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 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,

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

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 = $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

$P=\$85,000\left(P/A,4\%,5\right)\text{ - }\$10,000\left(P/G,4\%,5\right)$

$=\$85,000\left(4.45\right)\text{ - }\$10,000\left(8.55\right)$

$=\$378,250\text{ - }\$85,500$

$=\$292,750$.

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,

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

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 = $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

$P=\$85,000\left(P/A,4\%,5\right)\text{ - }\$10,000\left(P/G,4\%,5\right)$

$=\$85,000\left(4.45\right)\text{ - }\$10,000\left(8.55\right)$

$=\$378,250\text{ - }\$85,500$

$=\$292,750$.

Conclusion:

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