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ololo11 [35]
3 years ago
15

Lockheed Martin is increasing its booster thrust power in order to win more satellite launch contracts from European companies i

nterested in new global communications markets. A piece of earth-based tracking equipment is expected to require an investment of $13 million. Annual operating costs for the system are expected to start the first year and continue at $0.9 million per year. The useful life of the tracker is 8 years with a salvage value of $0.5 million. Calculate the Annual Worth for the system if the corporate MARR is currently 12% per year.
Business
1 answer:
Valentin [98]3 years ago
5 0

Answer:

Lockheed Martin needs to make $3,476,250 per year (during 8 years) to cover their costs and investment required for the project if their MARR is 12%.

Explanation:

required investment -$13,000,000

operating costs per year = -$900,000

8 years useful life, salvage value of $500,000

to calculate the annual worth we need to determine capital recovery of the project:

capital recovery = [-$13,000,000 x annuity factor PV (A/P, 12%, 8)] + [$500,000 x annuity factor (A/F, 12%, 8)] = (-$13,000,000 x .2013) + ($500,000 x .0813) = -$2,616,900 + $40,650 = -$2,576,250

this means that Lockheed Martin would need to earn $2,576,250 during 8 years just to recover their investment.

since the company will also incur in yearly operating costs, they must include them to determine the total annual worth of the project:

annual worth = -$2,576,250 - $900,000 = -$3,476,250

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A company is considering buying a new piece of machinery. A 10% interest rate will be used in the computations. Two models of th
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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  

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MACHINE 1

Operating cost:

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

C 18,000

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rate 0.1

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

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

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time   20.00  

rate  0.1

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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\\  

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time 20

rate 0.105126197

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

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<u>Machine II</u>

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\\  

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time 10

rate 0.1

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

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

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rate  0.1

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rate 0.1

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