Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
add comment

1 Answer

up vote 3 down vote accepted

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.)

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.