How do you solve for
If Machine A has a NPW of $29,282 and Machine B has a NPW of $40,847 over an 8-year period, what is the difference in the annual worth of Machine B over Machine A at a rate of 10% p. y. c. y?
Right answer: |
$2,168 |
Answers: |
$1,579 |
$1,789 |
|
$2,168 |
|
$2,328 |
$2,168
Working:
a. | Present Value of annuity of 1 | = | (1-(1+i)^-n)/i | Where, | |||||||
= | (1-(1+0.10)^-8)/0.10 | i | 10% | ||||||||
= | 5.3349262 | n | 8 | ||||||||
b. | Difference in NPW | = | $ 40,847 | - | $ 29,282 | ||||||
= | $ 11,565 | ||||||||||
c. | Difference in annual worth | = | $ 11,565 | / | 5.3349262 | ||||||
= | $ 2,168 | ||||||||||
Note: | |||||||||||
NPW is the present value of annual worth. | |||||||||||
Get Answers For Free
Most questions answered within 1 hours.