Business, 16.11.2019 05:31 jaeybird9

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?

Answers: 2

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Answers

Answer from: jayzelgaspar8005

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:

$C \times \frac{1-(1+r)^{-time} }{rate} = PV\\$

C18,000

time20

rate0.1

$18000 \times \frac{1-(1+0.1)^{-20} }{0.1} = PV\\$

PV$153,244.1470

Salvage value:

$\frac{Maturity}{(1 + rate)^{time} } = PV$

Maturity $20,000.0000

time 20.00

rate 0.1

$\frac{20000}{(1 + 0.1)^{20} } = PV$

PV 2,972.87

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

Total: 230,271.28

$PV \div \frac{1-(1+r)^{-time} }{rate} = C\\$

Present worth $(230,271.28)

time20

rate0.1

$-230271.28 \div \frac{1-(1+0.1)^{-20} }{0.1} = C\\$

C-$ 27,047.578

Fund to purchase in 20 years:

$FV \div \frac{(1+r)^{time} -1}{rate} = C\\$

FV $80,000.00

time20

rate0.1

$80000 \div \frac{(1+0.1)^{20} -1}{0.1} = C\\$

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

$PV \div \frac{1-(1+r)^{-time} }{rate} = C\\$

PV $230,271.28

time20

rate0.105126197

$230271.28 \div \frac{1-(1+0.105126197287798)^{-20} }{0.105126197287798} = C\\$

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:

$C \times \frac{1-(1+r)^{-time} }{rate} = PV\\$

C15,000

time10

rate0.1

$15000 \times \frac{1-(1+0.1)^{-10} }{0.1} = PV\\$

PV$92,168.5066

$C \times \frac{1-(1+r)^{-time} }{rate} = PV\\$

C20,000

time15

rate0.1

$20000 \times \frac{1-(1+0.1)^{-15} }{0.1} = PV\\$

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:

$\frac{Maturity}{(1 + rate)^{time} } = PV$

Maturity $152,121.5901

time 10.00

rate 0.1

$\frac{152121.590126167}{(1 + 0.1)^{10} } = PV$

PV 58,649.46

salvage value

$\frac{Maturity}{(1 + rate)^{time} } = PV$

Maturity $25,000.0000

time 25.00

rate 0.1

$\frac{25000}{(1 + 0.1)^{25} } = PV$

PV 2,307.40

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

$PV \div \frac{1-(1+r)^{-time} }{rate} = C\\$

PV $248,510.57

time25

rate0.1

$248510.57 \div \frac{1-(1+0.1)^{-25} }{0.1} = C\\$

C $ 27,377.930

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