## Replacement Answers 1a.

NAC 1 = 50000(A/P, .1, 1) + 10,000 = 50000*1.1 + 10000 = 65000

NAC 2 = 50000(A/P, .1, 2) + 10000 + 30000(A/G, .1, 2)
= 50000*0.5762 + 10000 + 30000*0.476= 28810 + 10000 + 14280 = 53090

NAC 3 = 50000(A/P, .1, 3) + 10000 + 30000(A/G, .1, 3)
= 50000*0.4021 + 10000 + 30000*0.937= 20105 + 10000 + 28110 = 58215

The economic life is two years. 1b.

NAC 1 = 50000(A/P, .1, 1) + 10,000 - 30000(A/F, .1, 1) = 50000*1.1 - 20000 = 35000

NAC 2 = 50000(A/P, .1, 2) + 10000 - 20000(A/F, .1, 2)
= 50000*0.5762 + 10000 - 20000*0.4762= 28810 + 10000 - 9524 = 29286

NAC 3 = 50000(A/P, .1, 3) + 10000 = 50000*0.4021 + 10000
= 20105 + 10000 = 30105

The economic life is two years. 2.a. What is the Net Annual Cost of the challenger?

NAC = 200,000(A/P, 0.1, 10)= 200,000_0.1627 = 32,540

b. What is the economic life of the defender?

For the defender:

The NAC for 1 year is (30,000)(A/P, 0.1, 1) + 20,000 + 10,000 = 63,000

The NAC for 5 years is (30,000)(A/P, 0.1, 5) + 20,000 + 10,000(A/F, 0.1, 5) = 28,541

There's no need to compute the NAC for the other years as they will obviously be greater than 28,541. The economic life is five years and the associated annual cost os \$28,541.

c. What is the best decision?

Since the NAC is greater than for the defender, do not make the replacement and keep the defender for five years. 3.

a. For a six year tax life, the SOYD is 6*7/2 = 21. The depreciation for the first three years is

1: 14,000*6/21 = 4000 2: 14,000*5/21 = 3333 3: 14,000*4/21 = 2667.

The book value after 3 years is 14,000 ­ 4000 ­ 3333 ­ 2667 = 4000

Selling the asset for \$10,000, results in a taxable gain of 6000. The tax is 2400, so the net receipts is 7600.

b.

The NAC for 1 year is (10,000 ­ 6000)(A/P, 0.1, 1) + 6000*0.1 + 1000 = 6000

The NAC for 2 years is (10,000 ­ 4000)(A/P, 0.1, 2) + 4000*0.1 + 1000 = 4857.20

The NAC for 3 years is (10,000)(A/P, 0.1, 3) = 5021.

The economic life of the old machine is two years. It's NAC at this life is less than that of the challenger. Keep the old machine. 4. For the old car: P = 3500 ­ 1000 = 2500, life = 1 year, S = 1000, Maint. = 1200

NAC_OLD = 1500(A/P, 0.1, 1) + 1000*0.1 + 1200 = 2950.0

NAC_NEW = (12,000­2000)(A/P, 0.1, 10) + 2000*0.1 + 500 = 2627

Buy the new car and sell the old one. 5. Existing Machine (Defender)

We use the NAW analysis here as the economic lives of the two options are different. PW and ROR analysis would not be valid for this problem due to the fact that identical replacement for the existing machine is not feasible.

 Year BTCF Depr. TI Tax ATCF 0 ­\$3500 ­\$3500 1-4 \$2000 \$1000 \$1000 \$500 \$1500

NAW = ­3500(A/P,10%,4) + 1500 = \$395.75

Challenger

 Year BTCF Depr. TI Tax ATCF 0 -\$10,000 -\$10,000 1-9 3000 \$800 \$2200 \$1100 1900 10 3000+2000 800 2200 1100 1900+2000

NAW= ­10,000(A/P,10%,10) + 1900 + 2,000(A/F,10%,10)= \$398.4

As the NAW of new machine is higher, we replace the existing machine with a new one. 6. Current book value of machine A = Cost ­ Depreciation to date

Book Value = 54,000 ­ (9/12)(54,000 ­ 0) = 13,500.

Long term capital gain if sold now = 30,000 ­ 13,500 = 16,500

Net return if machine A is replaced = 30,000 ­ 16,500*0.4 = 23,400

Machine A annual depreciation = (P ­ S)/n = (54,000 ­ 0)/12 = 4500 for 3 more years.

Alternative 1:

Keep A for 12 more years

 Year BTCF Depr. TI Tax ATCF 0 -30,000 -23,400 1-3 -7500 4500 -12,000 -4800 -2700 4-12 -7500 0 -7500 -3000 -4500

After tax annual cost:

NAC = [23,400 ­ 1800(P/A,10%,3)][(A/P,10%,12)] + 4500 = \$7278

The after tax cash flow in year 0 reflects the loss of income after capital gains tax from not selling machine A.

Alternative 2:

Buy machine B

Machine B annual depreciation = (P ­ S)/n = (42,000 ­ 0)/12 = \$3500

 Year BTCF Depr. TI Tax ATCF 0 -42,000 -42,000 1-12 -5000 3500 -8500 -3400 -1600

After tax annual cost = 42,000(A/P,10%,12) +1600 = \$7766

Choose the alternative with smaller annual cost => Keep Machine A 