# Thread: Interest - Holiday Fund

1. ## Interest - Holiday Fund

A man invests $60,000 in an account which earns interest at a compound rate fo 7% per annum. He wants to make a withdrawal of$W at the end of each year for hsi annual holiday expenses. The withdrawal is made immediately after that year's interest has been paid.

He intends to do this for 20 years, so that at ethe end of this time his account will have zero balance. He is trying to work out the amount of $W that he will have towards his holiday each year. (a) Firstly, he writes teh expression A1 in terms of W for the amount of money in his account immediately after he has made the first withdrawal. (b) Then he writes the expressions for A2 and A3, the amounts of money in his account immediately after his second and third withdrawals and simplifies them. (c) From this he deduces an expression for A20, and then finds how muhc he ahs towards his holiday each year for 20 years. Find the value of W, showing full justification. Okay, so for (a) I have A1 = 60,000*(1.07^1) - W, but i can't really decide what A2 or A3, or A20 is, and therefore i cannot find$W - could anyone walk em through how to do this? there is a similar problem i has ot this one, whihc i'd liek ot try unassisted. But i sort of need a template first ...I would hugely appreciate some help.

Thanks.

2. Hello, Sparta!

Are you looking for a derivation for the required formula?

A man invests $60,000 in an account which earns interest at a compound rate fo 7% per annum. He wants to make a withdrawal of$W at the end of each year for his annual holiday expenses.
The withdrawal is made immediately after that year's interest has been paid.

He intends to do this for 20 years, so that at the end of this time his account will be $0. He is trying to work out the amount of$W that he will have towards his holiday each year.

Your reasoning is correct. .Let $\displaystyle P$ = amount invested.

At the end of one year, the account has: $\displaystyle P(1.07)^1$ dollars.
He withdraws $W and the account has:$\displaystyle P(1.07) - W$dollars. At the end of year 2, the account has:$\displaystyle 1.07[P(1.07) - W] $dollars. He withdraws$W and the account has: $\displaystyle P(1.07)^2 - W(1.07) - W$ dollars.

At the end of year 3, the account has: $\displaystyle 1.07[P(1.07)^2 - W(1.07) - W]$ dollars.
He withdraws $W and the account has:$\displaystyle P(1.07)^3 - W(1.07)^2 - W(1.07) - W$dollars. . . . and so on . . . At the end of year 20, the account has: . .$\displaystyle 1.07)[P(1.07)^{19} - W(1.07)^{18} - W(1.07)^{17} - \:\cdots \:- W(1.07)^2 - W(1.07)]$dollars. He withdraws$W and the account has:
. . $\displaystyle P(1.07)^{20} - W(1.07)^{19} - W(1.07)^{18} - \:\cdots \:- W(1.07)^2 - W(1.07) - W$ dollars.

But this final balance will be zero dollars.

We have: . $\displaystyle P(1.07)^{20} \;=\;W\left[(1.07)^{19} + (1.07)^{18} + \cdots + (1.07)^2 + (1.07) + 1\right]$

The expression at the far right is a geometric series
. . with first term $\displaystyle a = 1$, common ratio $\displaystyle r = 1.07$, and 20 terms.

So we have: .$\displaystyle P(1.07)^{20} \;=\;W\cdot\frac{(1.07)^{20} - 1}{(1.07) - 1}$

. . Therefore: .$\displaystyle W \;=\;P\cdot\frac{(0.07)(1.07)^{20}}{(1.07)^{20}-1}$

For $\displaystyle P = \$60,000\!:\;\;W \;\approx\;\$5663.58$

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In general, the formula is: .$\displaystyle W \;=\;P\cdot\frac{i(1+i)^n}{(1 + i)^n - 1}$

. . where: .$\displaystyle \begin{Bmatrix}P & = & \text{principal invested} \\ i & = & \text{periodic interest rate} \\ n & = & \text{number of periods} \\ W & = & \text{periodic withdrawl}\end{Bmatrix}$

This formula is identical to the Amortization Formula
. . in which we are to pay off a debt.