How much should the smog control district pay to the firm now to provide for the first cost of the equipment and its maintenance for 10 years?

Question

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Answer 1

Question

Chapter 4, Problem 55P

To determine

How much should the smog control district pay to the firm now to provide for the first cost of the equipment and its maintenance for 10 years?

Expert Solution & Answer

Answer to Problem 55P

$11,292

Given information:

Maintenance cost for 1^st year (A1) = $75

Cost will increase by $150 each year, so G=$25

Time (n) = 10 years

Interest rate (i) = 6%.

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 = 75 for 10 years, an arithmetic series of G = 25 for 10 years, a single cash flow of $10,000 in the year 0, so the present worth of the above cash flow is

P= $75(P/A,6%,10) + $25(P/G,6%,10)

=$75(7.36) + $25(29.60)

=$552 + $740

=$1,292

Total present worth = $10,000+$1,292

=$11,292.

Conclusion:

Thus, the smog control district pays to the firm $11,292 now to provide for the first cost of the equipment and its maintenance for 10 years.

Explanation of Solution

Given information:

Maintenance cost for 1^st year (A1) = $75

Cost will increase by $150 each year, so G=$25

Time (n) = 10 years

Interest rate (i) = 6%.

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 = 75 for 10 years, an arithmetic series of G = 25 for 10 years, a single cash flow of $10,000 in the year 0, so the present worth of the above cash flow is

P= $75(P/A,6%,10) + $25(P/G,6%,10)

=$75(7.36) + $25(29.60)

=$552 + $740

=$1,292

Total present worth = $10,000+$1,292

=$11,292.

Conclusion:

Thus, the smog control district pays to the firm $11,292 now to provide for the first cost of the equipment and its maintenance for 10 years.

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A firm operates with the production function Q = K2 L. Q is the number of units of output per day when the firm rents K units of capital and employs L workers each day. The manager has been given a production target: to produce 8,000 units per day. She knows that the daily rental price of capital is $400 per unit and the wage rate is $200 day. a. What is the returns to scale of this production function? Show mathematically. b. Currently the firm employs 80 workers per day. What is the firm’s daily total cost if it rents just enough capital to produce at its target? c. Compare the marginal product per dollar spent on K and on L when the firm operates at the input choice in part (b). What does this suggest about the way the firm might change its choice of K and L if it wants to reduce the total cost in meeting its target? Explain your answer very clearly. d. In the long run, how much K and L should the firm choose if it wants to minimize the cost of producing 8,000 units of output a day?…

Answer 2

Question

Chapter 4, Problem 55P

To determine

How much should the smog control district pay to the firm now to provide for the first cost of the equipment and its maintenance for 10 years?

Expert Solution & Answer

Answer to Problem 55P

$11,292

Given information:

Maintenance cost for 1^st year (A1) = $75

Cost will increase by $150 each year, so G=$25

Time (n) = 10 years

Interest rate (i) = 6%.

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 = 75 for 10 years, an arithmetic series of G = 25 for 10 years, a single cash flow of $10,000 in the year 0, so the present worth of the above cash flow is

P= $75(P/A,6%,10) + $25(P/G,6%,10)

=$75(7.36) + $25(29.60)

=$552 + $740

=$1,292

Total present worth = $10,000+$1,292

=$11,292.

Conclusion:

Thus, the smog control district pays to the firm $11,292 now to provide for the first cost of the equipment and its maintenance for 10 years.

Explanation of Solution

Given information:

Maintenance cost for 1^st year (A1) = $75

Cost will increase by $150 each year, so G=$25

Time (n) = 10 years

Interest rate (i) = 6%.

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 = 75 for 10 years, an arithmetic series of G = 25 for 10 years, a single cash flow of $10,000 in the year 0, so the present worth of the above cash flow is

P= $75(P/A,6%,10) + $25(P/G,6%,10)

=$75(7.36) + $25(29.60)

=$552 + $740

=$1,292

Total present worth = $10,000+$1,292

=$11,292.

