Someone takes a loan of $S_1$ as an annuity at an annual interest rate $i$, paid back in monthly installments of $M$. Now I wonder how I can get the annual interest payments for each year $t$.

I can calculate the remaining debt after each year using

$$ S_{t}=\frac{M}{\frac{i}{12}} - \bigg(1+\frac{i}{12}\bigg)^{12}\times \bigg(\frac{M}{\frac{i}{12}}-S_{t-1}\bigg) $$

(e.g., wikipedia, adapted to have monthly payments). Using this, I can then calculate the interest numerically as $I_t = 12 \times M - (S_t - S_{t-1})$ but this approach fails in the last period when $M$ is paid only for a few months until $S_1$ is fully repaid. Therefore, I wonder whether anyone is aware of an equation to get annual interest payments analytically.


The loan $S=Ma_{\overline{n}|r}$ where $r=i/12$ in $n$ total months. If $S$ and $M$ are known we can easily find tne number of months to repay the loan $$ S=Ma_{\overline{n}|r}=M\frac{1-(1+r)^{-n}}{r}\quad\Longrightarrow\; n=\frac{\log\left(\frac{M}{M-rS}\right)}{\log(1+r)} $$ and the number of years is $m=n/12$.

At any time $k$ (month) the interest payment is $$ I_k=rMa_{\overline{n-k+1}|r}=M\left(1-v^{n-k+1} \right) $$ with $v=\frac{1}{1+r}$ and the principal payment is $$ P_k=M v^{n-k+1} $$ so that $M=P_k+I_k$ and the oustanding balance is $$ B_k=Ma_{\overline{n-k}|r} $$ If the outstanding balance at month $k-1$ is known, i.e. what is left to be paid back, and the payment $M$, we can find the interest payed at month $k$ as $$ I_k=rB_{k-1} $$ and $P_k=M-I_k$.

The monthly repayment of the loan is equivalent to the yearly repayment of the loan with effective annual interest $$j=\left(1+\frac{i}{12}\right)^{12}-1=(1+r)^{12}-1$$ for $m=n/12$ years: $$ S=Ma_{\overline{n}|r}=M\frac{1-\frac{1}{(1+r)^n}}{r}=M\frac{1-\frac{1}{(1+j)^m}}{r}=M\cdot\frac{j}{r}\cdot \frac{1-\frac{1}{(1+j)^m}}{j}=A a_{\overline{m}|j} $$ observing that $(1+r)^n=(1+r)^{12\times\frac{n}{12}}=(1+j)^m$ and the yearly installment is $A=M\times\frac{j}{r}$. Observe that $A$ is the future vale at the end of the year of 12 installment of value $M$ at monthly interest $r$: $$ A=M\times\frac{j}{r}=M\times\frac{(1+r)^{12}-1}{r}=M\cdot s_{\overline{12}|r} $$ and expressing $r=i/12$ we have that $$ A=M\times\frac{j}{r}=M\times\frac{j}{i/12}=12M\times\frac{j}{i} $$ that is $A>12M$: the installment that we have to pay annually is greater than the sum of the monthly installments in one year, because the annual effective interest $j$ is greater than the monthly interest $i$: $$ \frac{j}{i}=\frac{(1+i/12)^{12}-1}{i}>1\quad \text{for } i>0 $$

At any time $t$ (year) the interest payment is $$ I_t=jAa_{\overline{m-t+1}|j}=A\left(1-\nu^{m-t+1} \right) $$ with $\nu=\frac{1}{1+j}$ and the principal payment is $$ P_t=A \nu^{m-t+1} $$ so that $A =P_t+I_t$ and the oustanding balance is $$ B_t=Aa_{\overline{m-t}|j} $$ If the outstanding balance at year $t-1$ is known, i.e. what is left to be paid back, and the payment $A=Mj/r$ we can find the interest payed at year $t$ as $$ I_t=jB_{t-1} $$ and $P_t=A-I_t$.

  • $\begingroup$ Thanks for the detail! Two questions remain. 1) Is $a_{\overline{m-t}|j} = 1$? Combining the equations for $I_t$ and $P_t$ it seems that way. Also, isn't $P_t + I_t$ the total annual payment rather than the outstanding balance, or am I misunderstanding terms? 2) What if, as in my example, $n$ and $m$ are not known but only previous payments (interest & principal) and what is left to be paid back are known? $\endgroup$ – bumblebee Aug 29 '17 at 14:39
  • $\begingroup$ 1) there was a mistake. ..sorry. 2) I'll add it later $\endgroup$ – alexjo Aug 29 '17 at 15:00
  • $\begingroup$ 1) Alright, makes sense now. 2) thanks! $\endgroup$ – bumblebee Aug 30 '17 at 8:18
  • $\begingroup$ 2) details added $\endgroup$ – alexjo Aug 30 '17 at 9:53
  • $\begingroup$ Thanks, one more question though: Why is $12 \times M - P_t < I_t$`, that is why is what is actually paid in interest smaller than the calculated amount? I guess this comes down to the $j$ business: Shouldn't the effective interest rate be smaller than $i$ because the sum to be paid interest on decreases during the year? In fact, dividing $12 \times M - P_t $ by $B_{t-1}$ gives a lower rate than $i$. $\endgroup$ – bumblebee Aug 30 '17 at 11:21

In situation, where $S_t > 12M$, we can use this formula:

The fixed monthly payment $M$ for a loan of $S$ for $n$ months and a monthly interest rate $i$ is: $$M=S\frac{i(1+i)^n}{(1+i)^n-1}$$

After some transformations you obtain the number of months to pay this loan: $$n=\log _{i+1} \frac{M}{M-Si}$$ So the total value of loan to pay in the last year $t$ is equal to $$M\log _{i+1} \frac{M}{M-iS_{t-1}}$$


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