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Okay, so in my previous lecture I was talking
about the expansion of crossbars using cross
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bar basically building up a larger switches
using multiple of them in multistage interconnection
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configuration. We figured out that two stages
cannot give me strictly non blocking property,
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but three stages probably can and I will show
yes they can and they this requires much less
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cross point complexity compared to using a
single crossbar of the same size okay, we
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will actually show that.
But before we do that we would like to actually
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look into how we will estimate the blocking
probability of a system. And then of course,
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they are two kind of blocking estimations
blocking probability estimation that we need
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to also learn to differentiate and we need
to also learn how these two are virtually
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related to each other.
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So I will start with a very simple cross bar
which is M/N, so remember I am not taking
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my inputs are not same as the outputs okay.
So I am actually having M lines which are
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coming in and there are n lines which are
going out. And they were like the telephone
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communication which is happening the calls
will come and calls will be them connected
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to some outgoing line.
So if a call comes on this line this will
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be connected and call is true okay. So the
rate at which the calls will come here will
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be defined using a poison statistics that
actually means a point event on a time horizon,
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I will take time difference time duration
of T how many events are going to happen.
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So if I actually assume the calls do arrive
and they are instantaneously completed and
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instantaneously removed.
So even happens that is it, normally whenever
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a phone call will come it will be held up
for certain time so line will get occupied,
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new call cannot come, but if that is not happening
at what rate the call will be coming when
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call is not going on. So that we will define
using a poison statistics which is a commonly
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used model. So we define something called
? which is the number of calls per unit time
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I will call it call arrival rate okay or arrival
rate.
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And this also being used this technique is
also used for packet switching systems also.
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So in time T ?xT calls will be arriving on
an average okay. So the probability if this
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is a person statistics the probability that
N calls will be coming will be arriving in
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time T will be given by, now remember when
the call is arriving is being instantaneously
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served and line is made fries word new calls
can come in, because if you occupy the line
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new call cannot come in actually then I cannot
make this estimate.
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So ?TN E-?T/N!, so this will be the distribution
okay. So if you look for the probability of
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no call being arriving here probability of
one call probability of two calls arriving
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in time T and so on. The total sum of these
are all mutually exclusive events so when
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one call is arriving within time period T
two calls will not be arriving when two is
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arriving one will not be happening.
So these are all mutually exclusive events
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when I sum up all the probabilities this should
be equal to 1 which actually implies that
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summation
of ?TN E-?T/N! this should be equal to 1 which
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is obvious because, so n goes from 0 to 8
and this is nothing, but a series expansion
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for exponential. So this will be written as
E-?T which it turnout to be equal to 1. So
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this is a probability distribution and we
call it Poisson distribution actually okay.
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So that is what is happening when the line
when actually in real life when a call will
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come people will start talking so there will
be some time after which call will get cleared
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okay. So I define that the call duration is
exponentially distributed that is my assumption
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is exponentially distributed
and the mean value mean duration okay for
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every call is 1/ Á okay this also comes actually
can be derived from poisonous statistics.
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So before I move further I need to clarify
this so I had define ?Tn e- ?t / n! that is
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a arrival statistics but if I want to measure
to Britain between two consecutive arrivals
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what is the time duration let me find out
the statistics for this time okay so I want
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we can figure out a PDF but for finding out
a PDF, PDF is probability density function
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you can always take probability distribution
function and do that derivative of this
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and if the variable is T it has to be / T
it is derivative as to with respect to T.
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So I think this is very similar that your
x is less than X that gives you the your PDF
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probability distribution function you take
the density you will define something like
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probability density function, so the variable
X is going to be in range x + x2x + dx within
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this range the probability that x will be
lying within this is always given by px x
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dx okay that is what the density actually
means and when it is a probability distribution
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function this defines probability that x will
be having a value smaller than this X that
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is a PDF okay.
And this is incremental technically that is
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why you have to take derivative of this to
get this okay when I am actually I can use
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this same mechanism to estimate the exponential
distribution for the time so Poisson distribution
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is related to exponential distribution in
that sense the time between two consecutive
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arrivals actually is exponentially distributed
so this can be estimated by if the time is
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t < T the inter arrival time is smaller than
this is this will happen this probability
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is nothing but that probability that you have
more than one or more one arrivals in time
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T.
