Capital Recovery CostIn most engineering situations, the capital cost has two components I = the investment required at time 0, and S = the salvage value received at time N.I 0 SNCapital Recovery (CR) cost is the annual equivalent of the capital cost. CR is oftendescribed from the bank’s point of view and is a function of the MARR, i%.CR(i) = I(A/P, i, N) + S(A/F, i, N) Because of the similarities between the formulas for (A/P, i, N) and (A/F, i, N) , we can also calculate CR(i) usingCR(i) = (I – S)(A/P, i, N) + iS, orCR(i) = (I – S)(A/F, i, N) + iI ExampleYour company is planning to manufacture a new product which requires a machinethat the company does not now own. The cost of the machine is $10,000, its life is 4 years, and its salvage value (at the end of the 4th year) is $1000. With this machine, the new profit from each product is $12. What annual production makes the investment worthwhile? The MARR is 10%.SolutionWe must find the capital recovery costs. The data areI = $10,000 S = $1,000MARR = 10% Useful life N = 4CR = I(A/P, MARR, N) – S(A/F, MARR, N) = $10,000(A/P,10%,4) – $1000(A/F, 10%, 4) = $10,000(0.3155) – $1000(0.2155) = $3155 – $215.50= $2939.5Using the other approaches, we getCR = (I – S)(A/P,10%,4) + (0.1)S = $9,000 (0.3155) + (0.1)$1000= $2839.5 + $100 = $2939.5CR = (I – S)(A/F,10%,4) + (0.1)I = $9000 (0.2155) + (0.1)$10,000= $1939.5 + $1000 = $2939.5To be profitable, revenues must equal or exceed the capital recovery costs. Let X be the number of products produced per year.Revenues – CR = $12X – $2939.5  0. Solving for X, we get X  245