What is the equivalent present cost is for the first 5 years of repair work if interest is 4%?

Question

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Answer 1

Question

Chapter 4, Problem 56P

To determine

What is the equivalent present cost is for the first 5 years of repair work if interest is 4%?

Expert Solution & Answer

Answer to Problem 56P

$292,750

Given information:

Maintenance cost for 1^st year (A1) = $85,000

Cost will increase by $150 each year, so G=$10,000

Time (n) = 5 years

Interest rate (i) = 4%.

Concept used:

Arithmetic gradient present worth is given by,

(P/G,i,n)={[(1+i)n−in−1]/[i2(1+i)n]}

To find the Present Worth, at EOY 0, of a gradient series that begins EOY 1, use

P = A1 (P/A,i%,n) + G (P/G,i%,n)

To find the annual equivalent (A series) of a gradient series that begins EOY 1, use

A = A1 + G (A/G,i%,n).

Calculation:

In order to find out the present worth of the cash maintenance cash flow we use the given formula,

To find the Present Worth, at EOY 0, of a gradient series that begins EOY 1, use

P = A1 (P/A,i%,n) + G (P/G,i%,n)

Arithmetic gradient present worth is given by,

(P/G,i,n)={[(1+i)n−in−1]/[i2(1+i)n]}

The upper cash flow is equivalent to a uniform series of A = $85,000 for 5 years, an arithmetic series of G = 10,000 for 5 years. So the present worth of the above cash flow is

P= $85,000(P/A,4%,5) - $10,000(P/G,4%,5)

=$85,000(4.45) - $10,000(8.55)

=$378,250 - $85,500

=$292,750.

Conclusion:

Thus, the present worth of the repair cost for a 5 year period is $292,750.

Explanation of Solution

Given information:

Maintenance cost for 1^st year (A1) = $85,000

Cost will increase by $150 each year, so G=$10,000

Time (n) = 5 years

Interest rate (i) = 4%.

Concept used:

Arithmetic gradient present worth is given by,

(P/G,i,n)={[(1+i)n−in−1]/[i2(1+i)n]}

To find the Present Worth, at EOY 0, of a gradient series that begins EOY 1, use

P = A1 (P/A,i%,n) + G (P/G,i%,n)

To find the annual equivalent (A series) of a gradient series that begins EOY 1, use

A = A1 + G (A/G,i%,n).

Calculation:

In order to find out the present worth of the cash maintenance cash flow we use the given formula,

To find the Present Worth, at EOY 0, of a gradient series that begins EOY 1, use

P = A1 (P/A,i%,n) + G (P/G,i%,n)

Arithmetic gradient present worth is given by,

(P/G,i,n)={[(1+i)n−in−1]/[i2(1+i)n]}

The upper cash flow is equivalent to a uniform series of A = $85,000 for 5 years, an arithmetic series of G = 10,000 for 5 years. So the present worth of the above cash flow is

P= $85,000(P/A,4%,5) - $10,000(P/G,4%,5)

=$85,000(4.45) - $10,000(8.55)

=$378,250 - $85,500

=$292,750.

Conclusion:

Thus, the present worth of the repair cost for a 5 year period is $292,750.

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Answer 2

Question

Chapter 4, Problem 56P

To determine

What is the equivalent present cost is for the first 5 years of repair work if interest is 4%?

Expert Solution & Answer

Answer to Problem 56P

$292,750

Given information:

Maintenance cost for 1^st year (A1) = $85,000

Cost will increase by $150 each year, so G=$10,000

Time (n) = 5 years

Interest rate (i) = 4%.

Concept used:

Arithmetic gradient present worth is given by,

(P/G,i,n)={[(1+i)n−in−1]/[i2(1+i)n]}

To find the Present Worth, at EOY 0, of a gradient series that begins EOY 1, use

P = A1 (P/A,i%,n) + G (P/G,i%,n)

To find the annual equivalent (A series) of a gradient series that begins EOY 1, use

A = A1 + G (A/G,i%,n).

