Question

interest = 3%

Q2. The Department of Engineering is contemplating the purchase of a top-of-the- line PCB drilling machine to be used in its laboratories. The price of the machine is $5,000. The depreciation rate of the machine follows the SL method over its 10 years life. The market value of the machine at the end of its life is estimated to be $S. Annual fees paid by students to use the machine are estimated to be $500. What is the value for S that would make this purchase break even (using the PW method) if the Department would sell the machine at EOY 8? (Hint: the market value at EOY 8 is the book value of the machine at the time)

Answer 1

Given,

Interest rate = 3%

Price of the machine = $5,000

Market value at the end of life (10 years) = $S

Depreciation rate per year under SL depreciation = (Price - Market value at the end of life)/Useful life = ($5,000 - $S)/10 = 500 - 0.1S

Total depreciation up to the EOY 8 = 8 * Depreciation rate per year = 8 * (500 - 0.1S) = 4000 - 0.8S

Market value at the EOY 8 = Price - Total depreciation up to the EOY 8 = 5000 - (4000-0.8S) = 1000 + 0.8S

Annual fee paid by the students = $500

From the compound interest table, we obtain

(P/A, 3%, 8) = 7.020

(P/F, 3%, 8) = 0.7894

Present worth of annual fee = $500*(P/A, 3%, 8) = $500 * 7.020 = $3510

Present worth of the machine expenses = -$5000 + (1000+0.8S)*(F/P, 3%, 8) = -$5000 + (1000+0.8S)*0.7894 = -5000 + 789.4 + 0.6315S = -4210.6 + 06315S

In order to break-even, the net present worth of the machine = 0

=> PW of annual fee + PW of the machine expenses = 0

=> 3510 - 4210.6 + 0.6315S = 0

=> 0.6315S = 700.6

=> S = 1109.4

Ans: S = 1109.4