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Okay so let us move forward from where we
left in the last lecture so just to recall
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the last lecture was about the lesĺs approximation
which I did for three stage interconnection
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network which was clos configuration and the
of course we also did a cross point complicity
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of clos network so those two things we did
but leeĺs approximation is not precise it
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is error because we are approximating there
so then of course I suggested that we can
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do more precise calculations but this not
for time congestion but for the call congestions.
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So let us actually do it we are actually considering
here again three stage clos network only and
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we are going to this approaches.
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Unable to capture the image because lecture
writes and erases the content immediately.
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Attributed to Karnaughĺs approach and this
gives a beautiful result about the call lose
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or call congestion so in this case the PL
in fact remember PB when I am actually writing
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it is always the probability of switch being
in blocked state whenever it is PL is a probability
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when a call arrives it is going to be lose
it a conditional probability okay so PL will
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now began as so this we are not computing
so PL we can write as I will tell what this
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S is.
Now s is something which have now introduced
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this is known as this represents the state
the of the switch now what do you mean by
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state of a switch, so if I have only one single
switch what are the state for this one it
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can be either ON or OFF their only two possible
states okay if all the cross bar of 2/2 each
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one of them can actually be ON or OFF independently
technically there will be 24 possible states
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some of them are not valid states.
Okay so in fact and some of the states will
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correspond to multicasting scenario so I am
not looking into the multicasting scenario
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here I am actually looking into uni casting
case so that many possible states which can
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happen so normally if the number of cross
points are C in general total number of states
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which will be there will 2C and of course
which input and output are active where also
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be counted for ram actually assuming that
all inputs and all outputs are connected they
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are actually communicating.
So under that possibility there are 2C possible
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states okay so this s represents this but
there is going to be large number of states
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we need to actually take care of these so
we use concept of something non as equivalent
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states so what do we mean by equivalent state
if a switch is in one part of say A and if
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it is in state B the probability that you
are going to be in A or probability you are
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going to be B in is same.
The blocking which happens is going to be
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same so basically what you can do is the cross
switches consist of a smaller switches if
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I can move them around I can just do a renumbering
I can always go from a state A to state B
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by doing this transformation we will actually
do an example of this while looking at while
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since non blocking switch and it is prove
that time on will be actually doing this thing
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then these become equivalent states, so I
do not need all this possible states to be
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taken care of when trying to computing this.
I can represent it by much a smaller number
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of non equivalent states we also go further
ahead actually
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now how this s can be represented I am coming
to that but what does the P(s) means P(s)
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is the probability that your switch will be
in state s what do you may understand by ? s.
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? is the call arrival rate number of calls
which are arriving at any point of time okay
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so this represents that the call arrival rate
probability you are in this step average number
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of calls which are arriving per unit time
when you are in a state s now this is conditional
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loss probability when you are in a state s
what are the chances that a call will be loss
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when it arrives is that conditional loss probability
for an arrived call.
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And I submit over s so over a long range of
observation period this numerator will represents
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a number of calls which will be lost on an
average okay because it is being now average
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over all possible states taking care of the
probability of being in that state the arrival
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rates and the call loss probability now if
you come back to denominator this represents
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the total number of calls which have been
attempted so this also includes the calls
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which were not lost that total calls attempted
average over all possible states by using
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this probability total calls attempted.
So this is the ratio of the failed calls by
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total calls and this becomes an estimate of
call loss probability okay in general now
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question is how you will represent s so for
doing this I am going to actually take an
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assumptions so of the efficient known as the
we need to take assumptions so one of them
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is random root hunting now the way it happens
is whenever you are going to have switch which
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will consist of all these states all these
switches in three stages typically when a
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call as to be set up the input and output
ports are indentified then how the call will
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be rooted.
I will be setting up I will be taking up any
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middle state switch in setting it up I can
do it randomly actually so if I take another
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input ports another input ports where I can
take up any one of them randomly so when a
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call as to be set up the way the links are
chosen they will be taken care of they will
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be done randomly that is what the random root
hunting actually means, I will just pick up
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randomly we will try it out does not go through
I will make an attempt again till I get a
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free call.