Calculation:

In order to find out the present worth of the cash maintenance cash flow we use the given formula,

To find the Present Worth, at EOY 0, of a gradient series that begins EOY 1, use

P = A1 (P/A,i%,n) + G (P/G,i%,n)

Arithmetic gradient present worth is given by,

(P/G,i,n)={[(1+i)n−in−1]/[i2(1+i)n]}

The upper cash flow is equivalent to a uniform series of A = $85,000 for 5 years, an arithmetic series of G = 10,000 for 5 years. So the present worth of the above cash flow is

P= $85,000(P/A,4%,5) - $10,000(P/G,4%,5)

=$85,000(4.45) - $10,000(8.55)

=$378,250 - $85,500

=$292,750.

Conclusion:

Thus, the present worth of the repair cost for a 5 year period is $292,750.

Explanation of Solution

Given information:

Maintenance cost for 1^st year (A1) = $85,000

Cost will increase by $150 each year, so G=$10,000

Time (n) = 5 years

Interest rate (i) = 4%.

Concept used:

Arithmetic gradient present worth is given by,

(P/G,i,n)={[(1+i)n−in−1]/[i2(1+i)n]}

To find the Present Worth, at EOY 0, of a gradient series that begins EOY 1, use

P = A1 (P/A,i%,n) + G (P/G,i%,n)

To find the annual equivalent (A series) of a gradient series that begins EOY 1, use

A = A1 + G (A/G,i%,n).

Calculation:

In order to find out the present worth of the cash maintenance cash flow we use the given formula,

To find the Present Worth, at EOY 0, of a gradient series that begins EOY 1, use

P = A1 (P/A,i%,n) + G (P/G,i%,n)

Arithmetic gradient present worth is given by,

(P/G,i,n)={[(1+i)n−in−1]/[i2(1+i)n]}

The upper cash flow is equivalent to a uniform series of A = $85,000 for 5 years, an arithmetic series of G = 10,000 for 5 years. So the present worth of the above cash flow is

P= $85,000(P/A,4%,5) - $10,000(P/G,4%,5)

=$85,000(4.45) - $10,000(8.55)

=$378,250 - $85,500

=$292,750.

Conclusion:

Thus, the present worth of the repair cost for a 5 year period is $292,750.

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Answer 3

Question

Chapter 4, Problem 56P

To determine

What is the equivalent present cost is for the first 5 years of repair work if interest is 4%?

Expert Solution & Answer

Answer to Problem 56P

$292,750

Given information:

Maintenance cost for 1^st year (A1) = $85,000

Cost will increase by $150 each year, so G=$10,000

Time (n) = 5 years

Interest rate (i) = 4%.

Concept used:

Arithmetic gradient present worth is given by,

(P/G,i,n)={[(1+i)n−in−1]/[i2(1+i)n]}

To find the Present Worth, at EOY 0, of a gradient series that begins EOY 1, use

P = A1 (P/A,i%,n) + G (P/G,i%,n)

To find the annual equivalent (A series) of a gradient series that begins EOY 1, use

A = A1 + G (A/G,i%,n).

Calculation:

In order to find out the present worth of the cash maintenance cash flow we use the given formula,

To find the Present Worth, at EOY 0, of a gradient series that begins EOY 1, use

P = A1 (P/A,i%,n) + G (P/G,i%,n)

Arithmetic gradient present worth is given by,

(P/G,i,n)={[(1+i)n−in−1]/[i2(1+i)n]}

The upper cash flow is equivalent to a uniform series of A = $85,000 for 5 years, an arithmetic series of G = 10,000 for 5 years. So the present worth of the above cash flow is

P= $85,000(P/A,4%,5) - $10,000(P/G,4%,5)

=$85,000(4.45) - $10,000(8.55)

=$378,250 - $85,500

=$292,750.

Conclusion:

Thus, the present worth of the repair cost for a 5 year period is $292,750.

Explanation of Solution

Given information:

Maintenance cost for 1^st year (A1) = $85,000

Cost will increase by $150 each year, so G=$10,000

Time (n) = 5 years

Interest rate (i) = 4%.

Concept used:

Arithmetic gradient present worth is given by,

(P/G,i,n)={[(1+i)n−in−1]/[i2(1+i)n]}

To find the Present Worth, at EOY 0, of a gradient series that begins EOY 1, use

P = A1 (P/A,i%,n) + G (P/G,i%,n)

To find the annual equivalent (A series) of a gradient series that begins EOY 1, use

A = A1 + G (A/G,i%,n).

