Engineering Economics — break-even initial investment with unequal lives: 16-Civ-B8 Management of Construction, December 2014, Q4

Question 4: Engineering Economics — break-even initial investment 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 appraised at a discount rate of 10 per cent per year. Project A has an unknown initial investment P, and Project B a known initial investment of $70,000. Operating costs and revenues are end-of-year amounts; the major-maintenance charge falls at the end of every third year of service, which within each project's own life means a single disbursement at the end of year 3.

Quantity	Project A	Project B
Initial investment at year 0	P (unknown)	$70,000
Yearly operating cost	$1,500	$1,000
Yearly revenue	$15,000	$30,000
Net yearly cash flow (revenue less operating cost)	$13,500	$29,000
Major maintenance, end of year 3 of each cycle	$5,000	$3,000
Service life	4 years	3 years
Discount rate	10 per cent per year, no salvage value stated for either project

Find. The initial investment P for Project A at which the two alternatives are economically indifferent — that is, at which their present worths, compared over a common study period, are equal.

Figure 4.1 — Cash-flow diagrams over the 12-year least-common-multiple study period. The lives are 4 and 3 years, so Project A is repeated three times and Project B four times; downward arrows are disbursements (initial investment at each replacement, and major maintenance at year 3 of each cycle), upward arrows the net annual cash flow of revenue less operating cost. Comparing over a common horizon is what makes a present-worth comparison of unequal-life alternatives legitimate.

Approach. Because the lives differ, present worths taken over each project's own life are not comparable; adopt the repeated-lives assumption and compare over the least common multiple of 4 and 3 years, namely 12 years, then equate the two 12-year present worths and solve for P. The annual-worth comparison over each project's own life is carried out afterwards as an independent check, since under repeatability the two methods must agree exactly.

Reduce each project to its net annual cash flow. Revenue and operating cost are both uniform annual series, so combine them first: $$A_A=15{,}000-1{,}500=\$13{,}500\ \text{per year},\qquad A_B=30{,}000-1{,}000=\$29{,}000\ \text{per year}$$ The major maintenance is not part of this series — it is a single future amount at year 3 of each cycle and must be discounted separately.
Assemble the discount factors at 10 per cent. The factors needed are $$(P/A,10\%,4)=3.169865,\quad (P/A,10\%,3)=2.486852,\quad (P/F,10\%,3)=0.751315$$ $$(A/P,10\%,4)=0.315471,\quad (A/P,10\%,3)=0.402115$$ each computed as $(P/A,i,n)=\left[(1+i)^n-1\right]/\left[i(1+i)^n\right]$ and $(P/F,i,n)=(1+i)^{-n}$.
Present worth of Project B over one 3-year cycle. Applying the factors to the known cash flows, $$PW_B=-70{,}000+29{,}000\,(P/A,10\%,3)-3{,}000\,(P/F,10\%,3)$$ $$PW_B=-70{,}000+72{,}118.71-2{,}253.94=\boxed{-\$135.23}$$ Project B is therefore very slightly unattractive on its own account at a 10 per cent hurdle rate — it very nearly, but not quite, earns the required return. That near-balance is not an accident of the arithmetic; it is what makes the question well posed, because the break-even investment for A must land close to the present worth of A's own benefits.
Present worth of Project A over one 4-year cycle, with P carried symbolically. The benefit side is $$13{,}500\,(P/A,10\%,4)-5{,}000\,(P/F,10\%,3)=42{,}793.18-3{,}756.57=\$39{,}036.61$$ so that $$PW_A=-P+39{,}036.61$$
Repeat each project over the 12-year study period. Least common multiple of 4 and 3 is 12, so A is renewed at years 4 and 8 and B at years 3, 6 and 9. Each renewal reproduces the same cycle cash flow, discounted back to time zero, giving repetition factors $$R_A=1+(P/F,10\%,4)+(P/F,10\%,8)=1+0.683013+0.466507=2.149521$$ $$R_B=1+(P/F,10\%,3)+(P/F,10\%,6)+(P/F,10\%,9)=1+0.751315+0.564474+0.424098=2.739886$$ so the 12-year present worths are $PW_A^{12}=R_A\,PW_A$ and $PW_B^{12}=R_B\,PW_B$.
Evaluate Project B over the study period. Substituting the cycle value from step 3, $$PW_B^{12}=2.739886\times(-135.23)=-\$370.51$$
Equate and solve for the break-even investment. Setting $PW_A^{12}=PW_B^{12}$, $$2.149521\,(39{,}036.61-P)=-370.51$$ $$39{,}036.61-P=\frac{-370.51}{2.149521}=-172.37$$ $$P=39{,}036.61+172.37=\boxed{P=\$39{,}209}$$ so Project A is exactly as attractive as Project B when its initial investment is about $39,200. Below that figure A is preferred, above it B is preferred.
Independent check by annual worth. Under the repeated-lives assumption an annual-worth comparison over each project's own life is equivalent to a present-worth comparison over the least common multiple, so the two must agree. For B, $$AW_B=PW_B\,(A/P,10\%,3)=-135.23\times0.402115=-\$54.38\ \text{per year}$$ and setting $AW_A=(39{,}036.61-P)\,(A/P,10\%,4)$ equal to it gives $$39{,}036.61-P=\frac{-54.38}{0.315471}=-172.38\quad\Rightarrow\quad P=\$39{,}209$$ The two routes agree to the cent, which confirms both the factor values and the treatment of the repeated cycles.

