6 minute read

Notation:

  • $D$ : loan amount
  • $R$ : monthly installment
  • $I$ : nominal annual interest rate
  • $i = \frac{I}{12}$ : monthly interest rate, 30/360 convention
  • $N$ : total number of installments
  • $T = D + Z$ : total paid (loan + interests)
  • $Z$ : total interests paid

For the $n$-th installment:

  • $d_n$ : debt left after the $n$-th installment ($d_0=D$)
  • $p_n = R - i\,d_{n-1}$ : principal part of the $n$-th installment
  • $z_n = i\,d_{n-1}$ : interest part of the $n$-th installment

Here a summary of all the equations derived in this article:

\[\boxed{N = \lceil n^* \rceil} \quad \quad \quad \boxed{\;n^* = \frac{\ln{(R)} - \ln{(R-i\,D)}}{\ln{(i+1)}}}\] \[R_n = \begin{cases} R & \quad \forall \, n=1,...,N-1 \\[8pt] R_{\text{last}} \le R & \quad n={N} \end{cases} \quad \quad \quad \boxed{\;R_\mathrm{last} = \frac{R}{i}(i+1) + \left(D - \frac{R}{i}\right) (i+1)^N}\] \[\boxed{ \; R = \frac{i \, (1 + i)^{N}}{(1 + i)^{N} - 1}D = \frac{i}{1 - (1 + i)^{-N}} D } \quad \quad \quad \boxed{\; D = \frac{(1 + i)^{N} - 1}{i \, (1 + i)^{N}} R = \frac{1 - (1 + i)^{-N}}{i}\, R }\]

\(d_n = D\,(1+i)^n - R\,\frac{(1+i)^n - 1}{i}\) \(p_n = (R - iD)\,(i+1)^{n-1}, \qquad z_n = R - (R - iD)\,(i+1)^{n-1}\,\)

\[\boxed{ \;T =(N - 1)\,R + R_\mathrm{last} \approx n^* R } \quad \quad \quad \boxed{ \;Z = (N - 1)\,R + R_\mathrm{last} - D \approx n^* R - D } \quad \quad \quad P_\mathrm{tot} = D\]

Check: amortization condition

The debt decreases iff:

\[R > i\,D \quad\Longleftrightarrow\quad 12R > I\,D\]

If $R \le iD$, then $d_n \ge D$ for all $n$ and the loan is never repaid.

Debt left to be payed after $n$ installments

Each month the bank adds the interest on the current balance, then subtracts the installment:

\[\tag{1} \begin{cases} d_0 &= D \\[10pt] d_n &= d_{n-1} - (R - i\,d_{n-1}) & \quad\quad \forall n > 0\\[4pt] \end{cases}\]

With simple algebra:

\[\begin{aligned} \Rightarrow \quad d_n &= d_{n-1} - (R - i\,d_{n-1}) \quad\quad \forall n > 0 \\[6pt] & \quad\quad q := 1 + i\\[6pt] &= q\,d_{n-1} - R \\[6pt] &= q\,(q\,d_{n-2} - R) -R = q^2 d_{n-2} - R(1 + q) \\[6pt] &= q^2\,(q\,d_{n-3} - R) - R(1 + q) = q^3 d_{n-3} - R(1 + q + q^2) \\[6pt] &= \dots \\[6pt] &= q^n d_0 - R (1 + q + q^2 + ... + q^{n-1}) \\[6pt] & \quad\quad d_0 := D\\[6pt] &= q^n D - R \sum_{k=0}^{n-1} q^k \\[6pt] &= q^n D - R\,\frac{q^n - 1}{q - 1} \\[6pt] & \quad\quad q = 1 + i\\[6pt] &= D\,(1+i)^n - R\,\frac{(1+i)^n - 1}{i} \end{aligned}\] \[\boxed{\,d_n = D\,(1+i)^n - R\,\frac{(1+i)^n - 1}{i}}\]

From this equation, we can easily derive the relation between the total number of installments $N$, the loan $D$, the interest rate $i$ and the installment $R$ by imposing the condition $d_N=0$ (i.e. the debt must be repaid at the end of last installment).

\[\begin{align*} d_N = 0 \quad&\iff\quad 0 = \frac{R}{i} + \left(D - \frac{R}{i}\right) (i+1)^N \\[10pt] &\iff\quad \frac{R-i\,D}{i}(i+1)^N = \frac{R}{i} \\[10pt] &\iff\quad \boxed{(i+1)^N = \frac{R}{R-i\,D}} \end{align*}\]

We’ll derive all the expressions later.

