# Compound interest with penalty in withdrawal amount

Carl puts $10000$ into a bank account that pays an annual effective interest rate of $0.04$ for $10$ years. If a withdrawal is made during the first five and a half years, a penalty of $5\%$ of the withdrawal amount is made.Carl withdraws $K$ at the end of years $4$, $5$, $6$, $7$. The balance in the account at the end of year $10$ is $10000$. Calculate $K$.

From the effective interest rate, we get the compound interest rate to be $0.04$.

Then at $t=4$: $A(4)-K-0.15K=10000(1.04)^4-1.05K=11698.58-1.05K$

The, afterwards, I do not exactly know what to do.

• hope this may help interpretmaths.blogspot.com/2015/03/… - pure theory Feb 28 '16 at 7:18
• Is there penalty on the 6th and 7th years? Feb 28 '16 at 8:57
• @N.S.JOHN,there is no penalty on 6th and 7th yr, do you have any idea of the working?
– Tosh
Feb 28 '16 at 10:04

Let $i=4\%$, $p=5\%$, $n=10$, $C=10000$ and $S=10000$. At the end of the 4th year we have $C(1+i)^4-pK-K=P$ and at the end of the 5th year we have $P(1+i)-pK-K=Q$. At the end of the 6th year we have $Q(1+i)-K=M$ and at the end of 7th year $M(1+i)-K=N$. So the sum $N$ will produce $N(1+i)^{n-7}=S$ at the end of the $n$-th year. That is $$\Big(\underbrace{\{\underbrace{[\underbrace{(\underbrace{C(1+i)^4-(1+p)K}_{P})(1+i)-(1+p)K }_Q](1+i)- K}_M\}(1+i)-K}_N\Big)(1+i)^{n-7}=S\tag 1$$ and after some products

\scriptsize \begin{align*} S &=\left(\Big\{\Big[C(1+i)^4(1+i)-(1+p)K(1+i)-(1+p)K\Big](1+i)-K\Big\}(1+i)-K\right)(1+i)^{n-7}\\ &=\left(\Big\{C(1+i)^4(1+i)(1+i)-(1+p)K(1+i)(1+i)-(1+p)K(1+i)-K\Big\}(1+i)-K\right)(1+i)^{n-7}\\ &=\left(C(1+i)^4(1+i)(1+i)(1+i)-(1+p)K(1+i)(1+i)(1+i)-(1+p)K(1+i)(1+i)-K(1+i)-K\right)(1+i)^{n-7}\\ &=C(1+i)^4(1+i)(1+i)(1+i)(1+i)^{n-7}-(1+p)K(1+i)(1+i)(1+i)(1+i)^{n-7}-(1+p)K(1+i)(1+i)(1+i)^{n-7}-K(1+i)(1+i)^{n-7}-K(1+i)^{n-7}\\ &=C(1+i)^{4+1+1+1+n-7}-(1+p)K(1+i)^{1+1+1+n-7}-(1+p)K(1+i)^{1+1+n-7}-K(1+i)^{1+n-7}-K(1+i)^{n-7}\\ &=C(1+i)^{n}-(1+p)K(1+i)^{n-4}-(1+p)K(1+i)^{n-5}-K(1+i)^{n-6}-K(1+i)^{n-7}\\ &=C(1+i)^n-(1+p)K\Big[(1+i)^{n-4}+(1+i)^{n-5}\Big]-K\Big[(1+i)^{n-6}+(1+i)^{n-7}\Big] \end{align*}

or $$C(1+i)^n-(1+p)K\Big[(1+i)^{n-4}+(1+i)^{n-5}\Big]-K\Big[(1+i)^{n-6}+(1+i)^{n-7}\Big]=S\tag 2$$ and finally $$K=\frac{C(1+i)^n-S}{(1+p)\left[(1+i)^{n-4}+(1+i)^{n-5}\right]+\left[(1+i)^{n-6}+(1+i)^{n-7}\right]}$$

• I could not understand how you factorised it @alexjo
– Tosh
Feb 28 '16 at 17:08
• @Toshina11 Do you mean eq. 1 or eq. 2? Feb 28 '16 at 17:22
• I mean it is eq 2
– Tosh
Feb 28 '16 at 17:23
• @Toshina11 I've added some calculations Feb 28 '16 at 17:57
• I had an aha moment as you factorised it out so easily.
– Tosh
Feb 29 '16 at 1:46

Your expression for the value at $t=4$ is fine, though you have a typo of $0.15$ instead of $0.05$. If you define $A(4)$ as the value at the end of year $4$, it should just be $A(4)=10000\cdot 1.04^4-1.05K$ and the first equal sign is not correct. Now keep going $A(5)=1.04A(4)-1.05K$ and so on. Plug in $A(4)$ to this, the evaluate $A(6)$, but you will not have the factor $1.05$ because there is no penalty. Once you get to $A(7)$ you have $A(10)=A(7) \cdot 1.04^3=100000$ You will have a quadratic equation in $K$. Perhaps you are expected to use a spreadsheet and goal seek.

• Actually, you won't have a quartic equation in $K$... Feb 28 '16 at 17:59
• @alexjo: I missed that there was only a penalty for two years. Fixed. Feb 28 '16 at 23:01