SOLUTIONS TO CHAPTER 13 PROBLEMS
Sensitivity Analysis
131
If N  project life, i  interest rate, and A  net annual cash flow,
we have AW $100,000(A/P,i,N) + A + $20,000(A/F,i,N)
%
Change
50%
40%
30%
20%
10%
0%
10%
20%
30%
40%
50%
Net Annual Worth
129.
_ Alternative _ Expected Payoff _ Pr (Payoff 0)
A1
$110(0.4)  $20(0.6) = $32
0.4
A2
$40(0.4) + $85(0.6) = $35
0.6
A3
$0(0.4) + $50(0.6) = $30
1.0
The answers to parts (a) and (b) will vary. The more risk averse person would
choose A3. Most people
$24,000,000
$60,000,000
$5,000,000
$89,000,000
1314 a) Price of Property
Equipment Investment
Working Capital
Total Capital Invested
Annual Revenues
Annual Cleaning Costs
Annual Operation Costs
Annual Gross Income
($38/ton)(2,000,000 tons)
($5 / ton)(2,0
b) Project is good if AW >= 0. Let x =change in investment and y =change in net
annual receipts.
$15,000(1+x)(A/P,15%,10) + $2,500(1+y) >= 0
489.50  2,989.5x + 2500y >= 0
Project is good if y >= 0.1958 + 1.1958x
160
1312 Let X = annual Btu requirement
1315.
(continued)
The plot is shown below.
166
1316
a) AC(A) = $2,000(A/P,10%,6) + $3,500 = $3,959
AC(B) = $6,000(A/P,10%,3) + $1,000  $4,000(A/F,10%,3) = $2,204
Project B would be selected given the "assumed certain" estimates.
b) To find sensitivity
1317
(continued)
Alternative B (revised to include extra investment permissible to breakeven):
MACRS
Taxable
Depreciation Income
Cash Flow
for Income
Taxes

ATCF
42,731,490
Year
0
BTCF
42,731,490
1
12,400,000
2,136,574 14,536,574
5,814,630
6,585,37
1318
The following equations represent the present worths of the defender
(D) and the challenger (C) as a function of the effective income tax
rate, t:
PW(0) = 150,000 + 38450(t / 2)  578,000(1  t)(P/A,10%,8)
+50000(1  t)(P/F,10%,8) + 44600(t)(P/F,10
phase of a project is carefully monitored so that corrective actions may be taken if
warranted. Postaudit reviews are conducted to establish in retrospect why a given
project was a success or a failure in terms of attaining original investment objectives
1012.(a) The spreadsheet below is identical to example 103 in the text except that the timing of the working capital
infusion has been shifted from the end of year 1 to year 0 or immediately
Income
Tax
Rate
40.00%
MARR
20.00%
All dollars in thousands
Ye
821. (continued)
Set PW1= PW2 and solve for X.
$X + $0.1133X(P/A,20%,3) = $30,326
$0.761 X = $30,326
X = $39,836
Therefore, $39,836 could be spent for software with a 3 year upgrade agreement
(i.e., Option 1).
103
Solutions to Chapter 9 Problems
91
818.
Assume neither machine has a salvage value. Incremental Analsis: A
(A)
BTCF
(R$)
Year
0
50,000
(B)
Infl.
Adujustment
year1
(1.10)
(C)
BTCF
(A$)
(D)
Depre.
B
(E)
(F)
(G)
Taxable
Income
= C D
Cash Flow
for income
Taxes
= t x E
ATCF
A($)
=CD+F


917 Keep the Defender:
(a)
EOY
BTCF
Depr
0
 $14,000

0
0
0

1
6
6
TI
T (40%)
b
0
4,000
b
10,000
a
$12,400
$4,000
0
$1,600
0
a
$4,000
ATCF
Gain on disposal = $14,000  $10,000 = $4,000 (if sold now).
Loss on disposal for defender when sold 6 years
922 Assume study period = 6 years.
Keep the Defender: Assume MV = 0 at end of study period.
EOY
BTCF
0
$130,000
1
1,800
6
1,800
TI
0
a
1,800
25
a
Depr
$90,000
$4,000
T(40%)
b
ATCF
$36,000
$94,000
2,320
520
3,920
2,120
2,320
520
5,800
8,000
9,800
99 The repeatability assumption and the AW method (over one useful life cycle) are used in the comparison of the two
robots. The use of repeatability as a simplified modeling approach can be supported in this case.
The estimated annual expenses for the d
101. (continued)
(b)
Interest Rate
Purchase
Price
Tax Rate
Sale Price
15.00%
0
1
2
3
4
5
5
B
Rent per
Year
$ 500,000
30.00%
$500,000
A
Number of
unit
Units
C
D =AxBxC
E
%
occupancy
Gross Rent
Maintenance
$7,200
$7,200
$7,200
$7,200
$7,200
20
20
20
20
20
155.
Alternative II is dominated by Alternative V.
Thus II can be eliminated.
(a)
State of Nature
Alternative
(b)
A
B
C
D
Max
Min
18
18
10
14
18
10
5
26
10
14
26
5
IV
14
22
10
10
22
10
V
10
12
12
10
12
10
I
III
(c) Alternative
I
Best
E(Costs) for Equal P