Hydro-Forming Presses — Different Lives: 23-CS-1 Engineering Economics, December 2014, Q3

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.

Assumptions: the down payment is paid at $t=0$; the annual instalment, running cost and maintenance are end-of-year costs over each press's life; the $150,000 salvage is received at the end of that life.

(a) Necessary Assumption

Alternatives with different lives can only be compared over a common study period. The standard assumption is repeatability: each press is replaced at the end of its life by an identical press with the same costs, and the MARR stays the same, so each cash-flow cycle repeats until the two alternatives end together, at the least common multiple of the lives (LCM of 20 and 25 = 100 years). The other option is a study-period assumption: pick a finite horizon, such as the 20 years in part (e), and give any press still in service at that point an estimated salvage value. Annual Worth has repeatability built in, because one cycle's EAC is the EAC of every cycle.

(b) Annual Worth (i = 9%)

Press A (20 yr): $CR = 600{,}000(A/P,9\%,20)-150{,}000(A/F,9\%,20)=600{,}000(0.1095465)-150{,}000(0.0195465)=65{,}727.9-2{,}932.0=62{,}795.9$. Maintenance $=3{,}000+400(A/G,9\%,20)=3{,}000+400(6.7674)=5{,}707.0$.

EAC_A = 62{,}795.9 + 9{,}000 + 6{,}000 + 5{,}707.0 = 83{,}502.9 \approx \boxed{\$83{,}503/\text{yr}}

Press B (25 yr): $CR = 700{,}000(A/P,9\%,25)-150{,}000(A/F,9\%,25)=700{,}000(0.10180625)-150{,}000(0.01180625)=71{,}264.38-1{,}770.94=69{,}493.44$. Maintenance $=2{,}000+300(A/G,9\%,25)=2{,}000+300(7.8316)=4{,}349.48$.

EAC_B = 69{,}493.44 + 7{,}000 + 4{,}000 + 4{,}349.48 = 84{,}842.92 \approx \boxed{\$84{,}843/\text{yr}}

Since $EAC_A < EAC_B$ by about $1,340 a year, select Press A. Press B's lower operating costs and five extra years of life do not pay for its extra $100,000 down payment at a 9% MARR.

(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{,}503(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, as long as both use the same MARR and the same study period (for unequal lives, the same repeatability or study-period assumption). For any alternative, $PW = AW\,(P/A,i,N)$ over a common horizon $N$. That factor is positive and identical for every alternative, so it scales all of them equally and cannot change their order; part (c) shows this, with both PWs being the part (b) EACs multiplied by 11.1091. Apparent disagreements always come from inconsistent inputs. The usual culprit is comparing one life-cycle PW of a 20-year asset with one of a 25-year asset, which is not a common period.

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

Truncate Press B at 20 years with an unknown salvage $S_B$ at that point. Its instalment (7,000) and running cost (4,000) are unchanged, and its maintenance equivalent over 20 years is $2{,}000+300(A/G,9\%,20)=2{,}000+300(6.7674)=4{,}030.2$, so

EAC_B(20) = 700{,}000(0.1095465) - 0.0195465\,S_B + 7{,}000 + 4{,}000 + 4{,}030.2 = 91{,}712.8 - 0.0195465\,S_B

Setting $EAC_B(20) = EAC_A = 83{,}502.9$:

91{,}712.8 - 0.0195465\,S_B = 83{,}502.9 \;\Rightarrow\; S_B = \frac{8{,}209.9}{0.0195465} \approx \boxed{\$420{,}000}

Over a 20-year study period, Press B becomes the better choice only if it can be sold for about $420,000 or more after 20 years. That is a demanding requirement: it is 2.8 times the $150,000 salvage the paper expects at the end of either press's service life, and it would mean recovering 60% of B's $700,000 price after 20 of its 25 years. Unless there is firm evidence of an unusually strong resale market, Press A stays the recommendation under the study-period assumption as well.

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