Engineering Economics — Break-even Revenue for Alternatives with Unequal Lives: 16-Civ-B8 Management of Construction, May 2013, Q4

Question 4: Engineering Economics — Break-even Revenue for 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 different service lives, discounted at 10 % per year, with the unknown being Project A’s annual revenue.

Cash flows as tabulated in the question
Item	Project A	Project B
Initial investment	$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
Life	4 years	3 years
Discount rate	10 % per year

Find. The annual revenue R that makes Project A exactly as attractive as Project B on a like-for-like economic basis.

Figure 4.1 — cash-flow diagrams for the two projects. Downward arrows are disbursements and upward arrows receipts; the year-3 disbursement on each timeline combines the annual operating cost with the major maintenance charge.

Approach. Because the lives differ (four years against three), compare on equivalent uniform annual worth under the repeatability assumption, evaluate Project B’s annual worth numerically, express Project A’s in terms of R, and set the two equal.

Choose a comparison basis that is legitimate for unequal lives. Present worth may not be compared across a four-year and a three-year project directly, because the shorter alternative renders no service in year four. The two admissible remedies are a study period equal to the least common multiple of the lives (12 years) or a comparison of equivalent uniform annual worths. They give identical rankings whenever each alternative is assumed to repeat identically, and the annual-worth route is far shorter, so it is used here and verified against the 12-year present worth at the end.
Assemble the capital-recovery factors. Working at $i = 0.10$, $$\begin{aligned}(A/P,i,n) &= \frac{i(1+i)^n}{(1+i)^n-1}\\ (A/F,i,n) &= \frac{i}{(1+i)^n-1}\\ (P/F,i,n) &= (1+i)^{-n}\end{aligned}$$ which evaluate to $(A/P,10\%,3) = 0.402115$, $(A/F,10\%,3) = 0.302115$, $(A/P,10\%,4) = 0.315471$ and $(P/F,10\%,3) = 0.751315$.
Annualise Project B. Its major maintenance falls at the end of year three, which is also the end of its life and therefore the end of every repetition, so it is annualised with a sinking-fund factor rather than discounted and spread: $$AW_B = R_B - C_B - P_B\,(A/P,i,3) - F_B\,(A/F,i,3)$$ and substituting its own figures, $$AW_B = 20000 - 1000 - 80000\,(A/P,10\%,3) - 3000\,(A/F,10\%,3)$$ $$AW_B = 20000 - 1000 - 32169.18 - 906.34$$ so that $$\boxed{AW_B = -14075.53\ \text{per year}}$$ Project B does not pay for itself: at 10 % the revenue of $20,000 per year falls well short of recovering an $80,000 investment over only three years.
Annualise Project A in terms of the unknown revenue. Project A’s major maintenance falls at the end of year three of a four-year cycle, so it is discounted to time zero and then spread over four years: $$\begin{aligned} A_{\text{maint}} &= 5000\,(P/F,10\%,3)(A/P,10\%,4)\\ &= 5000(0.751315)(0.315471) = 1185.09\end{aligned}$$ and the full annual worth is $$\begin{aligned} AW_A &= R - 1500 - 90000\,(A/P,10\%,4) - 1185.09\\ &= R - 1500 - 28392.37 - 1185.09\end{aligned}$$ giving $$AW_A = R - 31077.46\ \text{per year}$$ The constant $31,077.46 is Project A’s equivalent uniform annual cost: capital recovery of $28,392.37, operating cost of $1,500 and annualised major maintenance of $1,185.09.
Impose the break-even condition and solve. The two projects are equally attractive when their annual worths coincide, $$AW_A = AW_B \;\Longrightarrow\; R - 31077.46 = -14075.53$$ so that $$\boxed{R = 17001.93 \approx 17000\ \text{per year}}$$ Rounding to the precision the question’s own data support, Project A becomes as attractive as Project B at an annual revenue of about $17,000.
Cross-check on the least-common-multiple study period. Repeating A three times and B four times over 12 years, and using $(P/A,10\%,12) = 6.813692$, both alternatives return the same present worth at the break-even revenue: $$PW = AW\,(P/A,10\%,12) = (-14075.53)(6.813692) = -95906.31$$ for each. The agreement confirms both the factor arithmetic and the equivalence of the two comparison methods.
Read the answer for what it is. At $R = 17002$ per year both alternatives lose about $14,076 per year in equivalent terms, so the break-even revenue is the point of indifference between two unattractive projects, not a threshold of profitability. Project A becomes genuinely worth doing only above $$R_{\min} = 31077.46\ \text{per year}$$ where its annual worth turns positive. Any revenue above $17,002 makes A the better of the two; below it, B. The sensitivity is one-for-one, since R enters the annual worth linearly.

