Dear Sir,
 
That was GREAT!!!. Thanks a lot for the solution. Once again it showed how powerful 'R' is otherwise I was breaking my head on Newton-Raphson method.
 
Thanks again Sir. That was really superb.
 
Madhavi

--- On Mon, 1/2/10, S Ellison <S.Ellison@lgc.co.uk> wrote:


From: S Ellison <S.Ellison@lgc.co.uk>
Subject: Re: [R] 'R' and 'Yield to Maturity'
To: "Craig P. Pyrame" <crappyr@gmail.com>, "Madhavi Bhave" <madhavi_bhave@yahoo.com>
Cc: r-help@r-project.org
Date: Monday, 1 February, 2010, 3:41 AM


If you know the likely range, uniroot would do it.

f.ytm<-function(ytm) 100 / (1+ytm)  +100 / ((1+ytm)^2) + 1100 / ((1 +
ytm)^3) -1010

uniroot(f.ytm, interval=c(0,25))  

#$root has the answer


>>> "Craig P. Pyrame" <crappyr@gmail.com> 01/02/2010 10:19 >>>
Madhavi Bhave wrote:
> Dear R helpers
>  
> I am calculating the 'Yield to Maturity' for the Bond with following
characteristics.
>  
> Its a $1000 face value, 3 year bond with 10% annual coupon and is
priced at 101. The yield to maturity can be calculated after solving the
equation - 
>  
> 1010 = [100 / (1+ytm)]  + [100 / (1+ytm)^2] + [ 1100 / (1 + ytm)^3]
>  
> This can be solved by trial and error method s.t. ytm = 9.601%.
>   

Why don't you use sage, for example:

sage: var('ytm');
sage: eqn = 1010 == (100/(1+ytm) + 100/(1+ytm)^2 + 1100/(1+ytm)^3);
sage: [solution.right().n() for solution in solve(eqn, ytm)][2]
0.0960070882662794

The third value is the only real solution.
There may be an R package for doing that, but I don't know one.

Best regards,
Craig


>  
> My query is (1) if there is any R package which will calcualte ytm or
(2) is there any method in 'R' which can solve the above equation.
>  
> Thanking you all in advance
>  
> Regards
>  
> Madhavi
>  
>  
>  
>
>
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