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,

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

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

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

Finding the present worth of the future salvage value,

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

Thus the overall present worth of the vehicle is,

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

Is this correct?

Re: Determining Present Worth w/ Interest

Re: Determining Present Worth w/ Interest

Quote:

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?

Re: Determining Present Worth w/ Interest

Quote:

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.

Re: Determining Present Worth w/ Interest

Quote:

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 ?