PLEASE PLEASE HELP been stuck on this for more then 2 hours!
A zero compound bond is a bond that is sold now at a discount and will pay its face value at the time when it matures, no interest payment are made.
A zero-coupon bond can be redeemed in 20 years for $10000. How much should you be willing to pay for it now if want a return of.
(a) 13% compounded daily.
For a return of 13% compounded daily, should pay $_____
(b) 13% compounded continuously?
for a return of 13% continuously should pay $_____

Question

PLEASE PLEASE HELP been stuck on this for more then 2 hours!
A zero compound bond is a bond that is sold now at a discount and will pay its face value at the time when it matures, no interest payment are made. 
A zero-coupon bond can be redeemed in 20 years for $10000. How much should you be willing to pay for it now if want a return of.
(a) 13% compounded daily.
     For a return of 13% compounded daily, should pay $_____
(b) 13% compounded continuously?
     for a return of 13% continuously should pay $_____

e.mccormick · Answer

You are dealing with two compound interst formulas.  First, is $P = C (1 + \frac{r}{n})^{nt}$ and the second is $P = C e^{rt}$ where
    P = future value
    C = initial deposit
    r = interest rate (expressed as a fraction: eg. 0.06)
    n = # of times per year interest is compounded
    t = number of years invested 
    e = The constant e.

In your case, they give you P=10,000, t=20, r=13%=0.13, and n=365 (well, that is roughlty dayly, there is also leap years so...  not sure if they care about that.)

e.mccormick · Answer

So you plug in what you know and solve for C, what you would pay for it now.

anonymous · Answer

I got 13457.51 is that right?

e.mccormick · Answer

You should get something LESS than the end value.  Basically yhey are saying, if you pay C, then C+compounded interest=10,000.

e.mccormick · Answer

a) is:
$$P = C (1 + \frac{r}{n})^{nt} \Rightarrow 10,000 = C (1 + \frac{.13}{365})^{(365)(20)}$$
b) is:
$$P = C e^{rt} \Rightarrow 10,000 = C e^{(.13)(20)}$$

anonymous · Answer

$$10000(1+\frac{ 0.13 }{ 365 })^{7300}$$ this is what I plug in

anonymous · Answer

Oh trying to find C

e.mccormick · Answer

Yah, I guessed where your mistake was.  The concept is that they are giving you the output and asking you to find the input.

anonymous · Answer

Thats what I been doing wrong the whole Time!! do you do the log to get rid of 7300? well how do you get rid of 7300

e.mccormick · Answer

No need to get rid of it.  Just use it.

$$10000=C (1 + \frac{.13}{365})^{(365)(20)}\Rightarrow \frac{10000}{(1 + \frac{.13}{365})^{7300}}=C$$

e.mccormick · Answer

So about $743.08.  Not much to get $10,000.  But with 20 years of interest, that does not surprise me.

anonymous · Answer

I just get 10000

e.mccormick · Answer

You need to solve for C, so you divide both sides by the known part that is normally multiplying C.  That is all I did.

anonymous · Answer

yeah i see what I did wrong!

e.mccormick · Answer

Now, do the same sort of thing to the pert one.  What do you get there?

anonymous · Answer

I got 742.74

e.mccormick · Answer

Same as me! 

I think all that happened was that after using principal to find interest or how much you get after so long, they tossed one at you with, "Here is the answer, find the question!" and it caught you off guard.

anonymous · Answer

lol thank you soo much! was stuck on this question for more then 2 hours an no one helped! again thank you!!!

e.mccormick · Answer

Yah, I saw it earlier.  I just took a break in my Linerar homework and saw it was still there!

anonymous · Answer

still thank you for coming back for it! ^_^

e.mccormick · Answer

np.  Don't have too much fun!

anonymous · Answer

haha try not to!