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"Consider an investment of $5,000 at 6% convertible semiannually. How much can be withdrawn each half−year to use up the fund exactly at the end of 20 years?"

To solve this problem, an equation of value would be set between 5000 and the unknown withdrawal amount multiplied by the present value annuity expression with t = 40 and i = 3%.

I don't understand how a withdrawal can be multiplied by the PV expression. I understand how a deposit can be multiplied by the PV expression because that money is being funneled into an account, not taken out. Can someone explain why withdrawals are treated the same as deposits?

Mel
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3 Answers3

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$50,000 =r \cdot \frac{1.03^{40}-1}{1.03-1}\cdot \frac{1}{1.03^{40}}$

$r \cdot \frac{1.03^{40}-1}{1.03-1}$ is the value of all the 40 withdrawals at t=40.

$r \cdot \frac{1.03^{40}-1}{1.03-1}\cdot \color{blue}{\frac{1}{1.03^{40}}}$ is the present value of all the 40 withdrawals. The blue factor discounts the expression above for 20 years (40 half-years). So we are at t=0.

And 50,000 on the RHS is the present value of the deposits (t=0).

Thus both sides are values at t=0. And the withdrawals and the deposits must be equal, if the account should have a value of 0 after 20 years.

You can also multiply the equation by $1.03^{40}$. In that case the RHS and the LHS show the values of the deposits and the withdrawals in 20 years.

OR just make the RHS equal to 0:

$\underbrace{50,000}_{\text{PV of the deposit}} -\underbrace{r \cdot \frac{1.03^{40}-1}{1.03-1}\cdot \frac{1}{1.03^{40}}}_{\text{PV of the withdrawals }}=0$

Here you can say the deposit is payed in and the withdrawals are cashed out.

callculus42
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Just think of a withdrawal of $x$ as a deposit of $-x$. The present value of a quantity of a future cash flow is the same, except for sign (signifying the direction of the flow), whether it is a deposit or a withdrawal.

Think of it this way: if an account today has $a+ b$ dollars, it will grow to $FV (a) +FV (b)$ in the future. But if you remove $b$ dollars today, then the remaining $a$ dollars will only grow to $FV (a)$ in the future. Isn't that the same as leaving $a+b$ in the account, and then waiting to withdraw $FV (b)$ in the future? In either case, you will have $FV (a)$ in the account in the future.

MPW
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6% yearly compounded semi-annually means that every six months, 3% of the amount in the fund is added to the fund. Then, some fixed withdrawal is made. The question is, what size of withdrawal will drain the account at the end of forty half-year periods?

Let's call the 3% interest (1.03) by the variable $i$, and the initial amount $P_0$, and the withdrawal $W$. At time $t = 0$, there is $P_0$ in the fund. At time $t = 1$, the amount is $P_1 = iP_0 - W$ At time $t = 2$ the amount is: $$P_2 = iP_1 - W$$ This is equal to $$P_2 = i(iP_0 - W) - W$$ $$P_2=i^2P_0 - iW - W$$

The subsequent elements of the series follow the same pattern: multiply by $i$, subtract $W$: $$P_3 = i^3P_0 - i^2W - iW - W$$ $$P_4 = i^4P_0 - i^3W - i^2W - iW - W$$ $$...$$

The $-W$ can be factored out, and the powers of $i$ reduced to a geometric series expressed as a summation: $$P_n = i^nP_0 - W\sum_{k = 0}^{n-1}{i^k}$$

Now let us turn our attention to $\sum$ part. We can attack this by using the closed formula for it: $$\sum_{k=0}^{n-1} i^k= \frac{1-i^{n}}{1-i}$$ which gets us to: $$P_n = i^nP_0 - W\frac{1-i^{n}}{1-i}$$

Now this can be solved by plugging in the known values ($n = 40; i = 1.03; P_0 = 5000; P_n = 0$) and isolating $W$.

Kaz
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