Calculation:

In order to find out the present worth of the cash maintenance cash flow we use the given formula,

To find the Present Worth, at EOY 0, of a gradient series that begins EOY 1, use

P = A1 (P/A,i%,n) + G (P/G,i%,n)

Arithmetic gradient present worth is given by,

(P/G,i,n)={[(1+i)n−in−1]/[i2(1+i)n]}

The upper cash flow is equivalent to a uniform series of A = $85,000 for 5 years, an arithmetic series of G = 10,000 for 5 years. So the present worth of the above cash flow is

P= $85,000(P/A,4%,5) - $10,000(P/G,4%,5)

=$85,000(4.45) - $10,000(8.55)

=$378,250 - $85,500

=$292,750.

Conclusion:

Thus, the present worth of the repair cost for a 5 year period is $292,750.

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Question

Chapter 4, Problem 56P

To determine

What is the equivalent present cost is for the first 5 years of repair work if interest is 4%?

Expert Solution & Answer

Answer to Problem 56P

$292,750

Given information:

Maintenance cost for 1^st year (A1) = $85,000

Cost will increase by $150 each year, so G=$10,000

Time (n) = 5 years

Interest rate (i) = 4%.

Concept used:

Arithmetic gradient present worth is given by,

(P/G,i,n)={[(1+i)n−in−1]/[i2(1+i)n]}

To find the Present Worth, at EOY 0, of a gradient series that begins EOY 1, use

P = A1 (P/A,i%,n) + G (P/G,i%,n)

To find the annual equivalent (A series) of a gradient series that begins EOY 1, use

A = A1 + G (A/G,i%,n).

Calculation:

In order to find out the present worth of the cash maintenance cash flow we use the given formula,

To find the Present Worth, at EOY 0, of a gradient series that begins EOY 1, use

P = A1 (P/A,i%,n) + G (P/G,i%,n)

Arithmetic gradient present worth is given by,

(P/G,i,n)={[(1+i)n−in−1]/[i2(1+i)n]}

The upper cash flow is equivalent to a uniform series of A = $85,000 for 5 years, an arithmetic series of G = 10,000 for 5 years. So the present worth of the above cash flow is

P= $85,000(P/A,4%,5) - $10,000(P/G,4%,5)

=$85,000(4.45) - $10,000(8.55)

=$378,250 - $85,500

=$292,750.

Conclusion:

Thus, the present worth of the repair cost for a 5 year period is $292,750.

Explanation of Solution

Given information:

Maintenance cost for 1^st year (A1) = $85,000

Cost will increase by $150 each year, so G=$10,000

Time (n) = 5 years

Interest rate (i) = 4%.

Concept used:

Arithmetic gradient present worth is given by,

(P/G,i,n)={[(1+i)n−in−1]/[i2(1+i)n]}

To find the Present Worth, at EOY 0, of a gradient series that begins EOY 1, use

P = A1 (P/A,i%,n) + G (P/G,i%,n)

To find the annual equivalent (A series) of a gradient series that begins EOY 1, use

A = A1 + G (A/G,i%,n).

Calculation:

In order to find out the present worth of the cash maintenance cash flow we use the given formula,

To find the Present Worth, at EOY 0, of a gradient series that begins EOY 1, use

P = A1 (P/A,i%,n) + G (P/G,i%,n)

Arithmetic gradient present worth is given by,

(P/G,i,n)={[(1+i)n−in−1]/[i2(1+i)n]}

The upper cash flow is equivalent to a uniform series of A = $85,000 for 5 years, an arithmetic series of G = 10,000 for 5 years. So the present worth of the above cash flow is

P= $85,000(P/A,4%,5) - $10,000(P/G,4%,5)

=$85,000(4.45) - $10,000(8.55)

=$378,250 - $85,500

=$292,750.

Conclusion:

Thus, the present worth of the repair cost for a 5 year period is $292,750.

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 4, Problem 56P

To determine

What is the equivalent present cost is for the first 5 years of repair work if interest is 4%?

Expert Solution & Answer

Answer to Problem 56P

$292,750

Given information:

Maintenance cost for 1^st year (A1) = $85,000

Cost will increase by $150 each year, so G=$10,000

Time (n) = 5 years

Interest rate (i) = 4%.

