Question 3 of 6: Engineering Economics — Comparing Two Projects of Unequal Life

Question 3: Engineering Economics — Comparing Two Projects of Unequal Life (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 cash flows tabulated below, appraised at a discount rate of 10 per cent per year. All amounts are in Canadian dollars, revenues and operating costs are end-of-year annuities, and the major maintenance is a lump sum falling at each five-year anniversary within the life.

Find. The present-value profit of each project and, since the lives are unequal, the ranking of the two on a properly comparable basis, together with a recommendation.

Approach. Net the annual revenue against the annual operating cost to get a single annuity, discount that annuity and the maintenance lump sums back to year zero to obtain each project’s present-value profit over its own life, and then convert both to equivalent annual worth (and cross-check on the 30-year least common multiple study period) so that the unequal lives are removed from the comparison.

1. Reduce each project to a net annual cash flow. Revenue and operating cost are both uniform annuities over the life, so they combine into one:$$A_{\text{net}} = R - C_{\text{op}}$$For Project A, $A_{\text{net,A}} = 12{,}500 - 1{,}500 = 11{,}000$ per year; for Project B, $A_{\text{net,B}} = 16{,}000 - 1{,}000 = 15{,}000$ per year. Project B earns almost 36 per cent more each year on a smaller investment, which is the first sign of where this is heading.
2. Write down the discount factors needed at 10 per cent. The series present-worth and single-payment present-worth factors are$$\begin{aligned}(P/A,i,n)&=\frac{1-(1+i)^{-n}}{i}\\ (P/F,i,n)&=(1+i)^{-n}\end{aligned}$$which give $(P/A,10\%,15)=7.6061$, $(P/A,10\%,10)=6.1446$, $(P/F,10\%,5)=0.6209$ and $(P/F,10\%,10)=0.3855$.
3. Discount Project A over its own 15-year life. Combining the capital outlay, the net annuity and the two overhauls,$$\begin{aligned}PW_A &= -60{,}000 + 11{,}000\,(P/A,10\%,15)\\ &\qquad - 5{,}000\left[(P/F,10\%,5)+(P/F,10\%,10)\right]\end{aligned}$$Substituting the factors gives$$PW_A = -60{,}000 + 83{,}666.87 - 5{,}032.32 = \boxed{+\,18{,}634.55}$$in Canadian dollars of present-value profit.
4. Discount Project B over its own 10-year life. The same construction with one overhaul gives$$PW_B = -50{,}000 + 15{,}000\,(P/A,10\%,10) - 3{,}500\,(P/F,10\%,5)$$$$PW_B = -50{,}000 + 92{,}168.51 - 2{,}173.22 = \boxed{+\,39{,}995.28}$$Both figures are positive, so both projects clear the 10 per cent hurdle rate and either would be acceptable on its own.
5. Recognise that these two numbers cannot yet be compared. $PW_A$ covers 15 years of service and $PW_B$ only 10, so the larger number does not automatically identify the better project; the shorter project leaves five years of the planning horizon unaccounted for. The standard remedy is to assume repeatability — that each project can be renewed at the same cost and performance when it wears out — and then compare either equivalent annual worths or present worths over a common study period.
6. Convert both to equivalent annual worth. Multiplying each present worth by its own capital-recovery factor,$$\begin{aligned}AW &= PW\,(A/P,i,n)\\ (A/P,i,n)&=\frac{i}{1-(1+i)^{-n}}\end{aligned}$$with $(A/P,10\%,15)=0.13147$ and $(A/P,10\%,10)=0.16275$ gives$$AW_A = 18{,}634.55\times 0.13147 = \boxed{+\,2{,}449.95\ \text{per year}}$$$$AW_B = 39{,}995.28\times 0.16275 = \boxed{+\,6{,}509.05\ \text{per year}}$$On a like-for-like annual basis Project B is worth 4,059.10 dollars a year more than Project A.
7. Cross-check on the least common multiple study period. The lives 15 and 10 have a least common multiple of 30 years, over which Project A is built twice (at years 0 and 15) and Project B three times (at years 0, 10 and 20):$$\begin{aligned}PW_A^{30}&=PW_A\left[1+(P/F,10\%,15)\right]\\ &=18{,}634.55\times 1.23939=23{,}095.51\end{aligned}$$$$\begin{aligned}PW_B^{30}&=PW_B\left[1+(P/F,10\%,10)+(P/F,10\%,20)\right]\\ &=39{,}995.28\times 1.53419=61{,}360.24\end{aligned}$$Project B leads by 38,264.73 dollars of present worth over the common horizon, and dividing either figure by $(P/A,10\%,30)$ reproduces the annual worths of step 6 exactly, which confirms the two methods agree.
8. Test the sensitivity of the ranking to the maintenance assumption. Charging an overhaul in the terminal year as well — years 5, 10 and 15 for A and years 5 and 10 for B — reduces the present worths to 17,437.59 and 38,645.88 dollars and the annual worths to 2,292.59 and 6,289.44 dollars per year. The gap narrows only slightly and the ordering is untouched, so the recommendation is robust to the ambiguity in the question statement.

The recommendation is therefore Project B. It is worth noting why it wins, because that is where the marks for judgement sit: B has both the smaller capital cost and the larger net annual return, so it dominates A on every measure once the life difference is removed. The only circumstance in which A could be preferred is if the repeatability assumption fails — for instance if the service is needed for exactly 15 years and no replacement for B will be available at year 10 — in which case the comparison must be redone over the actual required horizon with an explicit estimate of B’s replacement cost and of A’s salvage value.