It is estimated that the annual heat-loss cost in a small power plant is $5,200.

Question

It is estimated that the annual heat-loss cost in a small power plant is $5,200. Two mutually exclusive proposals have been formulated that will reduce the loss. Proposal A will reduce heat loss by 60% and cost $3,000. Proposal B will reduce heat loss by 55% and cost $2,500. If the interest rate is 8% and the plant will benefit from the reduction in heat loss for ten years, using the present worth on incremental investment, which proposal should be accepted? [Must show calculations to receive credit]

Explanation / Answer

Interest Rate = 8%

Life = 10 years

For proposal A,

Initial investment = Cost = $3000

Annual saving in heat loss cost = 60%*5200 = $3120

Incremental Cash Flows are as follows:

Present worth on incremental investment is the sum of dicounted cash flows = $17935.45

For proposal B

Initial investment = Cost = $2500

Annual saving in heat loss cost = 55%*5200 = $2860

Incremental Cash Flows are as follows:

Present worth on incremental investment is the sum of dicounted cash flows = $16690.83

Since proposal A has a higher present worth on incremental investment than proposal B, proposal A should be accepted.

Year Cash Flow Discounted Cash Flow 0.00 -3000.00 -3000.00 1.00 3120.00 2888.89 2.00 3120.00 2674.90 3.00 3120.00 2476.76 4.00 3120.00 2293.29 5.00 3120.00 2123.42 6.00 3120.00 1966.13 7.00 3120.00 1820.49 8.00 3120.00 1685.64 9.00 3120.00 1560.78 10.00 3120.00 1445.16 NPV 17935.45

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