HW Math 504. Spring 2008. CSUF
Problems 10.5 and 10.6
by Nasser Abbasi
Solution
Illustrating model diagram
for problem 10.5
In the above,
is the number of broken machines in the queue.
is maximum capacity of the operating room. The goal is to keep this room
filled to its capacity. In other words, to keep
machines in operations.
is the capacity of the spare room.
Need to determine
This can happen when one machine fails, but no server completes its service
meanwhile. Hence we do not need to consider the servers part in this analysis.
There are 2 cases to consider:
(there are
machines in
operations)
In the above, the last term
accounts for other possible conditions under which
can increase by one but which is considered to be less likely, such as 2
machines break down and one server completes its service.
In the above,
simplifies to zero when
is very small, hence the above equation becomes
Comparing the above with the Hence we see
we see
that
or in other words,
(there are less than
machines in
operations)
Hence we see that
or in other words,
Need to determine
,
this can happen when a server completes its job but no machine fails
meanwhile, Hence we only need to consider the servers. There are 2 cases to
consider:
(Queue is empty and not all servers at working on fixing machines at
hand)
Hence
,
or since this is a birth/death process, we write
(All servers at
busy)
Hence
,
Hence
Therefore, we summarize all the above as follows
Notice that
The following state transition diagram illustrates the above result, with
arrows leaving/entering states show the rate of arrival and departure on them
per the above result. To make the diagram easier to make, I assume the
following values:
Notice that
and
as
expected.
Now
compute the steady state distribution
(This is not asked for in this problem, but need to do this to solve problem
12.3 later on and implement it)
Starting with the balance equation, where to balance the rate out of a state, with the rate into a state. We have
Hence for state
we have
But
and
hence
For state
we have
but
,
hence
the above becomes
But from (1) we have
,
hence the above becomes
Continue this way, we obtain that
From the above, if we solve in terms of
we obtain that
and with the equation
we can now solve for all
as follows
Hence
Now that
is found, we can find the remaining
using (3)
Review
of the problem setup: Imagine there is a queue of length
.
Burned out bulbs enter the queue (with inter-arrival time which is a random
variable distributed as an exponential with rate
).
Bulbs continue to enter the queue until the queue is full, then at that moment
we imagine a single server processing the bulbs in the queue all at once and
immediately all
bulbs become operational again and the queue is now empty. This process
repeats again and again.
A stochastic process
is defined to have the Markov property if its transition to the next state
depends only on the current state and not on any earlier states. In other
words it satisfies the
following
In this problem
is the number of burned out bulbs in the queue at any time
.
When
then
can be viewed as a counting process (or pure birth process) or a Poisson
process (until the queue become full).
Therefore, The time between each successive events (where an event causes the count to increase by one) is a random variable with exponential distribution (we are also given this fact in the problem). But the exponential distribution is memoryless Note_1 by definition. Therefore it does not depend on clock time but only on the length of the time interval. Hence the process satisfies the Markov property.
A stochastic process
is defined to be
stationary Note_2
if its state transition
do not depend on when the transitions happen but only on the time interval
.
In other words, random process
is stationary if
For any
So, letting
,
the system is stationary
if
This process is clearly stationary, since it is a counting process (when
).
A counting Poisson process is stationary since it does not depend on clock
time as was argued in part (A). To show this more clearly, since this is a
counting process, then by definition of the Poisson
process
We see that the probability of
does not depend on
and depends only on the time interval
.
If this was a non-stationary process, then
would appear in the RHS above. I.e. the probability of the random variable
would depend on clock time, but we see from the above definition that it does
not.
A stochastic process is a pure jump process if the transition probabilities can be written as
and
as
In this problem
is the probability than no bulb burns out during an interval
.
This is given by the probability than no bulb burns out from the current
number of functional bulbs which is
.
Due to independence, we obtain
Applying Binomial expansion
,
to the above, and taking
we
obtain
Hence we can write
where
Now
is the probability that there will be
failed bulbs after
units of time given that there is already
failed bulbs. For this to occur, then we need to have
bulbs fail in
units of time. We can solve for the general case when
but
since we will let
it
is most likely that there will be only one event occur (one bulb fail) during
this time, and we can collect all other less likely probabilities in the
term. Hence we will only consider
in
the
following.
Therefore from
we see that
i.e.
Hence the
matrix (The rate matrix)
is
rate flow diagram for problem
10.6
The balance equation can be obtained from balancing the flow out rate of a
state
(which is given by
)
by all the flow in rate into the state which is given by
as illustrated below for the above problem
Hence we write
and for state
we have
Therefore we obtain
Hence we have for
Therefore we have
back substitute, we obtain
Hence
We notice that the last equation above, is the same as for the case
.
Hence we have one of the
equations duplicated. Hence we need one more equation to solve for the
unknowns
and for that we use
Therefore, the general expression for
is
Now since
,
then we write
So now that we know
from (2), we substitute it into (1) and solve for the remaining
But
Which is the difference between 2 partial sums of harmonic numbers. Let
,
then
hence (3) becomes
Hence
This is a small program which show the long term
for
using the above equation