If it does not happen certain time I will
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simply discover it is possible I can always
start from the top switch keep on doing it
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till I find out a free link certain connection
everybody does that so most of the time the
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upper state middle state switches will get
occupied lower one will not be so that ordered
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system or may be when I tried for the first
call I use this one next one this one the
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convert through here and when the next one
will come I will start from the next one so
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that can be happening in cyclic fashion but
I am talking random root hunting advantage
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of this is I can represent the state of the
switch by x and y they are independent of
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how the calls are setup at any point of time
when I observe the switch this is actually
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same it is not changing so time earlier time
state it does not matter actually.
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So that makes it far simpler for us to do
the analysis and I can actually take I need
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not take all the switches I am only worried
about only one of the input ports so one switch
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here and one switch here we need to take now
why we do it and number of links which are
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occupied here is x number of x which are occupied
here is y but remember I have other switches
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here also so where I look at this combination
so input and this input output pair that is
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independent of the earlier pair for example
this A,C which I was talking and BD which
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I am talking.
So these are independent and there will be
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in the similar situation so if I compute by
estimate for AC this is going to be true even
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for BD or any other possible pair so I am
also taking care of I am also making assumption
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that it is an uniform kind of routing which
is been used so when somebody wants to make
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a call it is going to make call to anybody
else with equally likely probability so it
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is a uniform loading condition so under that
condition the s now can be very well represented
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any a pair x, y okay and I need not bother
about anything else but accept one pair.
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So which is A and C here and this number is
x and this number is y and x and y will represent
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the state, so x can be 1 y can be 1 x can
be 0 so all possible pairs and they can take
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maximum values can be K where 1, 2 the number
of switches in the middle stage are K, okay.
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Under this condition we have to compute. So
I can now modify my expression this will become
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(x, y) and now I think I can compute.
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Unable to capture the image because lecturer
writes and erases the content immediately.
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So now if I draw this so this one gives you
x this one gives me y, probability of cal,
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arrival essentially this technically means
that a probability of (x, y ) that switch
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will be in a state (x, y) how this will be
decided. X and Y are independent Y because
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when actually I am setting up the calls all
the calls which are coming emanating from
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this switch are not terminating here there
might be terminating here also they might
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be terminating somewhere here also.
So this x becomes independent of this y this
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outputs might get connected to the difference
switches via this route. So y need not have
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any overlap with x so they can be independently
varying in this case, okay. So the probability
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of x and probability of y can be written as
the product of probability being in x and
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probability of being in y, that comes because
of the independence.
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Any two independent random variable they are
join probability is nothing but the product
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of the probability of each one of those independently
and that is what I have written, okay. So
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that is one important thing we have to estimate
what this p(x) and p(y) also. Now next look
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at this P(l, x, y) how this will be estimated?
So you have a switch here
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and you have 1, 2ůkn and of course these
arte A links and these are B links.
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And there x calls which are progressing here
there y calls which are progressing here,
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I have to find out the probability that call
vector get lost if I want to set up a connection
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between these free incoming and out coming
ports so this is a free incoming port this
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is a free outgoing port what are the chances
that when you randomly select a pair you would
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not be able to find out a path, okay. So that
is comes from a Jacobeans approximation this
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is also known as Jacobeans approximation.
So let us see how we actually figure it out,
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now if my x + y this is the x number of links
which are busy y number of links which are
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busy if x + y is less than K if this condition
is true then what is going to happen?
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You will always be able to find out a switch
in worst case x are occupied y are occupied
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and they are not connected to the same middle
stage switches in that case you can always
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find out something for which A link and B
link both are free and I can use that to set
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up a path between from input to output, so
blocking will not happen if this is true blocking
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will one can only happen when it is x + y
> = k.
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But when this x and y are actually being the
links which are occupied A type and B type
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are happening independently we have to find
out the chances that I may find out a path
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from input to output, okay. So let us see
how this will be done, so let me take an example
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so I have three links which are occupied three
connections and in total we have five so these
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three are occupied, three are occupied of
from this side out of these five.
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So these 3 can be like this so this one possible
combination. Now you see I cannot set up the
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path set up a connection between a free incoming
port and free outgoing port in this example.
