Question

A recycling company is looking to purchase a machine based on two options

	Machine A	Machine B
Initial cost AED	66,000	102,000
Annual cost/year AED	16,000	7,000
Salvage Value AED	8,000	22,000
Useful life in years	3	6

Using an interest rate of 12%, you must decide which machine is to be purchased by using a future worth (FW) analysis

-       A.       B.       C.       D.       E.       F.       G.	The FW of machine A is nearest to
-       A.       B.       C.       D.       E.       F.       G.	The FW of machine B is nearest to
-       A.       B.       C.       D.       E.       F.       G.	Which machine is to be chosen?

A.	-$305,935
B.	-$236,136
C.	Machine B
D.	-$215,023
E.	-$333,601
F.	Machine A
G.	-$338,713

Answer 1

Useful life of Machine A is 3 years

Useful life of Machine B is 6 years

Hence we need to repeat alternative A twice.

FW of MAchine A= FW of Benefits A-FW of Costs A

FW of Benefits A=8000(1.12)^3+8000=19239.42

FW of Costs A=66000(1.12)^6+16000(1.12)^5+16000(1.12)^4+16000(1.12)^2+16000(1.12)+16000+66000(1.12)^3=330361.72

FW of Machine A=19239.42-330361.72=-311,122.3

FW of Machine B= FW of Benefits B-FW of Costs B

FW of Benefits B=22000

FW of Costs B=102000(1.12)^6+7000(1.12)^5+7000(1.12)^4+7000(1.12)^2+7000(1.12)+7000=248301.74
FW of Machine B=22000-248301.74=-226301.74

Hence FW of Machine A is nearest to option A

Hence FW of Machine B is nearest to option B

Machine B is to be chosen over A