Acompany is considering buying a new piece of machinery. a 10% interest rate will be used in the computations. two models of the machine are available.

machine i

initial cost: $80,000

end -of -useful –life salvage value, s: 20,000

annual operating cost 18,000

useful life, in years 20

machine ii

initial cost: $100,000

end -of -useful –life salvage value, s: 25,000

annual operating cost: 15,000 first 10 years, 20,000 thereafter

useful life, in years: 25

(a) determine which machine should be purchased, based on equivalent uniform annual cost.

(b) what is the capitalized cost of machine i?

(c) machine i is purchased and a fund is set up to replace machine i at the end of 20 years. compute the required uniform annual deposit.

(d) machine i will produce an annual saving of material of $28,000. what is the rate of return if machine i is installed?

(e) what will be the book value of machine i after 2 years, based on sum -of -years' -digits depreciation?

(f) what will be the book value of machine ii after 3 years, based on double declining balance depreciation?

Machine I

capitalized cost:  230,271.28

EAC: $ 27,047.58

Machine II

EAC:  $ 27,377.930  

As Machine I cost per year is lower it is better to purchase that one.

Annual deposits to purchase Machine I in 20 years: $ 1,396.770  

return of machine I with savings of 28,000 per year: 10.51%

Explanation:

WE calculate the present worth of each machine and then calculate the equivalent annual cost:

MACHINE 1

Operating cost:

C18,000

time20

rate0.1

PV$153,244.1470  

Salvage value:

Maturity  $20,000.0000  

time   20.00  

rate  0.1

PV   2,972.87  

Total: -80,000 cost - 153,244.15 annual cost + 2,972.87 salvage value:

Total: 230,271.28

Present worth  $(230,271.28)

time20

rate0.1

C-$ 27,047.578  

Fund to purchase in 20 years:

FV  $80,000.00  

time20

rate0.1

C  $ 1,396.770  

IF produce a 28,000 savings:

we must solve using a financial calcualtor for the rate at which the capitalized cost equals 28,000

PV  $230,271.28  

time20

rate0.105126197

C  $ 28,000.000  

rate of 0.105126197 = 10.51%

Machine II

100,000 cost

25,000 useful life

15,000 operating cost during 10 years

20,000 for the next 15 years

Present value of the operating cost:

C15,000

time10

rate0.1

PV$92,168.5066  

C20,000

time15

rate0.1

PV$152,121.5901  

in the timeline this is at the end of the 10th year we must discount as lump sum for the other ten years:

Maturity  $152,121.5901  

time   10.00  

rate  0.1

PV   58,649.46  

salvage value

Maturity  $25,000.0000  

time   25.00  

rate  0.1

PV   2,307.40  

Total cost: 100,000 + 92,168.51 + 58,649.46 - 2,307.40 = $248,510.57

PV  $248,510.57  

time25

rate0.1

C  $ 27,377.930  

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answer; /// the above statement is (true);