EEL6507sp09L24

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EEL6507 Spring 2009, Lecture 24, Friday 2009/03/06 (Notes created by Nathan Blythe)

Continuing with Er / M / 1

This material follows that covered in the previous lecture.

The previous lecture concluded after finding P(z)= \frac{(1-z^r)(1-1/z_0)}{r (1-z)(1 - z/z_o)}, where z0 is the real root of rpzr + 1 − (1 + rp)zr + 1,z > 1.

Pulling out the constants we find P(z) = \frac{1 - 1/z_0}{r} \frac{(1 - z^r)}{(1 - z)(1 - z/z_0)}.

Computing the partial fraction expansion is straighforward:

\frac{1}{(1 - z)(1 - z/z_0)} = \frac{A}{1 - z} + \frac{B}{1-z/z_0}

\rightarrow A = \frac{1}{z - z/z_0}|_{z = 1} = \frac{1}{1 - 1/z_0}

\rightarrow B = \frac{1}{1 - z}|_{z = z_0} = \frac{1}{1 - z_0}

Then P(z) = \frac{1 - 1/z_0}{r} (1 - z^r) (\frac{1}{1 - 1/zo} \cdot \frac{1}{1 - z} + \frac{1}{1-z_0} \cdot \frac{1}{1-z/z_0} ). This simplifies to P(z) = (1 - z^r) (\frac{1/r}{1 - z} - \frac{1/(r z_0)}{1 - z/z_0}).

Let fj be the inverse Z-transform of the component (\frac{1/r}{1 - z} - \frac{1/(r z_0)}{1 - z/z_0}) in the above expression. Notice then that pj = fj − fj − r.

We determine fj to be:

f_j = \begin{cases} \frac{1}{r} (1 - z_0^{-j-1}), & j \geq 0\\ \\ 0, & j < 0 \end{cases}

There are then two cases for Pj. If 0 \leq j < r then fj is non-zero but fj − r is zero (by the above definition). If j > r, then both fj and fj − r are non-zero. These two cases yield the following result for Pj.

P_j = \begin{cases} \frac{1}{r} (1 - z_0^{-j-1}), & 0 \leq j < r\\ \\ \rho (z_0 - 1) z_0^{r-j-1}, & j \geq r \end{cases}

Probability distribution on the number of customers in the system

We wish to find an expression for the probability pk of the system containing k customers at any given time. We start by finding p0 as below. First we refer to our original queue state definition (see previous lecture) and state p0 = P(zero customers) = P(in queue state 0 through r - 1).

The math proceeds as follows.

p_0 = \sum_{j = 0}^{r - 1} \frac{1}{r} (1 - z_0^{-j-1}) = 1 - \frac{1}{r}\sum_{j = 0}^{r - 1} (\frac{1}{z_0})^{j + 1}

This yields p0 = 1 − ρ (which implies that our definition of ρ in the previous lecture was valid, as this is a known relationship between p0 and ρ).

Next we generalize p0 and state that pk = P(in queue state kr through (k + 1)r - 1). Mathematically, p_k = \sum_{j=kr}^{(k + 1)r - 1} P_j which expands to p_k = \rho (z_0 - 1) z_0^{r - 1} \sum_{j = kr}^{(k + 1)r - 1} (\frac{1}{z_0})^j.

The mathematics of solving this summation are not shown here; standard approaches or known forms work equally well.

After simplifying, we find p_k = \rho (z_0^r - 1)z_0^{-rk}. We write the complete solution as below.

p_k = \begin{cases} 1 - \rho, & k = 0 \\ \rho (z_0^r - 1) z_0^{-rk}, & k > 0 \end{cases}

Special case: r = 1

Consider r = 1. Then p0 = 1 − ρ and p_k = (z_0 - 1)\rho z_0^{-k}. But if r = 1 then <z_0> must be a root of (1 − ρz)(1 − z), from the results found in the previous lecture. Since z0 > 1 we know that z_0 = \frac{1}{\rho}. Substituting this into the previous result and simplifying we find that pk = (1 − ρ)ρk. Recall from lecture 12 that this is the identical result to the M/M/1 queue. Thus E1 / M / 1 is equivalent to M/M/1 (which agrees with the mechanism by which we developed the Er distribution and related mathematics)

Mean number of customers in the system

Finally we wish to find the mean number of customers in the system. This is the expected value of the distribution found above, which we find as \bar{N}=\sum_{k=1}^{\infty} k (z_0^r - 1) \rho (\frac{1}{z_0^r})^k. After some manipulation and applying summation rules we find that \bar{N} = \frac{\rho}{1-z_0^{-r}}.

Special case: r = 1

For r = 1 we have \bar{N} = \frac{\rho}{1 - \rho} which again agrees with the M/M/1 queue.

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