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But as I said we are using random route hunting
and this happens with the same probability
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as this there is a path available I can set
up the connection, there is possibility that
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this will happen with the same probability.
So essentially if I freeze this that many
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possible ways this now can these three connections
can actually be distributed across these five,
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in some of the ways you will find out you
will not be able to set up the path in some
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of the combinations you can set up the path,
so probability a call will be list or you
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cannot set up a connection, essentially now
can be written as total number of patterns.
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In which the blocking will happen divided
by total number of possible patterns which
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can exist, okay. So this we can compute for
this combination, now instead of this if this
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would have been this scenario and I would
have allowed all possible patterns here I
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would have got the same ratio it does not
matter is a same case. So I need not bother
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about x so I will just fix it x and this is
y the remaining balance will be k ľ x.
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And out of this y in how many ways this y
actually can be arranged among these case
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middle stage switches this will be k come
pictorial why, that is a total number of possible
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which you can arrange and of course now you
have x occupied versions and k ľ y unoccupied
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once, so k ľ 1 y which are unoccupied once
in how many ways they can be arranged across
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this occupied A links.
That is the time when the blocking will happen,
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okay.
So for example when you have this possibility
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these are blocked there is another possibility
this way this blocked if you do this is also
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blocked, so I am not trying to arrange the
free once across this busy stuff so in how
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many possible ways that can happen, there
was another number of blocked possibilities.
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So if I take this ratio for which will be
xck-y this should give me the probability
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that your call will be lost given (x, y).
This is what is known as Jacobeans approximation.
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So I can solve this and we will get x factorial
by k-y factorial, x-k + y factorial so in
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fact I can write this as x + y ľ k factorial
and then of course k factorial, y! (k ľ y)!
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So this one will cancel and that is what with
the probability x! y! x + yľk ! / k! now
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you can actually see x + y has to be greater
than k than only this make some sense. Otherwise
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blocking does not happen, okay. So this will
be the expression which we can use it for
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PL given (x, y).
Now to find out that probability that x calls
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will be here how to estimate this so I will
you can actually go to the some of the earlier
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lectures we are have done m/n composite switch
where I have done the state probabilities
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for a m/n composite switch now here important
thing is that you will might have here certain
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number of inputs I call them m and this is
actually k here so k = n so I can use same
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m/n composite switch result to find out P(x)
so let see how that will come so instead of
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this m/n this is now m/k switch okay.
Okay let because my notes I have actually
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n so let me keep it n and we will solve from
here at some point I will change the variables
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I will do the transformation for m/n probability
that this switch will be in state n where
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n means outgoing n number of links are occupied
and remember in this case n is greater than
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m okay greater than equal to m that is the
case which we are looking at so I can write
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down t(n) probability of being in the state
and outgoings are occupied if you recall from
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there this will be MCN.
?/Á ? is arrival rate Á is the departure
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rate or one over Á is the average call duration
okay and this will be raise power n I can
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00:21:55,880 --> 00:22:02,900
write any variable let me put i because k
have been we have been using there important
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00:22:02,900 --> 00:22:20,080
thing it was n MCN sorry this will be i now
one thing which we need to note that n is
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going to be larger when m so this has to be
change it m actually okay and once I change
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it m because that is into maximum number which
you can get you cannot actually have a state
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board than m.
So this is becomes a complete binomial so
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binomial expression is MCI ai bm ľ i? I I
goes from 0 to m this will be always return
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as a+ bm so this is the binomial expression
we can use this to solve this expression for
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Pn so I can actually at here one raise power
m- i so this becomes a and this becomes b
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so a+ b so 1+ ?/ Ám okay I can actually further
solve it I can write this as ?+Á/Ám and
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I can take it up so this will MCN let me write
it down at the bottom.
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Unable to capture the image because lecturer
writes and erases the content immediately.
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So we can write it as MCN so I just split
it into m and n parts the bottom one Ám ľ
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Án comes here ?n comes here now Á/ ?+ Á
there is nothing but one ľ ?/?+Á and I call
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this as e I will give the interpretation of
it what does it mean it technically means
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loading of the telephone line or line which
is coming here at the n input okay so if I
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do this so this becomes 1- a and this becomes
a MCN an that is what it will be the probability
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of x Px now whatever is true from this side
this is a symmetric system.
