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Once the stationary probability vector
is known,
we can obtain various performance measures of interest such as the
average queue length which we showed in the previous subsection.
There are additional measures of interest that one can compute
knowing
,
, and
.
In particular, the tail distribution of the number of jobs
in the system can be expressed as
![\begin{displaymath}
\mbox{{\bf\sf P}}\left[ Ql > x \right] \; = \; \sum_{k=x+1}...
... {({\bf I} - {\mathbf{R}})}^{-1} {\bf e} , \;\;\;\; x \geq 0 ,
\end{displaymath}](img314.gif) |
(3.5) |
with the corresponding expectation given by
![$\displaystyle {\rm E}{[Ql]} \, = \, {{\mbox{\boldmath$\pi$}}}^{(1)} \sum_{k=0}^...
...th$\pi$}}}^{(1)} {\mathbf{R}}{({\bf I} - {\mathbf{R}})}^{-2} {\bf e} . \;\;\;\;$](img315.gif) |
|
|
(3.6) |
The expected waiting time of a job in the system can
then be calculated using Little's law [51] and
Eq.(3.6), which yields
![\begin{displaymath}
{\rm E}{[W]} \; = \; {\lambda}^{-1} \left( \,
{{\mbox{\bol...
...\mathbf{R}}{({\bf I} -{\mathbf{R}})}^{-2} {\bf e} \, \right) .
\end{displaymath}](img316.gif) |
(3.7) |
Let
denote the spectral radius of the matrix
, which is
often called the caudal characteristic [68].
In addition to providing the stability condition for a QBD,
is indicative of the tail behavior of the stationary queue
length distribution.
Let
and
be the left and right eigenvectors corresponding to
normalized by
and
.
Under the above assumptions, it is known that [94]
which together with Eq.(3.4) yields
 |
(3.8) |
It then follows that
![\begin{displaymath}
\mbox{{\bf\sf P}}\left[ Ql > x \right] \; = \; \frac{ {{\mb...
...box{o}(\eta^x) , \;\;\;\; \mbox{as } \; x \rightarrow \infty ,
\end{displaymath}](img324.gif) |
(3.9) |
and thus
 |
(3.10) |
or equivalently
![\begin{displaymath}
\mbox{{\bf\sf P}}\left[ Ql > x \right] \; \sim \; \frac{ {{...
...a } \, \eta^x ,
\;\;\;\; \mbox{as } \; x \rightarrow \infty .
\end{displaymath}](img326.gif) |
(3.11) |
The caudal characteristic can be obtained without having to first solve
for the matrix
.
We define the matrix
,
for
.
Since the generator matrix
is irreducible, this matrix
is irreducible with nonnegative off-diagonal elements.
Let
denote the spectral radius of the matrix
.
Then, under the above assumptions,
is the unique solution in (0,1)
of the equation
.
A more efficient method is developed in [9].
Next: 3.2 Why does a
Up: 3.1 Matrix geometric solutions
Previous: 3.1 Matrix geometric solutions
Alma Riska
2003-01-13