A company has $N$ different items which are used for transactions. A transaction needs $m$ items with probability $1/M$, $1 \leq m \leq M$, for certain $M \leq N$. Given that a transaction needs $m$ items, every subset of $m$ requested items is equally likely. Applications for transactions come in according to a Poisson process with rate $\lambda$. The processing time of a transaction that is made up of $m$ items, is exponential distributed with rate $\mu$.
An item that is used in the treatment of a transaction, is not available for other transactions for the duration of processing. Applications for transactions for which not all items are available, will be lost.
This model can be described by a Markov chain with states $(n_1, ..., n_M)$, where $n_m =$ # current transactions of $m$ items, $1 \leq m \leq M$.
Determine the transition intensities of the Markov process.
This seems like a birth-death process. Let $q_{ij}$ denote the transition intensity from state $i$ to $j$.
I would say $q_{n_1 n_1} = \lambda$, $q_{n_M n_M} = \mu$ and $q_{n_m n_m} = -(\lambda + \mu)$, for $m \neq 1,M$.
Furthermore $q_{n_m n_{m+1}}= \lambda$ and $q_{n_{m} n_{m-1}}= \mu$.
But I strongly doubt about it. The probability that a transaction needs $m$ items is not given without any reason I guess. Should this be involved too?
I hope someone can give me some clearness about this.
You are right, this probability is not given without any reason. Since $n_i$ is the number of transactions involving $i$ items, the rate at which $n_i$ goes to $n_i+1$ is the rate at which orders involving $i$ products arrive. The total rate at which orders arrive is $\lambda$, and the probability that any order contains $i$ items is $1/M$, so that $q_{n_i,n_i+1}=\lambda/M$ for all $i$.
Furthermore, it says that the processing time of any order is at rate $\mu$. Therefore, if there are $n_i$ orders of $i$ products, each of them is processed at rate $\mu$, so the total rate from $n_i$ to $n_i-1$ is $\mu n_i$. Therefore, $q_{n_i,n_i-1}=\mu n_i$ for all $i$.
This is a bit abuse of notation since the whole state of the Markov chain is $(n_1,\dots n_M)$, and not $n_i$, but I hope you understand what I mean.