11-CS-1 Engineering Economics · December 2013
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
National Exams — December 2013 — 11-CS-1 Engineering Economics. Open book; non-communicating calculator permitted. Any four of the five questions constitute a complete paper; all questions are of equal value. Fully worked solutions to all five questions are given below; standard compound-interest factors are used and minor rounding is immaterial.
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.
Each offer, seen from ABC's side, is a receipt now, five annual disbursements, and a receipt at year 5 — for example Offer 1 is $+85{,}000$ at $t=0$, $-50{,}000$ at $t=1\ldots4$, and $-50{,}000+230{,}000=+180{,}000$ at $t=5$. A standalone rate of return cannot be used to rank these: because the client pays ABC first, each offer's present worth stays positive at every discount rate (the minimum over all $i>0$ is about $\$25{,}100$, $\$20{,}000$ and $\$30{,}500$ for Offers 1, 2 and 3), so none of the three has a rate of return at all. The rate-of-return method must therefore be applied incrementally, which is the correct procedure for mutually exclusive alternatives in any case.
Order the offers by the size of the commitment ABC takes on, i.e. by the present worth of its construction disbursements at 12%, using $(P/A,12\%,5)=3.60478$ and $(P/F,12\%,5)=0.56743$: Offer 1 $=50{,}000(3.60478)=\$180{,}239$, Offer 2 $=\$234{,}310$, Offer 3 $=\$288{,}382$. Offer 1 is the base; the increments are then:
Each increment takes cash in first and pays it out later, so it is a financing (borrowing) increment, not an investment. Solving each for the rate that zeroes its present worth:
For a borrowing cash flow the decision rule is reversed: the increment is worth taking only if its rate is below the MARR, because $i^{*}$ is the rate ABC effectively pays for the extra cash received at the start. Here the extra $25,000 offered by Offer 2 costs 39.8% and the extra $75,000 offered by Offer 3 costs 15.6%, both dearer than the 12% MARR, so both increments are rejected and ABC should select Offer 1.
Present worth confirms the same ranking, as it must: $PW_1 = 85{,}000 - 50{,}000(3.60478) + 230{,}000(0.56743) = +\$35{,}269$, $PW_2 = +\$23{,}221$, $PW_3 = +\$30{,}498$. Offer 1 has the greatest net worth, and the two increments rejected above are exactly the two that reduce it.
Watch the sign of the increment. Applying the ordinary rule "accept the increment if $i^{*}>$ MARR" to $\Delta(2-1)$ would have selected Offer 2 on a 39.8% figure, even though Offer 2 is worth $\$12{,}048$ less than Offer 1. Always check whether an increment's first cash flow is an outlay (invest: accept when $i^{*}>$ MARR) or a receipt (borrow: accept when $i^{*}<$ MARR) before applying the test.
A rate-of-return method is attractive when decision-makers want the result as a single percentage that can be compared directly against the MARR or cost of capital and communicated easily to management and investors, and when the exact MARR is uncertain (the ROR shows the break-even rate the project can withstand). It is most useful for judging a single project's acceptability; for choosing among alternatives it must be applied incrementally.
No. Present Worth, Annual Worth, and a correctly applied incremental ROR always give the same decision at a given MARR; Annual Worth would also select Offer 1. (No calculation needed—the methods are equivalent.)
No. The alternative with the highest individual rate of return need not be the best: a smaller commitment can show a higher percentage yet add less total value. Mutually exclusive alternatives must be compared by incremental ROR—accepting each increment of investment whose own return exceeds the MARR—which is equivalent to maximizing present worth.
