# Basic boostrapping question

Suppose I have three bonds:

Coupon bonds are paid semi-annually. Rates are continuous compounding.

I'm trying to bootstrap the zero rates for 0.5 years maturity using the 1 year zero coupon bond and the 1 year fixed rate coupon bond, but my eventual 0.5 year zero rate is higher than my 1 year zero rate.

My calculations are:

• 1-year zero rate: $$95 = 100 \times e^{-r}$$ $$r_{1y} = - \ln (0.95) = 5.129\%$$

• 6M zero rate: $$2.5 \times e^{-0.5 r_{6m}} + 102.5 \times e^{-r_{1y}} = 99.8$$ $$r_{6m} = - 2 \ln \left( \frac{99.8-102.5 e^{-r_{1y}} }{2.5} \right) = 6.118\%$$

I'm not sure if I'm doing it right, don't think a 0.5 year zero rate is supposed to be higher than a 1 year zero rate.

You think you make a mistake where you actually don´t make one. The exercise is just like it is. Resulting in $$r_{6m}>r_{12m}$$

The difference in your both answers, based on the same rounding, lays in the different basis for the logarithm.

$$r_{6m} = - 2 \log_e \left( \frac{99.8-102.5 e^{-r_{1y}} }{2.5} \right) = \textbf{6.118%}$$ $$r_{6m} = - 2 \log_{10} \left( \frac{99.8-102.5 e^{-r_{1y}} }{2.5} \right) = \textbf{2.6571%}$$

Calculation for the last interest rate base on $r_{6m}=0.06118%$ and $r_{12m}=0.05129%$ yields:

$$r_{18m}=-\frac{log_e\left(\frac{102.7-4\times(e^{0.5\times(-0.06118)}+e^{-0.05129})}{104}\right)}{1.5}=0.06019886318=\textbf{6.0199%}$$

Your answer is correct. You included .5 in the exponent and therefore got an annualized result. 6.118% divided by 2 is your bootstrapped 6 month spot rate.

I think your calculation is just wrong.

Starting from the edit, we have for the 6m zero rate:

• 6M zero rate: $$2.5 \times e^{-0.5 r_{6m}} + 102.5 \times e^{-r_{1y}} = 99.8$$ $$r_{6m} = - 2 \ln \left( \frac{99.8-102.5 e^{-r_{1y}} }{2.5} \right) = \textbf{2.6571%}$$