Under simultaneous, small changes of the parameters $a$ and $b$ , the present value of the liability becomes

$P_{L}(a\Delta_{a}\bm{r}+b\Delta_{b}\bm{r})=80^{\prime}000\mbox{e}^{-2.5a-2.5^{2}b}\simeq 80^{\prime}000(1-2.5a-6.25b)\,,$

where $\Delta_{b}=(1,2,2.5,5)^{\top}$ . Similarly, for the zero-coupon bonds,

$P_{1}(a\Delta_{a}\bm{r}+b\Delta_{b}\bm{r})$ $=1^{\prime}000\mbox{e}^{-a-b}\simeq 1^{\prime}000(1-a-b)\,,$
$P_{2}(a\Delta_{a}\bm{r}+b\Delta_{b}\bm{r})$ $=1^{\prime}000\mbox{e}^{-2a-4b}\simeq 1^{\prime}000(1-2a-4b)\,,$

$P_{3}(a\Delta_{a}\bm{r}+b\Delta_{b}\bm{r})=1^{\prime}000{\rm e}^{-5a-25b}\simeq 1^{\prime}000(1-5a-25b)\,,$

for small $a$ and $b$ . We require that

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

and that

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

for small $a$ and $b$ . Comparing coefficients, we get the system of equations,

\begin{cases}1^{\prime}000h_{1}+1^{\prime}000h_{2}+1^{\prime}000h_{3}&=80^{\prime}000\\ -1^{\prime}000h_{1}a-1^{\prime}000h_{2}(2a)-1^{\prime}000h_{3}(5a)&=-80^{\prime}000\cdot 2.5a\\ -1^{\prime}000h_{1}b-1^{\prime}000h_{2}(4b)-1^{\prime}000h_{3}(25b)&=-80^{\prime}000\cdot 6.25b\,.\end{cases}

Simplifying, we can write

\begin{pmatrix}1&1&1\\ 1&2&5\\ 1&4&25\end{pmatrix}\begin{pmatrix}h_{1}\\ h_{2}\\ h_{3}\end{pmatrix}=\begin{pmatrix}80\\ 200\\ 500\end{pmatrix}\,.

We solve the linear system using Gaussian elimination: Subtracting from the third and the second row once the first row, we get

\begin{pmatrix}1&1&1\\ 0&1&4\\ 0&3&24\end{pmatrix}\begin{pmatrix}h_{1}\\ h_{2}\\ h_{3}\end{pmatrix}=\begin{pmatrix}80\\ 120\\ 420\end{pmatrix}\,.

Next, subtracting from the third row three times the second row, we get

\begin{pmatrix}1&1&1\\ 0&1&4\\ 0&0&12\end{pmatrix}\begin{pmatrix}h_{1}\\ h_{2}\\ h_{3}\end{pmatrix}=\begin{pmatrix}80\\ 120\\ 60\end{pmatrix}\,,

which is in row echelon form. Thus $h_{3}=60/12=5$ , $h_{2}=120-4\cdot 5=100$ and $h_{1}=80-100-5=-25$ .

The portfolio positions are thus $(h_{1},h_{2},h_{3})=(-25,100,5)$ .