Question

A manager of Al-Ahram newspaper has an offer to buy a new printing machine which has initial cost of LE 29,000 and will have LE 4,000 salvage value after a life of 10 years. The estimated maintenance cost equals LE 800 in year 5, and the annual operating cost equals LE 12,000 for the first 4 years, increasing by 6% per year thereafter. a. Calculate the annual worth of the machine if the interest rate is 8% per year. b. If he has another bank that gives the interest rate at 5 % per year, Would the annual worth still the same? why

Answer 1

Present worth of geometric series = A *[1 - (1+g)^n / (1+i)^n] /(i-g)

a.

i = 8%

Present cost of annual operating cost from EOY 4 to EOY 10 at EOY 3

= 12000*[1 - (1+0.06)^7 / (1+0.08)^7] /(0.08-0.06)

= 12000*[1 - (1.06)^7 / (1.08)^7] /(0.02)

= 12000 *6.132309

= 73587.71

Present worth of machine = -29000 + 4000*(P/F,8%,10) - 800*(P/F,8%,5) - 12000*(P/A,8%,3) - 73587.71*(P/F,8%,3)

= -29000 + 4000*0.463193 - 800*0.680583 - 12000*2.577097 - 73587.71*0.793832

= -117033.14

Annual worth of machine = -117033.14 *(A/P,8%,10)

= -117033.14 * 0.149029

= -17441.33

b.

i = 5%

Present cost of annual operating cost from EOY 4 to EOY 10 at EOY 3

= 12000*[1 - (1+0.06)^7 / (1+0.05)^7] /(0.05-0.06)

= 12000*[1 - (1.06)^7 / (1.05)^7] /(-0.01)

= 12000 *6.860195

= 82322.34

Present worth of machine = -29000 + 4000*(P/F,5%,10) - 800*(P/F,5%,5) - 12000*(P/A,5%,3) - 82322.34*(P/F,5%,3)

= -29000 + 4000*0.613913 - 800*0.783526 - 12000*2.723248 - 82322.34 *0.863838

= -130963.31

Annual worth of machine = -130963.31 *(A/P,5%,10)

= -130963.31 *0.129505

= -16960.40

Annual worth is different when interest rate is changed