Let $p:\mathbb{P}^r\dashrightarrow\mathbb{P}^{r-k-1}$ with $0\le k<r$ be the projection $(a_0:\cdots:a_r)\longmapsto (a_{k+1}:\cdots:a_r)$. Such a rational map has base $K:=V(X_{k+1},\cdots,X_r)\subsetneq\mathbb{P}^r$, that is, the domain of $p$ is $\mathbb{P}^r\backslash K$. The blow-up, $X:=Bl_{K}(\mathbb{P}^r)$, of $\mathbb{P}^r$ along of $K$ is \begin{eqnarray} X &\overset{\pi}{\longrightarrow}& \mathbb{P}^r\\ ((a_0:\cdots:a_r),(b_{k+1}:\cdots:b_r)) &\longmapsto& (a_0:\cdots:a_r), \end{eqnarray} where $X=V(X_iY_j-X_jY_i\mid k+1\leq i,j,\le r)\subset \mathbb{P}^r\times\mathbb{P}^{r-k-1}$, here we have given homogeneous coordinates $X_i$ to $\mathbb{P}^r$ and $Y_i$ to $\mathbb{P}^{r-k-1}$. The morphim \begin{eqnarray} X &\overset{P}{\longrightarrow}& \mathbb{P}^{r-k-1}\\ ((a_0:\cdots:a_r),(b_{k+1}:\cdots:b_r)) &\longmapsto& (b_{k+1}:\cdots:b_r), \end{eqnarray} resolves $p$, i.e, $P\circ\pi^{-1}=p$. The exceptional divisor is given by $E=\pi^{-1}(K)=V(X_{k+1},\cdots,X_r)\subsetneq X$. We have a canonical immersion $(\pi,P): X \hookrightarrow\mathbb{P}^r\times\mathbb{P}^{r-k-1}$ which give us $P^*\mathcal{O}_{\mathbb{P}^{r-k-1}}(1)=\pi^*\mathcal{O}_{\mathbb{P}^r}(1)\otimes\mathcal{O}_{X}(-E)$, this allow us to express the bivariate Hilbert polynomial of $X$ as \begin{eqnarray} Hilb_{X}(a,b)&:=&h⁰(\mathcal{O}_{\mathbb{P}^r}(a)|_{X}\otimes\mathcal{O}_{\mathbb{P}^{r-k-1}}(b)|_{X})\\ &=&h⁰(\pi^*\mathcal{O}_{\mathbb{P}^r}(a)\otimes P^*\mathcal{O}_{\mathbb{P}^{r-k-1}}(b))\\ &=&h⁰(\pi^*\mathcal{O}_{\mathbb{P}^r}(a+b)\otimes\mathcal{O}_{X}(-bE)), \mbox{ for } a,b >>0. \end{eqnarray}
Unfortunately, I have no idea how to calculate $h⁰(\pi^*\mathcal{O}_{\mathbb{P}^r}(a+b)\otimes \mathcal{O}_{X}(-bE))$, or more generally, $Hilb_{X}(a,b)$.
I would be very happy if anyone could give me any idea how to calculate $h⁰(\pi^*\mathcal{O}_{\mathbb{P}^r}(a+b)\otimes \mathcal{O}_{X}(-bE))$, or more generally, $Hilb_{X}(a,b)$. References, a solution, absolutely anything which could help me is welcome!!