You are considering a hydraulic excavator. This machine will have an estimated service life of 10...

You are considering a hydraulic excavator. This machine will have an estimated service life of 10 years, with a salvage value of 10% of the investment cost. Its expected savings from annual operating and maintenance costs are estimated to be SR 70,000. To expect a 15% rate of return on investment, what would be the maximum amount that you are willing to pay for the machine: (Choose the closest answer)

a) SR 498,145

b) SR 320,104

c) SR 250,417

d) SR 475,139

e) SR 360,218

Solutions

Expert Solution

The life span of the machine = 10 years

Salvage value = 10% of investment cost

Annual Cash Savings = SR 70000

Interest rate = 15%

Let the Present worth of the machine = x

SR x would be the maximum amount that one would be willing to pay for the machine.

It shall be noted that the present worth is the sum of the PV of net cash inflow

The PV of net cash inflow = Net Cash Inflow/(1+15%)^n

where,

n = nth year

Hence, the present worth of the machine is:

Year	Cash Savings	Salvage Value	Net Cash Inflow	PV of Net Cash Inflow
0
1	70000		70000	60869.56522
2	70000		70000	52930.05671
3	70000		70000	46026.13627
4	70000		70000	40022.72719
5	70000		70000	34802.37147
6	70000		70000	30262.93171
7	70000		70000	26315.59279
8	70000		70000	22883.12417
9	70000		70000	19898.36884
10	70000	0.10x	70000 + 0.10x	17302.9294285306 + 0.10x/(1+15%)^10

This means,

x = 334010.9 + 17302.9294285306 + 0.10x/(1+15%)^10

x = 351313.8 + 0.247185*(0.10x)

x = 0.0247185x + 351313.8

x = 360217.8

x ~ 360218

Hence, the present worth of the machine is SR 360218, the maximum amount that one would be willing to pay for the machine.

Hence, the correct answer is e) SR 360,218

Rahul Sunny answered 1 year ago

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