Conclusion:

Thus, the smog control district pays to the firm $11,292 now to provide for the first cost of the equipment and its maintenance for 10 years.

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Answer 3

Question

Chapter 4, Problem 55P

To determine

How much should the smog control district pay to the firm now to provide for the first cost of the equipment and its maintenance for 10 years?

Expert Solution & Answer

Answer to Problem 55P

$11,292

Given information:

Maintenance cost for 1^st year (A1) = $75

Cost will increase by $150 each year, so G=$25

Time (n) = 10 years

Interest rate (i) = 6%.

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 = 75 for 10 years, an arithmetic series of G = 25 for 10 years, a single cash flow of $10,000 in the year 0, so the present worth of the above cash flow is

P= $75(P/A,6%,10) + $25(P/G,6%,10)

=$75(7.36) + $25(29.60)

=$552 + $740

=$1,292

Total present worth = $10,000+$1,292

=$11,292.

Conclusion:

Thus, the smog control district pays to the firm $11,292 now to provide for the first cost of the equipment and its maintenance for 10 years.

Explanation of Solution

Given information:

Maintenance cost for 1^st year (A1) = $75

Cost will increase by $150 each year, so G=$25

Time (n) = 10 years

Interest rate (i) = 6%.

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 = 75 for 10 years, an arithmetic series of G = 25 for 10 years, a single cash flow of $10,000 in the year 0, so the present worth of the above cash flow is

P= $75(P/A,6%,10) + $25(P/G,6%,10)

=$75(7.36) + $25(29.60)

=$552 + $740

=$1,292

Total present worth = $10,000+$1,292

=$11,292.

Conclusion:

Thus, the smog control district pays to the firm $11,292 now to provide for the first cost of the equipment and its maintenance for 10 years.

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Answer 4

Question

Chapter 4, Problem 55P

To determine

How much should the smog control district pay to the firm now to provide for the first cost of the equipment and its maintenance for 10 years?

Expert Solution & Answer

Answer to Problem 55P

$11,292

Given information:

Maintenance cost for 1^st year (A1) = $75

Cost will increase by $150 each year, so G=$25

Time (n) = 10 years

Interest rate (i) = 6%.

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 = 75 for 10 years, an arithmetic series of G = 25 for 10 years, a single cash flow of $10,000 in the year 0, so the present worth of the above cash flow is

P= $75(P/A,6%,10) + $25(P/G,6%,10)

=$75(7.36) + $25(29.60)

=$552 + $740

=$1,292

Total present worth = $10,000+$1,292

=$11,292.

Conclusion:

Thus, the smog control district pays to the firm $11,292 now to provide for the first cost of the equipment and its maintenance for 10 years.

Explanation of Solution

Given information:

Maintenance cost for 1^st year (A1) = $75

Cost will increase by $150 each year, so G=$25

Time (n) = 10 years

Interest rate (i) = 6%.

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 = 75 for 10 years, an arithmetic series of G = 25 for 10 years, a single cash flow of $10,000 in the year 0, so the present worth of the above cash flow is

P= $75(P/A,6%,10) + $25(P/G,6%,10)

=$75(7.36) + $25(29.60)

=$552 + $740

=$1,292

Total present worth = $10,000+$1,292

=$11,292.

Conclusion:

Thus, the smog control district pays to the firm $11,292 now to provide for the first cost of the equipment and its maintenance for 10 years.

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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 55P

To determine

How much should the smog control district pay to the firm now to provide for the first cost of the equipment and its maintenance for 10 years?

Expert Solution & Answer

Answer to Problem 55P

$11,292

Given information:

Maintenance cost for 1^st year (A1) = $75

Cost will increase by $150 each year, so G=$25

Time (n) = 10 years

Interest rate (i) = 6%.

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 = 75 for 10 years, an arithmetic series of G = 25 for 10 years, a single cash flow of $10,000 in the year 0, so the present worth of the above cash flow is

P= $75(P/A,6%,10) + $25(P/G,6%,10)

=$75(7.36) + $25(29.60)

=$552 + $740

=$1,292

Total present worth = $10,000+$1,292

=$11,292.

