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34P/N. Each component is treated individually and converted to present day values using the GPW factor and the P/A factor, respectively. The sum of these two present worth factors must equal P. 34P * P/A N Since there is only one unknown, the formulas can be readily solved. The results are indicated on the next page. Copyright © 2002 by The Fairmont Press. 54 e=0 e = 10% e = 14% e = 20% Plant Engineers and Managers Guide to Energy Conservation N=5 N = 10 N = 15 N =20 $P $P $P $P ——————————————————————————— 2869 4000 4459 4648 ——————————————————————————— 3753 6292 8165 9618 ——————————————————————————— 4170 7598 10,676 13,567 ——————————————————————————— 4871 10,146 16,353 23,918 ——————————————————————————— Figure 2-9 illustrates the effects of escalation.

The effect of escalation is not considered. 2. A 5 percent fuel escalation is considered. 3. A 10 percent fuel escalation is considered. 4. A 14 percent fuel escalation is considered. 5. A 20 percent fuel escalation is considered. Calculate for 5-, 10-, 15-, 20-year life. Assume straight-line depreciation over useful life, 34 percent income tax bracket, and no tax credit. 34P N Thus, the after-tax savings (AS) are comprised of two components. The first component is a uniform series of $660 escalating at e percent/ year.

In the case where the asset has a value after the end of its useful life, the annual cost becomes: AC = (P – L) Copyright © 2002 by The Fairmont Press. * A/P + iL (2-16) 48 Plant Engineers and Managers Guide to Energy Conservation where AC is the annual cost L is the net sum of money that can be realized for a piece of equipment, over and above its removal cost, when it is returned at the end of the service life. L is referred to as the salvage value. As a practical point, the salvage value is usually small and can be neglected, considering the accuracy of future costs.

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