Attempt at creating a formula relating debt, payments and interest

Question

I tried writing down a formula relating a given debt and interest to the periodic payments and number of payments.

So let's say someone starts off with a debt of $D$. The periodic interest is $r$ (for example, $0.6\% =0.006$ each month). Let's call the periodic payment $p$, and the number of payments $N$.

I based my equation off of the fact that the total payments must equal to the original debt plus the added interest. The total payments are clearly $Np$. The added interest each month is $r$ times the remaining debt, so after one period we would add $$rD$$ after another, we would add $$(D+rD-p)r=rD+r^2D-pr$$ next we would add $$(D+rD+r^2D-pr-p)r=rD+r^2D+r^3D-pr^2-pr$$ then $$(D+rD+r^2D+r^3D-pr^2-pr-p)r=rD+r^2D+r^3D+r^4D-pr^3-pr^2-pr$$ and so on, for each of the $N$ periods.

Adding all of these up, along with the initial debt $D$ will give us the total that needs to be payed. Noticing the similar terms in each expression, we get

$$\text{Total} = D(Nr+(N-1)r^2+(N-2)r^3+\dots +r^N)-p((N-1)r+(N-2)r^2+(N-3)r^3\dots+r^{N-1})$$

$$=D\sum_{i=1}^{N} (N+1-i)r^i-p\sum_{i=1}^{N-1} (N-i)r^i = D\sum_{i=1}^{N} (N+1)r^i -p\sum_{i=1}^{N-1} Nr^i - D\sum_{i=1}^{N} ir^i +p\sum_{i=1}^{N-1} ir^i$$

The first to sums are geometric. The second two can be done using:

$$\sum_{i=1}^{k} ir^i = r\cdot \sum_{i=1}^{k} \frac{\partial}{\partial r} r^i = r\cdot \frac{\partial}{\partial r} \sum_{i=1}^{k} r^i =r\cdot \frac{\partial}{\partial r} \frac{r^{k+1}-r}{r-1} = \frac{kr^{k+1}-kr^k-r^k+1}{(r-1)^2}$$

So overall we get

$$D(N+1)\frac{r^{N+1}-r}{r-1}-pN\frac{r^N-r}{r-1}-D\frac{Nr^{N+1}-Nr^N-r^N+1}{(r-1)^2}+p\frac{(N-1)r^{N}-(N-1)r^{N-1}-r^{N-1}+1}{(r-1)^2}$$

As mentioned above, this total should equal $Np$, therefore we can finally conclude

$$N=\frac{D}{p}(N+1)\frac{r^{N+1}-r}{r-1}-N\frac{r^N-r}{r-1}-\frac{D}{p}\frac{Nr^{N+1}-Nr^N-r^N+1}{(r-1)^2}+\frac{(N-1)r^{N}-(N-1)r^{N-1}-r^{N-1}+1}{(r-1)^2}$$

I have a few questions. Firstly, is my attempt correct (I have a feeling this shouldn't be this complicated...). Secondly, is there a way to solve this equation for $N$ (non-numerically)? If not, and I am correct, then how do people determine how long their loan will take them to pay off given their monthly payment?

Thanks in advance.

Answer 1

I think you will find it easier to work backwards so with $D_n$ being the amount outstanding with $n$ months remaining, starting with $D_0=0$, and you have $D_{n-1}=D_{n}(1+r)-p$ since you make a payment at the end of the month, i.e. $D_{n}=\frac{D_{n-1}+p}{(1+r)}$. For given $n$ and $r$, clearly $D_n$ and $p$ are proportional to each other, so using $\frac{D_n}{p}=\frac{\frac{D_{n-1}}{p}+1}{(1+r)}$ just gives a tangle of $r$s to deal with:

• $\frac{D_1}{p}=\frac{1}{1+r}$
• $\frac{D_2}{p}=\frac{1+(1+r)}{(1+r)^2}$
• $\frac{D_3}{p}=\frac{1+(1+r)+(1+r)^2}{(1+r)^3}$

and you can see a geometric series appearing in the numerator.

So you can then show, and prove by induction, that $\frac{D_n}{p} =\frac{1}{r}\left(1-\frac1{(1+r)^n}\right)$, i.e. $D_n =\frac{p}{r}\left(1-\frac1{(1+r)^n}\right)$ and $p= r\,D_n \left(1+\frac{1}{(1+r)^n-1}\right)$ and in answer to your question $$n=\dfrac{\log(p)-\log(p-rD_n)}{\log(1+r)}.$$

Answer 2

Let’s assume the payments are made at the end of each monthly period.

At the end of month $1$ you owe a total of $D(1+r)-p$

At the end of month $2$ you owe a total of $\left(D(1+r)-p\right)(1+r)-p$

At the end of month $3$ you owe a total of $\left(\left(D(1+r)-p\right)(1+r)-p\right)-p$ $$=D(1+r)^3-p\left(1+(1+r)+(1+r)^2\right)$$

Continuing to the end of the $N$th month, when you owe nothing, $$0=D(1+r)^N-p\left(1+(1+r)+…+(1+r)^{N-1}\right)$$ $$\implies D(1+r)^N=p\frac{\left((1+r)^N-1\right)}{(1+r)-1}$$

So your regular payment is $$p=\frac{Dr(1+r)^N}{(1+r)^N-1}$$

You can then rearrange this to get $$N=\frac{\log\frac{p}{p-Dr}}{\log(1+r)}$$

Answer 3

The fundamental thing is that whether you take time at start or the end, the debts and payments must balance out.

You can actually simplify the computations a lot by using a multiplication factor $m=1+r$

Suppose we take the final payment as the point where the time value of money is to be equated, then

$Dm^n = P(m^{n-1} + m^{n-2} +... + 1)$

and using the G.P. formula

$Dm^n = P\cdot\dfrac{m^n-1}{m-1}$

Actually your total payment valued at the final payment time is simply $Dm^n$ or $D$ at the time the debt was incurred, although the total payments spread over the amortisation period was $Pn$, because of the time value of money.

If you want to find how many months it will take to liquidate the debt, you will have to use logarithms, or use a spreadsheet to see how many months it takes to liquidate the debt.