MNET 414

HOME WORK ON CHAPTER 13

13-5.

Economic service life is the life that will give lowest Uniform Equivalent Annual Cost.

The bottling machine has two cost components (1) Capital recovery cost and (2) maintenance cost.

EUAC of Capital recovery = (P-S)*(A/P,I,n) + S*I = 40,000*(A/P, I, n). For any given interest rate, this EUAC will decrease indefinitely as ‘n’ increases.

EUAC for maintenance = 2500

So, total EUAC = 40,000*(A/P, I, n) + 2500

This total EUAC will be minimum when n = infinity.

Thus the economic life is infinity.

13.9.

(a) As the initial cost and salvage value are both $15,000,

EUAC of capital recovery = 15000*0.15 = 2250

This EUAC remains constant irrespective of the number of years it is used.

If you look at the maintenance cost, it has three components, P = 2000, A = 500, and G = 500

Thus EUAC of the maintenance cost = 2000(A/P,15%,n)+500+500(A/G,15%,n)

For n = 1: EUAC(Maintenance) =2000*(1.1500)+500+ 500*(0.000) = 2800.00

For n = 2: EUAC(Maintenance) =2000*(0.6151)+500+ 500*(0.465) = 1962.70

For n = 3: EUAC(Maintenance) =2000*(0.4380)+500+ 500*(0.907) = 1829.50

For n = 4: EUAC(Maintenance) =2000*(0.3503)+500+ 500*(1.326) = 1863.60

For n = 5: EUAC(Maintenance) =2000*(0.2983)+500+ 500*(1.723) = 1958.10

As the capital recovery EUAC is constant, based on the EUAC calculation of the maintenance cost, economic life of the new tank = 3 years.

(b) [This part requires the knowledge of marginal cost]

It may or may not be likely that this tank will be replaced after its economic life of 3 years.

This is because of the following:

The total marginal cost of this tank after 3 years =

Loss of market value + Forgone interest + maintenance cost of 4^th year

= 0 + 15000*0.15 + 2000 = $4250.00

The marginal cost is basically keeping the defender for one more year.

This marginal cost will increase at a rate of $500 every year (because G=500).

When this marginal cost of the defender exceeds the EUAC of a challenger, it is economic to replace the defender. Thus without the data about new tanks that will come in the future market, it is difficult to predict what will happen. If technology changes, for example, a plastic tank is capable of replacing the stainless steel tank with no or little requirement of insulation is available, then it might have a lower EUAC, and the tank may be economic to be replaced even earlier than its economic life.

If a challenger becomes available in the market that has a total EUAC less the total EUAC of this tank (which is 2250+1829.50) it should be replaced.

13-15.

The problem is asking you to do a before tax cash flow analysis, thus the depreciation information is not needed to solve the problem. Additionally, the past cost which was incurred 8 years ago and its present book value have no consequences in the decision at hand.

We are basically comparing the following two cash flows.

Year	Recondition	Buy New
0	-35,000	-75,000
1	0	+7,000
2	0	+7,000
3	0	+7,000
…	……	…….
10	+10,000	+7,000 +15,000

In the absence of any marginal cost data of the existing tank, we compute the EUAC over next 10 years period at 15% rate and select the one with lowest EUAC.

EUAC (existing) =

(35000-10000)(A/P,15%,10) + 10000*0.15

= 25000*0.1993+1500 = 6482.50

EUAC (New) =

(75000-15000)(A/P,15%,10) + 15000*0.15-7000

= 60000*0.1993+2250-7000 = 11958-4750 = 7208.

As EUAC for the existing is lower, then reconditioning is cheaper and economic.