Question

Consider the two mutually exclusive projects in the table below. Salvage values represent the not 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 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.)

Answer 1

Here we have to determine the present worth of the two projects for a time period of 15 years. The present worth of the project B1 can be determined using the following equation

PWB1 = -30,000–1,500(P/F,12%, 15)–30,000(P/F, 12%,5)– 30,000(P/F, 12%, 10)+8,500(P/F,12%,5)+8,500(P/F, 12%, 10)+ 8,500(P/F, 12%, 15)

PWB2 = -30,000–1,500(P/F,12%, 15)-21500(P/F,12%, 5)– 21,500(P/F, 12%, 10) + 8,500(P/F, 12%, 15)

= PWB1 = -30,000–1,500(P/F, 12%, 15)–21500(A/F, 12%,5)(P/A, 12%, 10)+ 8,500(P/F, 12%, 15)

0.12 PWB1 = -30,000 – 1,500 x = 1 - 1.12-15 -- 21500 X 1 - 1.12-10 0.12 0 .19 +8,500 x 1.12-15 1.125 - 1

On solving the above equation we get

PWB1 = - $ 57,785.48

Present worth of project B1 = - $ 57.8 thousand.

Now calculating the present worth of B2

PWB, = -14,000–1,500(P/A, 12%, 15)–12,000(A/F, 12%, 3)(P/A, 12%, 12)+ 2,000(P/F, 12%, 15)

= PWB2 = –14,000 – 1,500 x = 1 - 1.12-15 -- 12.000 0.12 0.12 1-1.12-12 1.123 – 1* 0.12 +2,000 1.12-15

On solving the above equation we get

Present worth of Project B2 = - $ 45.9 thousand.

Using the present worth of the two projects the present worth of B2 is less hence select Project B2.