Transcript DEPRECIATION-AFTER/BEFORE TAX RATE OF RETURN

DEPRECIATION-AFTER/BEFORE TAX
RATE OF RETURN
Exercise:An automobile manufacturer is buying some special tools
for 100.000 $. The corporation will pay 20.000 $ now and borrow
the remaining 80.000 $ which will be paid starting from the first
year with an interest rate on loan of 10% by 4 equal end of year
payments. The tools are being depreciated by double declining
balance using a 4 year depreciable life and a 6250 $ salvage value.
It is expected the tools will actually be kept in service for 6 years
and then sold for 6250 $. The income tax rate is 46%.
a)Calculate the before tax rate of return
b) Compute the after-tax rate of return by preparing an expanded
cash flow table.
c) What woul be the after-tax rate of return if the company did
not borrow 80.000 $ ( In other words total equity= 100.000 $)
Hint: Interest payments are tax deductible
.
Year
Before-Tax Cash Flow
($)
1
30.000
2
30.000
3
35.000
4
40.000
5
10.000
6
10.000
Solution: 80000 (A/P,10%,4) = 25240 =Total loan paid
annually
80000/4= 20000 (principal annual payment)
25240-20000 = 5240 (interest annual payment)
Year
BeforeTax Cash
Flow
2
Principal
payment
3
Interest
payment
4
Total
Loan
Payment
5=3+4
Net
Cash
Flow
before
tax 6=
2-5
Depreciation
7
(DDB= 2/4=
50%)
Taxes
8= [2(4+7)] x
0.46
0
-20000
1
30000
20000
5240
25240
4760
50000
-
4760
2
30000
20000
5240
25240
4760
25000
-
4760
3
35000
20000
5240
25240
9760
12500
7940
1820
4
40000
20000
5240
25240
14760
6250
13115
1645
5
10000
10000
4600
5400
6
10000
6250
10000
6250
4600
11650
20000
Net Cash Flow After
taxes
9= 6-8
-20000
• a) Trial error for estimating IRRBT
• Compound interest tables i=30%
• NPW= -20000+ 4760 (P/F, 30%,1) + 4760
(P/F,30%,2) + 9760 (P/F,30%,3) + 14760
(P/F,30%,4) + 10000 (P/F,30%,5) + 16250
(P/F,30%,6)= 2148
• For i= 35% NPW= -537,25
30 %
2148
X
0
35%
-537,25
(X-30) / -2148 = (35-30)/ -537,25-2148
IRRBT = 34%
• b) Calculating IRRAT
• Trial Error for i=12%
• NPW= -20000 + 4760 (P/F,12%,1) + 4760(P/F,12%,2) +
1820 (P/F,12%,3) + 1645 (P/F,12%,4) + 5400 (P/F,12%,5)
+ 11650 (P/F,12%,6)= -648,4
• Trial error for i= 10% NPW= 681
• Interpolation 1329.4 x -13294=1296.8
• X= 10.97 %
• If we had used the equation iAT= iBT x (1-T), we would
have found out iAT= 0.34x(1-0.46) = 0.1836 , but this is
inaccurate since this method is prevalent for capital
gains or losses obtained by the sale of non-depreciable
assets such as land or stocks
Replacement Analysis-Exercise
• The Traverse company bought 5 years ago a milling
machine Y for 35.000 TL. The salvage value at the end
of its useful life is 5000 TL and its useful life is 10 years.
The present market value of Y is 13.000 TL . If the
company makes a revision and modifies the machine Y
by spending 8000 TL, the machine Y will be available
for additional 10 years with an annual operating cost of
7500 TL and a salvage value of 3000 TL
• On the other hand, there is a second alternative :It
buys a new machine Z of 50.000 TL with a useful life of
20 years and annual maintenance costs of 4000 TL?
• Assess which of the alternatives should be selected?
Solution
DEFENDER
CHALLENGER
Present value
21.000(13.000 +8000)
50.000
Salvage value
3000
20.000
Annual operating cost
7500
4000
Useful life
10 years
20 years
MARR
40%
40%
EUACD= 21.000 (A/P,40%,10) – 3000(A/F,40%,10) + 7500= 16.158
TL
EUACC= 50.000 (A/P, 40%, 20) – 20.000 (A/F,40%,20) + 4000=
24.014 TL
In this case, the old machine must be used for 10 years by making
a modification since its equivalent uniform annual costs are
lower