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asambeis [7]
3 years ago
13

A remotely located air sampling station can be powered by solar cells or by running an electric line to the site and using conve

ntional power. Solar cells will cost $12,600 to install and will have a useful life of 4 years with no salvage value. Annual costs for inspection, cleaning, etc. are expected to be $1,400. A new power line will cost $11,000 to install, with power costs expected to be $800 per year. Since the air sampling project will end in 4 years, the salvage value of the line is considered to be zero. At an interest rate of 10% per year, which alternative should be selected on the basis of a future worth analysis?
Business
1 answer:
Vanyuwa [196]3 years ago
6 0

Answer:

Since the total future worth of running an electric line of $19,353.42 is less than the total future worth of solar cells is $24,132.22, it implies that it will be cheaper to run an electric line than to use solar cells. Therefore, running an electric line should be selected.

Explanation:

The future worth analysis refers to an act of determining what the the worth of present amount of money or stream of money invested at an interest rate will after in some period or years to come.

To determine which one to select between solar cells and running an electric line, the we need to calculate the future worth of both and compared as follows:

a. Calculation of future value of solar cells

<u>Calculation of future worth of $12,600 installation cost</u>

FW of $12,600 = PW of $12,600 * (1 + r)^n ................ (1)

Where;

FW of $12,600 = Future worth of $12,600 installation cost = ?

PW of $12,600 = Present worth of $12,600 installation cost = $12,600

r = interest rate = 10%, or 0.10

n = number of years = 4

Substitute the values into equation (1), we have:

FW of $12,600 = $12,600 * (1 + 0.10)^4

FW of $12,600 = $12,600 * 1.4641

FW of $12,600 = $18,447.66

<u>Calculation of future worth of annual costs for inspection, cleaning, etc. of $1,400</u>

The future worth of annual costs for inspection, cleaning, etc. of $1,400 can also be calculated using the formula for calculating the Future Value (FV) of an Ordinary Annuity as follows:

FW of $1,400 = M * (((1 + r)^n - 1) / r) ................................. (2)

Where,

FW of $1,400 = Future value of Annual costs for inspection, cleaning, etc. of $1,400 =?

M = Annual costs for inspection, cleaning, etc. = $1,400

r = interest rate = 10%, or 0.10

n = number of years = 4

Substitute the values into equation (2), we have:

FW of $1,400 = $1,400 * (((1 + 0.01)^4 - 1) / 0.01)

FW of $1,400 = $1,400 * 4.060401

FW of $1,400 = $5,684.56

<u>Calculation of total future worth of solar cells</u>

This is calculated by simply adding the FW of $12,600 and FW of $1,400 as follows:

Total future worth of solar cells = FW of $12,600 + FW of $1,400 = $18,447.66 + $5,684.56 = $24,132.22

Therefore, <u>the total future worth of solar cells is $24,132.22.</u>

b. Calculation of future value of running an electric line

<u>Calculation of future worth of $11,000 installation cost</u>

FW of $11,000 = PW of $11,000 * (1 + r)^n ................ (3)

Where;

FW of $11,000 = Future worth of $11,000 installation cost = ?

PW of $11,000 = Present worth of $11,000 installation cost = $11,000

r = interest rate = 10%, or 0.10

n = number of years = 4

Substitute the values into equation (3), we have:

FW of $11,000 = $11,000 * (1 + 0.10)^4

FW of $11,000 = $11,000 * 1.4641

FW of $11,000 = $16,105.10

<u>Calculation of future worth of expected annual power costs of $800</u>

The future worth of expected annual power costs of $800 can also be calculated using the formula for calculating the Future Value (FV) of an Ordinary Annuity as follows:

FW of $800 = M * (((1 + r)^n - 1) / r) ................................. (4)

Where,

FW of $800 = Future value of expected annual power costs of $800 =?

M = Expected annual power costs = $800

r = interest rate = 10%, or 0.10

n = number of years = 4

Substitute the values into equation (4), we have:

FW of $800 = $800 * (((1 + 0.01)^4 - 1) / 0.01)

FW of $800 = $800 * 4.060401

FW of $800 = $3,248.32

<u>Calculation of total future worth of running an electric line </u>

This is calculated by simply adding the FW of $11,000 and FW of $800 as follows:

Total future worth of running an electric line = FW of $11,000 + FW of $800 = $16,105.10 + $3,248.32 = $19,353.42

Therefore, the total future worth of running an electric line is $19,353.42.

c. Conclusion

Since the total future worth of running an electric line of $19,353.42 is less than the total future worth of solar cells is $24,132.22, it implies that it will be cheaper to run an electric line than to use solar cells. Therefore, running an electric line should be selected.

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Answer:

Actual Yiel to maturity is 9.3%

Explanation:

Yield to maturity is the annual rate of return that an investor receives if a bond bond is held until the maturity.

Face value = F = $1,000

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Number of payment = n = 5 years

Yield to maturity = [ C + ( F - P ) / n ] / [ (F + P ) / 2 ]

Yield to maturity = [ $40 + ( $1,000 - $785 ) / 5 ] / [ ( 1,000 + $785 ) / 2 ]

Yield to maturity = [ $40 + $43 ] / $892.5  = $83 /$892.5 = 0.0645 = 0.093%

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3 years ago
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Answer:

d. Idea development

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I hope this answered your question. If you have any more questions feel free to ask away at Brainly.

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3 years ago
The bonds issued by United Corp. bear a coupon of 6 percent, payable semiannually. The bond matures in 20 years and has a $1,000
pishuonlain [190]

Answer: 6.40%

Explanation:

Use Excel to calculate this by the formula;

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3 years ago
The use of debt in the firm's capital structure will increase ROE if the firm:_____.
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Answer: c. earns a higher return than the rate paid on debt.

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If the debt that the company incurs leads to the company making more money than they are paying as interest for the debt, then more money will be available as net income which would increase the Return on Equity.

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3 years ago
Mercury Inc. purchased equipment in 2019 at a cost of $400,000. The equipment was expected to produce 700,000 units over the nex
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Answer:

See explanation section

Explanation:

We know,

Annual depreciation rate under Units-of-production = Depreciable amount/Overall (expected) production

Given,

Purchase value = $400,000

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Expected production = 700,000 units

Depreciable Amount = $(400,000 - 50,000) = $350,000

Annual depreciation rate = $350,000/700,000

Depreciation rate = $0.50

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= $170,000

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Book value = $400,000 - $170,000 = $230,000

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Credit Equipment $400,000

7 0
3 years ago
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