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I'm working on the following problem at the moment while preparing for an exam.

Find the present value of payments of 200 every six months starting immediately and continuing through four years from the present, and 100 every six months thereafter through ten years from the present, if i^(2) = .06. (all quantities in dollars)

Since interest is given, and not discount I went ahead and calculated this as an annuity immediate instead of an annuity due.

$\displaystyle a=\frac{1-v^n}{i}=\frac{1-1.03^{-9}}{.03}= 7.786108922$

$\displaystyle a\times200=1557.221784$

So far so good. Here's where I think I might be doing something wrong. For the last six years of the annuity, I'm calculating the present value four years in the future and then discounting that another four years. Is this wrong?

$\displaystyle 100a= 100\times\frac{1-1.03^{-18}}{.03}=1063.495533$

So this is the present value of the last six years of the annuity at time four into the future. I discount it four years to get:

$\displaystyle 1063.495533\times\nu^8=839.5331944$

To answer the question, I get 2,396.75. The book gives 2,389.72, which is quite close. Can anyone account for the difference?

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I see a homework tag, but I also see that you said you were doing it while preparing for an exam, so that makes it sound like it's not homework??? Homework tag means it was assigned for homework. I guess I am justified in showing you how to do it because you were almost there any way. –  Graphth Sep 29 '12 at 1:10
    
I think your first error is that you forgot to fix the first part. You calculated the value of a series of payments, essentially at time -1 (6 months before the first) and I don't see anywhere where you accumulated it forward with a 1.03. Your second error is in the second part, $1.03^{-18}$ means you are calculating the value of 18 payments of 100 one period before the first one. There is nothing in this problem that would represent 18 payments of 100. It should be 12. Fix those two errors and see what you get. –  Graphth Sep 29 '12 at 1:14
    
There is some ambiguity. For the first chunk, should one interpret wording as saying there are $9$ payments? Or is it $8$? –  André Nicolas Sep 29 '12 at 1:15
    
@AndréNicolas 9 is what I interpreted it as and that gave the correct answer. So, that is what was meant I guess. But, I agree it's not that clear. I would probably say "with the last payment coming 4 years from now" or something. –  Graphth Sep 29 '12 at 1:17
    
I interpreted it as 9. –  Josh Infiesto Sep 29 '12 at 15:46

1 Answer 1

up vote 2 down vote accepted

My first piece of advice is to learn the ways to make these types of problems simpler. That's going to make it less likely you make a mistake and more likely that you can do it quickly. Here's a way on this problem. Learn this trick and use it in the future. (And, memorize $d = i/(1+i)$. You will need it many, many times. Don't make problems much harder because you didn't take 3 minutes to memorize this.)

Let's let our "period" be 6 months. Then our interest rate is 3% per period, as you have. My understanding of the question is we get a payment of 200 now, 1 period from now, up to 8 periods from now (4 years from now), for a total of 9 payments. Then, we get payments of 100 at times 9, 10, ..., 20, for a total of 12 of those.

Method 1: If we want the present value, this is equivalent to 21 payments of 100, starting immediately, plus another sequence of 9 payments of 100 starting immediately (the first 9 payments of each coincide and add up to 200). a double dot angle n = $\frac{1 - v^n}{d} = \frac{1 - v^n}{i/(1 + i)}$. Therefore, our total present value is

$$100 \left(\frac{1 - 1.03^{-21}}{.03/1.03} + \frac{1 - 1.03^{-9}}{.03/1.03}\right) = 2389.72$$

Method 2: The present value of the first 9 payments is:

$$200 \frac{1 - 1.03^{-9}}{.03/1.03}$$

The present value of the next 12 payments at time 8 (4 years from now, and 6 months before the first payment of 100) is:

$$100 \frac{1 - 1.03^{-12}}{.03}$$

so we need to discount this back 4 years and add to the other payments to get:

$$200 \frac{1 - 1.03^{-9}}{.03/1.03} + 100 (1.03)^{-8} \frac{1 - 1.03^{-12}}{.03}$$

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