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Lets say I have packets arrive to a terminal at Poisson rate $\lambda$ per hour and my terminal has an exponential service rate $\mu$ per hour (so the mean service time is $\frac{1}{\mu}$). So this is a classic M/M/1 queue. Now say that if my queue has 3 packets in it then it is full (so finite capacity) and new packets do not line up and are dismissed. I understand how to compute the steady state probabilities $P_0, P_1, P_2, P_3$ but I do not understand how to answer the following question: in the long run, how many packets does the terminal serve per hour?

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Suppose the steady state probabilities were $$(P_0,P_1,P_2,P_3) = (1,0,0,0)$$ then what would the average throughput be? Zero, right? Now note that that you know the average service required for each job is $1/\mu$ and consider the case where the steady state probabilities were $$(P_0,P_1,P_2,P_3) = (0,1,0,0).$$ This means there's always one job in the system, which gets served at rate $\mu$, so you server $\mu$ jobs per hour. Now consider the cases $$(P_0,P_1,P_2,P_3) = (0,0,1,0)$$ and $$(P_0,P_1,P_2,P_3) = (0,0,0,1).$$ What happens in these situations? Then return to your original problem using the actual values you have for $P_0,P_1,P_2$ and $P_3$.

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