# 6 - 27 by langkunxg

VIEWS: 0 PAGES: 2

• pg 1
```									Brown, Deidre
SE 413 Summer 2008

Wiki Solutions
Ch 6 problem 6 - 27

A piece of production equipment is to be replaced immediately because it no longer meets quality requirements for the end product. The two best alternatives
are a used piece of equipment (E1) and a new automated model (E2). The economic estimates for each are shown in the accompanying table.

Alternatives
E1                     E2
Capital investment                        14000                     65000
Annual expenses                           14000                      9000
Useful life (years)                           5                        20
Market value (at end of useful life)       8000                     13000

The MARR is 15% per year.

a. Which alternative is preferred, based on the repeatablity assumption? (6.5)                                  Alternative E1 because it is the least negative.

The least common multiple of the useful lives of Alternatives A and B is 20 years. Using the repeatability assumption and a 20-year study period, the first first
like (identical) replacement of Alternative E1 would occur at the end of year 5, the second one at the end of year 10, and the third one at the end of year 15. For
Alternative E2, there would not be a like replacement.

Four cycles of Alternative E1:
E1-1           E1-2       E1-3         E1-4
0                5         10            15           20 years

One cycle of Alternative E2:
E2
0                                                     20 years

Alternative E1:                        PW(15%) =           -14000 - 14000[(P/F,15%,5) + (P/F,15%,10) + (P/F,15%,15)] + 14000(P/A,15%,20) + 8000(P/F,15%,5) + 8000(P/F,15%,10) + 8000(P/F,15%,15) + 8000(P/F,15%,20)
PW(15%) =           -14000 - 14000[0.4972 + 0.2472 + 0.1229] - 14000(6.2593) + 8000(0.4972) + 8000(0.2472) + 8000(0.1229) + 8000(0.0611)
PW(15%) =           \$     (106,345)

AW(15%) =           PW(A/P,15%,20)                            AW Check:            AW =       -14000(A/P, 15%, 5) + 8000(A/F, 15%, 5) - 14000
AW(15%) =           61976.8(0.1598)                                                AW =       -14000 * 0.2983 + 8000 * 0.1483 - 14000
AW(15%) =           \$      (16,994)                                                AW =       \$    (16,990)

From Table C-15                        (P/F,15%,5) =               0.4972
(P/F,15%,10) =              0.2472
(P/F,15%,15) =              0.1229
(P/F,15%,20) =              0.0611
(P/A,15%,20) =              6.2593
(A/P,15%,20) =              0.1598
(A/P,15%,5) =               0.2983
(A/F,15%,5) =               0.1483

Alternative E2:                        PW(15%) =           -65000 - 9000(P/A,15%,20) + 13000(P/F,15%,20)
PW(15%) =           -65000 - 9000*(6.2594) + 13000 * (0.0611)
PW(15%) =           \$     (120,540)

AW(15%) =           PW(A/P,15%,20)                            AW Check:            AW =       -65000(A/P, 15%, 20) + 13000(A/F, 15%, 20) - 9000
AW(15%) =           -120846(0.1598)                                                AW =       -65000 * 0.1598 + 13000 * 0.0098 - 9000
AW(15%) =           \$      (19,262)                                                AW =       \$    (19,260)
From Table C-15                     (P/F,15%,20) =           0.0611
(P/A,15%,20) =           6.2593
(A/P,15%,20) =           0.1598
(A/F,15%,20) =           0.0098

b. Show, for the coterminated assumption with a five year study period and an imputed market value for Alternative B, that the AW of B remains
the same as it was in Part (a). [And obviously, the selection is the same as in Part(a)]. Explain why that occurs in this problem. (6.5)

First compute the PW at EOY 5 of the remaining CR amounts for E2:

PW(15%)CR =      [65000(A/P,15%,20) - 13000(A/F,15%,20)] * (P/A,15%,15)
PW(15%)CR =      [65000(0.1598) - 13000(0.0098)] * (5.8474)
PW(15%)CR =      \$      59,992

Next, compute the PW at EOY 5 of the original MV at end of useful life (20yrs):

PW(15%)MV =      13000(P/F,15%,15)
PW(15%)MV =      13000 * (0.1229)
PW(15%)MV =      \$       1,598

The estimated MV at the EOY 5 is:

MW5 =            PWCR + PWMV
MW5 =            59992 + 1598
MW5 =            \$      61,590

Now calculate PW at EOY 5 for E2:

PW(15%) =        -65000 - 9000(P/A,15%,5) + 61590(P/F,15%,5)
PW(15%) =        -65000 - 9000 * 3.3522 + 61590 * 0.4972
PW(15%) =        \$     (64,547)

AW(15%) =        PW(A/P,15%,5)                       AW Check:             AW =   -65000(A/P, 15%, 5) + 61590(A/F, 15%, 5) - 9000
AW(15%) =        -64547 * 0.2983                                           AW =   -65000 * 0.2983 + 61590 * 0.1483 - 9000
AW(15%) =        \$     (19,254)                                            AW =   \$   (19,256)

Note: This is equal to the AW of E2 in part (a), off a little due to rounding.
Therefore as in part (a), E1 is the best alternative with an AW = \$(16,990).

The answer in part (b) is the same in part (a) because all we did was adjusted the market value for the new study period.
The adjusted market value is equivalent to the original market value in part (a).

```
To top