Customers come in a very small pizza joint at an average rate of 0.5 per minute. Each customer spent time in ordering pizza for a random duration which is exponentially distributed with mean 3 minutes, independently of the duration of the other customers. Two employees are assigned to handle those customers. If a customer arrives when both employees are busy, the customer has to wait unless there are already two customers in waiting, in which case the new customer (the fifth one) is sent to another joint. When a customer leaves, one of the customers in waiting is immediately engaged to the newly free employee. 1. Write down the transition matrix of the Markov jump chain associated with the process and calculate the stationary distribution. Any hint on how to solve. I have got all the five states and transition diagram thought.
I am assuming that the process to be modelled is the number of customers in the shop (although this is not clear from the question). In that case, the states are 0,1,2,3,4. The only possible transitions are to move up one or down one. All the upward forces are 0.5 (as that is the arrival rate). If there is one customer in the shop, the transition rate to 0 is 1/3 (the exponential parameter). If there are 2 customers, then they are both being served, so the transition rate from 2 to 1 is 2*1/3. The forces from 3 to 2 and 4 to 3 are also 2/3 as there are only 2 servers. The transition matrix contains probabilities. If the process is now in state 0, the next state must be 1. So the (0,1) entry of the matrix is 1 (and all other values in this row are 0). if the process is now in state 1, the next state will be either 0 or 2. The total force out of state 1 is 0.5 + 1/3 = 5/6. The proportion of the total force that goes to state 1 is 0.5/(5/6) = 0.4. The other 0.6 of the total force goes to state 2. These are the probabilities in the second row of the matrix. The other rows work in a similar way.
