Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm studying on Kellison's Theory of Interest and I'm stuck on the exercise 20/a of the 1st chapter.

If the $i=0.1$ then $d = 0.0901$

$d_5=\frac{A_5-A_4}{A_5}$ when I insert $d$ into this equation, I reach to the;

$\frac{ (1/(1-5d)) - (1/(1-4d))}{1 / (1-5d) }$ which is simplified as $\frac d{1-4d}$. When I put the $0.0901$ value for $d$, I can't reach to the answer of $\frac1{15}$ with this solution.

But if I insert $i$ instead of $d$, I reach to $\frac i{1+5i}$ then the result is $\frac1{15}$, which is correct.

After this non-resulting work, I wanted to see the cash flow of this example. And I made a table in excel with $A_0=100$, $i = 0.1$ and $d= 0.0901$. And surprise surprise I couldn't find the same cash flow by using $i$ and $d$. At the 5th period, the simple interest accumulated value is 150, while the one with simple discount is $183.33$ (with the formula $\frac A {1 - nd}$). Actually, after the first period, the cash flow with the discount rate started to get higher than the one with the interest rate.

As a result I really don't understand how $i$ is converted into $d$.

Any help?

Note: Sorry for the messy equation syntax, I don't how to do it here properly.

share|cite|improve this question
Are you sure your cashflow $A_5$, $A_4$ are given in simple instead of compound discount? In the market, interest rates below/above 1 year is quoted in simple/compound conventions. In terms of mathematics, the approximation $exp(-rT) \approx 1 - rT$ is valid only for small $rT$. – achille hui Jan 31 '13 at 13:07
Hmm I'm confused more. Isn't the future value of A is A/(1-dt) with a simple discount rate of d? Is that correct only for below 1 year? – rakha Jan 31 '13 at 13:12
Could you post the actual question without any interpretation? – TheMathemagician Jan 31 '13 at 13:17
Find d5 if the rate of simple interest is 10%. – rakha Jan 31 '13 at 13:19
And what is d5? – TheMathemagician Jan 31 '13 at 13:29
up vote 1 down vote accepted

You are assuming the formula $d = \frac{i}{i+1}$ for simple interest when that formula is only valid for compound interest. Thus, your first step of determining $d = 0.0901$ is incorrect.

To answer your second question, asked in the comment, if $i$ and $d$ are equivalent rates of simple discount, then $1 + it = \frac{1}{1 - dt}$, so just solve for one or the other. Note, however, that the accumulation function for simple discount is only defined for $0 \leq t < \frac{1}{d}$, so they can only be equivalent over this interval.

We have $$1 = (1 + it)(1 - dt) = 1 - dt + it - idt^2.$$ Subtracting one from both sides gives $$-dt + it - idt^2 = 0.$$ Of course, this holds true when $t = 0$, so let's assume $t \neq 0$ and that allows us to divide both sides by $t$. This gives $$i - d = idt.$$ You can solve this for $i$ or $d$ easily now. Note that a constant simple interest rate $i$ will NOT lead to a constant discount rate $d$, which is clear because there is still a $t$ left in our equation. And, similarly, a constant simple discount rate will not lead to a constant simple interest rate. These facts are clear even before we solve this because $1 + it$ is linear when $i$ is constant and $\frac{1}{1 - dt}$ is not linear when $d$ is constant.

share|cite|improve this answer
Is there any formula of converting i to d (or d to i) if the case is simple interest? Or should I not convert them to each other at all in a simple interest question and use whatever is given? – rakha Feb 1 '13 at 9:26
@rakha I added that to my answer. Does this answer your question? Essentially, I wouldn't do this problem by finding $d$. It makes the problem much more difficult. – Graphth Feb 2 '13 at 17:53
Ok I got it. In a nutshell, there is no constant simple discount rate for all the periods equivalent to the simple interest rate. Thank you very much. – rakha Feb 2 '13 at 20:07

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.