07-Str-B2 · May 2018

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

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

Notes on this paper

Paper format: National Exams, May 2018 — 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) — precedence (activity-on-node) networks with start-to-start lags, forward and backward passes, total and free float, the late bar chart, and contractor cash-flow and overdraft analysis; these chapters carry Questions 1 and 5. Hendrickson, C. & Au, T., Project Management for Construction (2nd ed., Carnegie Mellon) — Chapter 5 (cost estimation and unit-cost data), Chapter 8 (bidding and contract award), Chapter 10 (fundamental scheduling procedures), Chapter 11 (advanced scheduling with lags) and Chapter 12 (cost control and financing of constructed facilities). Halpin, D.W. & Senior, B.A., Construction Management (4th ed., Wiley) — quantity take-off, crew productivity, bidding strategy, bonding and construction safety management. Peurifoy, R.L. & Schexnayder, C.J., Construction Planning, Equipment and Methods (9th ed., McGraw-Hill) — excavation production and the physical determinants of backhoe daily output, behind Question 2. R.S. Means, Building Construction Cost Data (annual) — the anatomy of a unit-price line: crew, daily output, unit, and bare material / labour / equipment / total columns. Sullivan, W.G., Wicks, E.M. & Koelling, C.P., Engineering Economy (17th ed., Pearson) — Chapters 5 and 6, present-worth analysis and the repeatability (least common multiple of lives) assumption for alternatives with unequal lives, used in Question 4. Canadian Construction Documents Committee, CCDC 2 — Stipulated Price Contract (2020), CCDC 220 Bid Bond, CCDC 221 Performance Bond and CCDC 222 Labour and Material Payment Bond, with the BC Builders Lien Act holdback provisions and the Master Municipal Construction Documents (MMCD) — the Canadian tendering and payment machinery behind Questions 3 and 5. WorkSafeBC Occupational Health and Safety Regulation (Parts 8, 11, 12, 13, 18 and 33), CSA Z259 fall-protection and CSA Z94.4 respirator series, and the Transportation Association of Canada Manual of Uniform Traffic Control Devices for Canada — the Canadian rule set behind Question 6.

Check — how the two printed figures on page 2 were read. Network (Question 1): nine activity boxes with seven links — A → D carrying the printed SS = 3 lag (the line leaves A's top-left corner, runs across the top of the sheet and drops into D's top-left corner), then D → H, B → E, B → F, C → G, F → I and G → I, all finish-to-start with zero lag. No further arrows are drawn; A, B and C are the only start activities and E, H and I the only finish activities. Foundation plan (Question 2): an 80 m × 70 m rectangle with a rectangular notch 25 m deep cut into the top edge, one internal trench running the full width 20 m up from the bottom (the 50 m and 20 m dimensions meet on it), and one internal trench dropping from the bottom of the notch to that line. The notch width is not dimensioned on the paper, and Step 1 below shows the total trench length does not depend on it, so nothing is assumed. All plan dimensions are taken as trench centrelines; reading them instead to the outside face of the trench would shorten the total by about 1.6 per cent and change no conclusion.

Question 4: Engineering Economics — present-worth comparison of alternatives 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 investment, annual operating cost, periodic major maintenance, annual revenue and service life tabulated above, discounted at 10 per cent per year.

Find. The present worth of each alternative on a common analysis period, and hence the more economical plan.

Cash-flow timelines for the two alternatives over one service life each, drawn to the same time axis so the difference in life is visible. Amounts are dollars.

Approach. Net the annual revenue against the annual operating cost to get one uniform series per project, discount that series plus the investment and the major-maintenance outlays over one service life, then place the two on a common footing by repeating each cycle to the least common multiple of the lives — 30 years — and confirm the ranking with an annual-worth check.

