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Okay, so now we will actually look into how
to differentiate between probability of call
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loss and probability of switch being in blocked
state these are two different things. And
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the ways of actually estimating these two
in the switches.
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Unable to capture the image because the lecturer
writes and erases the content immediately.
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So if there is a switch and there is a call
which arrives and when the call arrives it
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finds that switch is in the blocked state.
Now call will not be able to go through, so
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the probability that your call will not go
through is known as call loss because call
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has been lost. So if you make 1 million calls
out of which 10,000 calls you found that switch
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was in block the state at that point of time
call cannot go through.
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So that is a call loss or we call it a call
congestion okay. Secondly even if the call
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comes or call does not come does not matter
you look whether switch in which states, if
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switch is in the blocked state even if call
does not come it does not matter if there
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is no call coming in switch is in blocked
state there is no loss, there is no call is
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being lost actually in that case. So the second
possibility is we call it a, that switch is
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in the blocked state, so that we callas a
time congestion.
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So now question is how these two are going
to be related, one thing which is sure that
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then switch is in the blocked state that is
basically fraction of you are actually technically
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observing the time when for example, I estimated
in my previous lecture probability of switch
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being in blocked state what does that mean,
you observe the switch for a long time say
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a few years and you find out fraction of time
for which switch was in that state which is
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the blocked state.
So ratio of that particular time divided by
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the total observed time will give you the
time congestion that is why the time congestion
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word is being used, but the probability that
your switch is in blocked state does not mean
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call is blocked, only when the call will arrive
then only the blocking will happen for the
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calls. So call congestion usually will be
lower value having a lower value then the
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time congestion.
So that I think intuitively is clear, but
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let us build up a relationship between these
two. So if we define that probability that
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of call arrival is defined as PA probability
that a call is arriving and I also define
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PN(a) this is the conditional probability
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that call arrives when switch was in Blocked
state
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okay so I define call congestion as probability
of call loss PL I define the time congestion
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as PB being in the blocked state.
So now I actually have four variables and
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these need to be related to each other, remember
this is a conditional probability conditional
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probability when switch was in a blocked state
and a call arrives, so this call is going
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to be lost and when it probability that a
call arrives this is the probability that
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the call will be loosed okay condition on
that call arrives actually so P x L into P(a)
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this gives the number of calls which will
be getting lost over time okay remember when
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I am saying when a call arrives and it finds
switches in blocking state that is a conditional
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probability.
That is a call congestion and this will be
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on an average number of calls lost over long
time if my arrival rate is high more number
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of calls will be lost arrival rate is lower
less will be lost over time okay, so this
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should be equal to because both are going
to lead to the same thing this will lead to
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nothing but probability that is a conditional
probability that call will arrive and switches
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in block the state and probability that you
will be in block the state is PB this technically
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is a Base rule.
Okay what I am saying is that they are event
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and B switches in block the state and calls
arrive so this P(A,B) be that both things
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I happen is P (A|B) so probability that switches
in block B state probability that call arrives
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condition none switches in blocked state okay
so this represents right side it is also possible
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to write P(B| A) the call arrives and when
the call is arriving on that condition that
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is which was in block the state at that point
of time, is a technically basil which I have
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written here and this is what I am going to
use to estimate the relationship between PL
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and PB. So I can now write and we will do
it for m/m composites which only.
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I can right now PL(S) and of course now very
important observation which you can make,
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probability of call loss and probability of
blocking will be almost same, if my call arrivals
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probabilities are independent of switch state
if they are independent of switches state
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this and this has to be exactly same and they
will cancel both of them will be called loss
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probability or call congestion and time condition
will be same, okay.
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But that actually does not happen this is
going to be a true statement if my n is comparatively
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much smaller than m, m is very large. So the
arrival probabilities does not get impacted
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by the change in switch state they will be
almost same and both of them will be similar
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but when m and n are not n is not much smaller
not much smaller than m then this is a PL
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= PB cannot be taken we have to actually estimate
a relationship between that.
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So what is PN or K let me see how we can find
out, so when the switch is in state A then
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what is the arrival rate, I have to find out
that and the small elemental time ??T so I
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can put some ??T here. So that is a probability
of arrival when you are in state n in small
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time ??T, probability of arrival this will
be the average value of the arrival rate,
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with every state change there is going to
be different arrival date.
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Remember in the previous lecture when you
were in state zero, arrival rate was M? when
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you are in state one it was M -1? and so on,
I make an estimate average estimate depending
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on that probability of the events and I will
get this average value in totality, so that
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is what is going to be the call arrival rate
on an average basis which is independent of
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state states actually have been ironed out
by taking what we call a proud mystic average.
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So in this case ?T now can be estimated as
because I am taking a probabilistic average
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i can write that you I am in probability of
being in state k & state k what is the probability
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of what is the arrival rate so I am taking
a probabilistic average am I take all states
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goes from k to n so that will be PA.
So I can now solve it so i have already remember
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I am taking ratio ??T is will cancel these
are immaterial so PL will be ? n divided by
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? of I have to write down this thing
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okay so this is what will be the relation
now ?T there I will rate and let the end they
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will be related ?n will always be smaller
than ?T okay this of course you can make a
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drawing so when you were in state 0 m? was
the arrival rate as you keep on moving by
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state it becomes a - 1 number and you will
keep on reducing at this point it will be
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M -n ?.
And this monotonically decreasing as your
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state’s as you are actually going to from
0 to nth state so the average value will be
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somewhere here and there the smallest value
so this top one is the smallest value so this
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has to be smaller than the average and of
course with this you configure out that this
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is going to be smaller than this so PL is
always going to be less than PB this was also
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the intuitively the result which I explained
in the beginning the call loss probability
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will be smaller than the switch being in the
block state that probability from okay it
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will be smaller than that so let us solve
it so ?n essentially will be M -n so I am
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just going to put everything in this expression.
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Unable to capture the image because lecturer
writes and erases the content immediately.
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And we will solve it and this expression will
be
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that is your ?k probability that you are in
that state I am going to use the Tang set
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probity distribution will be given by MC
k ?/µk /again there has to be a summation
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okay so I can also now to place this PB in
the blocking probability x in fact I have
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to put again now this is independent of k
this and these two are same so I can cancel
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them actually once I cancel I will get I can
actually raise it now from here okay.
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So
this ? will cancel with this one this M factorial
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by n factorial m - n m factorial by K factorial
so
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I can cancel and I can make it - 1 so and
I can make m here and take 1 m here out i
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can do the similar exercise here this am will
cancel and i will have
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M -1 C n ? by µ n ? k goes from 0 to n, m
- 1 CK and of course you can see this is the
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same expression as the probability of being
in blocked state for M by n composites which
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except now instead of m I am going to have
M -1 here.
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So i can write this thing as nothing but probability
of blocking but for m - 1 inputs instead of
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using M by n if I use M -1 by n composites
which the probability of it being in block
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the state is equal to probability of call
loss for M by n switch so which actually implies
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that p l of m should be equal to PB (M -1)
that is how your call congestion and time
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congestions will be related to each other.