Compound interest

The total accumulated value, including the principal sum ${\displaystyle P}$ plus compounded interest ${\displaystyle I}$, is given by the formula:[2][3]

${\displaystyle P'=P\left(1+{\frac {r}{n}}\right)^{nt}}$

where:

P is the original principal sum

P' is the new principal sum

r is the nominal annual interest rate

n is the compounding frequency

t is the overall length of time the interest is applied (expressed using the same time units as r, usually years).

The total compound interest generated is the final value minus the initial principal:[4]

${\displaystyle I=P\left(1+{\frac {r}{n}}\right)^{nt}-P}$

Suppose a principal amount of $1,500 is deposited in a bank paying an annual interest rate of 4.3%, compounded quarterly.
Then the balance after 6 years is found by using the formula above, with P = 1500, r = 0.043 (4.3%), n = 4, and t = 6:

${\displaystyle P'=1\,500\times \left(1+{\frac {0.043}{4}}\right)^{4\times 6}\approx 1\,938.84}$

So the new principal ${\displaystyle P'}$ after 6 years is approximately $1,938.84.

Subtracting the original principal from this amount gives the amount of interest received:

${\displaystyle 1\,938.84-1\,500=438.84}$

Suppose the same amount of $1,500 is compounded biennially (every 2 years). (This is very unusual in practice.) Then the balance after 6 years is found by using the formula above, with P = 1500, r = 0.043 (4.3%), n = 1/2 (the interest is compounded every two years), and t = 6 :

${\displaystyle P'=1\,500\times \left(1+(0.043\times 2)\right)^{\frac {6}{2}}\approx 1\,921.24}$

So, the balance after 6 years is approximately $1,921.24.

The amount of interest received can be calculated by subtracting the principal from this amount.

${\displaystyle 1\,921.24-1\,500=421.24}$

The interest is less compared with the previous case, as a result of the lower compounding frequency.

Since the principal P is simply a coefficient, it is often dropped for simplicity, and the resulting accumulation function is used instead. The accumulation function shows what $1 grows to after any length of time.

Accumulation functions for simple and compound interest are

${\displaystyle a(t)=1+rt\,}$

${\displaystyle a(t)=\left(1+{\frac {r}{n}}\right)^{nt}}$

If ${\displaystyle nt=1}$, then these two functions are the same.

As n, the number of compounding periods per year, increases without limit, the case is known as continuous compounding, in which case the effective annual rate approaches an upper limit of e^r − 1, where e is a mathematical constant that is the base of the natural logarithm.

Continuous compounding can be thought of as making the compounding period infinitesimally small, achieved by taking the limit as n goes to infinity. See definitions of the exponential function for the mathematical proof of this limit. The amount after t periods of continuous compounding can be expressed in terms of the initial amount P₀ as

${\displaystyle P(t)=P_{0}e^{rt}.}$

As the number of compounding periods ${\displaystyle n}$ reaches infinity in continuous compounding, the continuous compound interest rate is referred to as the force of interest ${\displaystyle \delta }$.

In mathematics, the accumulation functions are often expressed in terms of e, the base of the natural logarithm. This facilitates the use of calculus to manipulate interest formulae.

For any continuously differentiable accumulation function a(t), the force of interest, or more generally the logarithmic or continuously compounded return is a function of time defined as follows:

${\displaystyle \delta _{t}={\frac {a'(t)}{a(t)}}={\frac {d}{dt}}\ln a(t)}$

This is the logarithmic derivative of the accumulation function.

Conversely:

${\displaystyle a(t)=e^{\int _{0}^{t}\delta _{s}\,ds}\ ,}$ (since ${\displaystyle a(0)=1}$; this can be viewed as a particular case of a product integral).

When the above formula is written in differential equation format, then the force of interest is simply the coefficient of amount of change:

${\displaystyle da(t)=\delta _{t}a(t)\,dt\,}$

For compound interest with a constant annual interest rate r, the force of interest is a constant, and the accumulation function of compounding interest in terms of force of interest is a simple power of e:

${\displaystyle \delta =\ln(1+r)\,}$ or

${\displaystyle a(t)=e^{t\delta }\,}$

The force of interest is less than the annual effective interest rate, but more than the annual effective discount rate. It is the reciprocal of the e-folding time. See also notation of interest rates.

A way of modeling the force of inflation is with Stoodley's formula: ${\displaystyle \delta _{t}=p+{s \over {1+rse^{st}}}}$ where p, r and s are estimated.

To convert an interest rate from one compounding basis to another compounding basis, use

${\displaystyle r_{2}=\left[\left(1+{\frac {r_{1}}{n_{1}}}\right)^{\frac {n_{1}}{n_{2}}}-1\right]{n_{2}},}$

where r₁ is the interest rate with compounding frequency n₁, and r₂ is the interest rate with compounding frequency n₂.

When interest is continuously compounded, use

${\displaystyle \delta =n\ln {\left(1+{\frac {r}{n}}\right)},}$

where ${\displaystyle \delta }$ is the interest rate on a continuous compounding basis, and r is the stated interest rate with a compounding frequency n.

The interest on loans and mortgages that are amortized—that is, have a smooth monthly payment until the loan has been paid off—is often compounded monthly. The formula for payments is found from the following argument.

