Solutions To Chapter 5 Problems

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```					                             Solutions to Homework #3

6-7    PWA(20%) = -\$28,000 + (\$23,000 - \$15,000)(P/A,20%,10) + \$6,000(P/F,20%,10)
= \$6,509

PWB(20%) = -\$55,000 + (\$28,000 - \$13,000)(P/A,20%,10) + \$8,000(P/F,20%,10)
= \$9,180

PWC(20%) = -\$40,000 + (\$32,000 - \$22,000)(P/A,20%,10) + \$10,000(P/F,20%,10)
= \$3,540

Select Alternative B to maximize present worth.

Note: If you were to pick the alternative with the highest total IRR, you would have
incorrectly selected Alternative A.

6-27    (a) Repeatability assumption
AWE1(15%) = - \$14,000 (A/P,15%,5) - \$14,000 + \$8,000 (A/F,15%,5)
= -\$16,990
AWE2(15%) = -\$65,000(A/P,15%,20) - \$9,000 + \$13,000 (A/F,15%,20)
= -\$19,260
Select Alternative E1 to minimize costs.
(b) Coterminated assumption (5-year study period)
AWE1(15%) = - \$16,990; unchanged from Part (a)
Imputed market value (MV5) for Alternative E2:
MV5 = [\$65,000(A/P,15%,20) - \$13,000 (A/F,15%,20)](P/A,15%,15)
+ \$13,000 (P/F,15%,15) = \$61,590
AWE2(15%) = - \$65,000(A/P,15%,5) - \$9,000 + \$61,590(A/F,15%,5)
= -\$19,256 (slight difference from Part (a) is due to rounding)
Select Alternative E1 to minimize costs. The reason AWE2 in Part b is the
same as in Part a is the annual expenses are the same over the 20-year period in
Part a as they are over the 5-year period in Part b.

6-38    Assume repeatabillity. Use method described in Section 6.5.1.
Rank order: DN,Alt. 1, Alt. 3, Alt. 2

IRR on  (1-DN) = 33% > 20%, so select Alt. 1

IRR on  (3-1): set AW1(i') = AW3(i') and solve for i'
- \$30,000(A/P, i',5) + \$12,000 +\$10,000(A/F, i',5)
= -\$40,000(A/P, i',6) + \$13,000 + \$10,000(A/F, i',6)

i' ≈ ½% << 20%, so select Alt. 1

IRR on  (2-1): set AW1(i') = AW2(i') and solve for i'
-\$30,000(A/P, i',5) + \$12,000 + \$10,000(A/F, i',5)
= -\$60,000(A/P, i',5) + \$23,500 + \$10,000(A/F, i',5)

i' ≈ 26.5% > 20%, so select Alternative 2

6-40 Design 1 PW1 = -\$100,000 + \$20,000 (P/A,10%,10) = \$22,891.34
Design 2 PW2 = -\$160,000 + \$27,000 (P/A,10%,10) + \$20,000 (P/F,10%,10)
= \$13,614.18
Design 3 PW3 = -\$200,000 + \$28,000 (P/A,10%,10) + \$40,000 (P/F,10%,10)
= -\$12,530.39
Design 4 PW4 = -\$260,000 + \$45,500 (P/A,10%,10) + \$10,000 (P/F,10%,10)
= \$23,433.23

Select Design 4 by slim margin.

IRR1 ≈ 15.1%, IRR2 ≈ 12%, IRR3 < 10%, IRR4 ≈ 12%;
Moral: Do not select Design 1 to maximize IRR!

7-7     From Table 7-2, the GDS recovery period is 3 years.
(a) Basis = \$195,000
d*3 = \$195,000 (0.3333 + 0.4445 + 0.1481) = \$180,550.50

(b)         d4 = 0.0741 (\$195,000) = \$14,449.50

(c)          BV2 = \$195,000 (1 - 0.3333 - 0.4445) = \$43,329

7-16    (a)    Income taxes = \$50,000 (0.15) + \$25,000 (0.25) + \$15,000 (0.34) = \$18,850
(b)    Depreciation + Expenses = \$220,000 - \$90,000 = \$130,000

7-26    Machine A:
(A)             (B)           (C) = (A) - (B)   (D) = -t (C)    (E) = (A) + (D)
EOY           BTCF           Depr                 TI            T (40%)            ATCF
0       -\$20,000             ---                ---              ---               -\$20,000
1-12        \$12,000         \$1,333.33         \$10,666.67          -\$4,266.67        \$7,733.33
12          \$4,000           ---                0                  0                  \$4,000

0.1468                                   0.0468
(a) AW = -\$20,000(A/P,10%,12) + \$7733.33 +\$4,000(A/F,10%,12) = \$4,984.53
8.9847
(b) PW = \$4,984.53(P/A,10%,24) = \$44,784.51

(c) IRR see part (c) below

Machine B:
(A)             (B)       (C) = (A) - (B)   (D) = -t (C)      (E) = (A) + (D)
EOY        BTCF             Depr           TI            T (40%)              ATCF
0    -\$30,000              ---           ---              ---                 -\$30,000
1-8     \$18,000             \$3,750       \$14,250               -\$5,700          \$12,300

0.1874
(a) AW = -\$30,000(A/P,10%,8) + \$12,300 = \$6,678
8.9847
(b) PW = \$6,678(P/A,10%,24) = \$60,000
Choose Machine B.

(c) IRR; use qualitative reasoning and ∆ (B-A); by observation from
differences in PWs, both IRRs > MARRAT = 10% and i'%B > i'%A
 Choose Machine B.