Reduce each project to its net annual cash flow. Revenue and operating cost are both uniform annual series, so they combine into one: $$A_A=23{,}500-12{,}500=\$11{,}000/\text{yr},\qquad A_B=26{,}000-11{,}000=\$15{,}000/\text{yr}$$ Project B earns the larger net return on the smaller investment, which is the first hint that its shorter life will not automatically disqualify it.
Place the major-maintenance outlays on the timeline. Major maintenance falls every five years during the service life; it is not incurred in the year the asset is retired and replaced, because the replacement carries new equipment. Project A (15-year life) is therefore overhauled at the end of years 5 and 10, and Project B (10-year life) at the end of year 5 only. This reading is tested for sensitivity in the callout below.
Discount one life cycle of Project A. With $i=0.10$, $(P/A,10\%,15)=7.6061$, $(P/F,10\%,5)=0.62092$ and $(P/F,10\%,10)=0.38554$: $$PW_A^{(15)}=-170{,}000+11{,}000(7.6061)-15{,}000(0.62092)-15{,}000(0.38554)$$ $$PW_A^{(15)}=-170{,}000+83{,}667-9{,}314-5{,}783=-\$101{,}430$$
Discount one life cycle of Project B. With $(P/A,10\%,10)=6.1446$ and the same five-year factor: $$PW_B^{(10)}=-150{,}000+15{,}000(6.1446)-13{,}000(0.62092)=-150{,}000+92{,}169-8{,}072=-\$65{,}903$$ Neither figure can be compared with the other yet: one buys fifteen years of service and the other ten.
Repeat each cycle to the least common multiple of the lives. The lives are 15 and 10 years, so the common analysis period is $\operatorname{lcm}(15,10)=30$ years — two cycles of A and three of B, each cycle identical to the first under the repeatability assumption. Displacing a cycle by $n$ years multiplies its present worth by $(P/F,10\%,n)$: $$PW_A^{(30)}=PW_A^{(15)}\left[1+(P/F,10\%,15)\right]=-101{,}430(1+0.23939)=\boxed{-\$125{,}712}$$ $$PW_B^{(30)}=PW_B^{(10)}\left[1+(P/F,10\%,10)+(P/F,10\%,20)\right]=-65{,}903(1+0.38554+0.14864)=\boxed{-\$101{,}108}$$
Check the ranking by annual worth. Annual worth handles unequal lives without the repetition step, so it is an independent confirmation rather than a restatement. With $(A/P,10\%,30)=0.106079$: $$AW_A=-125{,}712(0.106079)=-\$13{,}335/\text{yr},\qquad AW_B=-101{,}108(0.106079)=-\$10{,}725/\text{yr}$$ The same ordering appears, with Project B better by roughly $2,610 per year of service.
State the recommendation, and be honest about the sign. Both present worths are negative, so neither project returns 10 per cent on its investment; the arithmetic is easy to sanity-check without discounting at all, since Project A returns $11,000 a year against $170,000 invested, a simple payback of 15.5 years against a 15-year life, and Project B returns $15,000 a year against $150,000, a payback of exactly its ten-year life before any maintenance. If the appraisal is a genuine accept-or-reject decision, the correct answer at a 10 per cent hurdle rate is to reject both and look for a better use of the capital. If the decision is which of the two to build — the question asks for “the most economical plan” — then $$PW_B^{(30)}\gt PW_A^{(30)}\ \Longrightarrow\ \boxed{\text{choose Project B}}$$ by a margin of $24,604 in present worth over the 30-year comparison period.

Check — the major-maintenance timing assumption. The paper says only “every 5 years” and does not say whether the overhaul falls in the final year of each life. The solution above excludes it in the replacement year. Repeating the calculation with the overhaul also charged at year 15 for A and year 10 for B gives $PW_A^{(15)}=-\$105{,}021$ and $PW_B^{(10)}=-\$70{,}916$, hence $PW_A^{(30)}=-\$130{,}162$ and $PW_B^{(30)}=-\$108{,}798$. Project B still wins, by $21,364 instead of $24,604, so the recommendation is insensitive to the assumption. No salvage value is given for either project and none is assumed.

Question 4 — results at i = 10% per year
Quantity	Project A	Project B
Net annual cash flow	$11,000/yr	$15,000/yr
Service life	15 years	10 years
Present worth, one life cycle	−$101,430	−$65,903
Present worth over 30 years (LCM)	−$125,712	−$101,108
Equivalent annual worth	−$13,335/yr	−$10,725/yr
Recommendation	Project B is the more economical by $24,604 of present worth; at a 10% hurdle rate neither project is profitable in absolute terms

Topic. Present-worth comparison of mutually exclusive alternatives whose service lives differ, and the repeatability assumption that makes such a comparison legitimate.

Key relations. $(P/A,i,n)=\dfrac{1-(1+i)^{-n}}{i}$; $(P/F,i,n)=(1+i)^{-n}$; $(A/P,i,n)=\dfrac{i}{1-(1+i)^{-n}}$; comparison period $=\operatorname{lcm}$ of the lives; $PW^{(\text{LCM})}=PW^{(\text{cycle})}\sum_k (P/F,i,kn)$; $AW=PW\times(A/P,i,N)$.

Why this works. Present worth is only a valid ranking statistic when the alternatives deliver the same service over the same span — otherwise the longer-lived option is credited with benefits the shorter one was never asked to supply. Repeating each alternative to the least common multiple of lives restores that symmetry by assuming each asset is replaced in kind at identical cost, which is the standard repeatability assumption and is reasonable when the service is expected to continue indefinitely and technology and prices are stable. Annual worth reaches the same ranking by a different route: because $AW$ already expresses a whole cycle as a per-year equivalent, cycles of different lengths are directly comparable without repetition, which is why agreement between the two methods is a genuine check rather than a circular one.

Common pitfall. The dominant error is comparing $PW_A^{(15)}$ with $PW_B^{(10)}$ directly and concluding that B is better because its number is smaller in magnitude — true here by luck, but for the wrong reason and worth no marks. The second is discounting the maintenance as if it were an annual series rather than a set of discrete five-year outlays. The third is refusing to report a negative result: a present worth can legitimately be negative, and the professionally correct answer says so explicitly rather than quietly reversing a sign.

Source. Sullivan, Wicks & Koelling, Engineering Economy (17th ed.), Chapter 5 (evaluating a single project) and Chapter 6 (comparison of alternatives, including the repeatability and coterminated assumptions); Hendrickson & Au, Project Management for Construction, Chapter 6 on economic evaluation of facility investments.

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