Engineering Economics — maximum justifiable investment in a new surface: 16-Civ-B8 Management of Construction, December 2018, Q4

Question 4: Engineering Economics — maximum justifiable investment in a new surface (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. A maintenance stream that is reduced for ten years by resurfacing and then returns to its original level.

Given data — maintenance costs with and without the new surface
Period	Maintenance without resurfacing	Maintenance with the new surface	Annual saving
Years 1–5	$7,000 / yr	$2,500 / yr	$4,500 / yr
Years 6–10	$7,000 / yr	$4,000 / yr	$3,000 / yr
After year 10	$7,000 / yr	$7,000 / yr	nil
Interest rate	10 % per year, end-of-year cash flows

Find. The maximum first cost that the resurfacing can justify — the present worth of the maintenance savings it creates.

Figure 4.1 — cash-flow diagram of the maintenance savings. The upward arrows are the annual savings the new surface produces; the downward arrow at time zero is the investment those savings can justify. Nothing is saved after year 10, so the two maintenance streams are identical from year 11 onward and cancel exactly.

Approach. Form the incremental saving stream, discount it as two uniform series — the first starting immediately, the second deferred five years — and set the maximum investment equal to their combined present worth at 10 %.

Establish the saving stream. Only the difference between the two maintenance policies matters, since every other cash flow is common to both: $$A_1 = 7{,}000 - 2{,}500 = \$4{,}500\ \text{per year (years 1--5)}$$ $$A_2 = 7{,}000 - 4{,}000 = \$3{,}000\ \text{per year (years 6--10)}$$ After year 10 the two policies cost the same $7,000 a year, so the saving is zero and the analysis horizon closes at year 10. This is why no perpetuity term is needed — a point worth stating explicitly, because the question deliberately mentions what happens after year 10 to see whether the candidate recognises that it cancels.
Discount the first five years. The uniform series present-worth factor at 10 % for five years is $$\left(\frac{P}{A},10\%,5\right)=\frac{1-(1.10)^{-5}}{0.10}=3.790787$$ so the first block is worth $$P_1 = 4{,}500 \times 3.790787 = \$17{,}058.54$$
Discount the deferred second block. The second series also runs five years, but it begins at the end of year 6, so its own present-worth factor places it at the end of year 5 and it must then be brought back a further five years: $$P_2 = A_2\left(\frac{P}{A},10\%,5\right)\left(\frac{P}{F},10\%,5\right) = 3{,}000 \times 3.790787 \times 0.620921$$ $$P_2 = \$7{,}061.34$$
Sum for the maximum justifiable investment. Adding the two blocks, $$P_{\max} = 17{,}058.54 + 7{,}061.34 = \boxed{\$24{,}119.88}$$ At any first cost below this the resurfacing earns more than 10 %; at exactly this figure it earns precisely 10 % and the owner is indifferent; above it the work cannot be justified on maintenance savings alone.
Confirm the answer two independent ways. Writing the deferred block as a difference of factors, $$P_{\max} = 4{,}500\left(\frac{P}{A},10\%,5\right) + 3{,}000\left[\left(\frac{P}{A},10\%,10\right)-\left(\frac{P}{A},10\%,5\right)\right]$$ with $(P/A,10\%,10)=6.144567$ returns $24,119.88, and discounting the ten savings year by year at $(1.10)^{-n}$ returns the same figure. As a sense check, the equivalent uniform annual saving over the ten-year horizon is 24,119.88 / 6.144567 = $3,925.40 per year, which sits sensibly between the $4,500 and $3,000 blocks and closer to the earlier one, as discounting requires.

Check: end-of-year timing and the closed horizon. All maintenance is taken as an end-of-year cash flow, the standard convention for the factor tables, and the resurfacing cost is taken at time zero. The analysis horizon closes at year 10 because the saving is exactly zero thereafter; if instead the new surface were assumed to leave any residual benefit beyond year 10 — a longer life, a salvage value, deferred reconstruction — that benefit would have to be added, and the justifiable investment would rise. Nothing in the question supports such an addition, so none is made.

Final results — Question 4
Quantity	Result
Annual saving, years 1–5	$4,500 / yr
Annual saving, years 6–10	$3,000 / yr
Present worth of years 1–5	$17,058.54
Present worth of years 6–10	$7,061.34
Maximum justifiable investment	$24,119.88 (say $24,120)
Equivalent uniform annual saving over 10 years	$3,925.40 / yr

Topic. Present-worth analysis of an incremental cost saving, with a deferred uniform series, used to set a maximum justifiable first cost.

Key relations. $\left(\frac{P}{A},i,n\right)=\dfrac{1-(1+i)^{-n}}{i}$; $\left(\frac{P}{F},i,n\right)=(1+i)^{-n}$; a series starting in year $k+1$ is discounted by $\left(\frac{P}{A},i,n\right)\left(\frac{P}{F},i,k\right)$; the maximum justifiable investment is the present worth of the incremental benefits at the owner's minimum acceptable rate of return.

Why this works. The maximum justifiable investment is simply the break-even present worth: the amount at which the net present worth of the project is exactly zero and the internal rate of return is exactly the stated 10 %. Casting the problem as an incremental analysis is what makes it tractable, because every cash flow common to both alternatives — including the whole of the maintenance stream from year 11 to eternity — cancels before any discounting is done. The deferred series is handled with two factors rather than one because the $(P/A)$ factor always places its value one period before the first payment: applied to a series that starts in year 6 it produces a value at the end of year 5, which the $(P/F)$ factor then carries back to time zero.

Common pitfall. Using $(P/F,i,6)$ instead of $(P/F,i,5)$ on the deferred block, which shifts the whole second series a year too far and understates the answer by about $640. A second is treating the return to $7,000 after year 10 as a cost of the new surface and subtracting a perpetuity for it; both policies incur that same cost, so it cancels and must not appear. A third is answering with the undiscounted total saving, 5(4,500) + 5(3,000) = $37,500, which ignores the time value of money entirely and overstates the justifiable investment by more than half.

Source. Fraser et al., Global Engineering Economics: Financial Decision Making for Engineers, 5th Canadian ed., Ch. 3–4 (cash flow analysis, present worth, deferred annuities) and Ch. 5 (comparison methods); Hendrickson, Project Management for Construction, 2nd ed., Ch. 6.

Question 4 of 6: Engineering Economics — maximum justifiable investment in a new surface

Question 4: Engineering Economics — maximum justifiable investment in a new surface (20 marks)