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Suppose we have an M/G/1/c loss system, with equilibrium distribution $\;\pi,\;$ service times $\;S_i\;$ and arrival rate $\;\lambda.\;$ I'm trying to show that $\;(1-\pi_0)=(1-\pi_c)\lambda \mathbb{E}S_1.\;$ The question suggests use of Little's formula.

Now, I can see that $\;(1-\pi_c)\lambda\;\;$ is the effective arrival rate, and that $\displaystyle 1-\pi_0=\lim_{t\to \infty}\frac{\int_0^t Q_t dt}{t}$, where $\;Q_t\;$ is the number of items in the queue at time $\;t\;$, but I'm not sure where to go from here. It seems that we would need to know the expected sojourn time, or the expected wait time to use Little's formula?

Thank you.

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1 Answer 1

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Define $L_s$ to be the average number being served and $W_s$ to be the average service time. Then $L_s = 0 \pi_0 + 1 (1 - \pi_0) = 1 - \pi_0$, and $W_s$ is, in your notation, $\mathbb{E}S_1$. So you're trying to show $L_s = \bar{\lambda} W_s$, where $\bar{\lambda}$ is the effective arrival rate.

Remember that Little's formula holds for the queue as well as for the entire system. This is because the queue can be thought of as its own system, and Little's formula applies to subsystems within the larger system. (See, for example, the Wikipedia article on Little's law.) So not only is $L = \bar{\lambda} W$, but $L_q = \bar{\lambda} W_q$, where $L$ and $L_q$ are the average number in the system and in the queue, respectively, and $W$ and $W_q$ are the average wait time in the system and in the queue, respectively.

Since $L = L_q + L_s$, and $W = W_q + W_s$, you can just subtract Little's formula for the queue from Little's formula for the system to obtain $L_s = \bar{\lambda} W_s$.

(All of this, of course, refers to the long-term behavior of the queuing system.)

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