Stamping Presses — Different Lives: 23-CS-1 Engineering Economics, May 2015, Q3

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) Annual Worth (i = 9%)

Offer 1 (initial $1,050,000, 15 yr): $CR = 1{,}050{,}000(A/P,9\%,15)-210{,}000(A/F,9\%,15)=130{,}262-7{,}152=123{,}110$; maintenance $=8{,}000+1{,}000(A/G,9\%,15)=8{,}000+5{,}435=13{,}435$.

EAC_1 = 123{,}110 + 8{,}500 + 13{,}435 = \boxed{\$145{,}045/\text{yr}}

Offer 2 (initial $1,225,000, 20 yr): $CR = 1{,}225{,}000(A/P,9\%,20)-300{,}000(A/F,9\%,20)=134{,}194-5{,}864=128{,}330$; maintenance $=7{,}000+800(A/G,9\%,20)=7{,}000+5{,}414=12{,}414$.

EAC_2 = 128{,}330 + 7{,}500 + 12{,}414 = \boxed{\$148{,}244/\text{yr}}

Since $EAC_1 < EAC_2$, Offer 1 is better.

(b) Future Worth

The two lives (15 and 20 years) share a least common multiple of 60 years, so each offer is repeated (Offer 1 four times, Offer 2 three times) and its Future Worth taken at the end of year 60. Over that common period each alternative's cash flow is exactly its own uniform annual cost, so $FW = -EAC\,(F/A,9\%,60)$ with $(F/A,9\%,60)=\dfrac{1.09^{60}-1}{0.09}=\dfrac{176.031-1}{0.09}=1{,}944.79$:

FW_1 = -145{,}045(1{,}944.79) = \boxed{-\$282.1\text{ million}}

FW_2 = -148{,}244(1{,}944.79) = \boxed{-\$288.3\text{ million}}

Offer 1's future cost is the smaller of the two, by about $6.2 million in year-60 dollars, so Offer 1 is again better. The ranking is identical to part (a) because both measures are the same annual cost scaled by one positive factor.

(c) Do AW and FW Always Agree?

Yes, provided the same MARR and the same study period are used for both alternatives. Future Worth is Annual Worth multiplied by the single factor $(F/A,i,N)$, which is positive, and multiplying every alternative's figure by the same positive number cannot change their order. The two methods can appear to disagree only if the study periods are allowed to differ—for example, comparing Offer 1's Future Worth at year 15 with Offer 2's at year 20—which is not a valid comparison in the first place.

(d) Offer-2 Salvage for a 15-Year Study

Truncating Offer 2 to 15 years with unknown salvage $S_2$: $CR = 1{,}225{,}000(A/P,9\%,15)-S_2(A/F,9\%,15)=151{,}972-0.034059\,S_2$; operating 7,500; 15-yr maintenance $=7{,}000+800(A/G,9\%,15)=11{,}348$. Setting $EAC_2(15)=EAC_1=145{,}045$:

170{,}820 - 0.034059\,S_2 = 145{,}045 \;\Rightarrow\; S_2 \approx \boxed{\$757{,}000}

A salvage of about $757,000 would be required for Offer 2 over 15 years—about 62% of its $1,225,000 cost, a high but not impossible value. Below this, Offer 1 remains the better choice over a 15-year horizon.

(e) Necessary Assumption

The comparison of Annual Worths in part (a) rests on the repeatability assumption: each press is assumed to be replaced by an identical press at the end of its life, with the same costs, in the same economic environment, for as long as the service is needed. Under that assumption the comparison is effectively made over the least common multiple of the two lives (60 years here, used explicitly in part (b)), and each alternative's EAC over one life equals its EAC over every repeated cycle—which is why a 15-year and a 20-year alternative can be compared directly by Annual Worth. When repeatability is not credible (technology change, a fixed planning horizon), an explicit study period with an estimated truncated salvage value is used instead, as in part (d).

Question 3 of 5: Stamping Presses — Different Lives