
(a):
Calculate the annual worth.
(a):

Explanation of Solution
Defender: The first cost (FC) is $160,000. The annual operating cost (AC) is $7,000. The salvage value for the first year (SV1) is $50,000 and for the second year (SV2) is $40,000. The interest rate is 12% compounded monthly. Thus, the effective interest rate (i) is 1%
Challenger: The first cost (FC) is $210,000. The annual operating cost (AC) is $5,000. The salvage values for first year (SV1) is $100,000, second year (SV2) is $7,000, and third year (SV3) is $45,000. The interest rate is 12% compounded monthly. Thus, the effective interest rate (i) is 1%
The annual worth (AV) is calculated using the following formula:
Substitute the respective values in Equation (1) to calculate the annual worth (AV) of defender in 12-month period.
The annual worth of the defender for 12 months is -$17,2754.
Substitute the respective values in Equation (1) to calculate the annual worth (AV) of challenger in 12-month period.
The annual worth of the challenger for 12 months is -$15,774.
The challenger can be replaced if the annual cost of challenger is lower than the defender. Thus, the firm can replace the defender with challenger.
(b):
Calculate the annual worth for 2 years.
(b):

Explanation of Solution
Table 1 shows the annual worth of the defender and challenger for 2 years (24 months), which were obtained using Equation (1).
Table 1
Project | AV |
Defender | -13,048 |
Challenger | -12,290 |
The challenger can be replaced if the annual cost of challenger is lower than the defender. Thus, the firm can replace the defender with challenger.
(c):
Calculate the annual worth for 3 years.
(c):

Explanation of Solution
Table 1 shows the annual worth of the defender and challenger for 3 years (36 months), which were obtained using Equation (1).
Table 1
Project | AV |
Defender | -14,292 |
Challenger | -10,349 |
The challenger can be replaced if the annual cost of challenger is lower than the defender. Thus, the firm can replace the defender with challenger.
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Chapter 11 Solutions
EBK ENGINEERING ECONOMY
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