Okay and of course this actually means 1 - there
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is no arrival in time T okay so this I can
use this same expression Poisson thing 1-
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no arrival when n is 0 this will be e -?t
and this is nothing but a distribution function
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see here so I need to just take the derivative
to get the density is the probability that
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the value of x is lying going to lie between
x to nx + dx divided by this differential
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so this is the probability and so this px
x will be given I will be giving you the density
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function.
So let us take the derivative of this will
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give me now remember this the random variable
here is T so this time so I can actually replace
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it by a t also and if I take this if this
is a PDF the area under this should be equal
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to 1 which is actually true if I take ? e
- ?t dt now this t cannot be negative this
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can only be positive so it can take value
from 0 to 8 this will be equal to 1 so this
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actually can be proven now this time exponential
this is the exponential distribution.
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So I can represent that how many calls will
be served or how many calls will be finished
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essentially is a service rate the Á number
of call swill be serviced per unit time by
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the line that essentially service rate will
be decided by the call duration average call
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duration so that is also essentially being
assumed to be exponentially distributed okay
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so with this actually now I can analysis this
switch and try to estimate what is the blocking
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probability now before that when the blocking
is going to happen.
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If my number of outgoing links n is greater
than M will there be a blocking whenever a
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call will arrive call can always be connected
to an outgoing link because it is available
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and when you are talking anyway the call cannot
come at your line only on the free lines the
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call can come but you will always find an
outgoing link so there is no blocking here
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so blocking will always happen if n is going
to be < than M so when m is > than or equal
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to M then blocking will not happen blocking
will happen only in this case.
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So what is the blocking probability so I have
to find out that is all what is the probability
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that all the lines here are going to be occupied
we need to estimate that how to do this so
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we can actually use here Marko chains the
concept of that so the idea is that I will
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identify the state of this switch if only
one line is busy switches in one particular
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state this line can be anybody it does not
matter so for one line is busy switch in state1
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when no line is occupied switch is in state
0 and so on I can actually then been switch
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can be in state n when it is in a state n
m is larger than n.
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So there are some free lines call can still
come in but they cannot go through so switch
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is in blocked States so probability that this
switches in state n will give you the probability
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of the switch being in blocked state okay
this estimate or this number will be actually
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useful to us at some later point of time then
we will actually start doing estimation for
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a three-stage interconnection so when you
are switch so what I will do.
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Is I will represent these small round circles
will represent States so I am I can actually
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be instate 0 I can be in state 1 I can been
straight to and so on I can be in some state
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i and I can be in some state n in fact truly
speaking if this one is line is occupied.
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All over all free this is one particular state
next line is occupied all others are free
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is another state but they are equivalent states
because there is only one outgoing line which
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is occupied so I am actually not I have actually
have reduced by state space by merging all
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the equivalent states so there are only ten
non-equivalent states which are possible in
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the system so when you are in state 0 there
is no call I need to find out at what rate
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I will be transiting to state 1 so with ? number
of calls per unit time the calls are arriving.
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What is the probability that a call will arrived
in any one of these lines so there is only
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one line which I will consider probability
that one call will be coming in sometime elemental
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time ??T I am going to take it is very small
time and as Weill solve we will figure out
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this ??T is immaterial because it cancels
out when we will try to estimate the state
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probabilities so I can use the poison statistics
so this will give me ? into ??T1 is found
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one in fact I can estimate the probability
for to caller i will also in time ??T but
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??T I will take so small that probability
of having two calls.
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Will be still smaller actually so and then
of course I will have 1 factorial n in limit
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when this ??T goes to 0 this will turn out
to be nothing but ? into ??T so that is a
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probability or that is a instantaneous probability
in small time ??T that one call will be arriving
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on one line, there I am such lines so the
probability that one call will arrive will
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be given by m x ? ??T that is the rate at
which the calls will be arriving, so that
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is a transition rate for going from state
zero to state 1, okay. Now when you are in
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state 1 there is only one call which is going
through.
