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 1st 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,
To find the Present Worth, at EOY 0, of a gradient series that begins EOY 1, use
To find the annual equivalent (A series) of a gradient series that begins EOY 1, use
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
Arithmetic gradient present worth is given by,
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
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 1st 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,
To find the Present Worth, at EOY 0, of a gradient series that begins EOY 1, use
To find the annual equivalent (A series) of a gradient series that begins EOY 1, use
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
Arithmetic gradient present worth is given by,
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
Conclusion:
Thus, the present worth of the repair cost for a 5 year period is $292,750.
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