Check: treatment of the recurring major maintenance. The question says “major maintenance every 3 years” without saying whether the charge falls inside the stated life. It is taken here to occur at the end of year 3 of each repetition of each project, which is the only reading consistent with the repeatability assumption that licenses an annual-worth comparison. If instead Project B’s year-3 charge were dropped because it coincides with the end of its life, $AW_B$ would improve to $-13169.18$ per year and the break-even revenue would rise to $17,908.28 per year — a difference of about 5 %, which does not change any ranking. State whichever assumption is used, as Note 1 of the paper invites.

Break-even comparison at 10 % per year
Quantity	Project A	Project B
Capital recovery of first cost	$28,392.37 / yr	$32,169.18 / yr
Annual operating cost	$1,500 / yr	$1,000 / yr
Annualised major maintenance	$1,185.09 / yr	$906.34 / yr
Equivalent uniform annual cost	$31,077.46 / yr	$34,075.53 / yr
Annual revenue	$17,001.93 / yr (answer)	$20,000 / yr
Equivalent uniform annual worth	−$14,075.53 / yr	−$14,075.53 / yr
Present worth over the 12-year common study period	−$95,906.31	−$95,906.31

Topic. Comparing mutually exclusive alternatives with unequal service lives by equivalent uniform annual worth, and solving for a break-even parameter.

Key relations. $$AW = -P\,(A/P,i,n) - C_{\text{annual}} - F\,(A/F,i,n) + R$$ $$\begin{aligned}(A/P,i,n) &= \frac{i(1+i)^n}{(1+i)^n-1}\\ (A/F,i,n) &= (A/P,i,n)-i\end{aligned}$$ with break-even found by setting the two annual worths equal.

Why this works. An annual worth is a present worth already divided by the annuity factor for its own life, so it silently normalises alternatives of different duration onto a common per-year basis. That normalisation is only legitimate under the repeatability assumption — that each alternative would be renewed identically at the same cost and the same interest rate for as long as the service is needed — because it is exactly that assumption which makes one cycle representative of all of them. The 12-year least-common-multiple check demonstrates the equivalence: annualising and then re-annuitising over four cycles reproduces the same present worth, as it must.

Common pitfall. Comparing raw present worths of a four-year and a three-year project, which systematically favours the shorter one because it stops spending sooner. A second pitfall is annualising a one-off charge with the wrong factor: a payment at the end of a cycle takes $(A/F,i,n)$ directly, whereas a payment part-way through takes $(P/F,i,k)$ first and then $(A/P,i,n)$ — using $(A/F)$ for A’s year-3 charge would understate its annual cost by about $108 per year.

Source. Fraser et al., Global Engineering Economics, Ch. 4–5 (annual worth, comparison methods, repeatability and study periods); Newnan et al., Engineering Economic Analysis, Ch. 6.

Question 4 of 6: Engineering Economics — Break-even Revenue for Alternatives with Unequal Lives

Question 4: Engineering Economics — Break-even Revenue for Alternatives with Unequal Lives (20 marks)