I'm looking to achieve two things. Both might not be do-able in the same equasion but I'm happy to take advice.

I need to calculate compund interest on a loan. That in itself isn't an issue - I already have a formula that can do this.

Similarly, I need to calculate equal payments - again this is something I can do.

Here's the equasion that solves both of the above:


Representing a rate of 20% per month over 3 months.

However, I have a slightly more complex need:

Compund interest; Equal repayments over a fixed period (e.g 3 monthly payments); Varying interest periods.

As an example, credit cards work that way - they charge interest daily, but the number of days will vary dependent on the month and the way that weekend dates fall. However, it's not typical to repay a card over a fixed number of months like a loan, so I can't find an example to work through.

A credit card might have 3 periods of 30, 28 and 29 and I want to repay the full balance over over 3 periods. The repayments must be equal (or as equal as possible).

Can anyone give me advice, or help me solve the equasion to achieve this?

Thanks in advance.

  • $\begingroup$ Let $x$ be the unknown repayment. Figure out the balance after 30 days. Subtract $x$. Figure out the balance after the next 28 days. Subtract $x$. Figure out the balance after the 29 days. Set that equal to $x$. Solve for $x$. $\endgroup$ Mar 3, 2015 at 11:50
  • $\begingroup$ Thanks for the prompt reply Gerry - can you help me out here - follow the premise but I still don't know how I would go about solving x $\endgroup$
    – Johnnyboy
    Mar 3, 2015 at 12:14
  • $\begingroup$ Well, you can't solve for $x$ until you have written down the equation to solve, so I'd say you should do that, first. But I see you already have an answer that you have accepted. $\endgroup$ Mar 3, 2015 at 22:47

1 Answer 1


I will provide a general formula and then work out the example you have.

First some variable definitions: Let's call the amount of the original loan $L$, the rate of interest $r$, the number of pay periods $m$, the length of each pay period $y_1,\dots,y_m$ (the length $y_i$ here meaning how many times the interest is calculate in the $i$-th pay period), and finally the amount we pay each pay period $x$.

If $T$ is the amount left of our loan at the end of pay period $i-1$, then the total left on the loan at the end of pay period $i$ is $T(1+r)^{y_i}$. From this each month we subtract $x$ and eventually we need to end up with $0$.

Now we define equations $F_k$ representing how much we have to pay after pay interval $k$: $$F_0=L,$$ $$F_n=(1+r)^{y_n}F_{n-1}-x$$ We need to solve $F_m=0$ for $x$.

Now let make a helpful definition for the backwards partial sums of the $y$s: $$Y_n=\sum\limits_{i=n}^my_i$$ and solve for $x$ by expanding out the meaning of $0=F_m$:

$$\begin{align}0&=(1+r)^{y_m}((1+r)^{y_{m-1}}(\dots(1+r)^{y_2}((1+r)^{y_1}L-x)-x)\dots)-x)-x \\&=(1+r)^{y_m+\dots+y_1}L-(1+r)^{y_m+\dots+y_2}-(1+r)^{y_m\dots+y_3}x-\dots-x \\&=(1+r)^{Y_1}L-(1+r)^{Y_2}-(1+r)^{Y_3}x-\dots-(1+r)^{Y_{m}}x-x \\&=(1+r)^{Y_1}L-x\left(1+\sum\limits_{i=2}^m(1+r)^{Y_i}\right) \end{align}$$ Hence


In your example, we have $m=3$, $y_1=30$, $y_2=28$, and $y_3=29$. Note we have left the values of the original loan $L$ and the daily interest rate $r$ undefined. So we want now to solve $F_3=0$, and from above we have that the solution is $$x=\frac{(1+r)^{29+28+30}L}{1+(1+r)^{29+28}+(1+r)^{29}}=\frac{(1+r)^{87}L}{1+(1+r)^{57}+(1+r)^{29}}$$ If the original loan for example is $\$1000$, and the daily rate of interest is $1\%$, this would have you paying $\approx\$580$ per month for each of the three months.

  • $\begingroup$ Thanks for the great answer Peter - I can start working through to understand and and try to come up with a working solution. This is way out of my comfort zone so its going to take some time - can I ask what the ... signify? $\endgroup$
    – Johnnyboy
    Mar 3, 2015 at 14:36
  • 1
    $\begingroup$ @Johnnyboy the $\dots$ in mathematics generally means to repeat the pattern. So where I have $y_1+\dots+y_m$ I just mean adding all the terms $y_1$ through $y_m$. The $\dots$ which I have used in the first line of the solution of $0=F_m$ are harder to understand, but I mean them to repeat the pattern represented (I can't write out arbitrarily many "$($" symbols). If the pattern is confusing, you could try writing out the first few $F_1$, $F_2$, and $F_3$ and I think you'll see the picture. $\endgroup$ Mar 3, 2015 at 14:55
  • $\begingroup$ Yes, I get it Peter - thanks for the help. $\endgroup$
    – Johnnyboy
    Mar 3, 2015 at 14:59
  • $\begingroup$ Peter - just to let you know that I've been doing some testing and the formmula isn't entirely accurate. First I matched your example. Second, I tried simple example example r = 0.8%, 3 periods of 30 days. The results were significantly out - for instance the xample above should be £50.47 per period and I'm getting nearer £53 with the formula - any advice? $\endgroup$
    – Johnnyboy
    Mar 11, 2015 at 10:52
  • $\begingroup$ @Johnnyboy I agree with your evaluation of the formula I gave (about $\$53$ - more precisely $\$52.76$), but I disagree with the answer being $\$50.47$. If $50.47$ were the correct answer, then it should satisfy the equation $$((100\times1.008^{30}-x)\times1.008^{30}-x)\times 1.008^{30}-x=0$$ but when you plug in $x=50.47$ we get $8.88$, while when we plug in $52.76$ we get $-0.01$ - which improves if we add more accuracy. How did you arrive at $50.47$? $\endgroup$ Mar 11, 2015 at 15:12

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