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I'm preparing for an exam and one of the topics is pricing bonds where bootstrapping a yield curve is used. Among other things, it involves solving exponential equations which is a recurring problem. A sample equation may look like this:

$$ 106.065 = \frac{6.25}{e^{(0.0177 + 0.25r)\times{0.88}}} + \frac{106.25}{e^{(-0.0059 + 1.25r)\times{1.88}}} $$

or

$$ 106.065 = \frac{6.25}{(1.0177 + 0.25r)^{0.88}} + \frac{106.25}{(0.9941 + 1.25r)^{1.88}} $$

You can notice that both of these are the same thing, but the first one uses the continous compounding and the second one uses the ordinary yearly compound interest. Can I solve any of those on paper or at least approximate the value of $r$?

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  • $\begingroup$ You can get an approximate answer by trial and error. What's the issue? $\endgroup$
    – user124485
    Commented Jun 12, 2015 at 15:30
  • $\begingroup$ You can't solve them exactly. I'd post a derivation for the approximate values, but I'm not sure how mathematically inclined this is. You could just use Wolfram.... $\endgroup$
    – Zach466920
    Commented Jun 12, 2015 at 15:40
  • $\begingroup$ It´s not useful trying to solve the equation on paper. You can apply the newton-raphson-method to calculate the value of r. To get a solution in a appropiate time you should use Excel or a similar program for support. $\endgroup$ Commented Jun 12, 2015 at 16:01
  • $\begingroup$ Thanks, everyone. With your input I was able to confirm what I suspected. $\endgroup$
    – user247712
    Commented Jun 12, 2015 at 18:45

1 Answer 1

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You know that these equations will not show explicit solutions and numerical methods should be used.

The simplest would be Newton method which, starting from a "reasonable" guess $r_0$ will update it according to $$r_{n+1}=r_n-\frac{f(r_n)}{f'(r_n)}$$ We know that $r$ is small; so let us start at $r_0=0$.

For the first equation, the successive iterates would be $$r_1=0.029640$$ $$r_2=0.030716$$ $$r_3=0.030718$$ which is the solution for five significant figures.

For the second equation, the successive iterates would be $$r_1=0.029485$$ $$r_2=0.031144$$ $$r_3=0.031150$$ which is the solution for five significant figures.

Another way for a better estimate is to approximate the function by its simplest Pade approximant; built at $r=0$ this will give $r\approx 0.030705$ for the first equation and $r\approx 0.031140$ for the second equation.

Doing this on paper could be something hard; a good calculator should do it and any software so (Excel could be the easiest solution).

Since the procedures I described require derivatives (and, for these cases, they are awful) I suggest you compute them by finite differences. This will make life easier.

Edit

The first equation is much more simple than it looks. After some rearrangements, it write $$a = b e^{-\alpha r}+ c e^{-\beta r}$$ and the approximate solution given by the simplest Pade approximant is $$r=\frac{2 (-a+b+c) (\alpha b+\beta c)}{\alpha ^2 b (a+b-c)+\beta ^2 c (a-b+c)+4 \alpha \beta b c}$$ For the case considered $$a=106.065$$ $$b=\frac{6.25}{e^{0.0177\times 0.88}}\approx 6.153404241$$ $$c=\frac{106.25}{e^{-0.0059\times 1.88}}\approx 107.4350853$$ $$\alpha=0.25\times0.88=0.22$$ $$\beta=1.25\times 1.88=2.35$$ Applying the last formula, the result is $r\approx 0.0307045 $.

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