n}}{(1+i)^{n}-1}$; $(P/F,i,n) = (1+i)^{-n}$; $AW = R - C_{oper} - P\,(A/P,i,n) - M\,(P/F,i,k)(A/P,i,n)$; indifference when $AW_A = AW_B$.

Why this works. Present worth is only meaningful when two alternatives are measured over the same span of time, because a four-year project and a three-year project simply do not buy the same thing. The classical repair is to replicate both over the least common multiple of their lives, twelve years here, which assumes each is repeated identically. Annual worth reaches the same answer in one step: converting a life-cycle to a uniform yearly equivalent already assumes indefinite identical repetition, so two annual worths are directly comparable no matter how the lives differ. Capital recovery is the heart of it — the factor $(A/P,i,n)$ turns the purchase price into the yearly rent that would exactly repay it with interest over the asset’s life — and a non-uniform charge such as the three-yearly overhaul is handled by discounting it to time zero first and then annualising the whole present worth.

Common pitfall. Comparing present worths over the two different lives without a common horizon, which here would favour Project B purely for being shorter. A second is annualising the overhaul over its own three-year cycle with $(A/F,10\%,3)$ and adding that to a four-year annual worth, mixing two different annualisation periods in one expression. A third is to stop at the arithmetic and not report that both annual worths are negative at the breakeven revenue: the projects are equally attractive there, but neither is attractive.

Source. Sullivan, Wicks & Koelling, Engineering Economy (17th ed.), Ch. 5 (Evaluating a Single Project) and Ch. 6 (Comparison and Selection Among Alternatives), §6.4 on the coterminated and repeatability assumptions; Halpin & Senior, Construction Management, chapter on the time value of money.

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