NivaarExam PrepOfficial exam papers ↗

11-CS-1 Engineering Economics · May 2017

Question 3 of 5: Stamping Presses — Different Lives

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

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 3: Stamping Presses — Different Lives (25 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.

(a) Necessary Assumption

Repeatability—each press replaced identically at end of life (or compared over the LCM period, 100 years). Annual Worth builds this in.

(b) Annual Worth (i = 9%)

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 $=6{,}000+800(6.76745)=6{,}000+5{,}414=11{,}414$.

$$EAC_X = 125{,}592 + 18{,}000 + 12{,}000 + 11{,}414 = \boxed{\$167{,}006/\text{yr}}$$

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 $=4{,}000+600(7.83160)=4{,}000+4{,}699=8{,}699$.

$$EAC_Y = 138{,}987 + 14{,}000 + 8{,}000 + 8{,}699 = \boxed{\$169{,}686/\text{yr}}$$

Since $EAC_X < EAC_Y$, select Press X.

(c) Present Worth

Over the common (LCM = 100-year) period, $PW = EAC\times(P/A,9\%,100)$ for each; the same factor applies, so Press X is again preferred.

(d) Do PW and AW Always Agree?

Yes, with the same MARR and a consistent study period.

(e) Press Y Salvage for a 20-Year Study

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 $EAC_Y(20)=EAC_X=167{,}006$:

$$183{,}425 - 0.0195465\,S_Y = 167{,}006 \;\Rightarrow\; S_Y \approx \boxed{\$840{,}000}$$

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.