07-Str-B2 · December 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, December 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 that the paper can be used for revision whichever five a candidate chooses.

Reference texts: Hegazy, T., Computer-Based Construction Project Management (Prentice Hall) — precedence (activity-on-node) networks with lags, forward and backward passes, total and free float, earned-value progress measurement and the time–cost trade-off curve; these chapters carry Questions 1 and 3. Hendrickson, C. & Au, T., Project Management for Construction (2nd ed., Carnegie Mellon) — Chapter 5 (cost estimation and unit-cost data), Chapter 10 (fundamental scheduling procedures), Chapter 11 (advanced scheduling with lags) and Chapter 12 (cost control, monitoring and accounting), including percent-complete and earned-value reporting. Halpin, D.W. & Senior, B.A., Construction Management (4th ed., Wiley) — labour productivity, crew balance, construction contracts and bonding, and construction safety management. Peurifoy, R.L. & Schexnayder, C.J., Construction Planning, Equipment and Methods (9th ed., McGraw-Hill) — crew productivity and the physical determinants of daily output, behind Question 3. R.S. Means, Building Construction Cost Data (annual) — the structure of a unit-price line: crew, daily output, labour-hours per unit, bare material / labour / equipment, and total including overhead and profit. 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. 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 220 Bid Bond, CCDC 221 Performance Bond, CCDC 222 Labour and Material Payment Bond and CCDC 40 — Rules for Mediation and Arbitration, together with the BC Builders Lien Act holdback provisions — the Canadian contractual machinery behind Questions 2 and 5. 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, 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.

Check — what the printed R.S. Means line in Question 3 actually contains. On line 04810‑3000 the printed cells are labour-hours 0.092, bare material $3.62, bare labour $2.93, bare total $6.55 and total including O&P $8.45. The CREW, DAILY OUTPUT, UNIT and EQUIPMENT cells are blank on the paper — they are not faint, they carry no ink at all. The self-consistency of the printed row ($3.62 + $2.93 = $6.55) confirms the money columns were read correctly.

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 whose whole cash flow is set out in the table above, to be compared at a discount rate of $i = 10\ \%$ per year. Project A costs $30,000 to build, earns $12,500 and costs $1,500 to operate each year, needs a $5,000 overhaul every three years, and lasts six years. Project B costs $25,000, earns $10,000 and costs $1,000 to operate each year, needs a $3,000 overhaul every two years, and lasts four years. No salvage value is stated for either.

Find. The present-worth profit of each project and, on a defensible common basis given that the lives differ, which of the two is the more economical.

Figure 4.1 — Project A: net annual cash flow of $11,000 for six years against an initial outlay of $30,000, with the $5,000 overhaul falling at the end of year 3.

Figure 4.2 — Project B: net annual cash flow of $9,000 for four years against an initial outlay of $25,000, with the $3,000 overhaul falling at the end of year 2.

Approach. Reduce each project to a net annual cash flow, discount that uniform series and the discrete overhaul back to time zero, and subtract the investment to get the present-worth profit of one life cycle. Because the lives differ, compare the two over the least common multiple of the lives — twelve years — assuming each project can be repeated on identical terms, and confirm the ranking with the equivalent annual worth, which gives the same conclusion without the repetition arithmetic.

