23-CS-1 Engineering Economics · May 2017
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
National Exams — May 2017 — 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 follow; 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.
Repeatability (identical replacement): each press is assumed to be replaced at the end of its service life by an identical press with the same costs and the same $300,000 salvage, so that the two alternatives can be compared over a common period — formally the least common multiple of the lives, $\text{LCM}(20,25)=100$ years (five cycles of X, four of Y). Annual Worth builds this assumption in automatically, because the equivalent annual cost of one cycle is the same in every cycle, which is why AW needs no explicit 100-year schedule. If repeatability is not credible (technology change, a fixed contract horizon), a stated study period must be used instead, with truncated (estimated) salvage values — which is exactly what part (e) does.
Assumption stated for parts (b)–(e): the “Annual Installment” is treated as a fixed end-of-year cost paid every year of the press's service life, in addition to the down payment.
Press X (20 yr), with $(A/P,9\%,20)=0.1095465$, $(A/F,9\%,20)=0.0195465$ and $(A/G,9\%,20)=6.76745$: $CR = 1{,}200{,}000(0.1095465)-300{,}000(0.0195465)=131{,}456-5{,}864=125{,}592$; maintenance $=6{,}000+800(6.76745)=6{,}000+5{,}414=11{,}414$.
Press Y (25 yr), with $(A/P,9\%,25)=0.1018063$, $(A/F,9\%,25)=0.0118063$ and $(A/G,9\%,25)=7.83160$: $CR = 1{,}400{,}000(0.1018063)-300{,}000(0.0118063)=142{,}529-3{,}542=138{,}987$; maintenance $=4{,}000+600(7.83160)=4{,}000+4{,}699=8{,}699$.
Since $EAC_X < EAC_Y$, select Press X.
Present worths must be compared over the common 100-year period. First find the present worth of one cycle of each press at its own start (costs positive):
Here $36{,}000 = 18{,}000+12{,}000+6{,}000$ and $26{,}000 = 14{,}000+8{,}000+4{,}000$ are the uniform annual costs (installment + running + base maintenance), multiplied by $(P/A,9\%,20)$ and $(P/A,9\%,25)$; the gradients use $(P/G,9\%,20)$ and $(P/G,9\%,25)$, and the salvage $(P/F,9\%,20)$ and $(P/F,9\%,25)$. Under repeatability X is bought at $t=0,20,40,60,80$ and Y at $t=0,25,50,75$, so:
Press X has the lower present cost, by $29,773, so Press X is again preferred. As a check, $PW = EAC\times(P/A,9\%,100)$ with $(P/A,9\%,100)=11.109102$ gives $167{,}005.8(11.109102)=1{,}855{,}284$ and $169{,}685.8(11.109102)=1{,}885{,}057$ from the unrounded equivalent annual costs — the same figures.
Yes, provided the same MARR and the same (consistent) study period are used for both alternatives. The two measures are related by $AW = PW\,(A/P,i,n)$, and $(A/P,i,n)$ is strictly positive, so multiplying every present worth by it cannot reorder the alternatives — parts (b) and (c) illustrate this. They can only appear to disagree when the analyst uses inconsistent horizons, for instance comparing single-cycle present worths over each press's own life ($1,524,520 for 20 years against $1,666,753 for 25 years), which is not a valid comparison because the two figures buy different lengths of service.
Truncating Y to 20 years with unknown salvage $S_Y$: $CR = 1{,}400{,}000(0.1095465)-S_Y(0.0195465)=153{,}365-0.0195465\,S_Y$; instalment 14,000; running 8,000; 20-yr maintenance $=4{,}000+600(6.76745)=8{,}060$. Setting $EAC_Y(20)=EAC_X=167{,}006$:
A salvage of about $840,000 would be required for Press Y over 20 years—almost exactly 60% of its $1,400,000 cost, an implausibly high resale value. So over a 20-year horizon Press X remains the better choice for any realistic Y salvage.
