WEBVTT
Kind: captions
Language: en
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Mean number of inter-arrival time.
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So that is nothing but the mu by lambda.
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Now we can define the traffic intensity that
is nothing but Rho.
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Rho is equal to arrival rate divided by the
service rate.
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From above you can get Rho is nothing but
lambda by mu.
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It can be shown that the DTMC is a positive
recurrent when Rho is less than 1.
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We have already made it it is irreducible
and a periodic now we are giving the condition
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for a positive recurrent when Rho is less
than 1 given DTMC is a positive recurrent.
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If Rho equal to 1 it is a null recurrent.
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If Rho is greater than 1 then it will be a
transient.
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That means when Rho is less than 1 all the
states are positive recurrent.
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Therefore the DTMC is said to be a positive
recurrent.
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Similarly when Rho is 1 all the states are
a null recurrent therefore the DTMC will be
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a null recurrent and similarly for Rho is
greater than 1.
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So our interest is to find out the limiting
distribution.
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So when Rho is less than 1 the DTMC is irreducible
a periodic and positive recurrent along with
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the condition b naught is greater than 0 and
b naught plus b1 is less than 1. with this
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condition it is reducible.
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With Rho is less than 1 it is a positive recurrent
we can easily verify it is a periodic.
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Hence , the limiting distribution exists and
it is unique and that will be independent
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of initial stat i.
00:02:08.429 --> 00:02:18.860
Therefore, Pij will be a limit n tends to
infinity the probability of i to j in n steps.
00:02:18.860 --> 00:02:25.379
So define the limiting distribution probability
vector as a Pi it's consists of Pi naught,
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Pi1, Pi2 and so on where Pij's are defined
in this form limit n tends to infinity P of
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i to j in n steps.
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We determine the limiting distribution by
solving Pi is equal to Pip and the summation
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of Pi's is equal to 1.
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So the first one is a homogeneous equation
including this normalizing condition you have
00:02:55.970 --> 00:02:59.980
a system of non-homogeneous equation.
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So all this non-homogenous - so all these
system of non-homogeneous equation to get
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the limiting probabilities.
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We get the limiting probabilities are one
minus Psi, Psi power j where Psi is the unique
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root of the equation Z is equal to the Laplace
stieltjes transform of the inter-arrival distribution
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with the variable mu minus mu Z in the interval
0 to 1.
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That means first you have to solve the equation
Z is equal to Psi of mu minus mu Z and you
00:03:47.200 --> 00:03:51.739
know what is a Psi of mu minus mu Z from the
beta Z.
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Solve the equation in the interval 0 to 1.
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So the unique root you have to substitute
as a Psi then substitute Psi in the Pij and
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that will be the - since the unique root is
in the interval 0 to 1 therefore the Pij 1
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minus Psi Psi power j that will form a probability
mass function for the limiting Rho, limiting
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distribution.
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Note that the queue length distribution found
just before arriving customer is geometric
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distribution with the parameter Psi.
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The limiting distribution which we got it
that is just before arriving customer the
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queue side the queue length distribution found
just before an arriving customer which is
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a geometric distribution with the parameter
Psi.
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The best computational method for determination
of the limiting distribution seems to be their
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direct matrix multiplication to get P power
n for increasing values of n until the Rhos
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can be considered to be reasonable identical.
00:05:13.050 --> 00:05:24.150
Whenever for a larger n P power n has the
identical rows it means the limiting distribution
00:05:24.150 --> 00:05:26.080
exists.
00:05:26.080 --> 00:05:35.080
Therefore, the best computation method is
find the P power n for larger n until the
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rows can be considered to be a reasonably
identical.
00:05:39.700 --> 00:05:45.580
It is important to note that the embedded
Markov chain analysis gives the properties
00:05:45.580 --> 00:05:51.950
of the number of -- number in the system at
arrival epochs.
00:05:51.950 --> 00:05:58.150
The Q of t for t greater than or equal to
0 that is a discrete state continuous time
00:05:58.150 --> 00:06:05.530
stochastic process whereas Qn for n greater
- n is equal to 0, 1, 2, and so on that is
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a embedded time homogeneous discrete time
Markov chain and that is Qn is nothing but
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Q of tn minus 0 that is nothing but the number
of customers in the system just before the
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nth arrival.
00:06:29.900 --> 00:06:36.090
Hence, it is important to note that the embedded
Markov chain analysis gives the properties
00:06:36.090 --> 00:06:46.810
of the number in the system at arrival epochs
not the departure epochs or not at the arbitrary
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time instants.
00:06:47.810 --> 00:06:54.090
It gives the embedded Markov chain analysis
gives the properties of the number in the
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system at only at the arrival epochs.
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As pointed out the under - the discussion
of MI/M/1 queue the limiting distribution
00:07:08.941 --> 00:07:14.000
of the number of customers in the system at
arrival epochs at departure epochs and at
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arbitrary points in time are the same only
when the arrivals occurs as the Poisson process.
00:07:23.430 --> 00:07:30.470
So in the MI/G/1 queue the limiting distribution
of the number of customers in the system at
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arrival epochs, departure epochs and arbitrary
time points are same because the arrival occurs
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in the Poisson process for inter arrival follows
independent exponential distribution with
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the same parameter whereas in the GM GIM/1/
queue model the embedded Markov chain results
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gives the limiting distribution of the number
of customers in the system at the arrival
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epochs only.
