Cash accumulation equation

The cash accumulation equation is an equation which calculates how much money will be in a bank account, at any point in time. The account pays interest, and is being fed a steady trickle of money.

Compound interest

We will approach the development of this equation by first considering the simpler case, that of just placing a lump sum in an account and then making no additions to the sum. With the usual notation, namely

$y\,\!$	= the current sum (dollars)
$P\,\!$	= principal (dollars)
$i\,\!$	= force of interest (per year)
$t\,\!$	= time (years)

the equation is

${\displaystyle y=Pe^{it}}$ (1)

and so the sum of money grows exponentially. Differentiating this we derive

${\displaystyle {\frac {dy}{dt}}=iPe^{it}}$ (2)

and applying the definition of y from eqn (1) to eqn (2), yields

${\displaystyle {\frac {dy}{dt}}=iy}$ (3)

Note that eqn. (1) is a particular solution to the ordinary differential equation in eqn. (3), with y equal to P at t=0.

Cash infeed

Having achieved this we are ready to start feeding money into the account, at a rate of $F\,\!$ dollars/year. This is effected by making a small change to eqn (3) as follows

${\displaystyle dy=iy\,dt+F\,dt}$

and accordingly we need to solve the equation

${\displaystyle t=\int {\frac {dy}{iy+F}}}$

From a table of integrals, the solution is

${\displaystyle t={\frac {1}{i}}\ln(iy+F)+k}$

where $k\,\!$ is the constant of integration. The initial sum deposited was $P\,\!$ so we know one point on the curve :

${\displaystyle (t,y)=(0,P)\,\!}$

and making this substitution we find that

${\displaystyle k=-{\frac {1}{i}}\ln(iP+F)}$

Using this expression for $k\,\!$, and recalling that

${\displaystyle \ln(a)-\ln(b)=\ln \left({\frac {a}{b}}\right)}$

gives us the solution :

${\displaystyle it=\ln \left({\frac {iy+F}{iP+F}}\right)}$

This is the neatest form of the cash accumulation equation, as we are calling it, but it not the most useful form. Using the exponential instead of the logarithmic function, the equation can be written out like this :

${\displaystyle y=Pe^{it}+{\frac {F}{i}}(e^{it}-1){\mbox{ , }}i\neq 0}$ (4)

First special case

From this new perspective, eqn (1) is just a special case of eqn (4) - namely with ${\displaystyle F=0\,}$.

Second special case

For completeness we will consider the case ${\displaystyle i=0\,\!}$, and specifically the expression

${\displaystyle {\frac {e^{it}-1}{i}}{\mbox{ , }}i=0\,}$

One way of evaluating this is to write out the Maclaurin expansion

${\displaystyle e^{it}=1+it+{\frac {(it)^{2}}{2!}}+\cdots }$

At a glance we can subtract $1\,\!$ from this series and divide by $i\,\!$, to find out that

${\displaystyle {\frac {e^{it}-1}{i}}=t{\mbox{ , }}i=0\,}$

With this result the cash accumulation equation now reads

${\displaystyle y=P+Ft{\mbox{ , }}i=0\,}$

Thus the cash sum just increases linearly, as expected, if no interest is being paid.

Third special case

The only other special case to mention is ${\displaystyle F=-iP\,\!}$. Upon making this substitution, eqn (4) becomes simply

${\displaystyle y=P\,}$

Evidently $F\,\!$ is negative, and money is being withdrawn rather than deposited. Specifically, the interest is being withdrawn as fast as it is being earned.

An alternative interpretation of this special case is that $P\,\!$ is negative - the account is overdrawn - and money is being fed in at a rate which just meets the interest charges. A force of interest value is always positive.

External links

• an extended version of this article
• try out the equation with you own values