07-Str-B2 · Undated paper

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 2019 — 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 that appear in the answer book are marked. All six are worked below so the paper serves as a complete revision set whichever five a candidate elects.

Reference texts: Hegazy, T., Computer-Based Construction Project Management (Prentice Hall) — activity-on-node networks, the forward and backward passes, total and free float, resource profiles and levelling, the time–cost trade-off, and the earned-value formulation with the 20/80 progress convention; these chapters carry Questions 1 and 5. Hendrickson, C. & Au, T., Project Management for Construction (2nd ed., Carnegie Mellon) — Chapter 8 (construction contracts, delivery systems and the allocation of risk), Chapter 10 (fundamental scheduling procedures) and Chapter 12 (cost control, monitoring and accounting), behind Questions 1, 3 and 5. Halpin, D.W. & Senior, B.A., Construction Management (4th ed., Wiley) — delivery-system comparison, bid evaluation and responsibility determination, and construction safety management, behind Questions 3 and 6. Sullivan, W.G., Wicks, E.M. & Koelling, C.P., Engineering Economy (17th ed., Pearson) — Chapters 4, 5 and 6: the uniform-series present-worth factor, single-payment factors, and the comparison of alternatives with unequal lives by repeated study period or by annual worth; this is Question 4. AACE International, Recommended Practice 29R-03, Forensic Schedule Analysis, together with the Society of Construction Law Delay and Disruption Protocol (2nd ed., 2017) — the delay-analysis methods and the excusable / compensable / concurrent taxonomy in Question 2. Canadian Construction Documents Committee, CCDC 2 Stipulated Price Contract (2020), CCDC 5B Construction Management Contract — for Services and Construction, and CCDC 23 A Guide to Calling Bids and Awarding Contracts — the Canadian contract and tendering machinery behind Questions 2 and 3. WorkSafeBC Occupational Health and Safety Regulation (B.C. Reg. 296/97), especially Part 20 (Construction, Excavation and Demolition) and Part 6 (Substance Specific Requirements — asbestos and lead), with the federal Transportation of Dangerous Goods Regulations — the Canadian regulatory frame for Question 6.

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 printed on page 3, at a discount rate of i = 10 % per year. Note that the lives are unequal — 15 years against 10 — and that no salvage value is stated for either project.

Cash-flow data as printed (all amounts in dollars)
Item	Project A	Project B
Initial investment (year 0)	$170,000	$150,000
Yearly operating cost	$12,500	$11,000
Major maintenance (every 5 years)	$15,000	$13,000
Yearly revenue	$23,500	$26,000
Life	15 years	10 years

Find. The present worth of each alternative over a fair common study period, and hence which project is the more economical.

Approach. Collapse each project's yearly revenue and operating cost into a single net annual benefit, discount that uniform series with the series present-worth factor, subtract the discounted major-maintenance events and the initial investment to get one life-cycle present worth, then — because the lives differ — place the two on a common basis by repeating each over the 30-year least common multiple, with an annual-worth calculation as an independent cross-check.

Figure 4.1 — cash-flow diagram for Project A. Upward arrows are the net annual benefit; downward arrows are the initial investment and the two major-maintenance events.

Figure 4.2 — cash-flow diagram for Project B, on the same convention. The shorter life means only one intermediate overhaul.

Reduce each project to a net annual benefit. Revenue and operating cost are both uniform annual series over the whole life, so they combine directly: $$A_A=23{,}500-12{,}500=\$11{,}000/\text{yr},\qquad A_B=26{,}000-11{,}000=\$15{,}000/\text{yr}$$ Project B has both the smaller investment and the larger net annual benefit, which already suggests the answer; the arithmetic below confirms it and quantifies the margin.
Fix the timing of the major maintenance. “Every 5 years” over a 15-year life places overhauls at the ends of years 5 and 10, and over a 10-year life at the end of year 5. An overhaul is not booked in the final year of the life, because the asset is retired at that point and would not be refurbished on the day it is disposed of. This assumption is stated explicitly in the callout below and tested there.
Compute the required factors at 10 %. From $(P/A,i,n)=\dfrac{1-(1+i)^{-n}}{i}$ and $(P/F,i,n)=(1+i)^{-n}$: $$(P/A,10\%,15)=7.6061,\qquad (P/A,10\%,10)=6.1446$$ $$(P/F,10\%,5)=0.6209,\quad (P/F,10\%,10)=0.3855,\quad (P/F,10\%,15)=0.2394,\quad (P/F,10\%,20)=0.1486$$
Present worth of Project A over its 15-year life. Discounting the net annual benefit and subtracting the investment and the two overhauls, $$PW_A=-170{,}000+11{,}000\,(P/A,10\%,15)-15{,}000\left[(P/F,10\%,5)+(P/F,10\%,10)\right]$$ $$PW_A=-170{,}000+83{,}667-15{,}000(0.6209+0.3855)=-170{,}000+83{,}667-15{,}097$$ $$\boxed{PW_A=-\$101{,}430}$$
Present worth of Project B over its 10-year life. Identically, with one overhaul, $$PW_B=-150{,}000+15{,}000\,(P/A,10\%,10)-13{,}000\,(P/F,10\%,5)$$ $$PW_B=-150{,}000+92{,}169-8{,}072$$ $$\boxed{PW_B=-\$65{,}903}$$ These two figures cannot yet be compared, because they buy different amounts of service — fifteen years against ten. Comparing them directly would be the single most common error on this question.
Put the alternatives on a common study period. The least common multiple of 15 and 10 is 30 years, over which Project A is repeated twice and Project B three times. Each repetition is the same cash flow displaced in time, so it is discounted by the single-payment factor for its start: $$PW_A^{30}=PW_A\left[1+(P/F,10\%,15)\right]=-101{,}430\,(1+0.2394)=\boxed{PW_A^{30}=-\$125{,}712}$$ $$PW_B^{30}=PW_B\left[1+(P/F,10\%,10)+(P/F,10\%,20)\right]=-65{,}903\,(1+0.3855+0.1486)=\boxed{PW_B^{30}=-\$101{,}108}$$ Project B's present worth is the higher (less negative) by $125,712 − $101,108 = $24,603 over the 30-year period.
Cross-check by annual worth. Annual worth needs no common study period at all, because converting each life-cycle present worth to a uniform annual equivalent over its own life already normalises for duration: $$AW_A=PW_A\,(A/P,10\%,15)=-101{,}430\times 0.13147=\boxed{AW_A=-\$13{,}335/\text{yr}}$$ $$AW_B=PW_B\,(A/P,10\%,10)=-65{,}903\times 0.16274=\boxed{AW_B=-\$10{,}725/\text{yr}}$$ Project B is better by $2,610 per year. Converting the 30-year present worths back to annual figures reproduces exactly these two values, which is the arithmetic check that the repeated-study-period and annual-worth routes agree.
Conclude. On both the 30-year present-worth basis and the annual-worth basis, Project B is the more economical: $$PW_B^{30}=-\$101{,}108\;>\;PW_A^{30}=-\$125{,}712\quad\Longrightarrow\quad\boxed{\text{select Project B}}$$ The margin is substantial — about 20 % of Project A's 30-year cost — so the conclusion is not sensitive to small changes in the assumptions.

