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11-CS-1 Engineering Economics · December 2014

Question 3 of 5: Hydro-Forming Presses — Different Lives

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

Notes on this paper

National Exams — December 2014 — 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: Hydro-Forming 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 is replaced identically at the end of its life (or compared over the LCM period, 100 years). Annual Worth builds this in.

(b) Annual Worth (i = 9%)

Press A (20 yr): $CR = 600{,}000(A/P,9\%,20)-150{,}000(A/F,9\%,20)=65{,}726-2{,}931=62{,}794$. Maintenance $=3{,}000+400(A/G,9\%,20)=3{,}000+400(6.7676)=5{,}707$.

$$EAC_A = 62{,}794 + 9{,}000 + 6{,}000 + 5{,}707 = \boxed{\$83{,}501/\text{yr}}$$

Press B (25 yr): $CR = 700{,}000(A/P,9\%,25)-150{,}000(A/F,9\%,25)=71{,}264-1{,}771=69{,}493$. Maintenance $=2{,}000+300(A/G,9\%,25)=2{,}000+300(7.8316)=4{,}349$.

$$EAC_B = 69{,}493 + 7{,}000 + 4{,}000 + 4{,}349 = \boxed{\$84{,}843/\text{yr}}$$

Since $EAC_A < EAC_B$, select Press A.

(c) Present Worth

Present Worth needs a common study period, so take the least common multiple of the two lives, LCM$\,=\,$100 years (five cycles of Press A, four of Press B). Under the repeatability assumption of part (a) each press's cost stream repeats identically, so its annual cost over the whole 100 years is still the EAC found in part (b), and

$$PW = EAC\,(P/A,9\%,100),\qquad (P/A,9\%,100)=\frac{1-1.09^{-100}}{0.09}=11.1091$$
$$PW_A = 83{,}501(11.1091) = \boxed{\$927{,}600}\ \text{(cost)};\qquad PW_B = 84{,}843(11.1091) = \boxed{\$942{,}500}\ \text{(cost)}$$

$PW_A < PW_B$, so Press A is again preferred — the same decision as Annual Worth, by a present-worth margin of about $14,900 over the 100-year period. Note that the single factor $(P/A,9\%,100)$ multiplies both EACs, so it cannot reverse the ranking; that is the algebraic reason for the answer to part (d).

(d) Do PW and AW Always Agree?

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

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

Truncating B to 20 years with unknown salvage $S_B$: $CR = 700{,}000(A/P,9\%,20)-S_B(A/F,9\%,20)=76{,}680-0.019543\,S_B$; instalment 7,000; running 4,000; 20-yr maintenance $=2{,}000+300(A/G,9\%,20)=4{,}030$. Setting $EAC_B(20)=EAC_A=83{,}501$:

$$91{,}710 - 0.019543\,S_B = 83{,}501 \;\Rightarrow\; S_B \approx \boxed{\$420{,}000}$$

Over a 20-year study period, a salvage value of about $420,000 or more for Press B would make it the better choice (this is below its $700,000 cost, so it is a feasible requirement).