Question

Johnson Farms is planning to install one of two mechanical devices to reduce costs. Both cost $1000 and last five years with no salvage value. Device A can be expected to result in $300 savings annually. Device B will provide cost saving of $400 the first year but will decline by $50 annually, making the second year saving to $350 and third year savings $300 and so forth, with interest at 7% per year, calculate PW of each device. Which device should the firm purchase? Why?

Answer 1

MARR, i	8%	p.a.
device fixed cost	$1,000
useful life, n	5	years
Both the devices here cost $1000, and the decision criterion will be to choose the device that maximizes the present worth of the benefits.
The Present worth of cost of each device is $1000.
Present worth of benefits:
Device A = 400(P/A,8%,5)
(here $400 is the saving from the device every year)
Using standard COMPOUND INTEREST TABLES: (i=8%,n=5)
PW of benefits for A = 400(3.993)
	$1,597.20
PW of Device A = PW of benefits for A - PW of cost
	$597.20
Present worth of benefits:
Device B = 400(P/A,8%,5)-50(P/G,8%,5)
Using standard COMPOUND INTEREST TABLES: (i=8%,n=5)
PW of benefits for B = 400 (3.993) - 50(7.372)
PW of benefits for B =	$1,228.60
PW of Device B = PW of benefits for B - PW of cost
	$228.60
PW of each device:
	PW
Device A	$597.20
Device B	$228.60
Hence, device A should be chosen as it has higher present worth.