Question

You invest in a piece of equipment costing $30,000. The equipment will be used for two years,

at the end of which time the salvage value of the machine is expected to be

1/3 of its original

value

. The expected annual net savings in operating costs will

be $25,000 during the first year

and $40,000 during the second year.

a)

If your interest rate is 10%, what would be the

Net Annual Worth (or Equivalence) of the

project?

b)

The machine will be used for 5,000 hours during the first year and 8,000 hours during the

second year.

If your interest rate is 10%, what would be the equivalent net savings per

machine

-

hour?

Hint:

Assume C as unit saving per hour ($/h) so that the

first

-

year

saving will be equal to 5000 (h)

x C ($/h) and the second 40,000 (h) x C ($/h). Use the NAW

(Net Annual

Worth) calculated

in 2.a

and the time value of money of both savings transformed into annuity to calculate C.

Answer 1

a) Investment = 30000

first year saving = 25000

Second year saving = 40000

i = 10%

n = 2 yrs

To find annuity value from present worth we use factor (A/P, i,n) = i((1 + i)^n)/((1 + i)^n-1)

To find annuity value from Future worth we use factor (A/F, i,n) = i/((1 + i)^n-1)

We will consider 25000 saving in both year and extra saving of 15000 in 2nd year

Now Net Annual worth = -30000 *(A/P, 10%, 2) + 25000 + 15000 * (A/F, 10%, 2)

= -30000 * [i((1 + i)^n)/((1 + i)^n-1)] + 25000 + 15000 * [i/((1 + i)^n-1)]

= -30000 * [0.1*((1 + 0.1)^2)/((1 + 0.1)^2 - 1)] + 25000 + 15000 * [0.1/((1 + 0.1)^2 - 1)]

= -30000 * [0.1*(1.1^2)/(1.1^2 - 1)] + 25000 + 15000 * [0.1/(1.1^2 - 1)]

= -30000 * [0.1*1.21/(1.21- 1)] + 25000 + 15000 * [0.1/(1.21 - 1)]

= -30000 * 0.5761904 + 25000 + 15000 * 0.4761904

= 14857.14

B) Let C be the aveage cost saved then

Savings in first year = 5000 * C

Savings in second year = 8000 * C

we will consider saving of 5000*C for both years and extra 3000*C for second year then,

Annual worth = -30000 *(A/P, 10%, 2) + 5000*C + 3000 *C * (A/F, 10%, 2)

We know values of factors from above, therefore,

Annual worth = -30000 *0.5761904 + 5000*C + 3000 *C * 0.4761904

this should be equal to annual worth calculated above

Now,

-30000 *0.5761904 + 5000*C + 3000 *C * 0.4761904 = 14857.14

C*(5000+3000 * 0.4761904) = 14857.14 + 30000*0.5761904

C = 32142.852/6428.5712

C = 4.9999 ~ $5/hr