Interest and Discount

When interest `i` is paid at the end of the period, the accumulation
function is:

`a`(`t`) = (1 + `i`)^{t}

When discount `d` is paid at the beginning of the period, the accumulation
function is:

`a`(`t`) = (1 `-` `d`)^{-t}

Interest: | 1 | 1 + i |
|||

| |
___________________ |
___________________ |
| | ||

Discount: | 1 - d |
1 |

The first important relation is found by accumulating 1 `-` `d`:

(1 `-` `d`)(1 + `i`) = 1

This is the important identity: `a`^{-1}(`t`)`a`(`t`) = 1

`v` = 1 `-` `d`

The discount factors are equal

`d` = 1 `-` `v`

The second important relation is:

(1 `-` `d`)(1 + `i`) = 1 `-` `d` + `i`(1 `-` `d`) = 1
`d` = `i`(1 `-` `d`)
`d` = `iv`

Discount-amount equals discounted interest-amount

The third important relation compares the events depicted in the above diagram:

`d` = `i`(1 `-` `d`) = `i` `-` `id`
`i` `-` `d` = `id`

Difference in earnings equals interest-rate times difference in principal

As an alternative method, accumulate 1 in separate portions of
1 `-` `d` and `d`:

[(1 `-` `d`) + `d`](1 + `i`) = 1 + `i`
1 + `d`(1 + `i`) = 1 + `i`
`v` + `d` = 1
`d` = 1 `-` `v`

The first relation

1 + `d`(1 + `i`) = 1 + `i`
`d`(1 + `i`) = `i`

Accumulated discount-amount equals interest-amount

`d` = `iv`

The second relation

`d`(1 + `i`) = `i`

Accumulated discount-amount equals interest-amount

`d` + `id` = `i`
`i` `-` `d` = `id`

The third relation

Dividing the third relation by `id` yields:

1 |
- |
1 |
= 1 |

d |
i |

Copyright © 2004 The Stevens Computing Services Company, Inc. All rights reserved.