Question

Question 1 A. Using Ms Excel, find out which alternative should be selected on the basis of the Present Worth method, if the rate of interest is 8% per year. • Alternative 1: Initial purchase price = $2500000, Annual operating cost $45000 at the end of 1st year and increasing by $3000 in the subsequent years till the end of useful life, Annual income = $120000, Salvage value = $550000, Useful life = 3 years. Alternative 2: Initial purchase price = $3000000, Annual operating cost = $30000, Annual income $150000 for first three years and increasing by $5000 in the subsequent years till the end of useful life, Salvage value = $800000, Useful life = 6 years. Alternative 3: Initial purchase price = $2700000, Annual operating cost $35000 for first 5 years and increasing by $2000 in the successive years till the end of useful life, Annual income = $140000, Expected salvage value = $650000, Useful life = 9 years. B. Using Ms Excel, Compare the previuos alternatives in Question 1 on the basis of the future worth analysis, would the the most economical one at the interest rate of 8 % per year be different to the answer in Q1-A?. C. Using Ms Excel, find the the best Equivalent Annuity Cash Flow of the most economical one at the interest rate of 8 % per year. D. Draw by your hand Cash flow diagram. E. Write the standard notation of each solution above.

Answer 1

present worth of alternative 1, PWL PW, = -2, 500,000+120,000(2, 8%,3)– 45,000(%, 8%, 3)-30002,8%,3). 550,000 ,8%,3) PW1 = -2,500,000 + 120,000 * 2.577 – 45,000 * 2.577 – 3,000 * 2.445 + 550,000 * 0.7938 PW1 = -2,500,000+ 193, 275 – 7,335 + 436,590 PW1 = -1,877, 470

present worth of alternative 2, PW2 PW2 = -3,000,000+150,000(7,8%, 2)+(150,000(%, 8%,4)+5000€, 8%,4)ę,8%, 2) 30,000(?,8%, 6) + 800,000 (5,8%, 6) PW2 = -3,000,000+150,000* 1.783+(150,000*3.312+5000 * 4.650.8573- 30,000 * 4.623 + 800,000 * 0.6302 PW2 = -3,000,000+267450+ [496, 800+ 23250]0.8573–138,690+504, 160 PW2 = -3,000,000 + 267, 450 + 445, 838.865 – 138, 690 + 504,160 PW2 = -1,921, 241.135

present worth of alternative 3, PW3 PW3 = -2, 700,000+140, 000(, 8%,9)–350,000(%, 8%, 4)-2,8%, 4)[350,000(%, 8%,5)+ 2,000 (5,8%, 5)] + 650,000, 8%,9) PW3 = -2, 700,000 + 874,580 – 1, 159, 200 – 0.735[1, 397,550 + 14, 744] + 650,000 * 0.5002 PW3 = -2,984, 620 – 1,038,036.09 + 325, 130 PW3 = -3,697,526.09
using present worth analysis, Alternative 1 would be preferred because it has the highest present worth compared to other alternatives

7,3) = (-1,877, 470)*0.388 = equivalent annual worth, AW1 -728, 458.36

) = (-1,921, 241.135) * equivalent annual worth, AW2 0.2163 = -415,564.4575

6,9) = (-3,697,526.09) * equivalent annual worth, AW3 = 0.1601 = -591, 973.927

using equivalent annual worth analysis, alternative 2 is preferred because it has highest equivalent annual worth compared to other alternatives

FW1 = PW, F.,8%, 3) = (-1,877, 470) * 1.26 = –2, 365,612.2

FW, = PW25,8%,6) = (-1,921, 241.135) * 1.587 = -3,049, 009.681

FW3 = PW35,8%,9) = (-3,697, 526.09) * 1.999 = -7,391, 354.654

using future worth analysis, Alternative 1 would be preferred because it has the highest future worth compared to other alternatives

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