Concept used:

Arithmetic gradient present worth is given by,

(P/G,i,n)={[(1+i)n−in−1]/[i2(1+i)n]}

To find the Present Worth, at EOY 0, of a gradient series that begins EOY 1, use

P = A1 (P/A,i%,n) + G (P/G,i%,n)

To find the annual equivalent (A series) of a gradient series that begins EOY 1, use

A = A1 + G (A/G,i%,n).

Calculation:

In order to find out the present worth of the cash maintenance cash flow we use the given formula,

To find the Present Worth, at EOY 0, of a gradient series that begins EOY 1, use

P = A1 (P/A,i%,n) + G (P/G,i%,n)

Arithmetic gradient present worth is given by,

(P/G,i,n)={[(1+i)n−in−1]/[i2(1+i)n]}

The upper cash flow is equivalent to a uniform series of A = $85,000 for 5 years, an arithmetic series of G = 10,000 for 5 years. So the present worth of the above cash flow is

P= $85,000(P/A,4%,5) - $10,000(P/G,4%,5)

=$85,000(4.45) - $10,000(8.55)

=$378,250 - $85,500

=$292,750.

Conclusion:

Thus, the present worth of the repair cost for a 5 year period is $292,750.

Explanation of Solution

Given information:

Maintenance cost for 1^st year (A1) = $85,000

Cost will increase by $150 each year, so G=$10,000

Time (n) = 5 years

Interest rate (i) = 4%.

Concept used:

Arithmetic gradient present worth is given by,

(P/G,i,n)={[(1+i)n−in−1]/[i2(1+i)n]}

To find the Present Worth, at EOY 0, of a gradient series that begins EOY 1, use

P = A1 (P/A,i%,n) + G (P/G,i%,n)

To find the annual equivalent (A series) of a gradient series that begins EOY 1, use

A = A1 + G (A/G,i%,n).

Calculation:

In order to find out the present worth of the cash maintenance cash flow we use the given formula,

To find the Present Worth, at EOY 0, of a gradient series that begins EOY 1, use

P = A1 (P/A,i%,n) + G (P/G,i%,n)

Arithmetic gradient present worth is given by,

(P/G,i,n)={[(1+i)n−in−1]/[i2(1+i)n]}

The upper cash flow is equivalent to a uniform series of A = $85,000 for 5 years, an arithmetic series of G = 10,000 for 5 years. So the present worth of the above cash flow is

P= $85,000(P/A,4%,5) - $10,000(P/G,4%,5)

=$85,000(4.45) - $10,000(8.55)

=$378,250 - $85,500

=$292,750.

Conclusion:

Thus, the present worth of the repair cost for a 5 year period is $292,750.

Answer 6

Chapter 4, Problem 56P

To determine

What is the equivalent present cost is for the first 5 years of repair work if interest is 4%?

Expert Solution & Answer

Answer to Problem 56P

$292,750

Given information:

Maintenance cost for 1^st year (A1) = $85,000

Cost will increase by $150 each year, so G=$10,000

Time (n) = 5 years

Interest rate (i) = 4%.

Concept used:

Arithmetic gradient present worth is given by,

(P/G,i,n)={[(1+i)n−in−1]/[i2(1+i)n]}

To find the Present Worth, at EOY 0, of a gradient series that begins EOY 1, use

P = A1 (P/A,i%,n) + G (P/G,i%,n)

To find the annual equivalent (A series) of a gradient series that begins EOY 1, use

A = A1 + G (A/G,i%,n).

Calculation:

In order to find out the present worth of the cash maintenance cash flow we use the given formula,

To find the Present Worth, at EOY 0, of a gradient series that begins EOY 1, use

P = A1 (P/A,i%,n) + G (P/G,i%,n)

Arithmetic gradient present worth is given by,

(P/G,i,n)={[(1+i)n−in−1]/[i2(1+i)n]}

The upper cash flow is equivalent to a uniform series of A = $85,000 for 5 years, an arithmetic series of G = 10,000 for 5 years. So the present worth of the above cash flow is

P= $85,000(P/A,4%,5) - $10,000(P/G,4%,5)

=$85,000(4.45) - $10,000(8.55)

=$378,250 - $85,500

=$292,750.

Conclusion:

Thus, the present worth of the repair cost for a 5 year period is $292,750.