Conclusion:

Thus, the smog control district pays to the firm $11,292 now to provide for the first cost of the equipment and its maintenance for 10 years.

Explanation of Solution

Given information:

Maintenance cost for 1^st year (A1) = $75

Cost will increase by $150 each year, so G=$25

Time (n) = 10 years

Interest rate (i) = 6%.

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 = 75 for 10 years, an arithmetic series of G = 25 for 10 years, a single cash flow of $10,000 in the year 0, so the present worth of the above cash flow is

P= $75(P/A,6%,10) + $25(P/G,6%,10)

=$75(7.36) + $25(29.60)

=$552 + $740

=$1,292

Total present worth = $10,000+$1,292

=$11,292.

Conclusion:

Thus, the smog control district pays to the firm $11,292 now to provide for the first cost of the equipment and its maintenance for 10 years.

Answer 6

Chapter 4, Problem 55P

To determine

How much should the smog control district pay to the firm now to provide for the first cost of the equipment and its maintenance for 10 years?

Expert Solution & Answer

Answer to Problem 55P

$11,292

Given information:

Maintenance cost for 1^st year (A1) = $75

Cost will increase by $150 each year, so G=$25

Time (n) = 10 years

Interest rate (i) = 6%.

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 = 75 for 10 years, an arithmetic series of G = 25 for 10 years, a single cash flow of $10,000 in the year 0, so the present worth of the above cash flow is

P= $75(P/A,6%,10) + $25(P/G,6%,10)

=$75(7.36) + $25(29.60)

=$552 + $740

=$1,292

Total present worth = $10,000+$1,292

=$11,292.

Conclusion:

Thus, the smog control district pays to the firm $11,292 now to provide for the first cost of the equipment and its maintenance for 10 years.

Explanation of Solution

Given information:

Maintenance cost for 1^st year (A1) = $75

Cost will increase by $150 each year, so G=$25

Time (n) = 10 years

Interest rate (i) = 6%.

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 = 75 for 10 years, an arithmetic series of G = 25 for 10 years, a single cash flow of $10,000 in the year 0, so the present worth of the above cash flow is

P= $75(P/A,6%,10) + $25(P/G,6%,10)

=$75(7.36) + $25(29.60)

=$552 + $740

=$1,292

Total present worth = $10,000+$1,292

=$11,292.

Conclusion:

Thus, the smog control district pays to the firm $11,292 now to provide for the first cost of the equipment and its maintenance for 10 years.

Answer 7

Chapter 4, Problem 55P

To determine

How much should the smog control district pay to the firm now to provide for the first cost of the equipment and its maintenance for 10 years?

Answer 8

Expert Solution & Answer

Answer to Problem 55P

$11,292

Given information:

Maintenance cost for 1^st year (A1) = $75

Cost will increase by $150 each year, so G=$25

Time (n) = 10 years

Interest rate (i) = 6%.

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 = 75 for 10 years, an arithmetic series of G = 25 for 10 years, a single cash flow of $10,000 in the year 0, so the present worth of the above cash flow is

P= $75(P/A,6%,10) + $25(P/G,6%,10)

=$75(7.36) + $25(29.60)

=$552 + $740

=$1,292

Total present worth = $10,000+$1,292

=$11,292.

Conclusion:

Thus, the smog control district pays to the firm $11,292 now to provide for the first cost of the equipment and its maintenance for 10 years.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Given information:

Maintenance cost for 1^st year (A1) = $75

Cost will increase by $150 each year, so G=$25

Time (n) = 10 years

Interest rate (i) = 6%.

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 = 75 for 10 years, an arithmetic series of G = 25 for 10 years, a single cash flow of $10,000 in the year 0, so the present worth of the above cash flow is

P= $75(P/A,6%,10) + $25(P/G,6%,10)