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Then same should be true for py also okay
because a technically when I am talking about
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py that is because of the calls arriving it
all other input ports which are terminated
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at the outgoing line which I am currently
observing and that should be same as this
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is the input line n all outgoing lines to
which this call is being rotate px is because
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00:26:10,620 --> 00:26:18,040
of that so both are symmetric systems so py
as be equal to px and they both are independent
193
00:26:18,040 --> 00:26:38,380
so I can write down py also as MCN an1 ľ
a m- n so px and py.
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00:26:38,380 --> 00:26:48,760
We have found out we have found out Pl given
x and y which is this using karnaughĺs approximation
195
00:26:48,760 --> 00:26:53,470
so we have got this now we have to find out
this and we have to ultimately put it into
196
00:26:53,470 --> 00:26:59,860
the expression for getting my answer so in
this case.
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Unable to capture the image because lecturer
writes and erases the content immediately.
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Now x ?x is we talking about that is the call
arrival probability so if I talk about the
199
00:27:11,660 --> 00:27:23,330
switch here x links are busy their total number
of m inputs so what is going to happen is
200
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how many input links are free here it should
be m ľ x so the call arrival rate per line
201
00:27:34,340 --> 00:27:51,790
is ? in fact I can call it ?x so that will
be the arrival probability similarly on the
202
00:27:51,790 --> 00:28:00,120
other side m outputs are there one of them
is free which I am trying to connect.
203
00:28:00,120 --> 00:28:09,830
Links which are basis y so how many are the
free links available here is m ľ y so this
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00:28:09,830 --> 00:28:24,900
is the arrival probability here so I can write
this ?x ?y this ?x is this, this ?y so I can
205
00:28:24,900 --> 00:28:40,410
write ?x ?y so
this can be written as ?? m ľ y I am also
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00:28:40,410 --> 00:28:48,730
got my third term so I have to just actually
now use all these three to compute my expression
207
00:28:48,730 --> 00:28:56,340
so let us write down the expression and solve
it so P(L) so I have to write down summation
208
00:28:56,340 --> 00:29:03,060
since a = x, y I am going to put I have to
do summation over x and y.
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00:29:03,060 --> 00:29:10,080
This actually means I need to do summation
of with one with x and one with y okay so
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00:29:10,080 --> 00:29:14,260
forward timing let me put it this feel at
some point I will just into double summation
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00:29:14,260 --> 00:29:23,620
oh maybe now itself I can do it, it does not
matter so sorry this is I am using ? so it
212
00:29:23,620 --> 00:29:31,140
will be ? now be very carefully what I am
actually doing is I am doing now transformation
213
00:29:31,140 --> 00:29:39,010
of variables the middle state which is here
rk n number here the number of inputs are
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00:29:39,010 --> 00:29:46,140
n number of inputs are n we have to be careful
because I have use this fall n.
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00:29:46,140 --> 00:29:51,200
Slightly earlier when I am doing transformation
and n is now going to mean something else
216
00:29:51,200 --> 00:29:58,700
okay so it will be n ľ x so this is just
to be consistent with my notes so I do not
217
00:29:58,700 --> 00:30:07,380
end up in making some mistake n ľ y while
solving these will also understand certain
218
00:30:07,380 --> 00:30:16,780
things which we can observe in the process.
So this was nc something so it will become
219
00:30:16,780 --> 00:30:44,480
ncx and then of course you have x! y!/k!x+y
this coming from Jacobian approximation denominator
220
00:30:44,480 --> 00:30:58,900
we consist of except PL rest everything will
be there and you can also observe very careless
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00:30:58,900 --> 00:31:03,950
about this ? this is ultimately it concerns
out this is simply proportionality constant
222
00:31:03,950 --> 00:31:39,400
does not matter actually here.
Yeah I made a mistake here
223
00:31:39,400 --> 00:31:52,050
okay so now it is a correct expression. Now
look at A, A is written as ?/ ?+Á what does
224
00:31:52,050 --> 00:31:59,440
it mean. So on the trunk line or on telephone
line ? is arrival rate and remember I had
225
00:31:59,440 --> 00:32:06,550
actually mentioned in one of my previous lectures
that ? is because of poison distribution.