Equivalent derivation from fixed-point theorem

The fixed point of $d \mapsto q d - R$ is $d^* = R/i$, and the deviation from it grows geometrically:

\[\begin{aligned} d_{n-1} \mapsto d_{n} &= q d_n - R \\[8pt] d_{n-1} = d_{n} &\iff d_n=d^*:=\frac{R}{q-1}=\frac{R}{i} \\[8pt] d_{n} - d^* &= d_{n} - \frac{R}{i} \\[6pt] &= q \, d_{n-1} - R - \frac{R}{i} \\[6pt] &= q \left( d_{n-1} - \frac{1}{\cancel{i+1}}\frac{\cancel{(i+1)}R}{i}\right) \\[8pt] &= q \left(d_{n-1} - d^*\right) \\[10pt] &= q^2 \left(d_{n-2} - d^*\right) \\[8pt] &= ... \\[10pt] &= q^n \left(d_{0} - d^*\right) \\[8pt] \end{aligned}\] \[\;\Longrightarrow\; \boxed{d_n - d^* = q^n \left(d_{0} - d^*\right)}\] \[\;\Longrightarrow\; \boxed{d_n = \frac{R}{i} + \left(D - \frac{R}{i}\right) (i+1)^n}\]

Total number of installments

To get the number of installments needed to repay a debt, we can set $d_n = 0$ and get the value $n^*$:

\[\begin{align*} d_n = 0 \quad&\iff\quad 0 = \frac{R}{i} + \left(D - \frac{R}{i}\right) (i+1)^n \\[10pt] &\iff\quad \frac{R-i\,D}{i}(i+1)^n = \frac{R}{i} \\[10pt] &\iff\quad (i+1)^n = \frac{R}{R-i\,D} \\[10pt] &\iff\quad n \, \ln{(i+1)} = \ln{\left(\frac{R}{R-i\,D}\right)} \\[10pt] &\iff\quad n = n^* :=\frac{\ln{(R)} - \ln{(R-i\,D)}}{\ln{(i+1)}} \\[10pt] \end{align*}\]

In the previous expression, $n^{*}$ is real-valued => the (integer) number of installments is $N = \lceil n^{*} \rceil$. However, the last installment, that pays the remaining debt plus one month of interest on it, is smaller than (or equal to) the standard $R$:

\(\boxed{\,n^* = \frac{\ln{(R)} - \ln{(Ri\,D)}}{\ln{(i+1)}}}\) \(\mathrm{Total \; number \; of \; installments:} \quad \boxed{N = \lceil n^* \rceil}\)

\[R_n = \begin{cases} R & \quad \forall \, n=1,...,N-1 \\[8pt] R_{\text{last}} \le R & \quad n={N} \end{cases}\] \[\boxed{ R_\mathrm{last} := (i+1)\,d_{N-1} = \frac{R}{i}(i+1) + \left(D - \frac{R}{i}\right) (i+1)^N}\]

Installment and Loan from the other terms

Recalling the equation we derived by setting $d_N=0$, we can get all the inverses:

\[\begin{align*} d_N = 0 \quad&\iff\quad 0 = \frac{R}{i} + \left(D - \frac{R}{i}\right) (i+1)^N \\[10pt] \quad&\iff\quad R + (iD - R) (i+1)^N = 0 \\[10pt] \quad&\iff\quad \left[1-(i+1)^N\right]R + i(i+1)^N D = 0 \\[10pt] \end{align*}\]

