e355t1q3

e355t1q3 - salvage value 10% of initial cost, maintenance...

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E355-Spring08-Test1-Sol 1.3 P I - S Salvage A Savings N Life (yrs) Note: All units are in '000 dollars. a) Draw the cash flow diagram for each alternative. [4 point] Alternative A; Alternative B; 1 3 6 9 10 1 8 b) Calculate the EUAC for both alternatives. [6 point] EUAC = A + P(A/P, i, N) - S(A/F, i, N) Alt. EUAC A = P (A/P,i,n) +A - S(A/F,i,n) = 25,000 $ X (A/P,12%,10) + -5000 + 3500 x (A/F,12%,3) - 2500 x (A/F,12% ,10) = 25000 x 0.177 + -5000 + 3500 x 0.2963 - 2500 x 0.057 = 319.55 $ EUAC B = 1,000 $ + 30,000 $ (A/P,12%,8) - ( 3,000 $ ) X (A/F,12%,8) EUAC B = 1,000 $ + 30,000 $ 0.2013 - ( 3,000 $ ) X 0.0813 EUAC B = 6795.10 K c) Which alternative should be chosen? Why? [2 point] Choose Alternative A due to the lower EUAC. 10 8 2,500 $ 3,000 $ 5,500 $ 3,500 $ 5,000 $ 4,500 $ A contractor is evaluating two different building insulation materials urethane foam and fiberglass. The initial cost of the foam will be $25,000 with
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Unformatted text preview: salvage value 10% of initial cost, maintenance costs is estimated at $3500 every three year and savings in utility bills are estimated at $5,000 per year. The fiberglass may be installed for $30,000 and has salvage value 10% of initial cost. There will be maintenance cost of $5500 and savings in energy costs are expected to be only $4500. The foam is expected to last 10 years, whereas the fiberglass will have to be replaced in 8 years. 25,000 $ 30,000 $ Alternative A Alternative B I = $ 25k S = $ 2500 O & M = $ 3500 savings= $ 5000 I = $ 30K S = 3000 O & M = 5500 savings =4500 E355-Spring08-Test1-Sol Page 1 of 1 3/3/2008...
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