07-Str-B2 · May 2016

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 2016 — 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 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 networks with lags, total and free float, project overhead versus general overhead, and the bar-chart/S-curve control method behind Questions 1, 3 and 6; Hendrickson, C. & Au, T., Project Management for Construction (2nd ed., Carnegie Mellon) — Chapters 5 (cost estimation), 10 (scheduling) and 12 (cost control, monitoring and accounting), the source of the earned-value quantities used in Question 6; Halpin, D.W. & Senior, B.A., Construction Management (4th ed., Wiley) — competitive bidding, unbalanced bids, indirect-cost structure and construction safety; Sullivan, W.G., Wicks, E.M. & Koelling, C.P., Engineering Economy (17th ed., Pearson) — Chapters 5 and 6, present-worth analysis and the repeatability assumption for alternatives with unequal lives, used in Question 4; Peurifoy, R.L. & Oberlender, G.D., Estimating Construction Costs (6th ed., McGraw-Hill) — job overhead versus general overhead; Canadian Construction Documents Committee, CCDC 2 — Stipulated Price Contract (2020) and CCDC 23 — A Guide to Calling Bids and Awarding Contracts — bid-call practice, bid security and award criteria for Question 2; Ron Engineering (M.J.B. Enterprises line of cases) as summarised in Goldsmith, I. & Heintzman, T.G., Goldsmith on Canadian Building Contracts (5th ed., Thomson Reuters) — the Contract A/Contract B doctrine that governs a Canadian public bid call; WorkSafeBC, Occupational Health and Safety Regulation (Parts 4, 8, 11, 13, 18, 19 and 20) and the BC Workers Compensation Act, together with CSA Z259 (fall protection), CSA Z94.4 (respirators) and CSA W117.2 (welding safety) — the Canadian rule set behind Question 5.

Question 1 (network). Every activity letter and duration is printed inside its box and reads cleanly. The link routing was traced at high magnification: Start feeds A, D and G; a riser from the right edge of D feeds B; the horizontal link D → E carries the only labelled lag on the sheet, FS 6; a riser from the right edge of G feeds E; G also feeds H, H feeds I, A feeds B, B feeds C, E feeds F; and C, F and I terminate at End. Every unlabelled arrow is an ordinary finish-to-start link with zero lag, which is the only reading consistent with the drawing.

Question 6 (bar chart). Every percentage label falls on a week boundary, so the printed figures are cumulative percent complete at each week end. Planned: A 20/60/100 in weeks 1–3; B 10/80 in weeks 1–2, finishing in week 3; C 20/70 in weeks 3–4, finishing in week 5. Actual: A 10/50/90 in weeks 1–3, its bar closing in week 4; B 70 at week 2, its bar closing in week 3; C 50 at week 3, its bar closing exactly on the week-4 gridline. A bar that closes is read as 100 % complete from that week end onward, which is the standard convention and the only reading that lets part (c) be answered at all.

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, discounted at $i=10\,\%$ per year.

	Project A	Project B
Initial investment (year 0)	$60,000	$70,000
Yearly operating cost	$2,500	$1,000
Major maintenance	$12,000 every 3 years	$17,000 every 4 years
Yearly revenue	$12,000	$16,000
Life	9 years	12 years
Discount rate	10 % per year

Find. The present-value profit of each alternative and hence the more economical plan, on a basis that is fair to two projects of different lives.

Figure 4.1 — Project A, one 9-year cycle. Net annual cash flow of $9,500 ($12,000 revenue less $2,500 operating cost) in years 1–9, against a $60,000 investment at year 0 and $12,000 major maintenance in years 3 and 6.

Figure 4.2 — Project B, one 12-year cycle. Net annual cash flow of $15,000 ($16,000 revenue less $1,000 operating cost) in years 1–12, against a $70,000 investment at year 0 and $17,000 major maintenance in years 4 and 8.

Approach. Reduce each project to a net annual cash flow, discount that stream and the discrete maintenance outlays back to year 0 with the uniform-series and single-payment present-worth factors, subtract the initial investment, and then place the two answers on a common footing — either by repeating each cycle over the 36-year least common multiple of the two lives, or equivalently by converting each present worth to an annual worth.

