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Browser
08-20-2002, 12:52 PM
Given that (S 2n) - Accumulated Due value with 2n payments = 300 and (An) = 10 (Simple PV formula) find i.

The hint is to change the accumulated value to a present vlaue with 2n payments. I seem to be getting stuck on good ole Algebra. any help would be great

MathGuy
08-20-2002, 01:19 PM
Clarification Request:

You state that the first annuity is an annuity due, accumulated 2n periods, so the accumulated value is:

S<sub>2n</sub> = (1 + v + ... + v<sup>2n-1</sup>) * (1 + i)<sup>2n</sup> = 300

(This is correct right?) Anyway, the clarification is whether the second annuity you mention (A<sub>n</sub> = 10) is due or immediate. Thanks.

Browser
08-20-2002, 01:26 PM
The first is due, the second is immediate.

Bama Gambler
08-20-2002, 01:31 PM
Mathguy,

How did you use subscripts in your post?

Thanks,
Bama Gambler

MathGuy
08-20-2002, 01:38 PM
Assuming it's what I think it is:

A<sub>n</sub> = 1 + v + ... + v<sup>n-1</sup> = 10
vA<sub>n</sub> = v + v<sup>2</sup> + ... + v<sup>n</sup> = 10v

A<sub>n</sub> - vA<sub>n</sub> = 1 - v<sup>n</sup> = 10 - 10v

1 - v<sup>n</sup> = 10(1 - v) [Formula I]

We'll save this for later.

Next,

S<sub>2n</sub> = (1 + v + ... + v<sup>2n-1</sup>) * (1 + i)<sup>2n</sup> = 300

If we "de-accumulate", which means we just discount back 2n periods using the discount factor v, we get:

1 + v + ... + v<sup>2n-1</sup> = 300v<sup>2n</sup>
1 + v + ... + v<sup>n-1</sup> + v<sup>n</sup> + ... + v<sup>2n-1</sup> = 300v<sup>2n</sup>

but, the bolded terms are the same as A<sub>n</sub> = 10 from above:

10 + v<sup>n</sup> + ... + v<sup>2n-1</sup> = 300v<sup>2n</sup>
10 + v<sup>n</sup>(1 + v + ... + v<sup>n-1</sup>) = 300v<sup>2n</sup>

And there he is again!

10 + v<sup>n</sup>(10) = 300v<sup>2n</sup>
0 = 300v<sup>2n</sup> - 10v<sup>n</sup> - 10
0 = 300(v<sup>n</sup>)<sup>2</sup> - 10v<sup>n</sup> - 10

And use the handy dandy quadratic formula, with v<sup>n</sup> = x:

v<sup>n</sup> = (10 + sqrt(100 - 4(300)(-10)))/(2(300)) = (10 + 110)/600 = 120/600 = .20.

Now, we take Formula I from way back at the beginnning and plug in v<sup>n</sup> = .20:

1 - v<sup>n</sup> = 10(1 - v)
1 - .20 = 10(1 - v)
.8 = 10(1 - v)
.08 = 1 - v
v = .92
(1 + i)<sup>-1</sup> = .92
(1 + i) = .92<sup>-1</sup>
1 + i = 1.087
i = .087

MathGuy
08-20-2002, 01:40 PM
Subscripts: use the html tags &lt;sub&gt; and &lt;/sub&gt; around what ever you'd like to have as a subscript. (Same for sup).

If the second is immediate I'll have to think a little more.

Browser
08-20-2002, 02:00 PM
With my inproved notational abilites the problem reads:

..
S<sub>2n</sub> = 300 (Due)

And

a<sub>n</sub>=10 (Immediate)

find i

MathGuy
08-20-2002, 02:10 PM
This solution will be similar to my previous one:

a<sub>n</sub> = v + ... + v<sup>n</sup> = 10
va<sub>n</sub> = v<sup>2</sup> + ... + v<sup>n+1</sup> = 10v

v - v<sup>n+1</sup> = 10 - 10v FORMULA I

The "de-accumulation" of S<sub>2n</sub> works the same as above, and we get to:

1 + (v + v<sup>2</sup> + ... + v<sup>n</sup>) + v<sup>n+1</sup> + ... + v<sup>2n-1</sup> = 300v<sup>2n</sup>
1 + (10) + v<sup>n+1</sup> + ... + v<sup>2n-1</sup> = 300v<sup>2n</sup>
11 +v<sup>n+1</sup> + ... + v<sup>2n-1</sup> = 300v<sup>2n</sup>
11 + v<sup>n</sup>(v + ... + v<sup>n-1</sup>) = 300v<sup>2n</sup>
11 + v<sup>n</sup>(10 - v<sup>n</sup>) = 300v<sup>2n</sup>
11 - 10v<sup>n</sup> - 2v<sup>2n</sup> = 300v<sup>2n</sup>
0 = 302v<sup>2n</sup> - 10v<sup>n</sup> - 11

Again, using the quadratic formula, we get v<sup>n</sup> = .2081

And, using Formula I from above:

v - v<sup>n+1</sup> = 10 - 10v
v(1 - v<sup>n</sup>) = 10 - 10v
v(1 - .2081) = 10 - 10v
.7919v = 10 - 10v
10.7919v = 10
v = 10/10.7919
1/(1+i) = .9266
1 + i = 1.079
i = .079

Browser
08-20-2002, 02:18 PM

c3 taker
08-20-2002, 03:34 PM
And, using Formula I from above:

v - v<sup>n+1</sup> = 10 - 10v
v(1 - v<sup>n</sup>) = 10 - 10v
v(1 - .2081) = 10 - 10v
.7919v = 10 - 10v
10.7919v = 10
v = 10/10.7919
1/(1+i) = .9266
1 + i = 1.079
i = .079

Rather than use formula 1 you could also just say that since a<sub>n</sub> = 10 = (1 - v<sup>n</sup>) / i
then i = (1 - .2081) / 10 = .079

same premise, I find it's easier to remember this way (if you know the formula for an annuity-immediate, which I assume you won't do very well on interest theory without) :D

RiSK kid
08-20-2002, 04:45 PM
Subscripts: use the html tags <sub> and </sub> around what ever you'd like to have as a subscript. (Same for sup).

Amazing, I totally forgot that we could use HTML. This is gonna make the questions a lot more readable.
Thanks

MathGuy
08-21-2002, 08:30 AM
C2 taker: Formulas are for babies. I prefer the all-out guerilla-math style: Everything on first principles!!! For Interest Theory, this is particularly useful because the entire topic is light and fluffy, but "the man" weighs it down with all sorts of formulas and stuff. If you can write down 1 + v + ... + v<sup>n</sup>, and you truly understand what v means, you can answer any question about annuities that comes up. Sure it makes for a crazy four hours come exam time, but it means that there is no studiying required.

c3 taker
08-21-2002, 09:17 AM
C2 taker: Formulas are for babies. I prefer the all-out guerilla-math style: Everything on first principles!!! For Interest Theory, this is particularly useful because the entire topic is light and fluffy, but "the man" weighs it down with all sorts of formulas and stuff. If you can write down 1 + v + ... + v<sup>n</sup>, and you truly understand what v means, you can answer any question about annuities that comes up. Sure it makes for a crazy four hours come exam time, but it means that there is no studiying required.

I understand the concept of 1 + v + ... + v<sup>n</sup>, I just think it's easier to remember a few formulas, but maybe I'm just frightened by your "all-out guerilla-math style" :D