Installment from $D$, $I$, $N$

\[\boxed{ R = \frac{i \, (1 + i)^{N}}{(1 + i)^{N} - 1}D = \frac{i}{1 - (1 + i)^{-N}} D }\]

Loan from $R$, $I$, $N$

\[\boxed{ D = \frac{(1 + i)^{N} - 1}{i \, (1 + i)^{N}} R = \frac{1 - (1 + i)^{-N}}{i}\, R }\]

Interest rate from $D$, $R$, $N$: no closed form

Substitute $x = 1/q$ into the present-value condition:

\[g(x) \equiv R\,(x + x^2 + \dots + x^N) - D = 0\]

This is a polynomial of degree $N$ in $x$. For $N \ge 5$ there is no general solution in radicals (Abel–Ruffini), and $N$ is usually in the hundreds. The same equation defines the APR / internal rate of return.

Existence and uniqueness of a positive rate: $g$ is strictly increasing on $x > 0$, with $g(0) = -D < 0$ and $g(x) \to \infty$. So there is exactly one positive root $x^$. It satisfies $x^ < 1$, i.e. $i > 0$, iff

\[g(1) = R\,N - D > 0\]

Numerically, in terms of $i$: the present value

\[\mathrm{PV}(i) = R\,\frac{1 - (1 + i)^{-N}}{i}, \qquad \lim_{i \to 0} \mathrm{PV}(i) = R\,N\]

is strictly decreasing in $i$. Bisection on $\mathrm{PV}(i) = D$ therefore converges to the unique root, in $\lceil \log_2(\text{width}/\text{tol}) \rceil \approx 43$ steps for width $1$ and tolerance $10^{-13}$.

For small $i$, evaluate $1 - (1 + i)^{-N}$ as -expm1(-N*log1p(i)) to avoid cancellation.

Principal part and Interest part

Recalling the definitions of the principal part $p_n$ and interest part $z_n$ of each installment

\[\begin{align*} z_n &:= i\,d_{n-1} \\[10pt] p_n &:= R - i\,d_{n-1} = d_{n-1} - d_n \end{align*}\]

we can get their expressions by inserting the fixed-point form of $d_{n-1}$ into $p_n$:

\[\begin{align*} p_n &= R - i\,d_{n-1} \\[6pt] &=\cancel{R} - i\left[\cancel{\frac{R}{i}} + \left(D - \frac{R}{i}\right) q^{n-1}\right] \\[8pt] &= (R-iD)\,q^{n-1} \\[10pt] z_n &= i\,d_{n-1} \\[6pt] &= R - (R - iD) q^{n-1} \end{align*}\] \[\boxed{\,p_n = (R - iD)\,(i+1)^{n-1}, \qquad z_n = R - (R - iD)\,(i+1)^{n-1}\,}\]

The principal part grows geometrically, so the interest part shrinks.

Total paid and total paid interest

We can get the total paied $T$ and the total interest paid $Z$ easily (the total principal paid $P_\mathrm{tot}$ is trivial):

\[\boxed{ T =(N - 1)\,R + R_\mathrm{last} \approx n^* R}\] \[\boxed{ Z = (N - 1)\,R + R_\mathrm{last} - D \approx n^* R - D }\] \[P_\mathrm{tot} = D\]

Treating $N$ as continuous gives $Z \approx R\,n^* - D$, with an error below $R$.

Function definitions

Analytical definitions

Note: for numerical reasons, it’s better to rewrite some terms as follows and use the numerical functions $\mathrm{log1p}(x)$ and $\mathrm{expm1}(x)$:

\[\begin{align*} \ln(1+i) &= \mathrm{log1p}(i) \\[6pt] 1-(1+i)^{-N} &= - \left[(1+i)^{-N} -1 \right] = - \left[e^{-N\,\ln(1+i)} -1 \right] = -\mathrm{expm1}(-N \, \mathrm{log1p}(i))\\[10pt] \mathrm{log1p}(x) :&= \ln(1+x) \\[6pt] \mathrm{expm1}(x) :&= e^x -1 \end{align*}\]

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