Reduce each project to its net annual flow. Revenue and operating cost both recur every year, so they combine into one uniform series: $$ A_A=12{,}000-2{,}500=\$9{,}500\ \text{per year},\qquad A_B=16{,}000-1{,}000=\$15{,}000\ \text{per year} $$ The major maintenance is not part of this series; it falls in specific years only and must be discounted individually.
Write down the factors needed at 10 %. The uniform-series and single-payment present-worth factors are $$ (P/A,i,n)=\frac{1-(1+i)^{-n}}{i},\qquad (P/F,i,n)=(1+i)^{-n} $$ which give $(P/A,10\%,9)=5.7590$, $(P/A,10\%,12)=6.8137$, $(P/F,10\%,3)=0.7513$, $(P/F,10\%,6)=0.5645$, $(P/F,10\%,4)=0.6830$, $(P/F,10\%,8)=0.4665$ and $(P/F,10\%,9)=0.4241$.
Place the maintenance events. Project A's major maintenance falls "every 3 years" over a 9-year life, so it occurs at the end of years 3 and 6; the year-9 occurrence coincides with the end of the project and the replacement of the asset, so it is not charged. By the same logic Project B's maintenance at 4-year intervals over a 12-year life falls at years 4 and 8, not year 12. This is the standard treatment and it is applied consistently to both projects, so it cannot bias the comparison.
Present worth of Project A over its own 9-year life. Discounting the net series and the two maintenance outlays and subtracting the investment: $$ PW_A=-60{,}000+9{,}500\,(P/A,10\%,9)-12{,}000\left[(P/F,10\%,3)+(P/F,10\%,6)\right] $$ $$ PW_A=-60{,}000+9{,}500(5.759024)-12{,}000(0.751315+0.564474) $$ $$ PW_A=-60{,}000+54{,}710.73-15{,}789.46 $$ $$ \boxed{PW_A = -\$21{,}078.74} $$ Project A does not recover its investment at a 10 % required return: the present value of everything it earns is some $21,079 short of what it costs.
Present worth of Project B over its own 12-year life. Repeating the calculation with B's data: $$ PW_B=-70{,}000+15{,}000\,(P/A,10\%,12)-17{,}000\left[(P/F,10\%,4)+(P/F,10\%,8)\right] $$ $$ PW_B=-70{,}000+15{,}000(6.813692)-17{,}000(0.683013+0.466507) $$ $$ PW_B=-70{,}000+102{,}205.38-19{,}541.85 $$ $$ \boxed{PW_B = +\$12{,}663.53} $$ B clears the 10 % hurdle with roughly $12,664 of present-value profit to spare.
Put the two on a common study period. A present worth computed over 9 years cannot be compared directly with one computed over 12 — the longer project has more time in which to earn. The standard remedy is the repeatability assumption: analyse both over the least common multiple of the lives, $\operatorname{lcm}(9,12)=36$ years, so that A is repeated four times and B three times. Each repetition is the same cycle displaced in time, so the present worth of the repeated stream is the single-cycle present worth multiplied by a geometric series of shift factors: $$ PW_{A,36}=PW_A\left[1+(P/F,10\%,9)+(P/F,10\%,18)+(P/F,10\%,27)\right]=PW_A(1.680234) $$ $$ PW_{B,36}=PW_B\left[1+(P/F,10\%,12)+(P/F,10\%,24)\right]=PW_B(1.420156) $$ Substituting the single-cycle results: $$ PW_{A,36}=-21{,}078.74\times 1.680234=-\$35{,}417.22 $$ $$ \boxed{PW_{B,36}=12{,}663.53\times 1.420156=+\$17{,}984.18} $$
Cross-check with annual worth. Annual worth is immune to the unequal-life problem because it already expresses each project as a per-year figure, so it is the natural independent check: $$ AW_A=\frac{PW_A}{(P/A,10\%,9)}=\frac{-21{,}078.74}{5.759024}=-\$3{,}660.12\ \text{per year} $$ $$ AW_B=\frac{PW_B}{(P/A,10\%,12)}=\frac{12{,}663.53}{6.813692}=+\$1{,}858.54\ \text{per year} $$ Dividing the 36-year present worths by $(P/A,10\%,36)=9.676508$ reproduces exactly these two annual worths, which confirms the least-common-multiple arithmetic.
State the decision. Project B is superior on every measure computed: it is the only one with a positive present-value profit over its own life, it is the only one with a positive annual worth, and over the common 36-year study period it is ahead of A by $$ PW_{B,36}-PW_{A,36}=17{,}984.18-(-35{,}417.22)=\$53{,}401.40 $$ $$ \boxed{\text{Select Project B; Project A is not economically justified at }i=10\,\%} $$ The result is not close, and the reason is visible in the raw data: B earns $15,000 net per year against A's $9,500 for only $10,000 more capital, and it pays its larger maintenance charge less often.

