it contains $\ell$ amount of water at time $t=0$. Thus, $\tau_{\ell}^{-}$ represents the first time after $t=0$ that the tank goes empty. We will compute the quantity $$\begin{aligned}
\label{eq:computeTarget}
\Pi_{c}^{-}(i,j)=\mathbb{E}^{i}\left(e^{-c\tau_{0}^{-}}{\mathbbm{1}}_{\{X_{\tau_{0}^{-}}=j\}}\right),\quad i\in\mathbf{E}^{+},\, j\in\mathbf{E}^{-}.\end{aligned}$$ Towards this end, we further assume that the tank has either an aggregate water inflow at rate $v^{+}$ or an aggregate water outflow at rate $v^{-}$. In other words, $$\begin{aligned}
\mathbf{E}^{+}=\{e_+\},\quad\mathbf{E}^{-}=\{e_-\},\quad v(e_+)=v^{+},\quad\text{and}\quad v(e_-)=v^{-}.\end{aligned}$$ Moreover, we assume that the time-inhomogeneous Markov chain $X$ has the generator $$\begin{aligned}
\mG_{t}=\left\{\begin{array}{lll}
& \mG_{1},\quad & s_{0}\le t <s_{1},\\
& \mG_{2},\quad &s_{1}\le t <s_{2},\\
& \mG_{3},\quad &t\ge s_{2},
\end{array}\right.\end{aligned}$$ where $0<s_{1}<s_{2}$.

We take the following inputs: $c=0.5, v(e_+)=2, v(e_-)=-3,s_{1}=2,s_{2}=8$, $$\begin{aligned}
  \mG_{0}=
  \kbordermatrix{
    & e_+ & e_- \\
    e_+ &-2 & 2\\
   e_- &1 & -1
  }, \quad
    \mG_{1}=
    \kbordermatrix{
      & e_+ & e_- \\
      e_+ &-3 & 3\\
     e_- &2 & -2
    },\quad
      \mG_{2}=
      \kbordermatrix{
        & e_+ & e_- \\
        e_+ &-5 & 5\\
       e_- &3 & -3
      }.\end{aligned}$$ The following table compares our result and execution time with Monte-Carlo simulation (10000 paths).\

[llr]{}\
Method & Wiener-Hopf & Monte-Carlo\
$\Pi_{c}^{-}(e_+,e_-)$ & $0.6501$ & $0.6462$\
Execution time & $0.15\,s$ & $3.12\,s$\

One can also compute $\Pi_{c}^{+}(e_-,e_+)$, if it is the quantity of interest in the 