Question

A company 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?

Answer 1

Answer:

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:

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

C 18,000

time 20

rate 0.1

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

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)

time 20

rate 0.1

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

C -$ 27,047.578  

Fund to purchase in 20 years:

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

FV  $80,000.00  

time 20

rate 0.1

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

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  

time 20

rate 0.105126197

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

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:

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

C 15,000

time 10

rate 0.1

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

PV $92,168.5066  

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

C 20,000

time 15

rate 0.1

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

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  

time 25

rate 0.1

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

C  $ 27,377.930