# Need help and clarification in Exam FM problem, future value.

The problem that I am working on is the following.

Jim began saving money for this retirement by making monthly deposits of 200 into a fund earning 6% interest compounded monthly. The first deposit occurred on 1/1/1985. He became unemployed and missed making deposits 60 through 72. He then continued making monthly deposits of 200. How much did Jim accumulate in his fund on 12/31/1999?

My thought process is this.

He missed twelve payments in the middle, so we can pretend that he paid the entire 180 months but he spent 200 twelve times from payment 60 to 72. So,

$$200 \ddot{s}_{\overline{180}\rceil .5\%} -200\ddot{s}_{\overline{12}\rceil .5\%}(1.005)^{108} =FV$$

which gives me 54,205 as an answer.

But the answer is supposedly 53,840.

What am I doing wrong?

• If I'm reading correctly, missing payments 60 through 72 means missing 13 payments, not 12. There shouldn't otherwise be anything wrong with the procedure you're using. – mardat Feb 2 '15 at 9:41

The effective monthly rate is $j = 0.06/12 = 0.005$.

Next, we have to figure out the times at which payments are made: there are $15$ years from 1/1/1985 to 1/1/2000, so the first payment has had $15 \cdot 12 = 180$ periods to accrue interest. Thus, in general, the $k^{\rm th}$ deposit has had $180 - k + 1$ months to accrue interest, and if the $60^{\rm th}$ deposit had been made, it would have had $180 - 60 + 1 = 121$ months to accrue interest. Similarly, had the $72^{\rm nd}$ deposit been made, it would have had $180 - 72 + 1 = 109$ months to accrue interest. So the cash flow can be written out as $$200\left((1+j)^{180} + (1+j)^{179} + \cdots + (1+j)^{122} + (1+j)^{108} + \cdots + (1+j)^1 \right).$$ Note that the final deposit, occurring on 12/1/1999, has had one month to accrue interest until 12/31/1999.

Thus we can now write the accumulated value in a few ways; one way is $$200\left((1+j)^{121} \ddot s_{\overline{59}\rceil j} + \ddot s_{\overline{108}\rceil j}\right) = 53839.8336.$$ Another way is $$200\left(\ddot s_{\overline{180}\rceil j} - (1+j)^{108} \ddot s_{\overline{13}\rceil j}\right) = 53839.8336.$$

The number we are looking for is $$\sum_{i=1}^{180} a_i\cdot 1.005^{i-1},$$ where $a_i = 0$ if $60\le i \le 72$ and $200$ otherwise. Therefore the sum splits as $$\sum_{i=1}^{180} a_i\cdot 1.005^{i-1} = 200\left(\sum_{i=1}^{61}\cdot 1.005^{i-1} + \sum_{i=73}^{180}\cdot 1.005^{i-1}\right).$$ Now these are geometric sums, which have known closed form. Can you solve the problem from here on?

• where are you getting 188 from? – hyg17 Jan 29 '15 at 20:06
• 14 years times 12 months makes, oopsie, 168.... – Marc Jan 29 '15 at 20:55
• But it's from Jan 1 to Dec 31 though. I'm sure it's 180 months. – hyg17 Jan 29 '15 at 22:30
• Good point, my calendar knowledge is worthless XD The idea of the exercise doesn't change though. – Marc Jan 29 '15 at 22:43
• I really appreciate your help, but in finance, "annual interest of 6% convertible monthly" usually means "nominal interest" where the "effective interest" is 6%/12=0.5%. Either way, I will try solving it as a geometric series, which of course, I am comfortable with. – hyg17 Jan 29 '15 at 22:47