7-39 Use a study period of 3 years

Quotation I
EOY         BTCF BTCF          Depr.   Depr.     Book Gain (Loss)                       in           Cash Flow
Capital Operating Fact.              Value On Disp.                        Ord. Inc         for IT (Cap)
0       \$(180,000)                            \$180,000
1            ---   \$(28,000) 0.2000 \$(36,000) \$144,000                                \$(64,000)
2            ---   \$(28,000) 0.3200 \$(57,600) \$ 86,400                                \$(85,600)
3       \$50,000    \$(28,000) 0.0960 \$(17,280) \$ 69,120 \$ (19,120)                     \$(45,280) \$ 7,648
PW of ATCF, Quotation I: \$(143,174)
IT = Income Taxes

Quotation II
EOY        BTCF BTCF           Depr.   Depr.     Book Gain (Loss)                       in           Cash Flow
Capital Operating Fact.              Value On Disp.                        Ord. Inc         for IT (Cap)
0       \$(200,000)                            \$200,000
1            ---   \$(17,000) 0.2000 \$(40,000) \$160,000                                \$(57,000)
2            ---   \$(17,000) 0.3200 \$(64,000) \$ 96,000                                \$(81,000)
3       \$60,000    \$(17,000) 0.0960 \$(19,200) \$ 76,800 \$ (16,800)                     \$(36,200) \$ 6,720
PW of ATCF, Quotation II: \$(136,848)
IT = Income Taxes
Accept Quotation II.

8-12   MARR = ic = 25% per year; Assume f = 8% per year; Let k = 0
Note that the estimated cash flows are in R\$ except for the contract maintenance
agreement (\$3,000 / year). However, the PW of \$3,000 per year at ic = 25% per year
is equal to the PW of the R\$ equivalent at ir. Therefore, the PW of the cash flows as
a function of N is:
PW = -\$50,000 + \$18,000 (P/A, ir , N) - \$3,000 (P/A, 25%, N)
and,
0.25 - 0.08
ir                  01574 or 15.74% per year
.
1.08
By trial and error we have:               N         PW
3      -\$15,257
4        -6,455
5         1,230
The life of the computer system must be at least 5 years for it to be economically
justified.

8-27   f = 4.5% per year; ic (after-tax) = 12% per year; t = 40%; b = 0
increase rate = 6 % per year (applies to annual expenses, replacement costs, and
market value)
Analysis period = 20 years; Useful life = 10 years
MACRS (GDS) 5-year property class
Capital investment (and cost basis, B)                   = -\$260,000
Market value (at end of year 10) in year 0 dollars       = \$50,000
Annual expenses (in year 0 dollars) = -\$6,000
Annual property tax = 4% of capital investment (does not inflate)
Assume like replacement at end of year 10.
Annual Property                                                       ATCF     ATCF*
EOY   Expenses Taxes            BTCF       Depr.         TI       T(40%)      (A\$)      (R\$)
0                            -\$260,000                                    -\$260,000 -\$260,000
1     -\$6,360 -\$10,400         -16,760    52,000 -68,760 27,504               10,744    10,281
2      -6,742 -10,400          -17,142    83,200 -100,342 40,137              22,995    21,057
3      -7,146 -10,400          -17,546    49,920 -67,466 26,986                9,440     8,272
4      -7,575 -10,400          -17,975    29,952 -47,927 19,171                1,196     1,003
5      -8,029 -10,400          -18,429    29,952 -48,381 19,352                  923       741
6      -8,511 -10,400          -18,911    14,976 -33,887 13,555               -5,356    -4,113
7      -9,022 -10,400          -19,422         0 -19,422    7,769           -11,653     -8,563
8      -9,563 -10,400          -19,963         0 -19,963    7,985           -11,978     -8,423
9     -10,137 -10,400          -20,537         0 -20,537    8,215           -12,322     -8,292
10    -10,745 -10,400          -21,145         0 -21,145    8,458           -12,687     -8,170
10                              89,542             89,542 -35,817             53,725    34,595
10                            -465,620                                     -465,620 -299,826
11    -11,390      -18,625     -30,015    93,124 -123,139 49,256              19,241    11,856
12    -12,073      -18,625     -30,698   148,998 -179,696 71,878              41,180    24,282
13    -12,798      -18,625     -31,423    89,399 -120,822 48,329              16,906     9,540
14    -13,565      -18,625     -32,190    53,639 -85,829 34,332                2,142     1,157
15    -14,379      -18,625     -33,004    53,639 -86,643 34,657                1,653       854
16    -15,242      -18,625     -33,867    26,820 -60,687 24,275               -9,592    -4,743
17    -16,157      -18,625     -34,782         0 -34,782 13,913             -20,869     -9,875
18    -17,126      -18,625     -35,751         0 -35,751 14,300             -21,451     -9,713
19    -18,154      -18,625     -36,779         0 -36,779 14,712             -22,067     -9,562
20    -19,243      -18,625     -37,868         0 -37,868 15,147             -22,721     -9,421
20                             160,357            160,357 -64,143             96,214    39,894

* ATCF(R\$) = ATCF(A\$)  1/(1.045)k

012  0.045
.
ir =               = 0.0718 or 7.18% per year
.
1045

20                                  20
PW =   
k0
ATCFk (A\$) (P/F, 12%, k) =      ATCF
k0
k   (R\$) (P/F, 7.18%, k) = -\$359,665

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