# Problem about relationships between interest factors

1. Jan 19, 2014

### s3a

1. The problem statement, all variables and given/known data
The problem:
Derive (3.3) without reference to (2.3) or (2.1).

(2.1):
F/P = (1 + i)^n

(2.3):
F/A = [(1 + i)^n – 1]/i

(3.3):
(F/A, i%, n) = (F/A, i%, n_1) + (F/P, i%, n_1) + (F/P, i%, n_1 + 1) + . . . + (F/P. i%, n – 1) (n > n_1)

The solution (which also includes the problem) is attached as TheProblemAndSolution.png.

2. Relevant equations
(F/P, i%, n) and (F/A, i%, n)

3. The attempt at a solution
Here are my questions for the two things that I don't get for this problem.:

1. How exactly is (1) determined?

2. How do I go from Fig. 3-2 to (F/A, i%, n – n_1) (F/P, i%, n_1)? In other words, what the solution says is “obvious” is not obvious to me.

#### Attached Files:

• ###### TheProblemAndSolution.png
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2. Jan 20, 2014

### maajdl

These (F/A, i%, n) notations are awful.
You do not provide the exact meaning of (F/A, i%, n) .
Formulas (2.1) and (2.3) do not use the same notations as formula (3.3).
And furthermore, formulas (2.1) and (2.3) should not be the basis of the derivation!

Therefore, it is impossible to answer your question without guessing what you exactly want.
In addition, this exercise seems to depend on the context of this book and I do not have a copy of it.

I suggest you to go back to the definitions of the (F/A, i%, n) and (F/P, i%, n).

by the way, what are these F, A, P?

3. Jan 22, 2014

### s3a

Hello, and sorry for the delay.

I used to feel that way, until I programmed an application that calculated things for me easily (and that notation is very programming friendly). :)

(2.1):
(F/P, i%, n) = F/P = (1 + i)^n

(2.3):
(F/A, i%, n) = F/A = [(1 + i)^n – 1]/i

I had just listed them, with the goal in mind that whoever helped me knew what not to use.

I gave the algebraic definitions above. As for the theoretical definitions and meaning of variables from my book (Schaum’s Outline of Engineering Economics), they are as follows (with minor modifications).:

(F/P, i%, n):
Suppose that a given sum of money, P, earns interest at a rate i, compounded annually. The total amount of money, F (which I think is referred to as the “future worth”), which will have accumulated from an investment of P dollars after n years is given by F = P(1 + i)^n. The ratio F/P = (1 + i)^n is called the single-payment, compound-amount factor. The fuller notation for this ratio is (F/P, i%, n).

(F/A, i%, n):
Let equal amounts of money, A, be deposited in a savings account (or placed in some other interest-bearing investment) at the end of each year. If the money earns interest at a rate i, compounded annually, how much money will have accumulated after n years? To answer this question, we note that after n years, the first year’s deposit will have increased in value to F_1 = A(1 + i)^(n – 1). Similarly, the second year’s deposit will have increased in value to F_2 = A(1 + i)^(n – 2) and so on. The total amount accumulated will thus be the sum of a geometric progression.: F = F_1 + F_2 + . . . + F_n = A(1 + i)^(n – 1) + A(1 + i)^(n – 2) + . . . + A = A[(1 + i)^n – 1]/i. This implies that F/A = [(1 + i)^n – 1]/i. The ratio F/A = [(1 + i)^n – 1]/i is called the uniform-series, compound-amount factor. The extended notation for this ratio is (F/A, i%, n).

If I haven’t already done so, I can provide you with whatever else it is you need to know.

4. Jan 22, 2014

### haruspex

You're putting in an amount P in each of n-n1 years. The first amount will accumulate interest for n-1 years, the second for n-2, etc. That gives the sum (1).
An alternative way to think of those first n-n1 payments is in two stages:
- an F/A process of regular payments over n-n1 years, giving a future value (F/A, i%, n – n1);
- the investment of that resulting sum for n1 years, amplifying it by the factor (F/P, i%, n1)

5. Jan 26, 2014

### s3a

Thanks, haruspex, I get it now! :D