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Cost data related to two alternative systems (A and B) is given below
Cost data related to two alternative systems (A and B) is given below. Interest rate is 10% per year: A B First cost, $ -25,000 - 150,000 Refurbishment cost -2,500 at the end of year 3, $ Annual cost, $ per year -7,000 -4,500 Salvage Value, s 3,000 50,000 Life, years 6 1. [ 4 pts. ] Which one of the value is closest to the annual worth for alternative A? $-12,425 $-14,780 $-14.425 $-12.780 2. [ 4 pts. ] Which one of the value is closest to the annual worth for alternative B? $-17,500 $-19,500 $-16,500 $-14,500 3. [ 4 pts. ] Which one of the value is closest to the capitalized cost of Machine B? $-175.000 $-181,250 $-206,250 $-195,000
Expert Solution
Answer:
1. -$12,780
2. -$14,500
3. -$195,000
Explanation:
1. Annual Worth of Alternative A
interest rate i = 10%
AW = -$25,000 * (A/P, 10%, 6) - $2,500 * (P/F, 10%, 3) * (A/P, 10%, 6) - $7,000 + $3,000 * (A/F, 10%, 6)
Compute (A/P, i%, n) using [i * (1 + i)n] / [(1 + i)n - 1]
(A/P, 10%, 6) = [0.1 * (1 + 0.1)6] / [(1 + 0.1)6 - 1] = [0.1 * (1.1)6] / [(1.1)6 - 1]
= [0.1 * 1.7716] / [1.7716 - 1] = 0.1772 / 0.7716 = 0.2296
Compute (P/F, i%, n) using (1 + i)(-n)
(P/F, 10%, 3) = (1 + 0.1)(-3) = (1.1)(-3) = 0.7513
Compute (A/F, i%, n) using i / [(1 + i)n - 1]
(A/P, 10%, 6) = 0.1 / [(1 + 0.1)6 - 1] = 0.1 / [(1.1)6 - 1]
= 0.1 / [1.7716 - 1] = 0.1 / 0.7716 = 0.1296
AW = (-$25,000 * 0.2296) - ($2,500 * 0.7513 * 0.2296) - $7,000 + ($3,000 * 0.1296)
= -$5,740.18 - $431.27 - $7,000 + $388.82 = -$12,782.63
Annual Worth of Alternative A is -$12,782.63
1. Annual Worth of Alternative B
interest rate i = 10%
Annual Worth of fist cost = - $150,000 * 0.1 = -$15,000
Annual cost = -$4,500
Annual Salvage value = $50,000 * 0.1 = $5,000
Annual Worth = -$15,000 - $4,500 + $5,000 = -$14,500
Annual Worth of Alternative B is -$14,500
3. Capitalized cost of Alternative B
Capitalized cost = first cost + (Annual cost / interest rate)
= -$150,000 - ($4,500 / 0.1) = -$150,000 - $45,000
= -$195,000
Capitalized cost of Alternative B is -$195,000
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