Question

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

Note- We need the MARR/Interest Rate to calculate PW. It is not given, so assuming it as 8%.

Detailed cashflows are shown below. Explanation follows-

Year Initial Cost Benefit Depreciation Selling Value 0 -5000 1 500 2 500 3 500 4 5 (5000-S)/10 (5000-5)/10 (5000-5)/10 (5000-5)/10 (5000-5)/10 (5000-5)/10 (5000-S)/10 (5000-5)/10 5000-8*(5000-S)/10 (5000-S)/10 (5000-5)/10 500 500 500 500 500 6 7 8 9 10

For the break even to happen, the net benefit in terms of PW should be equal to the initial cost.

As given, the final value after 10 years is S and the depreciation method is Straight Line. Meaning that each year the machine will depreciate by (5000-10)/S. So, after 8 years, its value would be 5000-8*(5000-S)/10. The total benefit at the end of year 8 is 500+(5000-8*(5000-S)/10)

Now, total NPV of benefits is

500 500 5000-5 10 500 1.08 + 500 1.083 十 十 十 500 1.08 500 + 5000-8x 1.088 十 500 500 1.085 1.08? 十 1.08 1.084

At break even, this would be equal to initial cost of 5000. So at break even

50005 5000 : 500 1.08 500 500 + 5000-8 10 十 500 1.08? 十 500 1.084 十 500 1.085 十 500 500 1.086 1.087 1.08 1.088

Solving for S (ideal to use a equation solver tool such as mathematica or wolfram etc), we get

S=3670.42