Question

A large manufacturing company is now evaluating three mutually exclusive production systems. The following table describes the data:

System	A	B	C
Initial investment (millions)	$2,005	$8,800	$19,000
Annual receipts less expenses (millions)	$  600	$1,500	$ 2,300
Salvage value (millions)	$15	$100	$330
Useful life (years)	5	10	20

Apply repeatability assumption and use the PW method to select the best system. The MARR is 10% per year, and the system chosen must provide service for 20 years. (correct your final answer to 3 decimal places)

Answer 1

System	A	B	C
Initial investment (million $)	$2,005	$8,800	$19,000
Annual receipts less expenses (million $)	$ 600	$1,500	$ 2,300
Salvage value (million $)	$15	$100	$330
Useful life (years)	5	10	20

MARR = 10%

Study Period = 20 years

It otherwise means

System A is to be repeated 4 times along with all cash flows.

System B is to be repeated 2 times along with all cash flows.

System C has the life of 20 years. So, no need to repeat the cash flows.

PW of System A

PW = -2005 – 2005 (P/F, 10%, 5) – 2005 (P/F, 10%, 10) – 2005 (P/F, 10%, 15) + 600 (P/A , 10%, 20) + 15 (P/F, 10%, 5) + 15 (P/F, 10%, 10) + 15 (P/F, 10%, 15) + 15 (P/F, 10%, 20)

PW = -2005 – 2005 (0.62092) – 2005 (0.38554) – 2005 (0.23939) + 600 (8.51356) + 15 (0.62092) + 15 (0.38554) + 15 (0.23939) + 15 (0.14864)

PW = 626.12

PW of System B

PW = -8,800 – 8,800 (P/F, 10%, 10) + 1,500 (P/A, 10%, 20) + 100 (P/F, 10%, 10) + 100 (P/F, 10%, 20)

PW = -8,800 – 8,800 (0.38554) + 1,500 (8.51356) + 100 (0.38554) + 100 (0.14864)

PW = 631

PW of System C

PW = -19,000 + 2,300 (P/A, 10%, 20) + 330 (P/F, 10%, 20)

PW = -19,000 + 2,300 (8.51356) + 330 (0.14864)

PW = 630.23

PW of A = 626.12

PW of B = 631

PW of C = 630.23

On the basis of PW analysis, Select the System B

System B has highest PW, so select system B.