SOLUTIONS TO CHAPTER 8 PROBLEMS
Dealing With Price Changes in Capital Investment Analysis.
8-1. (a)
(0)
0
P=$30,000(P/A,5%,15)=$311,391
(b) ir = 1.9%–or 0.019
1.05
0.050.03 −=
−
8-2. R$ )0(
n= 1 A$n = 2 f = 3%
8-3. R$ )0(
40 = $1,000,000(P/F,3%,20) = $553,700
8-4. (a) Labor2007 = $1,000,000(F/P,10%,5) = $1,610,510
18-5. If you don’t purchase the lathe now,
95
8–6. 0=-$8,870+$1,000(P/A,ic,10)+$10,000(P/F,ic,10)
ir= =
06.1
8-7. (a) EW2005 = $10,000 + $1 0,000(P/F, 12%,1) + $12,000(P/F,12%,2) = $28,495
(b) 2005: (2005)
2005
R$ = $10,000(P/F,8%,0) = $10,000
8-8. Alternative Present Worth of Costs ( (0)
0
P)
I $12,000 = $12,000
II $7,000 + $6,000(P/F,5%,6) = $11,477
8–9. Assuming your son was just born (deposit will be made four years hence).
Actual value of
account at end = A$18 = $50,000(F/P,6%,18) = $142,715
year eighteen
96
8-10. Using is = 5% + 10% = 15%
(a) Alternative A:
(0)
0
P= $110,000(P/A,15%,4) + $2,000(P/G,15%,4) = $321,630
Alternative B:
8-11. R$ )0(
10 = $400M(1.75) = $700M
8-12. ir = 08.1
08.025.0 −= 0.1574 or 15.74 %
Find minimum N (integer) such that
97
8-13. ic = (1.02)(1.04) – 1 = 0.061 or 6.1%
8-14. (a) 0 = -$28,000+ 5,000(P/A, ir,15)
By trial and error, ir ≅ 15.9%
1.159
8–15 Annual Cost of energy
(A(0)) = 50.562,1$
ft
$2.50
Btu10
1ft
0.8
Btu10500
36
36
=
×
5% 4.545%or 0.04545
1.10
0.100.15
ir≅=
−
=
(0)
0
PW = $1,562.5(P/A,4.545%,12) = $14,211
If the approximate ir = 5% were used,
(0)
0
P would = $13,849
EOY,k Amount
Deposited (A$) k
FW of Deposit @ EOY5
(A$)5 = A($)k * (F/P,6.1%,5-k)
1 $1
,
000 1
,
262
2
$
1
,
080 1
,
286
3 $1
,
166 1
,
310
4 $1,253 1,334
5 $1
,
361 1
,
360
Total (A$)5 = FW5 = $6,552
98
8-16. PW(0) = –$2,500 – $100(P/A,9%,10) – $500(P/F,9%,8) – $680(P/A,9%,10)
= -$7,757
18-17. (a)
Year
(A)
BTCF
(R$)
(B)
INFL.
Adjustment
(1.10)Year
(C)
BTCF
(A$)
(D)
Lease
Payment
(A$)
(E)
Taxable
Income:
= C+D
(F)
Cash Flow
for Income
Taxes
= -t x E
(G)
ATCF
(A$)
=C+D+F
99
8-18. Assume neither machine has a salvage value. Incremental Analsis: A Æ B
(A) (B) (C) (D) (E) (F) (G)
Year
BTCF
(R$)
Infl.
Adujustment
(1.10)year-1
BTCF
(A$)
Depre. Taxable
Income
= C -D
Cash Flow
for income
Taxes
= -t x E
ATCF
A($)
=C-D+F
0 -50,000 – – – – – -50,000
1 500 1.000 500 $10,000 -9,500 4,750 5,250
2 500 1.100 550 16,000 -15,450 7,725 8,275
100
8-19.
Yeark
(A)
BTCF
(R$)
(B)
Inflation
Adjustment
(1.06)Year
(C)
BTCF
(A$)
(D)
Property
Tax
(A$)
(E)
Total
BTCF
(A$)
(F)
Depre.
(G)
Taxable
Income:
E-F
(H)
Cash
Flow for
Income
Taxes
= -txG
(I)
ATCF
(A$)
= E+H
(J)
ATCF
(R$)
=(I) x
(P/F,4.5%,Year)
0 –
$
260
,
000 1.000 –
$
260
,
000 — –
$
260
,
000 – – –
$
260
,
000 –
$
260
,
000
1 -6,000 1.060 -6,360 -$10,400 – 16,760 $52,000 -$ 68,760 $27,504 10,744 10,281
2 -6
,
000 1.124 -6
,
742 -10
,
400 – 17
,
142 83
,
200 -100
,
342 40
,
137 22
,
995 21
,
057
3 -6
,
000 1.191 -7
,
146 -10
,
400 – 17
,
546 49
,
920 -67
,
466 26
,
986 9
,
440 8
,
272
4 -6,000 1.262 -7,572 -10,400 – 17,972 29,952 -47,924 19,170 1,198 1,005
5 -6,000 1.338 –8,028 -10,400 – 18 428 29 952 -48,380 19,352 924 741
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8-20. dk=($150,000 – 0) / 3 =$50,000
(A) (B)
Revenues Expenses
A($) A($)
(C)
BTCF
(A$)
= A + B
(D)
Depre.
(E)
Taxable
Income
= C – D
(F)
Cash Flow
for Income
Taxes
= -t x
E
(G)
ATCF
(A$)
= C + F
0 -$150,000 -$150,000 — — — -$150,000
8-21.
Option 1: Software with 3 year upgrade agreement:
(A)
BTCF
(B)
Depre.
(C)
Taxable
Income:
= A – B
(D)
Cash Flow
for Income
taxes
=-txC
(E)
ATCF
(A$)
A+D
0 –$X — — — –$X
102
8-21. (continued)
Set PW1= PW2 and solve for X.
103