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.
Best Answer
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.