Question

Consider the two mutually exclusive projects in the table below. Salvage values represent the net proceeds (after tax) from disposal of the assets if they are sold at the end of each year. Both projects B1 and B2 will be available (or can be repeated) with the same costs and salvage values for an indefinite period. B Click the icon to view the additional data about the mutually exclusive projects. Click the icon to view the interest factors for discrete compounding when MARR = 12% per year. (a) Assuming an infinite planning horizon, which project is a better choice at MARR = 12%? Use 15 years as the common analysis period. The present worth for project B1 is $ thousand. (Round to one decimal place.) x i More Info The present worth for project B2 is $thousand. (Round to one decimal place.) Which project is a better choice? Choose the correct answer below. Project B1 O Project B2 B2 Salvage Value n 0 (b) With a 10-year planning horizon, which project is a better choice at MARR = 12%? 1 B1 Cash Flow Salvage Value - $18,000 - 1,700 9,000 - 1,700 7,500 - 1,700 4,500 - 1,700 2,500 - 1,700 1,500 Cash Flow - $16,000 -2,400 - 2,400 - 2,400 2 8,000 5,500 3,500 The present worth for project B1 is $thousand. (Round to one decimal place.) The present worth for project B2 is $thousand. (Round to one decimal place.) 3 4 5 Which project is a better choice? Choose the correct answer below. Print Done Project B1 Project B2

Answer 1

a.

Analysis period = 15 yrs

NPW of B1 = -18000 -1700*(P/A,12%,15) - (18000-1500)*((P/F,12%,5) + (P/F,12%,10)) + 1500*(P/F,12%,15)

= -18000 -1700*6.810864 - (18000-1500)*(0.567427 + 0.321973) + 1500*0.182696

= -43979.5 ~ -44.0 thousand

NPW of B2 = -16000 -2400*(P/A,12%,15) - (16000-3500)*((P/F,12%,3) + (P/F,12%,6) +(P/F,12%,9) + (P/F,12%,12)) + 3500*(P/F,12%,15)

= -16000 -2400*6.810864 - (16000-3500)*(0.711780 + 0.506631 + 0.360610 + 0.256675) + 3500*0.182696

= -54652.84 ~ -54.7 thousand

As present cost of B1 is less, it should be selected

b.

Analysis period = 10 yrs

NPW of B1 = -18000 -1700*(P/A,12%,10) - (18000-1500)*(P/F,12%,5) + 1500*(P/F,12%,10)

= -18000 -1700*5.650223 - (18000-1500)*0.567427 + 1500*0.321973

= -36484.97 ~ -36.5 thousand

NPW of B2 = -16000 -2400*(P/A,12%,10) - (16000-3500)*((P/F,12%,3) + (P/F,12%,6) +(P/F,12%,9)) + 8000*(P/F,12%,10)

= -16000 -2400*5.650223 - (16000-3500)*(0.711780 + 0.506631 + 0.360610) + 8000*0.321973

= -46722.51 ~ -46.7 thousand

As present cost of B1 is less, it should be selected