ch2 (part2)arithmetic gradient
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Ch2 (part2)arithmetic gradientTRANSCRIPT
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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