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Remember it is like I told you that exponential
distribution and Poisson distribution are
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related to each other, so it is like there
was only one call and this call is going to
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be serviced and there is only one server and
the service rate is Á so in time ??T Á x
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??T en e- Á ?? T by those many calls will
be served so probability that n calls will
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be served is Á ?? T en e- Á ?? T/n! since
there is only one server only one call can
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be served so in time ?? T when limit when
?? T goes to 0.
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n service to be given so when I am looking
for n = 1 if this becomes Á x ?? T that is
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a service rate. If the two lines which are
occupied in time ?? T it will be 2 Á ?? T
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that will be the service rate. So I can actually
come from one back using Á x ?? T when you
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are in state 1well line is occupied from this
side. The free incoming lines are M - 1 so
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the rate h way at which the calls can arrive
is. M - 1 x ? ?? T and where you are in state
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2 there are two calls which are there.
With a small time ?? T the probability of
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chances this basically is the probability
transition probability, okay. In small time
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?? T remember is a probability expression
which have been approximated as ? ?? T there
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is a probability expression, so chances are
that it will be 2 x Á x ?? T and now important
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thing is that if I want to if this is a system
which is in a steady state condition.
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So the probability of being in certain state
will not change it will remain it is a constant
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actually and something can always remain constant
if I take any closed surface in this is known
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as Markov chain
this is called Markov chain and this is being
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used extensively for queuing theory for analyzing
the Q Q's performance, so this is also technically
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a queue, so if I take any closer surface like
this.
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So the rate at which you are going to move
out of the surface and rate at which you are
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coming into the surface has to be equal if
the switch is in steady state, which implies
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that probability that you are going to be
instate 0 is p0 and that is a probability
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of transitioning going from 0 to 1 so that
should be a transition rate now, should be
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this and this should become p1 Á??T. Now
interestingly this ??T gets cancelled because
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it is common on both sides you actually make
any closed surface here and build up a balanced
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equation this is known as balance equation.
So the rate at which you will go out of the
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surface and read rate at which you will come
into the surface has to be equal, okay. Because
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this system is in steady state under that
condition ?? T will always cancel out, okay.
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00:19:14,139 --> 00:19:19,601
So ultimately we normally in when we build
Markov chains we do not write this ?? T because
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of this condition, when I put ?? T it is a
probability, probability of the transition
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taking place.
But when I am NOT putting ?? T this value
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can actually be more than one, remember probability
cannot take a value more than one that is
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why ?? T was multiplied ideal it should be
probability is a transition probability going
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from one state to another state what is a
chance it will happen, so that is what we
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do in finite state machines, okay. So ??T
now actually can be removed. So ultimately
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what how this whole chain will look like,
so chain will now look like if you go from
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here M ľ 2? 3 Á.
So I have written these are basically now
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become transition rates not the transition
probabilities actually okay so the probability
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are proportional to this so proportionality
constant d T is be removed out so balance
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equations it does not matter actually.
So now I can actually use a balanced equation
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to solve it my idea is that I want to figure
out what is the probability of being in state
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I, I want to estimate that okay so once I
know the probability of being in state I,
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I can find out the probability of being in
state n which is the blocking probability
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of the system so let us write down the balanced
equation I can write down in many ways okay
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so one of the possible ways is this.
So I can actually take a closed surface around
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every state and then write down the balanced
equation so balanced equation around 0 will
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be p0 the outgoing rate is m? this should
be equal to incoming rate which is p1 Á okay
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so which gives me p1 as I can solve it m ? by
Á p0 I can look at the second surface and
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write down my equation this will turn out
to be p0 m ? that is incoming rate one incoming
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rate is from this side to ÁP2 the outgoing
rate is Á p 1 + p1 M - 1 ? okay.
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So I just need to have a relation between
p2 and p 0that is what I want I want to represent
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all the state probabilities in terms of p0
okay once I have that I can use some fundamental
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axiom of probability to figure out what is
p0 and henceforth I can actually estimate
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all state probabilities so I need to put out
p 1so let me put the value of P1 there so
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that I can get p 2so this p0 also I need to
move on that side so this will turn out to
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be m ? p0 this term comes on the site will
become p0 m ?.