An exact formula for the monthly payment (${\displaystyle c}$) is

${\displaystyle c={\frac {Pr}{1-{\frac {1}{(1+r)^{n}}}}}}$

or equivalently

${\displaystyle c={\frac {Pr}{1-e^{-n\ln(1+r)}}}}$

where:

${\displaystyle c}$ = monthly payment

${\displaystyle P}$ = principal

${\displaystyle r}$ = monthly interest rate

${\displaystyle n}$ = number of payment periods

This can be derived by considering how much is left to be repaid after each month.
The Principal remaining after the first month is

${\displaystyle P_{1}=(1+r)P-c,}$

that is, the initial amount plus interest less the payment.
If the whole loan is repaid after one month then

${\displaystyle P_{1}=0}$, so ${\displaystyle P={\frac {c}{1+r}}}$

After the second month ${\displaystyle P_{2}=(1+r)P_{1}-c}$ is left, so

${\displaystyle P_{2}=(1+r)((1+r)P-c)-c}$

If the whole loan was repaid after two months,

${\displaystyle P_{2}=0}$, so ${\displaystyle P={\frac {c}{1+r}}+{\frac {c}{(1+r)^{2}}}}$

This equation generalises for a term of n months, ${\displaystyle P=c\sum _{j=1}^{n}{\frac {1}{(1+r)^{j}}}}$. This is a geometric series which has the sum

${\displaystyle P={\frac {c}{r}}\left(1-{\frac {1}{(1+r)^{n}}}\right)}$

which can be rearranged to give

${\displaystyle c={\frac {Pr}{1-{\frac {1}{(1+r)^{n}}}}}={\frac {Pr}{1-e^{-n\ln(1+r)}}}}$

Spreadsheet formula

In spreadsheets, the PMT() function is used. The syntax is:

PMT( interest_rate, number_payments, present_value, future_value,[Type] )

See Excel, Mac Numbers, LibreOffice, Open Office, Google Sheets for more details.

For example, for interest rate of 6% (0.06/12), 25 years * 12 p.a., PV of $150,000, FV of 0, type of 0 gives:

= PMT(0.06/12, 25 * 12, -150000, 0, 0)

= $966.45

A formula that is accurate to within a few percent can be found by noting that for typical U.S. note rates (${\displaystyle I<8\%}$ and terms ${\displaystyle T}$=10–30 years), the monthly note rate is small compared to 1: ${\displaystyle r<<1}$ so that the ${\displaystyle \ln(1+r)\approx r}$ which yields a simplification so that

${\displaystyle c\approx {\frac {Pr}{1-e^{-nr}}}={\frac {P}{n}}{\frac {nr}{1-e^{-nr}}}}$

which suggests defining auxiliary variables

${\displaystyle Y\equiv nr=IT}$

${\displaystyle c_{0}\equiv {\frac {P}{n}}}$.

Here ${\displaystyle c_{0}}$ is the monthly payment required for a zero–interest loan paid off in ${\displaystyle n}$ installments. In terms of these variables the approximation can be written

${\displaystyle c\approx c_{0}{\frac {Y}{1-e^{-Y}}},}$

The function ${\displaystyle f(Y)\equiv {\frac {Y}{1-e^{-Y}}}-{\frac {Y}{2}}}$ is even:

${\displaystyle f(Y)=f(-Y)}$

implying that it can be expanded in even powers of ${\displaystyle Y}$.

It follows immediately that ${\displaystyle {\frac {Y}{1-e^{-Y}}}}$ can be expanded in even powers of ${\displaystyle Y}$ plus the single term: ${\displaystyle Y/2.}$

It will prove convenient then to define

${\displaystyle X={\frac {1}{2}}Y={\frac {1}{2}}IT}$

so that

${\displaystyle c\approx c_{0}{\frac {2X}{1-e^{-2X}}}}$

which can be expanded:

${\displaystyle c\approx c_{0}\left(1+X+{\frac {X^{2}}{3}}-{\frac {1}{45}}X^{4}+...\right)}$

where the ellipses indicate terms that are higher order in even powers of ${\displaystyle X}$. The expansion

${\displaystyle P\approx P_{0}\left(1+X+{\frac {X^{2}}{3}}\right)}$

is valid to better than 1% provided ${\displaystyle X\leq 1}$.

For a $10,000 mortgage with a term of 30 years and a note rate of 4.5%, payable yearly, we find:

${\displaystyle T=30}$

${\displaystyle I=0.045}$

which gives

${\displaystyle X={\frac {1}{2}}IT=.675}$

so that

${\displaystyle P\approx P_{0}\left(1+X+{\frac {1}{3}}X^{2}\right)=\$333.33(1+.675+.675^{2}/3)=\$608.96}$

The exact payment amount is ${\displaystyle P=\$608.02}$ so the approximation is an overestimate of about a sixth of a percent.

Given an principal (initial) deposit and a recurring deposit, the total return of an investment can be calculated via the compound interest gained per unit of time. If required, the interest on additional non-recurring and recurring deposits can also be defined within the same formula (see below).[5]

${\displaystyle P}$ = Principal deposit

${\displaystyle r}$ = Rate of return (monthly)

${\displaystyle M}$ = Monthly deposit, and

${\displaystyle t}$ = Time, in months

${\displaystyle M{\frac {(1+r)^{t}-1}{r}}+P(1+r)^{t}}$

If two or more types of deposits occur (either recurring or non-recurring), the compound interest earned can be represented as

${\displaystyle M{\frac {(r+1)^{t}-1}{r}}+P(r+1)^{t}+k{\frac {(r+1)^{t-x}-1}{r}}+C(r+1)^{t-y}}$

where C and k are non-recurring and recurring deposits, respectively, and x and y are the differences in time between a new deposit and whichever variable ${\displaystyle t}$ is modeling.