Engineering Economic Analysis

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Rate of Return Analysis 213

If an electromagnet is installed on the input conveyor of a coal-processing plant, it will pick up

scrap metal in the coal. The removal of this metal will save an estimated $1200 per year in costs

associated with machinery damage due to metal. The electromagnetic equipment has an estimated

useful life of 5 years and no salvage value. Two suppliers have been contacted: Leaseco will provide

the equipment in return for three beginning-of-year annual payments of $1000 each; Saleco will

provide the equipment for $2783. If the MARR is 10%, which supplier should be selected?

SOLUTION

Sinceboth suppliers will provide equipment with the same useful life and benefits, thisis afixed-

output situation. In rate of return analysis, the method of solution is to examine the differences

between the alternatives. By taking Saleco - Leaseco,we obtain an increment of inves-tment..

Compute the NPW at various interest rates on the increment of investment represented by the

difference between the alternatives.

*Each year the cash flow is multiplied by(P/F, i, n).

AtO%:(P/F,0%,n)= 1 for all values ofn

At 00%:(P/F,00%,0) = 1

(Pj F,00%,n) = 0 for all other values ofn

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Difference Between Alternatives:

Year Leaseco Saleco Saleco-Leaseco

(^0) -$1000 -$2783 -$1783
1
{ -1000

• 1200 + 1200 +1000
  2
  { -1000


• 1200 + 1000.


• 1200
  (^3) + 1200 + 1200 0
  (^4) +1200 + 1200 0
  (^5) + 1200 +1200 0
  ,. ..' .,., -' .........,.g
  pw*
  Year Cash Flow:
  n Saleco-Leaseco At 0%. At 8% At 20% At 00%
  (^0) -$1783 -$1783 -$173 -$1783 -$1783
  (^1) + 1000 + 1000 +926 +833 0
  (^2) + 1000 + 1000 +857 +694 0
  3 0 0 0 0 0
  (^400000)
  (^500000)
  NPW= +217 0 -256 -1783