Question

For the below ME alternatives, which machine should be selected based on the PW analysis. MARR=10%.

	Machine A	Machine B	Machine C
First cost, $	15000	30000	10,360
Annual cost, $/year	8,320	6,000	4,000
Salvage value, $	4,000	5,000	1,000
Life, years	3	6	2

Answer the below questions :

C- PW for machine C =

Answer 1

SOLUTION :

r = MARR = 10% = 0.1

=> (1 + r) = 1.1

PV of costs of Machine A :

= 18000 + 8320/ 1.1 + 8320/1.1^2 + 8320/1.1^3 - 4000/1.1^3 

= 42682.49 ($) 

PV of costs of Machine B :

= 30000 + 6000 (1.1^6 - 1)/(0.1*1.1^6) - 5000/1.1^6

= 53309.19 ($) 

PV of costs of Machine C :

= 10360 + 4000/1.1 + 4000/1.1^2 - 1000/1.1^2

= 16475.70 ($)

Therefore, as per PV of costs , Machine C has lowest cost. So Machine C should be selected.

(ANSWER