Check: treatment of Project B's major maintenance. The table gives major maintenance "every 3 years" while Project B's life is exactly 3 years, so the charge falls precisely at the end of B's service life. The solution above takes the table at its word and charges it, which is the reading consistent with the repeated-lives model in which the maintenance recurs at years 3, 6, 9 and 12. If instead the charge is assumed not to be incurred in the year the asset is retired, Project B's cycle present worth becomes $-70{,}000+72{,}118.71=+\$2{,}118.71$, its annual worth becomes $851.96 per year, and the break-even initial investment for Project A falls to $36,336. An exam answer should state which reading it has adopted; both are defensible, and the assumption is exactly the kind the paper's own instruction 1 invites the candidate to record.

Break-even analysis at 10 per cent per year, 12-year study period
Quantity	Value
Net annual cash flow, Project A	$13,500 per year
Net annual cash flow, Project B	$29,000 per year
Present worth of Project B, one 3-year cycle	−$135.23
Annual worth of Project B	−$54.38 per year
Present worth of Project A's benefits, one 4-year cycle	$39,036.61
Repetition factors over 12 years (A, B)	2.149521, 2.739886
Present worth of Project B over 12 years	−$370.51
Break-even initial investment for Project A	$39,209, say $39,200
Decision rule	Choose A if its initial investment is below $39,209; choose B if above
Sensitivity (excluding B's year-3 maintenance)	$36,336

Topic. Present-worth comparison of mutually exclusive alternatives with unequal service lives, and solving for a break-even parameter.

Key relations. $(P/A,i,n)=\dfrac{(1+i)^n-1}{i(1+i)^n}$, $(P/F,i,n)=(1+i)^{-n}$, $(A/P,i,n)=\dfrac{i(1+i)^n}{(1+i)^n-1}$. Over a study period equal to the least common multiple $L$ of the lives, a project of life $n$ repeated $L/n$ times has $PW^{L}=PW^{n}\sum_{k=0}^{L/n-1}(1+i)^{-kn}$. Break-even is the parameter value at which $PW_A^{L}=PW_B^{L}$, equivalently $AW_A=AW_B$.

Why this works. Present worth measures the value added by a project over the horizon it occupies, so comparing a four-year present worth with a three-year one compares different amounts of service and is meaningless. The least-common-multiple device removes the mismatch by giving both alternatives the same twelve years of service, which is legitimate provided the repeated-lives assumption holds — that each project can be renewed on identical terms until the horizon is reached. Annual worth achieves the same thing more economically by expressing each project as an equivalent uniform annual amount, which is automatically comparable regardless of life; that is precisely why the two methods must give the same answer, and why running both is a genuine check rather than a duplicate calculation. Once the horizon is settled, solving for the break-even investment is straightforward because the unknown enters linearly: P appears once, at time zero of each cycle, so the equation never needs iteration.

Common pitfall. By far the most common error is equating the present worths taken over each project's own life — four years against three — which silently credits Project B with a year of service it does not deliver and gives a materially wrong break-even. The second is mishandling the repeated initial investment: candidates often discount only the first outlay and forget that A must be bought again at years 4 and 8 and B at years 3, 6 and 9. A third, subtler slip is netting revenue against operating cost but then also discounting the maintenance as if it were part of that uniform series; it is a single future amount and needs a $(P/F)$ factor of its own.

Source. Fraser, Jewkes et al., Global Engineering Economics: Financial Decision Making for Engineers, Canadian ed., Ch. 4 (present worth) and Ch. 6 (comparison of alternatives with unequal lives, repeated-lives and study-period assumptions); Hendrickson & Au, Project Management for Construction, Ch. 6 (economic evaluation of facility investments).

Question 4 of 6: Engineering Economics — break-even initial investment with unequal lives

Question 4: Engineering Economics — break-even initial investment with unequal lives (20 marks)