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13-13 The Clap Chemical Company needs a large insulated stainless steel tank to

ID: 1118307 • Letter: 1

Question

13-13 The Clap Chemical Company needs a large insulated stainless steel tank to expand its plant. A recent 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. Installing the tank will cost $900%and remov- ing it will cost $5000. The outside of the tank is covered with heavy insulation that requires consid- erable maintenance. This will cost $3500 in year l and increase by $1000 per year (a) Based on a 12% before-tax MARR, what life of the insulated tank has the lowest EUAC?

Explanation / Answer

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.

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