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For a M/M/1 queue, calculating the estimated number of jobs $n$ in the queue is given by:

$$E[n] = \sum_{i=1}^{\infty} p_i i = \sum_{i=1}^{\infty} \rho^i (1-\rho) i .$$

The final result for a M/M/1 queue is:

$$E[n] = \frac{\rho}{(1-\rho)}.$$

How is it possible to derive this last step from the formulas above?

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  • $\begingroup$ Do you want to a proof of the identity $\sum_{i=1}^{\infty} i \rho^i (1-\rho) = \frac{\rho}{1-\rho}$? $\endgroup$
    – Srivatsan
    Jan 30, 2012 at 13:45
  • $\begingroup$ lack of dependence on $n$ is a bit confusing $\endgroup$
    – SBF
    Jan 30, 2012 at 13:51
  • $\begingroup$ The following related posts: 28331, 20418, 11464. They do not answer your exact question but the methods can be modified in a straightforward way. Also 60660 is relevant. $\endgroup$
    – Srivatsan
    Jan 30, 2012 at 13:57
  • $\begingroup$ @Srivatsan: yes, thats what I want. $\endgroup$
    – ptikobj
    Jan 30, 2012 at 14:39
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    $\begingroup$ @Ilya: $n$ is a random variable denoting the number of jobs in the system (I have seen $N$ used more commonly), and $E[n]$ is the expected number of jobs in the system. The right hand side gives an expression for this expectation in terms of some other parameter governing the system (the "load factor" $\rho$). Naturally, the right hand side shouldn't involve $n$. $\endgroup$
    – Srivatsan
    Feb 1, 2012 at 8:37

2 Answers 2

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Let $$\begin{align} f(\rho)&=\sum_{i=1}^\infty i \rho^i (1-\rho)\\ &=(1-\rho)\sum_{i=1}^\infty i \rho^i \\ &=(1-\rho)\rho\sum_{i=1}^\infty i \rho^{i-1} \\ &=(1-\rho)\rho\sum_{i=1}^\infty \frac{d}{d\rho} (\rho^i) \\ &=(1-\rho)\rho\frac{d}{d\rho}\left(\sum_{i=1}^\infty \rho^i\right) \\ &=(1-\rho)\rho\frac{d}{d\rho}\left(\frac{\rho}{1-\rho}\right) \\ &=(1-\rho)\rho\frac{1}{(1-\rho)^2} \\ &=\left(\frac{\rho}{1-\rho}\right) \\ \end{align}$$

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To prove $\sum_{i=1}^{\infty} i \rho^i (1-\rho) = \frac{\rho}{1-\rho}$ (which is meaningful at least for the $0 \le \rho \lt 1$ required for this problem):

Let $f( \rho )=\sum_{i=1}^{\infty} i \rho^i (1-\rho)$.

Let $g( \rho )= f( \rho )-\rho f( \rho )$ and so $g( \rho ) = \sum_{i=1}^{\infty} \rho^i (1-\rho)$ as each term almost cancels.

Then $g( \rho )-\rho g( \rho ) = \rho (1-\rho)$ when all the other terms cancel, so $g( \rho ) = \rho$.

So $f( \rho )-\rho f( \rho ) = \rho $, so $f( \rho ) =\frac{\rho}{1-\rho}$.

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  • $\begingroup$ @did I don't know what happened there $\endgroup$
    – Henry
    Aug 17, 2012 at 21:27

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