Final Results — Question 4
Quantity	Project A	Project B
Net annual cash flow	$9,500/yr	$15,000/yr
PW of net annual flows	$54,710.73	$102,205.38
PW of major maintenance	−$15,789.46	−$19,541.85
Initial investment	−$60,000	−$70,000
Present-value profit over own life	−$21,078.74 (9 yr)	+$12,663.53 (12 yr)
Present-value profit over 36-year LCM	−$35,417.22	+$17,984.18
Annual worth	−$3,660.12/yr	+$1,858.54/yr
Decision: adopt Project B. It is ahead of Project A by $53,401.40 in present worth over the common 36-year study period, and it is the only alternative that earns more than the 10 % required return.

Check — two modelling assumptions, both stated on the exam paper's own terms. First, "major maintenance every 3 years" over a 9-year life is charged at years 3 and 6 only, on the reasoning that the year-9 event coincides with the end of the study period and the disposal of the asset; the same rule puts B's charges at years 4 and 8. Charging the terminal-year event as well would give $PW_A=-26{,}167.91$ and $PW_B=+7{,}246.80$ dollars, and Project B still wins by a wide margin, so the decision is insensitive to this choice. Second, no salvage value is stated for either project, so both are taken as zero.

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

Key relations. $(P/A,i,n)=\dfrac{1-(1+i)^{-n}}{i}$; $(P/F,i,n)=(1+i)^{-n}$; $PW=-I_0+A\,(P/A,i,n)-\sum M_k\,(P/F,i,k)$; for repeated cycles $PW_{\text{LCM}}=PW_{\text{cycle}}\sum_{m=0}^{r-1}(P/F,i,mn)$; and $AW=PW/(P/A,i,n)$, which is invariant to the number of repetitions.

Why this works. Present worth converts every future dollar to a common date so that streams of different shapes can be added, but it is only a fair basis for choosing between alternatives if they occupy the same span of time. Two devices restore that fairness. The least-common-multiple method assumes each alternative is repeated identically until both end together, which is defensible whenever the service will genuinely be needed indefinitely and the cost structure is expected to repeat. Annual worth achieves the same thing more cheaply: because the annual worth of a repeated cycle equals the annual worth of one cycle, the repetition never has to be modelled at all. The two methods must agree, and checking that they do — as done in step 7 — is the cheapest available audit of the whole calculation. Recurring but non-annual charges such as the major maintenance are handled with single-payment factors rather than folded into the uniform series, because a $(P/A)$ factor assumes a payment in every period.

Common pitfall. The dominant error is comparing $PW_A$ over 9 years with $PW_B$ over 12 years and declaring a winner, which silently rewards the longer-lived project; here the conclusion happens to survive, but on closer alternatives it would not. The second is averaging the major maintenance into an annual figure — $12,000 ÷ 3 = $4,000 per year — and discounting it with $(P/A)$; that misplaces the money earlier than it is actually spent and understates the project's worth. The third is forgetting to net the operating cost against revenue before applying $(P/A)$, and the fourth is being inconsistent about the terminal maintenance event between the two projects, which is the one choice here that could genuinely tilt a close comparison.

Source. Sullivan, Wicks & Koelling, Engineering Economy (17th ed.), Chapter 5 (present-worth and annual-worth methods) and Chapter 6 (comparing alternatives, the repeatability assumption and the coterminated assumption); Hendrickson & Au, Project Management for Construction, Chapter 6 (economic evaluation of facility investments).

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