Since at present $a=b=0$ , the present value of the liability is

$P_{L}(0)=80^{\prime}000\,,\qquad\mbox{(in \$)}\,.$

The present values of the bonds are $P_{1}(0)=P_{2}(0)=P_{3}(0)=1^{\prime}000$ , as the zero-rates are all zero. A change in the parameter $a$ corresponds to a parallel shift:

$\Delta_{a}\bm{r}=(1,1,1,1)^{\top}\,,$

where the corresponding times are $(1,2,2.5,5)$ in years. Thus under a small change in $a$ , the present value of the liability due in 2.5 year changes to

$P_{L}(a\Delta_{a}\bm{r})=80^{\prime}000\mbox{e}^{-2.5a}\simeq 80^{\prime}000(1-2.5a)\,,$

using a first order Taylor approximation for small $a$ . For the zero-coupon bonds we find in the same way

$P_{1}(a\Delta_{a}\bm{r})$ $=1^{\prime}000\mbox{e}^{-a}\simeq 1^{\prime}000(1-a)\,,$
$P_{2}(a\Delta_{a}\bm{r})$ $=1^{\prime}000\mbox{e}^{-2a}\simeq 1^{\prime}000(1-2a)\,,$
$P_{3}(a\Delta_{a}\bm{r})$ $=1^{\prime}000\mbox{e}^{-5a}\simeq 1^{\prime}000(1-5a)\,,$

see also equation (3.10) in the book. To determine the hedging portfolio, we require that the present values match, i.e.,

$P_{L}(0)=\sum_{i=1}^{3}h_{i}P_{i}(0)\,,$

and that

$P_{L}(a\Delta_{a}\bm{r})\simeq\sum_{i=1}^{3}h_{i}P_{i}(a\Delta_{a}\bm{r})\,.$

Alternatively,

$P_{L}(0)=\sum_{i=1}^{3}h_{i}P_{i}(0)\,,\qquad\frac{\partial}{\partial a}P_{L}(a\Delta_{a}\bm{r})\bigg|_{a=0}=\frac{\partial}{\partial a}\sum_{i=1}^{3}h_{i}P_{i}(a\Delta_{a}\bm{r})\bigg|_{a=0}\,.$

Following the hint, we set $h_{1}=0$ . Hence, comparing coefficients, we obtain the system

\begin{pmatrix}1^{\prime}000&1^{\prime}000\\ -2^{\prime}000a&-5^{\prime}000a\end{pmatrix}\begin{pmatrix}h_{2}\\ h_{3}\end{pmatrix}=\begin{pmatrix}80^{\prime}000\\ -80^{\prime}000\cdot 2.5\cdot a\end{pmatrix}

Simplifying, we find

\begin{pmatrix}1&1\\ 2&5\end{pmatrix}\begin{pmatrix}h_{2}\\ h_{3}\end{pmatrix}=\begin{pmatrix}80\\ 200\end{pmatrix}\,.

Solving the system, we find $h_{2}=\frac{200}{3}$ and $h_{3}=\frac{40}{3}$ .