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.

  • $\begingroup$ 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 $rT$. $\endgroup$ – achille hui Jan 31 '13 at 13:07
  • $\begingroup$ 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? $\endgroup$ – rakha Jan 31 '13 at 13:12
  • $\begingroup$ Could you post the actual question without any interpretation? $\endgroup$ – TheMathemagician Jan 31 '13 at 13:17
  • $\begingroup$ Find d5 if the rate of simple interest is 10%. $\endgroup$ – rakha Jan 31 '13 at 13:19
  • $\begingroup$ And what is d5? $\endgroup$ – TheMathemagician Jan 31 '13 at 13:29

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.

  • $\begingroup$ 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? $\endgroup$ – rakha Feb 1 '13 at 9:26
  • $\begingroup$ @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. $\endgroup$ – GeoffDS Feb 2 '13 at 17:53
  • $\begingroup$ 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. $\endgroup$ – rakha Feb 2 '13 at 20:07

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