Stamping Presses — Different Lives: 11-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

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

Question 3 of 5: Stamping Presses — Different Lives