Reduce each project to a net annual cash flow. Revenue and operating cost both recur every year of the life, so they combine into one uniform series: $$A_A = 12{,}500 - 1{,}500 = \$11{,}000\ \text{per year}, \qquad A_B = 10{,}000 - 1{,}000 = \$9{,}000\ \text{per year}.$$ Project A therefore earns more per year in absolute terms, but it also costs more to build and carries the larger overhaul, so nothing can be concluded yet.
Place the overhauls on the time line. A $5,000 overhaul "every 3 years" over a six-year life falls at the end of year 3; a $3,000 overhaul "every 2 years" over a four-year life falls at the end of year 2. In each case the overhaul that would coincide with the end of the life is not performed, because the asset is retired at that instant and a major overhaul on a machine about to be disposed of buys nothing. This is the standard reading and it is used throughout below; Step 8 tests what happens if the opposite convention is adopted.
Write the discount factors. At 10 % per year the factors needed are $$(P/A,10\%,6) = \frac{1 - 1.10^{-6}}{0.10} = 4.35526, \qquad (P/A,10\%,4) = \frac{1 - 1.10^{-4}}{0.10} = 3.16987,$$ $$(P/F,10\%,3) = 1.10^{-3} = 0.75131, \qquad (P/F,10\%,2) = 1.10^{-2} = 0.82645.$$
Compute the present-worth profit of Project A over its six-year life. Discounting the uniform series and the single overhaul and subtracting the investment, $$PW_A = -30{,}000 + 11{,}000\,(P/A,10\%,6) - 5{,}000\,(P/F,10\%,3)$$ $$PW_A = -30{,}000 + 11{,}000(4.35526) - 5{,}000(0.75131) = -30{,}000 + 47{,}907.87 - 3{,}756.57$$ $$\boxed{PW_A = \$14{,}151\ \text{over 6 years}}$$
Compute the present-worth profit of Project B over its four-year life. By the same route, $$PW_B = -25{,}000 + 9{,}000\,(P/A,10\%,4) - 3{,}000\,(P/F,10\%,2)$$ $$PW_B = -25{,}000 + 9{,}000(3.16987) - 3{,}000(0.82645) = -25{,}000 + 28{,}528.79 - 2{,}479.34$$ $$\boxed{PW_B = \$1{,}049\ \text{over 4 years}}$$ Both projects are profitable at 10 %, but only just in the case of B.
Put the two on a common study period. The two present worths cover six years and four years respectively and cannot be compared directly — a longer-lived project has more time in which to earn. Under the repeatability assumption each project is replaced at the end of its life on identical terms, and the two are compared over the least common multiple of the lives, $\mathrm{LCM}(6,4) = 12$ years. Project A is then built twice, at $t = 0$ and $t = 6$; Project B three times, at $t = 0$, 4 and 8: $$PW_A^{(12)} = PW_A\left(1 + 1.10^{-6}\right) = 14{,}151.30\,(1 + 0.56447) = \$22{,}139,$$ $$PW_B^{(12)} = PW_B\left(1 + 1.10^{-4} + 1.10^{-8}\right) = 1{,}049.45\,(1 + 0.68301 + 0.46651) = \$2{,}256.$$
Select the more economical project. Comparing the two on the identical twelve-year basis, $$PW_A^{(12)} - PW_B^{(12)} = 22{,}139 - 2{,}256 = \$19{,}883$$ $$\boxed{\text{Project A is the more economical, by } \$19{,}883 \text{ of present-worth profit over 12 years}}$$ The equivalent annual worth confirms it without any repetition arithmetic, since an annuity is unaffected by how many cycles are laid end to end: $$AW_A = PW_A\,(A/P,10\%,6) = 14{,}151.30 \times 0.229607 = \$3{,}249\ \text{per year},$$ $$AW_B = PW_B\,(A/P,10\%,4) = 1{,}049.45 \times 0.315471 = \$331\ \text{per year}.$$ Annualising the twelve-year present worths reproduces exactly these figures, $22{,}139 \times 0.146762 = \$3{,}249$ and $2{,}256 \times 0.146762 = \$331$, which is the arithmetic check on the whole comparison.
Test the sensitivity of the answer to the overhaul convention. If instead the overhaul is charged in the final year as well — A paying $5,000 at $t = 6$ and B paying $3,000 at $t = 4$ — then $$PW_A' = 14{,}151.30 - 5{,}000(0.56447) = \$11{,}329, \qquad PW_B' = 1{,}049.45 - 3{,}000(0.68301) = -\$1{,}000 .$$ Project A still wins, and Project B now loses money outright. The recommendation is therefore robust to the ambiguity in the wording, which is the point worth stating in an exam answer: identify the assumption, price both readings, and show that the decision does not depend on it.

