Is there a formula for this interest over a period

Question

Say I have £100 overdraft and my balance is currently £0. when I go into my overdraft I am charged 1.25 % at the end of each day, but I have to divide the rest of the money equally for the next 10 days to see what I can realistically afford.

Is there a way formula to work out the most optimum daily spend for each day without trial and error?

for example, if 0% was deducted each day, clearly £10 each day would be optimal leaving you with -£100

with interest being deducted each day it would make this harder to work out

the point is trying to work out, how much money someone can afford to spend each day, if it was an equal amount through to the last day

Answer 1

Let the number of days be $D=10$, the total money available be $M=100$, and the interest fee be $r=0.0125$. The daily spending is $s$. The total money spent after $D$ days is \begin{align*} M & =\bigg(\Big(\big(s(1+r)+s\big)(1+r)+s\Big)+\cdots\bigg)(1+r) \\ & = s(1+r)^{D} + s(1+r)^{D-1} + \cdots + s(1+r) \\ & = s\frac{(1+r)^{D+1} - (1+r)}{r} \end{align*} Therefore, the optimal daily spending is \begin{equation*} s=\frac{Mr}{(1+r)^{D+1} - (1+r)} \end{equation*} which in your case is $s=9.33363693543$.

Answer 2

Say the amount of money I have at day $0$ is $M_0$. Over the course of $N$ days, I want to spend an equal amount of money $f$ each day such that at the end of the $N$ days, I have $0$ left. However, each day after I spent my money, I loose a certain amount $x$ of money. What is the value of $f$ ?

The case $x=0$ is already taken care, so let us consider that $x\neq 0$

We can write a sequence corresponding to our problem:

-day $0$: $M_0$

-day $1$: $M_1 = (M_0-f) - x(M_0-f) = (1-x)(M_0-f)$

-...

-day $N$: $M_N = (1-x)(M_{N-1}-f) = (1-x)M_{N-1} - f(1-x)$

This sequence can be solved and we find: \begin{equation} M_N = M_0(1-x)^N + f \frac{1-x}{x}[(1-x)^N - 1]. \end{equation} At the end of the $N$ days, we want $M_N=0$, such that it can be solved for $f$, we have: \begin{equation} f=-M_0 \frac{x(1-x)^{N-1}}{(1-x)^N - 1} \end{equation}

If the money is taken before we can spend some, the sequence is a little bit modified:

-day $0$: $M_0$

-day $1$: $M_1 = (1-x)M_0 - f$

-...

-day $N$: $M_N = (1-x)M_{N-1} - f$

Again solving the sequence gives: \begin{equation} M_N = M_0(1-x)^N + f \frac{1}{x}[(1-x)^N - 1]. \end{equation} At the end of the $N$ days, we want $M_N=0$, such that it can be solved for $f$, we have: \begin{equation} f=-M_0 \frac{x(1-x)^{N}}{(1-x)^N - 1} \end{equation}

For the specific example asked, we have $N=10$, $x=1.25\%$ and $M_0=100$, we can compute: \begin{align} f_{after} \approx 9.4435\\ f_{before} \approx 9.3255 \end{align}

Answer 3

Suppose you withdraw $x$ every day.

At the end of day 1, your overdraft is $x\times1.0125$.

At the end of day 2, your overdraft is $(x\times1.0125+x)\times1.0125$ $$=x\times(1.0125+1.0125^2)$$

Continuing this pattern, at the end of day $10$, your overdraft is $$x\times(1.0125+1.0125^2+...+1.0125^{10})$$

The bracketed term is a geometric series whose sum is $$1.0125\left(\frac{1.0125^{10}-1}{1.0125-1}\right)$$

Your overdraft is $100$ and this equals the above expression, so $$x=\frac{100\times0.0125}{1.0125(1.0125^{10}-1)}=9.33363...$$