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So this cancels with this and this Á will
become Á2 this will be 2 so ultimately you
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will have
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alternatively I could actually have taken
a surface like this so incoming rate has to
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be equal to outgoing rate and you would have
got this equation directly okay so this would
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have become a relation between p2and p1 and
p1 I could have replaced from the previous
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one to get this particular equation.
So doing it this way for this surface when
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I am taking to enclosing this P 1 M - 1 ? should
be equal to p2 2Á you so which becomes p2
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00:25:20,800 --> 00:25:39,629
is equal to M - 1 / 2 ? by Á 1 which is
I can actually keep on doing it so ultimately
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00:25:39,629 --> 00:25:56,509
I can find out what is the probability of
being in state I so this will be
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00:25:56,509 --> 00:26:16,020
probability of being in state I will be given
by MM - 1 M - I + 1 okay.
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00:26:16,020 --> 00:26:31,090
Remember their inner brackets okay and of
course I* I ľ 1* 1 and a hair it will be
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00:26:31,090 --> 00:27:11,149
? by Á this power I p 0 now whatever what
is this expression they think right so this
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00:27:11,149 --> 00:27:22,740
particular expression is nothing but a combinatorial
it is MC IM ! m- I! I effect Orion the lower
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00:27:22,740 --> 00:27:32,379
part is I !l and this part is what gives you
the numerator here so I can write P I s M
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00:27:32,379 --> 00:27:45,080
commute oriole I ? / Á I into p 0 so for
every state the probability now can be represented
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00:27:45,080 --> 00:27:50,530
in terms of p0 .So now let us solve for what
the p0 will look like.
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00:27:50,530 --> 00:27:52,780
Unable to capture the image because lecture
writes and erases the content immediately
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00:27:52,780 --> 00:27:59,970
So to find out p 0 remember all the states
which always has to be in one of the states
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00:27:59,970 --> 00:28:05,429
so sum of all state probabilities has to be
equal to 1 so technically sum of all mutually
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00:28:05,429 --> 00:28:11,649
exclusive events all possible of them and
the probability of those events been happening
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00:28:11,649 --> 00:28:15,610
when you sum up all the things together that
should be equal to 1 that is a fundamental
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00:28:15,610 --> 00:28:28,190
axiom of probability so I will use that so
p0 + p 1 + p 2 and so on + P I and PN should
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00:28:28,190 --> 00:28:53,530
be = 1 so which I can right now p 0 as 1 +
MC 1 ? by Á 1this should be equal to 1 which
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00:28:53,530 --> 00:29:00,420
gives me nothing but p 0 is this is not a
closed form solution if n would have been
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00:29:00,420 --> 00:29:05,750
equal to M then I could have got a closed
form solution here I cannot okay.
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00:29:05,750 --> 00:29:28,139
So p0 into ? i going from 0 to n MC n ? / Á
i is equal to 1 so p 0 is 1 divided by ? i
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00:29:28,139 --> 00:29:42,139
= 0 to n okay so you can now find out what
is going to be probability of being in state
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00:29:42,139 --> 00:30:08,029
I mci in fact I should write a different index
here so I can put j because I have used I
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00:30:08,029 --> 00:30:13,990
here so I have to use this so this is the
probability now what is the probability of
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00:30:13,990 --> 00:30:20,799
being in blocked state so what is the blocked
state when the switch is instate n and when
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00:30:20,799 --> 00:30:33,580
I do that probability of switch being in blocked
state we call it also as a PB okay.
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00:30:33,580 --> 00:30:38,769
Probability of being in blocked state that
is the way I will be representing this so
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00:30:38,769 --> 00:30:58,210
this will be MC n ? / Á that is power n ? i
goes from 0 to n but that is a blocking probability
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00:30:58,210 --> 00:31:07,779
for am by n composites which we call it m
by n composites which
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00:31:07,779 --> 00:31:57,399
and this also is known as AND set
distribution okay.