If the cost of a new machine remained constant during this time, at what date would it be
most economical to replace the machine with a duplicate? Use a 7 percent interest rate.
Calculation Procedure:- Compute the present worth of all payments on the asset
Let P' denote the present worth (i.e., the value at the date of purchase) of all expenditures
ascribable to an asset. In the capital-recovery annual-cost equation, substitute P' for P and
set c = O to obtain the following alternative equation: A = (P' - L)(CR) + Li 1. Using I 1 = 7
percent, compute the present worth of the operating costs. Or,
Year
1 P W = ($2300)(0.9346) = $2150
2 P W = ($2500)(0.8734) = $2184
3 P W = ($3300)(0,8163) = $2694
4 P W = ($4800)(0.7629) = $3662
5 P W = ($6800)(0.7130) = $4848- Determine the present worth for each life span
Take the sum of the installed cost, $10,000, and the present worth of the operating cost
found in step 1. Or,
Life, years
0 P' = $ 10,000 + $0 = $ 10,000
1 P' = $10,000 + $2,150 = $12,150
2 P' = $12,150 + $2,184 = $14,334
3 P' = $14,334 ± $2,694 = $17,028
4 P' = $17,028 + $3,662 = $20,690
5 P' = $20,690 + $4,848 = $25,538- Apply the annual-cost equation developed in step 1
Life, years
1 A= $6,150(1.07000)+ $6,000(0.07) = $7,001
2 A = $10,334(0.55309) + $4,000(0.07) = $5,996
3 A = $13,828(0.38105) + $3,200(0.07) = $5,493
4 A = $18,190(0.29523) + $2,500(0.07) = $5,545
5 A = $23,538(0.24389) + $2,000(0.07) = $5,881Inspect these annual costs to determine when the minimum annual cost occurs. Since the
annual cost is a minimum when TV = 3, the asset should be retired at the end of the third
year.