Question 4 of 6: Engineering Economics

Question 4: Engineering Economics (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 unequal service lives, appraised at a discount rate of 10% per year. Major maintenance recurs on the stated cycle within each project's own life, so Project A incurs it in years 3 and 6 and Project B in years 2 and 4.

Find. The present-value profit of each project and, on a defensible basis given the unequal lives, the more economical plan.

Approach. Reduce each project to a net annual operating surplus, discount that surplus with the uniform-series present-worth factor, subtract the discounted major-maintenance payments and the initial investment to obtain present worth, then convert both present worths to annual worths so that projects of six and four years can be compared on a common basis, and cross-check on a twelve-year least-common-multiple study period.

1. Reduce each project to a net annual operating surplus. Revenue less the yearly operating cost gives a uniform annual amount that can be discounted with a single factor: $$\begin{aligned} A_A &= 12{,}500 - 1{,}500 = 11{,}000 \text{ per year} \\ A_B &= 10{,}000 - 1{,}000 = 9{,}000 \text{ per year} \end{aligned}$$ Major maintenance is deliberately kept out of this surplus, because it is not a uniform annual amount and must be discounted separately as individual future payments.
2. Assemble the factors at 10%. The uniform-series present-worth factor and the single-payment present-worth factor are $$\begin{aligned} \left(\frac{P}{A}, i, n\right) &= \frac{(1+i)^n - 1}{i\,(1+i)^n} \\ \left(\frac{P}{F}, i, n\right) &= \frac{1}{(1+i)^n} \end{aligned}$$ At 10% these give $(P/A, 10\%, 6) = 4.3553$ and $(P/A, 10\%, 4) = 3.1699$, with $(P/F, 10\%, 2) = 0.8264$, $(P/F, 10\%, 3) = 0.7513$, $(P/F, 10\%, 4) = 0.6830$ and $(P/F, 10\%, 6) = 0.5645$.
3. Compute the present worth of Project A. Discounting the six-year surplus and the two maintenance payments, then deducting the investment: $$PW_A = -30{,}000 + 11{,}000\,(4.3553) - 5{,}000\,(0.7513 + 0.5645)$$ The surplus is worth $47{,}907.87$ and the maintenance costs $6{,}578.94$ in present terms, so $$PW_A = -30{,}000 + 47{,}907.87 - 6{,}578.94 = \boxed{+11{,}328.92 \text{ dollars}}$$
4. Compute the present worth of Project B. The same construction over four years: $$PW_B = -25{,}000 + 9{,}000\,(3.1699) - 3{,}000\,(0.8264 + 0.6830)$$ Here the surplus is worth $28{,}528.79$ and the maintenance $4{,}528.38$, giving $$PW_B = -25{,}000 + 28{,}528.79 - 4{,}528.38 = \boxed{-999.59 \text{ dollars}}$$ Project B's present-value profit is negative. That is not an arithmetic error and the sign must not be flipped: over its own four-year life, B fails to recover its $25,000 investment at a 10% required return. Its undiscounted total is a positive $5,000, so the shortfall is entirely a discounting effect, and B's internal rate of return is 8.06%, below the 10% hurdle.
5. Put the unequal lives on a common basis with annual worth. Present worths computed over six and four years are not directly comparable, because the six-year project is being credited with two extra years of earning. Assuming each project is repeatable on the same terms, converting to annual worth with the capital-recovery factor removes the difference: $$\begin{aligned} AW &= PW \times \left(\frac{A}{P}, i, n\right) \\ \left(\frac{A}{P}, i, n\right) &= \frac{i\,(1+i)^n}{(1+i)^n - 1} \end{aligned}$$ With $(A/P, 10\%, 6) = 0.2296$ and $(A/P, 10\%, 4) = 0.3155$: $$AW_A = 11{,}328.92 \times 0.2296 = +2{,}601.20 \text{ per year}$$ $$AW_B = -999.59 \times 0.3155 = -315.34 \text{ per year}$$ Project A is better by $2,916.55 per year, and it is the only one of the two that earns more than its cost of capital.
6. Cross-check on the least-common-multiple study period. Six and four years share a least common multiple of twelve, so A is repeated twice and B three times over a common twelve-year horizon: $$\begin{aligned} PW_A^{12} &= 11{,}328.92\left(1 + 0.5645\right) = 17{,}723.81 \\ PW_B^{12} &= -999.59\left(1 + 0.6830 + 0.4665\right) = -2{,}148.64 \end{aligned}$$ The ranking is unchanged, and annuitising each of these over twelve years reproduces the annual worths of the previous step exactly — which is the arithmetic identity that justifies using annual worth as a shortcut for the repeatability assumption in the first place.
7. State the recommendation. On every basis examined — present worth over each project's own life, annual worth, present worth over the common twelve-year horizon, and internal rate of return — Project A is the more economical plan. Project A returns 22.83% against the 10% hurdle and produces a present-value profit of $11,328.92 over six years; Project B returns 8.06% and should be rejected outright rather than merely ranked second, because at the stated discount rate it destroys value.