Question

A project has a first cost of $12,000 net annual benefits of $2,300, and a salvage value of $3,300 at the end of its 10-year useful life.  The project will be replaced identically at the end of 10 years, and again at the end of 20 years.  What is the present worth of the entire 30 years of service if the interest rate is 9%?

Solution:

PW of 10 years = -12,000 + 2,300(P/A, 9%, 10) + 3,300(P/F, 9%, 10)

                          = -12,000+2,300(6.418)+3,300(0.4224) = $4,155.32

PW of 30 years = (PW of 10 years) [1+ (P/F, 9%, 10) + (P/F, 9%, 20)]

                           = $4,155.32 [1+0.4224+0.1784] = $6,651.84

why did he put [1+ (P/F, 9%, 10) + (P/F, 9%, 20)]

why is the 1 ?

can we solve it in another way ??

Answer 1

Present worth of the project = -12000 - 12000(P/F, 9%, 10) - 12000(P/F, 9%, 20) + 2300(P/A, 9%, 30) + 3300(P/F, 9%, 10) + 3300(P/F, 9% + 20) + 3300(P/F, 9%, 30)

This is because currently we pay 12000 as initial cost. Then we repurchase in 10th year so again the purchase cost is 12000 and then it is repeated in 20th year. Similarly the salvage value fist appears at the end of 10th, 20th and 30th year

PW = -12000 - 12000 x 0.4224 - 12000 x 0.1784 + 2300 x 10.2737 + 3300 x 0.4224 + 3300 x 0.1784 + 3300 x 0.0754

= 6651.37

This is the required present worth.