Check — two points the candidate should state on the answer paper.

Both present worths are negative. Neither project earns the 10 % discount rate: Project A returns $11,000 a year against a $170,000 investment (simple payback 15.5 years, longer than its own life), and Project B returns $15,000 against $150,000 (payback exactly 10 years, again the whole life). Undiscounted, A loses $35,000 and B loses $13,000 over a life cycle, so the negative sign is not a discounting artefact. The question asks which is the most economical, and that comparison remains perfectly valid between two loss-making alternatives — it is the correct form of answer where the service must be provided regardless. Should the projects be genuinely optional, the correct recommendation is to accept neither at a 10 % cost of capital, and that should be said in one sentence.

The maintenance-timing assumption. Taking overhauls at years 5 and 10 for A and year 5 for B is the reading adopted above. If instead an overhaul is also booked in the terminal year of each life, the present worths become $−105,021 for A and $−70,916 for B, giving annual worths of $−13,808 and $−11,541. Project B still wins, by $2,266 per year. The ranking is therefore insensitive to the assumption, which is exactly what should be demonstrated rather than asserted.

Question 4 — results at i = 10 % per year
Quantity	Project A	Project B
Net annual benefit	$11,000/yr	$15,000/yr
Life	15 years	10 years
Major maintenance booked at	years 5 and 10	year 5
Present worth, one life cycle	$−101,430	$−65,903
Present worth over 30-year LCM	$−125,712	$−101,108
Annual worth	$−13,335/yr	$−10,725/yr
Decision: Project B is the more economical — better by $24,603 in 30-year present worth, or $2,610 per year. Both alternatives have negative present worth, so neither recovers a 10 % cost of capital.

Topic. Present-worth comparison of mutually exclusive alternatives with unequal lives, including a repeating non-annual cost.

Key relations. $(P/A,i,n)=\dfrac{1-(1+i)^{-n}}{i}$; $(P/F,i,n)=(1+i)^{-n}$; $(A/P,i,n)=\dfrac{i}{1-(1+i)^{-n}}$. For a project repeated k times over a common period, $PW_{\text{total}}=PW_{\text{cycle}}\sum_{m=0}^{k-1}(P/F,i,mn)$; and the annual-worth shortcut $AW=PW_{\text{cycle}}\,(A/P,i,n)$ is invariant under repetition, which is why it needs no least common multiple.

Why this works. Present worth is only a valid ranking criterion between alternatives that deliver the same service over the same period. Two projects with different lives do not, so either the shorter must be repeated until the horizons match (the LCM method) or both must be converted to a rate per year (the annual-worth method). The two are algebraically the same operation viewed from opposite ends, which is why they must agree — and their agreement is a free check on the arithmetic.

Common pitfall. Comparing $−101,430 with $−65,903 and concluding that B wins “by $35,527”. The conclusion happens to be right but the reasoning is wrong and the margin is wrong by a factor of about 1.5, because the two figures buy different amounts of service. The second trap is booking the major maintenance in the terminal year, and the third is netting revenue against operating cost after discounting rather than before — harmless here, since both are uniform over the same span, but wrong the moment the two series have different durations.

Source. Sullivan, Wicks & Koelling, Engineering Economy (17th ed.), Chapter 4 (Money-Time Relationships and Equivalence) and Chapter 6 (Comparison and Selection Among Alternatives), specifically the repeatability assumption and the coterminated / repeated study-period methods for unequal lives.

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