We use cookies to enhance your experience on our website. By continuing to use our website, you are agreeing to our use of cookies. You can change your cookie settings at any time. Find out more

### More Interest Formulas

Go to questions covering topic below

Suppose that there is a series of "n" payments uniformly spaced, but differing from one period to the next by a constant multiple. The change or "gradient" multiple from one period to the next is denoted "g." There will, of course, also be an interest rate “i” that applies.

g = gradient percent (as a decimal in calculations)

i = interest percent (as a decimal in calculations)

n = number of periods

A1 = payment at EOY 1

The Present Worth (P) is given by:

P = A1 ( P/ A1, g, i, n ) = A1 [ 1 – ( 1 + g ) n ( 1 + i ) –n ] / ( i – g )

Example 1: A1 = \$100; g = 15%; i = 9%; n = 5

P = A1 ( P/ A1, g, i, n ) = 100 [ 1 – ( 1 + 0.15 ) 5 ( 1 + 0.09 ) –5 ] / ( 0.09 – 0.15 )

= 100 [ 1 – 2.0114 * 0.6499] / (-0.06)

= 100 [-0.3072] / (-0.06)

= 512.1

Example 2: A1 = \$100; g = 5%; i = 8%; n = 10

P = A1 ( P/ A1, g, i, n ) = 100 [ 1 – ( 1 + 0.05 ) 5 ( 1 + 0.08 ) –5 ] / ( 0.08 – 0.05 )

= 100 [ 1 – 1.628894627 * 0.463193488] / (0.03)

= 100 [ 0.2455] / (0.03)

= 818.3554

### More Interest Formulas

Question 1

Question 1.

Suppose that the maintenance for a piece of equipment costs \$300 EOY1 and increases by 15% every year for 5 years. The value of money is 9%. What is the equivalent present cost over the time horizon?

Choose an answer by clicking on one of the letters below, or click on "Review topic" if needed.

A P = \$2022.71

B P = \$1175.16

C P = \$1536.22

D P = \$1166.90

Review topic