Check — assumptions declared. (i) The overhaul that would fall on the disposal date is not performed, since the asset is retired at that instant; Step 8 shows the ranking is unchanged if it is. (ii) No salvage value is stated for either project and none is assumed. (iii) The comparison uses the repeatability assumption over the twelve-year least common multiple; the equivalent-annual-worth route in Step 7 gives the same answer and is the preferred presentation when repeatability is doubtful. (iv) "Present value profit" is read as net present worth of revenues less all costs, which is what the data support.

Question 4 — final results (i = 10 % per year)
Quantity	Project A	Project B
Net annual cash flow	$11,000 /yr	$9,000 /yr
PW of the annual series	$47,907.87	$28,528.79
PW of the major overhaul	−$3,756.57 (year 3)	−$2,479.34 (year 2)
Present-worth profit, one life cycle	$14,151 (6 yr)	$1,049 (4 yr)
Present-worth profit over 12 years	$22,139	$2,256
Equivalent annual worth	$3,249 /yr	$331 /yr
PW profit if the final-year overhaul is also charged	$11,329	−$1,000
Decision	Select Project A — higher present-worth profit on every consistent basis, by $19,883 over the twelve-year study period

Topic. Present-worth comparison of mutually exclusive alternatives with unequal lives, using the repeatability (least common multiple) assumption and the equivalent-annual-worth cross-check.

Key relations. $(P/A,i,n) = \dfrac{1-(1+i)^{-n}}{i}$; $(P/F,i,n) = (1+i)^{-n}$; $(A/P,i,n) = \dfrac{i}{1-(1+i)^{-n}}$; one-cycle present worth $PW = -I + A\,(P/A,i,n) - M\,(P/F,i,t)$; repeated-cycle present worth $PW^{(N)} = PW \sum_{k=0}^{N/n-1}(1+i)^{-kn}$; equivalent annual worth $AW = PW\,(A/P,i,n)$.

Why this works. Present worth answers the question "how much would I have to have today, at this interest rate, to reproduce this cash flow" — so the alternative with the larger present worth leaves the owner better off, provided both cover the same span of time. That proviso is the whole reason for the least-common-multiple device: comparing a six-year present worth with a four-year one silently rewards the project that happens to be measured over more years. Repeating each project until the two horizons coincide removes the bias without inventing any new data. Equivalent annual worth reaches the same conclusion by the opposite route: it spreads each cycle's present worth uniformly over that cycle's own length, and because an infinitely repeated annuity is unchanged by repetition, the annual figures are directly comparable as they stand. The two methods must agree, and checking that they do is a free error trap on the arithmetic.

Common pitfall. The dominant error is comparing $PW_A$ over six years against $PW_B$ over four years and declaring A the winner for the wrong reason — here the conclusion happens to be right, but on other data the shorter project wins once its cycles are counted properly, so the shortcut is unsafe. The second is mis-placing the recurring overhaul: charging it at years 3 and 6 for A without saying so, or spreading it as an annual amount of $5,000/3, both of which change the answer. The third is discounting the annual figures one year at a time and losing a term, when the uniform-series factor does the job in one line. A fourth is forgetting that a repeated cycle's present worth must be discounted from the start of that cycle, not from its end.

Source. Sullivan, Wicks & Koelling, Engineering Economy (17th ed.), Chapter 5 (§5.3 present worth, §5.6 the capitalised-worth and annual-worth methods) and Chapter 6 (§6.4–6.5, comparing alternatives with unequal lives by repeatability or by a coterminated study period). Hendrickson & Au, Project Management for Construction, Chapter 6 (economic evaluation of facility investments).

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