MHB Find the maximum amount of the equal annual withdrawal

[Joe_1234]

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]

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I would let \(A_n\) be the amount in the account at the end of year \(n\), and \(W\) be the amount withdrawn at the end of year \(n\).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:

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

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

 

[MarkFL]

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

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

And so our homogeneous solution will take the form:

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

Our particular solution will take the form:

$$p_n=B$$

Substitution into the recursion yields:

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

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

And so by the principle of superposition, we obtain:

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

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

And so our closed form is:

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

To finish the problem, we want to set:

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

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

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

Using the value \(A_0=25000\), we then find:

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

 

[Joe_1234]

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

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

And so our homogeneous solution will take the form:

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

Our particular solution will take the form:

$$p_n=B$$

Substitution into the recursion yields:

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

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

And so by the principle of superposition, we obtain:

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

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

And so our closed form is:

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

To finish the problem, we want to set:

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

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

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

Using the value \(A_0=25000\), we then find:

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

Tnx Sir