I am having trouble understanding how to solve problems with varying annuities. There is this problem I was given as a homework which I can't figure out.
Barry presently has 2.9 million dollars in an account paying a nominal rate of 6.6 percent convertible quarterly. He plans to start making quarterly withdrawals from the account when he retires, the first coming in exactly 20 years. If he would like to be able to make 88 withdrawals (with the last emptying the account) and the withdrawals will increase by 1.2 percent from one to the next, how large is his first withdrawal?
In the solution, I used the formula for payments varying in geometric progression, where R is the amount of the first withdrawal: $$ PV =R \frac{1-(\frac{1+k}{1+i})^n}{i-k} $$
Interest rate is convertible quarterly, so the annual effective interest rate is
$$(1+\frac{0.066}{4})^4-1=0.06765$$
For PV I have
$$2,900,000(1.06765)^{4*19}=10,059,099.498 $$
After putting everything into the formula and solving for R, I get $$ 10,059,099.498=R \frac{1-(\frac{1+0.012}{1+0.06765})^{88}}{(0.06765-0.012)} $$ $$ R=\frac{10,059,099.498}{17.8078}=564,870.42 $$
But that is not correct answer.
I also tried to put $(4*20)$ instead of $(4*19)$ or $\frac{88}{4}$ instead of $88$ and so on, but the answer is still incorrect.
Maybe I should use another approach, but I don't have other ideas on how to solve it.
Any help would be appreciated. Thank you.
Write out the cash flow. Let $PV = 2.9 \times 10^6$ be the present value, $K$ be the amount of the first withdrawal, and let $j = i^{(4)}/4 = 0.066/4$ be the effective quarterly rate of interest. Then $$PV = Kv^{20} + (1.012)Kv^{21} + (1.012)^2 Kv^{22} + \cdots + (1.012)^{87} K v^{107},$$ where $v = 1/(1+j)$ is the effective quarterly present value discount factor. Equivalently, we may write this as $$(1.012)^{20} PV = (v')^{19} K \left(v' + (v')^2 + (v')^3 + \cdots + (v')^{88}\right)$$ where $v' = 1.012v$ can be regarded as an adjusted quarterly present value discount factor. Consequently, $$K = \frac{(1.012)^{20} PV}{(v')^{19} a_{\overline{88}\rceil j'}} = \frac{1.012PV}{v^{19}} \frac{j'}{1 - (v')^{88}}$$ where $j' = 1/v' - 1$. All that is left is to substitute the values into this expression, which gives $K = \$55096.90$.