Explanation of Solution

Given information:

Maintenance cost for 1^st year (A1) = $85,000

Cost will increase by $150 each year, so G=$10,000

Time (n) = 5 years

Interest rate (i) = 4%.

Concept used:

Arithmetic gradient present worth is given by,

(P/G,i,n)={[(1+i)n−in−1]/[i2(1+i)n]}

To find the Present Worth, at EOY 0, of a gradient series that begins EOY 1, use

P = A1 (P/A,i%,n) + G (P/G,i%,n)

To find the annual equivalent (A series) of a gradient series that begins EOY 1, use

A = A1 + G (A/G,i%,n).

Calculation:

In order to find out the present worth of the cash maintenance cash flow we use the given formula,

To find the Present Worth, at EOY 0, of a gradient series that begins EOY 1, use

P = A1 (P/A,i%,n) + G (P/G,i%,n)

Arithmetic gradient present worth is given by,

(P/G,i,n)={[(1+i)n−in−1]/[i2(1+i)n]}

The upper cash flow is equivalent to a uniform series of A = $85,000 for 5 years, an arithmetic series of G = 10,000 for 5 years. So the present worth of the above cash flow is

P= $85,000(P/A,4%,5) - $10,000(P/G,4%,5)

=$85,000(4.45) - $10,000(8.55)

=$378,250 - $85,500

=$292,750.

Conclusion:

Thus, the present worth of the repair cost for a 5 year period is $292,750.

Answer 7

Chapter 4, Problem 56P

To determine

What is the equivalent present cost is for the first 5 years of repair work if interest is 4%?

Answer 8

Expert Solution & Answer

Answer to Problem 56P

$292,750

Given information:

Maintenance cost for 1^st year (A1) = $85,000

Cost will increase by $150 each year, so G=$10,000

Time (n) = 5 years

Interest rate (i) = 4%.

Concept used:

Arithmetic gradient present worth is given by,

(P/G,i,n)={[(1+i)n−in−1]/[i2(1+i)n]}

To find the Present Worth, at EOY 0, of a gradient series that begins EOY 1, use

P = A1 (P/A,i%,n) + G (P/G,i%,n)

To find the annual equivalent (A series) of a gradient series that begins EOY 1, use

A = A1 + G (A/G,i%,n).

Calculation:

In order to find out the present worth of the cash maintenance cash flow we use the given formula,

To find the Present Worth, at EOY 0, of a gradient series that begins EOY 1, use

P = A1 (P/A,i%,n) + G (P/G,i%,n)

Arithmetic gradient present worth is given by,

(P/G,i,n)={[(1+i)n−in−1]/[i2(1+i)n]}

The upper cash flow is equivalent to a uniform series of A = $85,000 for 5 years, an arithmetic series of G = 10,000 for 5 years. So the present worth of the above cash flow is

P= $85,000(P/A,4%,5) - $10,000(P/G,4%,5)

=$85,000(4.45) - $10,000(8.55)

=$378,250 - $85,500

=$292,750.

Conclusion:

Thus, the present worth of the repair cost for a 5 year period is $292,750.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Given information:

Maintenance cost for 1^st year (A1) = $85,000

Cost will increase by $150 each year, so G=$10,000

Time (n) = 5 years

Interest rate (i) = 4%.

Concept used:

Arithmetic gradient present worth is given by,

(P/G,i,n)={[(1+i)n−in−1]/[i2(1+i)n]}

To find the Present Worth, at EOY 0, of a gradient series that begins EOY 1, use

P = A1 (P/A,i%,n) + G (P/G,i%,n)

To find the annual equivalent (A series) of a gradient series that begins EOY 1, use

A = A1 + G (A/G,i%,n).

Calculation:

In order to find out the present worth of the cash maintenance cash flow we use the given formula,

To find the Present Worth, at EOY 0, of a gradient series that begins EOY 1, use

P = A1 (P/A,i%,n) + G (P/G,i%,n)

Arithmetic gradient present worth is given by,

(P/G,i,n)={[(1+i)n−in−1]/[i2(1+i)n]}

The upper cash flow is equivalent to a uniform series of A = $85,000 for 5 years, an arithmetic series of G = 10,000 for 5 years. So the present worth of the above cash flow is

P= $85,000(P/A,4%,5) - $10,000(P/G,4%,5)