Consider the following inventory model. You are managing an inventory of bikes. Demand for bikes arrives
as a rate 1 Poisson process. Whenever a demand arrives, it is always a demand for exactly one bike.
At any given time t, you have two types of inventory - inventory on hand $H_t$, and inventory ordered but not
yet arrived $O_t$. When a demand arrives, if you have at least one bike on hand (i.e. $H_t \ge 1$), you meet the
demand with the inventory on hand, i.e. you give the bike to the customer and $H_t$ is reduced by 1. If you
have no inventory on hand and a demand comes, AND you have either 0 or 1 bikes in inventory ordered
but not yet arrived (i.e. $O_t \le 1$), you lose that customer’s demand, and place an order for a new bike. A
newly ordered bike takes an i.i.d. Expo(2) amount of time to arrive. If you have no inventory on hand
and a demand comes, and you have at least 2 bikes in inventory ordered but not yet arrived, you lose that
customer’s demand, and do not place any new orders. Suppose that a demand arrives at some time $t\gg 0$.
What is the probability that this customer receives a bike?
I tried to model this problem by CTMC with states of double-index. That is, state $(i,j)$ for $i\ge 0$ and $0\le j\le 2,$ where $i$ indicates for number of $H_t$ and $j$ indicates for number of $O_t$. And let $A$ be generating matrix of this Markov chain and find $\eta A=0$ with all entries of $\eta$ are positive and $\sum_{(i,j)}\eta_{(i,j)}=1$. Then the answer should be $1-\sum_{j=0}^2\eta_{(0,j)}$. But the system of $\eta A=0$ is too complicated to find a solution. I wonder if there is some other way to look at it.