equipment. this vehicle and its economic life. ent. 13-13 The Clap Chemical Comp
ID: 1113125 • Letter: E
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equipment. this vehicle and its economic life. ent. 13-13 The Clap Chemical Company needs a large insulated stainless steel tank to expand its plant. A recently closed brewery has offered to sell their tank for $15,000 delivered. The price is so low that Clap believes it can sell the tank at any future time and recover its $15,000 investment. ned replace- rchased and declared sal life of 10 vith a more ing installa- open market ld machine for the old Installing the tank will cost $9000 and removing it will cost $5000. The outside of the tank is cov- ered with heavy insulation that requires considerable maintenance. Insulation Year Maintenance Cost $2000 500 1000 1500 2000 2500 0 year has a drop by for the next Deter- for each of will 2 250. 4 $40,000. day, or for or $2000 at nnual oper- 6 years will and $4700. service for for (a) Based on a 12% before-tax MARR, what life of the insulated tank has the lowest EUAC? (b) Is it likely that the insulated tank will be replaced by another tank at the end of the period with the lowest EUAC? Explain.Explanation / Answer
PART-1) As the initial cost and salvage value are both $15,000, thus P = S = $15,000
Thus, EUAC of capital recovery = 15000*0.15 = 2250
The maintenance cost has 3 components, P = 2000, A = 500, and G = 500.
EUAC of the maintenance cost = 2000(A/P,15%,n)+500+500(A/G,15%,n)
When n = 1:EUAC (Maintenance) =2000*(1.1500) + 500 + 500*(0.000) = 2800.00
When n = 2:EUAC (Maintenance) =2000*(0.6151) + 500 + 500*(0.465) = 1962.70
When n = 3:EUAC (Maintenance) =2000*(0.4380) + 500 + 500*(0.907) =1829.50
When n = 4:EUAC (Maintenance) =2000*(0.3503) + 500 + 500*(1.326) =1863.60
When n = 5:EUAC (Maintenance) =2000*(0.2983) + 500 + 500*(1.723) =1958.10
Total EUAC = $2,250 + EUAC of Maintenance
Thus, to minimize Total EUAC, select the alternative with minimum EUAC of maintenance. As here the capital recovery EUAC is constant which is based on the EUAC calculation of the maintenance cost thus economic life of the new tank = 3 years.
PART- 2) The stainless steel tank will be compared always with the best available replacement or challenger. When the challenger is superior, then the defender tank probably will be replaced. However there will cost be a substantial amount for the removal of the existing tank from the plant when it is sold it to someone else, and then buy and install another one.
In this scenario the total marginal cost of this tank after 3 years =Loss of market value + Forgone interest + maintenance cost of fourthyear = 0 + (15000*0.15) + 2000 = $4250.00
The marginal cost is will be keeping the defender for one more year. This will be increasing at a rate of $500 each year (given G=500).
When the challenger becomes available in the market then will have total EUAC less the total EUAC of this tank (which equals 2250+1829.50) then should be replaced.
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