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 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,

$(P/G,i,n)=\{[{(1+i)}^n-in-1]/[i^2{(1+i)}^n]\}$

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

$P=A1 \left(P/A,i\%,n\right)+G\left(P/G,i\%,n\right)$

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

$A=A1 +G\left(A/G,i\%,n\right)$.

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

$P=A1 \left(P/A,i\%,n\right)+G\left(P/G,i\%,n\right)$

Arithmetic gradient present worth is given by,

$(P/G,i,n)=\{[{(1+i)}^n-in-1]/[i^2{(1+i)}^n]\}$

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

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

$=\$75\left(7.36\right)+\$25\left(29.60\right)$

$=\$552+\$740$

$=\$1,292$

$Totalpresentworth=\$10,000+\$1,292$

$=\$11,292$.

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,

$(P/G,i,n)=\{[{(1+i)}^n-in-1]/[i^2{(1+i)}^n]\}$

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

$P=A1 \left(P/A,i\%,n\right)+G\left(P/G,i\%,n\right)$

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

$A=A1 +G\left(A/G,i\%,n\right)$.

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

$P=A1 \left(P/A,i\%,n\right)+G\left(P/G,i\%,n\right)$

Arithmetic gradient present worth is given by,

$(P/G,i,n)=\{[{(1+i)}^n-in-1]/[i^2{(1+i)}^n]\}$

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

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

$=\$75\left(7.36\right)+\$25\left(29.60\right)$

$=\$552+\$740$

$=\$1,292$

$Totalpresentworth=\$10,000+\$1,292$

$=\$11,292$.

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.