I've been preparing for the FM exam coming up next month and am having some trouble understanding the solution given to Problem 24.15 in the book "A Basic Course in the Theory of Interest and Derivatives Markets." The question is as follows:
You want to accumulate \$1,000,000. You intend to do this by making deposits of \$5,000 into an investment account at the end of each month, until your account balance equals \$1,000,000. The account earns 8% convertible quarterly. Determine the number of monthly deposits you will need to make to achieve your goal.
The book gives a solution of 127, while I found 128 as my answer. I worked the problem as follows:
$\$1,000,000 = 12 * \$5,000 * \frac{1.02^{t/3} - 1}{i^{(12)}}$
$i^{(12)} = [1.03^{1/3} - 1] * 12 = 0.0794725$
At the end of it all, we are left with:
$1.02^{t/3} = 2.32454$
$t = 3*ln(2.32454)/ln(1.02) = 127.7895$
Since the number of monthly payments would have to be an integer, I am under the impression that it should be 128. I would understand the solution being 127 if the sequence of 127 payments would accumulate to be \$1,000,000 by the time t = 128, but that is not the case; the accumulated value of the annuity for t = 127 is approximately \$990,877.25, and accumulating a month of interest to this amount leaves us with only \$997,439.54 in the account (short of our million dollar goal).
If there's something I missed here I would greatly appreciate the assistance - want to make sure I don't miss it again on the exam. Thanks!
$\frac{dP}{dt} = .02P + 15000$
$\frac{50\cdot dP}{P + 750000} = dt$
$50\cdot \ln|P + 750000| = t + C$
$\ln|P + 750000| = .02(t + C)$
$P + 750000 = e^{.02C}\cdot e^{.02t}$
When $t = 0, P = 0$
$e^{.02C} = 750000$
$P = 750000e^{.02t} - 750000$ .....is the specific solution
When $P = 1000 000, t= ?$
$1000 000 = 750000e^{.02t} - 750000$
$.02t = \ln (\frac{1750000}{750000}) = \ln (2.3333333)$
$t = 42.36489$ quarters
$ = 42.36489\cdot 3 = 127.09$ months