# Find the maximum amount of the equal annual withdrawal

#### Joe_1234

##### New member
P25,000 is deposited in a savings account that pays 5% interest, compounded semi-annually. Equal annual withdrawals are to be made from the account, beginning one year from now and continuing forever. The maximum amount of the equal annual withdrawal is closest to
A. P625
B. P1000
C. P1250
D. P1625

#### MarkFL

Staff member
Hello, and welcome to MHB!

I would let $$A_n$$ be the amount in the account at the end of year $$n$$, and $$W$$ be the amount withdrawn at the end of year $$n$$. Then we may model this problem with the following recursion:

$$\displaystyle A_{n+1}=1.05A_{n}-W$$

Can you give the homogeneous solution (based on the root of the characteristic equation)?

#### MarkFL

Staff member
I made an error in my above post, where I used annual compounding. We want:

$$\displaystyle A_{n+1}=1.025^2A_{n}-W=\left(\frac{41}{40}\right)^2A_{n}-W$$

And so our homogeneous solution will take the form:

$$\displaystyle h_n=c_1\left(\frac{41}{40}\right)^{2n}$$

Our particular solution will take the form:

$$\displaystyle p_n=B$$

Substitution into the recursion yields:

$$\displaystyle B-\left(\frac{41}{40}\right)^2B=-W$$

$$\displaystyle \left(\frac{9}{40}\right)^2B=W\implies B=\left(\frac{40}{9}\right)^2W$$

And so by the principle of superposition, we obtain:

$$\displaystyle A_{n}=h_n+p_n=c_1\left(\frac{41}{40}\right)^{2n}+\left(\frac{40}{9}\right)^2W$$

$$\displaystyle A_0=c_1+\left(\frac{40}{9}\right)^2W\implies c_1=A_0-\left(\frac{40}{9}\right)^2W$$

And so our closed form is:

$$\displaystyle A_{n}=\left(A_0-\left(\frac{40}{9}\right)^2W\right)\left(\frac{41}{40}\right)^{2n}+\left(\frac{40}{9}\right)^2W$$

To finish the problem, we want to set:

$$\displaystyle \lim_{n\to\infty}A_n=0$$

In order for that to happen (for the limit to be finite), we need:

$$\displaystyle W=\left(\frac{9}{40}\right)^2A_0$$

Using the value $$A_0=25000$$, we then find:

$$\displaystyle W=\frac{10125}{8}=1265.625$$

#### Joe_1234

##### New member
I made an error in my above post, where I used annual compounding. We want:

$$\displaystyle A_{n+1}=1.025^2A_{n}-W=\left(\frac{41}{40}\right)^2A_{n}-W$$

And so our homogeneous solution will take the form:

$$\displaystyle h_n=c_1\left(\frac{41}{40}\right)^{2n}$$

Our particular solution will take the form:

$$\displaystyle p_n=B$$

Substitution into the recursion yields:

$$\displaystyle B-\left(\frac{41}{40}\right)^2B=-W$$

$$\displaystyle \left(\frac{9}{40}\right)^2B=W\implies B=\left(\frac{40}{9}\right)^2W$$

And so by the principle of superposition, we obtain:

$$\displaystyle A_{n}=h_n+p_n=c_1\left(\frac{41}{40}\right)^{2n}+\left(\frac{40}{9}\right)^2W$$

$$\displaystyle A_0=c_1+\left(\frac{40}{9}\right)^2W\implies c_1=A_0-\left(\frac{40}{9}\right)^2W$$

And so our closed form is:

$$\displaystyle A_{n}=\left(A_0-\left(\frac{40}{9}\right)^2W\right)\left(\frac{41}{40}\right)^{2n}+\left(\frac{40}{9}\right)^2W$$

To finish the problem, we want to set:

$$\displaystyle \lim_{n\to\infty}A_n=0$$

In order for that to happen (for the limit to be finite), we need:

$$\displaystyle W=\left(\frac{9}{40}\right)^2A_0$$

Using the value $$A_0=25000$$, we then find:

$$\displaystyle W=\frac{10125}{8}=1265.625$$
Tnx Sir