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Gradient Exampels

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Arithmetic Gradient Factors

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Arithmetic Gradient Problem

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• Two computations must be made and added: the first for the present worth of the base amount PA and a second for the present worth of the gradient PG. The total present worth PT occurs in year 0. This is illustrated by the partitioned cash flow diagram.

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A)PT = 500(P/A,5%, IO) + 100(P/G,5%,10) = 500(7.7217) + 100(31.652) = $7026.05 ($7,026,050)

B) Here, too, it is necessary to consider the gradient and the base amount separately. The total annual series AT is found by AT = 500 + 100(A/G,5%,10) = 500 + 100(4.0991) = $909.91 per year ($909,910)

• And AT occurs from year 1 through year 10.• If PT is known AT can be calculated directly

AT = PT(A/P,5%,1O) = 7026.05(0.12950) = $909.87 ($909,870)

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Geometric Gradient• It is common for cash flow series, such as

operating costs, construction costs, and revenues, to increase or decrease from period to period by a constant percentage, for example, 5% per year. This uniform rate of change defines a geometric gradient series of cash flows. In addition to the symbols i and n used thus far, we now need the term g

• g = constant rate of change, in decimal form, by which amounts increase or decrease from one period to the next

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Geometric Gradient

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Geometric Gradient• In summary, the engineering economy relation

and factor formulas to calculate Pg in period t = 0 for a geometric gradient series starting in period 1 in the amount A I and increasing by a constant rate of g each period are

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Geometric Gradient Problem

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Geometric Gradient Problem