# Interest Theory- Annuity Withdrawals

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1. Jul 30, 2014

### uestions

1. The problem statement, all variables and given/known data

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 How much can be withdrawn each half−year to use up the fund exactly at the end of 20 years? 2. Relevant equations the present value annuity-immediate equation equation of value relating 5000 to the above equation 3. The attempt at a solution withdrawal is unknown 5000 = withdrawal * present value of annuity I have a more urgent question: why can a withdrawal value be multiplied by the present value function when the withdrawal is being taken out? My thinking is the present value function can only be multiplied by deposits because deposits will be affected by interest. This question bothers me more than solving the problem. 2. Jul 31, 2014 ### Ray Vickson Two answers: (1) Withdrawals are the opposite of deposits, so the present value of a series of payments is numerically equal to the present value of the same series of withdrawals. In other words, it is true just because of the math. However, I suspect you agree with the math but are still a bit mystified by why it works, so here is another answer. (2) We can work out a detailed explanation, step-by-step (but for a much smaller example). Look at the PV of two withdrawals of$1 at times t = 0 and 1 (with 1-period interest = r). The PV is $PV_2$ (the '2' standing for two withdrawals)
$$PV_2 = 1 + \frac{1}{1+r}$$
We start with $PV_2$ dollars in the bank. After the initial $1 withdrawal at time t = 0 the bank account contains$$1/(1+r)$. Because of interest earned, this grows to $$(1+r) \times 1/(1+r) = 1$ at time t = 1, when our second withdrawal of$1 empties the bank account.

Now let's try it again for three $1 withdrawals at times t = 0,1,2. The PV is $PV_3$: $$PV_3 = 1 + \frac{1}{1+r} + \frac{1}{(1+r)^2}$$ We start with$$PV_3$ in the bank. At t = 0 we withdraw $1 and so are left with a balance of $$B_1 = \frac{1}{1+r} + \frac{1}{(1+r)^2}$$ at time t = 1 (just after the withdrawal). This grows to $B_2 = (1+r)B_1$ at time t = 1; here, $$B_2 = 1 + \frac{1}{1+r}$$ and we still have two$1 withdrawals to go. But that case was already treated in the previous example; that is $B_2 = PV_2$, and we already know from before that two more $1 withdrawals will empty the bank account. So, again, the PV represents the total effects of withdrawals plus interest earned throughout the payment period---in such a way that we end up with exactly$0 at the end.

For larger problems having more than three payments you can just do something similar. So, for 4 payments, the PV that remains after the first withdrawal and after earning interest is just $PV_3$, which has already been analyzed. Similarly, 5 payments, after the first withdrawal and interest earned, becomes the 4-payment case, etc., etc.