Question

Problem 1. (20 Pts) Two storage structures, given code names Y and Z, are being considered for a military base. The military uses a 6% percent/year expected rate of return and a 24-year life for decisions of this type. The relevant characteristics for each structure are shown below.

Structure Y Structure Z

First Cost 24225 24225

Estimated Life 12 years 24 years

Estimated Salvage Value None $1,800

Annual Maintenance Cost $1,000 $720

a) What is the future worth of each machine? b) What is the decision rule for determining the preferred machine based on future worth ranking? c) Which structure should be recommended? d) Plot a graph for both alternatives future worth changes with expected ARR (annual rate of return) if expected annual rate of return varies from 6 % to 9% ?

Answer 1

Structure Y
Year		Cost	PVF (6%)	PV
0	First cost	24225	1	24225
1-12	Annual maintenance cost	1000	8.3838	8383.8
12	Estimated salvage value	0	0.497	0
Future Worth				32608.8

Structure Z
Year		Cost	PVF (6%)	PV
0	First cost	24225	1	24225
1-24	Annual maintenance cost	720	12.55036	9036.257
24	Estimated salvage value	-1800	0.246979	-444.561
Future Worth				32816.7

structure y is perfered on basis of future worth ranking

Structure Y

Annual worth = 32608.8 / 8.3838 = $3890

Structure Z

Annual worth = 32816.70 / 12.55036 = $ 2614.8

Structure Z Should be recommended having annual cost less than structure Y.

Structure Y
Year		Cost	PVF (9%)	PV
0	First cost	24225	1	24225
1-12	Annual maintenance cost	1000	7.160725	7160.725
12	Estimated salvage value	0	0.355535	0
Future Worth				31385.73

Structure Y
Year		Cost	PVF	PV
0	First cost	24225	1	24225
1-12	Annual maintenance cost	720	9.706612	6988.76
12	Estimated salvage value	-1800	0.126405	-227.529
Future Worth				30986.23