# Thread: Determining Present Worth w/ Interest

1. ## Determining Present Worth w/ Interest

A bakery is thinking of purchasing a small delivery truck that has a first cost of $18,000 and its to be kept in service for 6 years, at which time the salvage value is expected to be$2500. Maintenance and operating costs are estimated at $2500 the first year and will increase at a rate of$200 per year. Determine the present worth of this vehicle, using an interest rate of 12 percent.

My attempt:

Finding the regular annuity for such an arithmetic series,

$A^{''} = 2500 + 200(\frac{A}{G}, 12\%, 6) = \2934.41$

Finding the present worth of maintenance costs given such an annuity, (NOTE: A here is A'' found previously)

$P = 2934.41(\frac{A}{P},12\%,6) = \12064.56$

Finding the present worth of the future salvage value,

$P = 2500(\frac{P}{F},12\%,6) = 1266.58$

Thus the overall present worth of the vehicle is,

$\18000 + \12064.56 - \1266.58 = \28797.98$

Is this correct?

$28797.97 is correct ! 3. ## Re: Determining Present Worth w/ Interest Originally Posted by jegues .....that has a first cost of$18,000...
Can you explain what that means?
This style of problem usually states a purchase price, then the positive and negative
subsequent flows resulting from this purchase; then, is the purchase price advantageous?

4. ## Re: Determining Present Worth w/ Interest

Originally Posted by Wilmer
Can you explain what that means?
This style of problem usually states a purchase price, then the positive and negative
subsequent flows resulting from this purchase; then, is the purchase price advantageous?
I'm assuming it means the purchase price.

5. ## Re: Determining Present Worth w/ Interest

Originally Posted by jegues
I'm assuming it means the purchase price.
Ok; then 28797.97 is definitely correct.
Can you calculate this:
if \$K is the savings each year due to the purchase, what is K in order to break even ?