00:07:59.810 --> 00:08:06.410
That is not same as the limiting distribution
at the departure epochs, that is also not
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same as the arbitrary time points.
00:08:12.680 --> 00:08:18.560
Now we are finding - now we are going to discuss
the limiting distribution at the arrival - at
00:08:18.560 --> 00:08:25.590
the arbitrary time points.
00:08:25.590 --> 00:08:34.319
Prabhu and Bhat derived the results of limiting
distributions at a arbitrary time t.
00:08:34.319 --> 00:08:41.090
Let Pj is the probability that limit t tends
to infinity the probability Qt is equal to
00:08:41.090 --> 00:08:44.740
j.
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So this is nothing to do with the embedded
Markov chain Qn.
00:08:49.699 --> 00:08:56.100
We are finding limit at t tends to infinity
probability that Qt is equal to j where Qt
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is a number of customers in the system at
arbitrary time t.
00:09:06.029 --> 00:09:14.589
The limiting distribution exists whenever
the Rho is less than 1 and probability that
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no customer in the system in a long-run are
the limiting in a long-run that P naught is
00:09:20.529 --> 00:09:27.350
equal to 1 minus Rho and the Pj is equal to
Rho times 1 minus Psi Psi power j minus 1
00:09:27.350 --> 00:09:31.571
for j is equal to 1, 2 and so on.
00:09:31.571 --> 00:09:37.400
So this is the limiting distribution at arbitrary
time.
00:09:37.400 --> 00:09:45.009
To determine the distribution of waiting time
of a customer we need a distribution of a
00:09:45.009 --> 00:09:51.750
number of customers in the system at the time
of that customers arrive if you want to find
00:09:51.750 --> 00:09:55.180
out the distribution of waiting time.
00:09:55.180 --> 00:10:06.370
Replace Rho by Psi in the MM1 results since
the queue length the distribution is same.
00:10:06.370 --> 00:10:13.199
Our interest is to find out the waiting time
distribution since the limiting time distribution
00:10:13.199 --> 00:10:22.829
at arrival epochs is the same as the limiting
distribution of MM1 queue model.
00:10:22.829 --> 00:10:36.279
Therefore, you can replace Rho in the MM1
results by Psi to get queue length distribute,
00:10:36.279 --> 00:10:38.959
to get the waiting time distribution.
00:10:38.959 --> 00:10:46.779
Therefore, by replacing Rho by Psi in the
moment results of waiting time distribution
00:10:46.779 --> 00:10:57.029
you get the waiting time distribution for
GI/M/1 queue as the CDF of waiting time distribution
00:10:57.029 --> 00:11:06.199
is 1 minus Psi times e power minus mu times
1 minus Psi times t for t greater than or
00:11:06.199 --> 00:11:07.199
equal to 0.
00:11:07.199 --> 00:11:12.579
For t is less than 0 it will be 0.
00:11:12.579 --> 00:11:28.290
So this is the waiting time distribution for
GI/M/1 queueing system.
00:11:28.290 --> 00:11:31.610
First we found the limiting distribution at
the arrival epochs.
00:11:31.610 --> 00:11:34.660
Next we find the limiting distribution.
00:11:34.660 --> 00:11:42.199
Next we discuss the limiting distribution
at the arbitrary time points from the Prabhu
00:11:42.199 --> 00:11:45.330
and Bhat results.
00:11:45.330 --> 00:11:49.189
Then we have discussed the waiting time distribution.
00:11:49.189 --> 00:11:57.959
Now we are discussing the moments of queue
Psi.
00:11:57.959 --> 00:12:03.709
The mean and variance of waiting time, time
spent distribution and its mean and variance
00:12:03.709 --> 00:12:12.300
can be obtained.
00:12:12.300 --> 00:12:18.850
Once we know the limiting distribution at
the arrival epochs as well as you know the
00:12:18.850 --> 00:12:25.680
waiting time distribution you can find the
mean and variance of waiting time by adding
00:12:25.680 --> 00:12:39.399
mean, you can find the mean of a time spent
also.
00:12:39.399 --> 00:12:48.060
You cannot use the mean value approach because
the arrivals are not Poisson process.
00:12:48.060 --> 00:12:54.540
So you can find out the average time spent
in the system that is E of t will be average
00:12:54.540 --> 00:13:02.559
number of customers seen by at the arrival
epochs that is ELa multiplied by 1 by mu plus
00:13:02.559 --> 00:13:13.069
the average service time that is 1 by mu will
give the average time spent in the system.
00:13:13.069 --> 00:13:19.769
Since the arrivals are not Poisson processes,
not Poisson process we cannot use the mean
00:13:19.769 --> 00:13:25.920
value approach.
00:13:25.920 --> 00:13:32.240
By simplification you can get already we know
what is the average - what is the distribution
00:13:32.240 --> 00:13:41.680
of number of customers seen at the arbitrary,
sorry, already we know the limiting distribution
00:13:41.680 --> 00:13:43.610
at the arrival epochs.
00:13:43.610 --> 00:13:46.550
So we can find the mean from those results.
00:13:46.550 --> 00:13:55.939
Then multiply by 1 by mu plus 1 by mu will
give the average time spent in the system.
00:13:55.939 --> 00:14:03.300
Then you can find the variance of time spent
in the system also.
00:14:03.300 --> 00:14:11.670
The first this one the mean and variance of
waiting times since you know the waiting time
00:14:11.670 --> 00:14:14.399
distribution you can find the mean and the
variance.