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00:32:06,550 --> 00:32:11,470
So if I look at on time scale the events which
are happening or the calls which are arriving
227
00:32:11,470 --> 00:32:20,190
? calls per second which are arriving.
So ? represented 1/? represents the inter
228
00:32:20,190 --> 00:32:25,700
arrival time will be exponential distributed
with mean value of 1/?. So the gap between
229
00:32:25,700 --> 00:32:35,210
two calls is 1/?, 1/Á simply is also in the
same fashion I had that 1/ Á is the call
230
00:32:35,210 --> 00:32:40,970
duration when the call starts is going to
be occupied for some time before again the
231
00:32:40,970 --> 00:32:46,370
line goes to free and the new call came coming
that call duration that also exponential distributed
232
00:32:46,370 --> 00:32:53,500
with average 1/ Á
So or we can say the calls are been processed
233
00:32:53,500 --> 00:33:00,590
very fast the Á number of calls per second
or per unit time will be processed okay. So
234
00:33:00,590 --> 00:33:05,610
because exponential distribution and Poisson
distributions are interrelated. So I can actually
235
00:33:05,610 --> 00:33:24,420
modify this and write do the inversion I can
actually put this as 1/Á/1/?+1/Á okay, so
236
00:33:24,420 --> 00:33:33,080
I can do this. So once I do this and this
is nothing but as good as A you can actually
237
00:33:33,080 --> 00:33:49,470
proved this one, so this term has to be same.
So this is on a timing scale when a call arrives
238
00:33:49,470 --> 00:33:56,460
this is the call duration, so average duration
for this is 1/Á and when the next call will
239
00:33:56,460 --> 00:34:03,730
arrive that gap is 1/? and then, the next
call will be arriving. So what I am saying
240
00:34:03,730 --> 00:34:11,060
is the busy duration divided by busy plus
free duration on the line that is what is
241
00:34:11,060 --> 00:34:19,200
A. So A becomes technically a trunk utilization
and its value will be between 0 and 1 okay,
242
00:34:19,200 --> 00:34:27,329
this value will be between 0 and 1.
So it is busy period divided by this period
243
00:34:27,329 --> 00:34:37,599
between two calls busy + free periods so that
is what the A is, so it is line loading actually.
244
00:34:37,599 --> 00:34:50,389
So let us now start solving it. So we can
see it the ? and ? will cancel, so that is
245
00:34:50,389 --> 00:34:57,609
very good let us see now what we have to do
next I can explain the Cx so that I can solve
246
00:34:57,609 --> 00:35:14,109
these n-y and n-x thing. So let me do that,
I can write as n!/x!n-x! I can do same thing
247
00:35:14,109 --> 00:35:21,829
for this
and same thing can here also.
248
00:35:21,829 --> 00:35:34,960
So
249
00:35:34,960 --> 00:35:48,470
I can take this n-x and n-x! and I can cancel
it out this will become n-x-1, I can do the
250
00:35:48,470 --> 00:36:08,089
same thing with this, same I can do on this
side. But this is not full combinatorial,
251
00:36:08,089 --> 00:36:23,190
so let me take n out and make it n-1, same
I can take here n2 so this n2 can come out
252
00:36:23,190 --> 00:36:29,309
of the summation and we can cancelled out.
So this is what I am going to have so let
253
00:36:29,309 --> 00:36:43,519
us solve it for that n2 I can remove from
here and what I need to have here is ax so
254
00:36:43,519 --> 00:36:47,789
this is the combinatorial which I am using
if I want to make it a full combinatorial
255
00:36:47,789 --> 00:37:00,289
sum full binomial sum it has to be n-1-x.
So it has to be n-1-x, it has to be n-1-y
256
00:37:00,289 --> 00:37:08,369
which actually implies I have to take 1-a2
out which I can take out of the summation
257
00:37:08,369 --> 00:37:15,589
it independent of x and y. Now these two are
this summations now can be taken and I can
258
00:37:15,589 --> 00:37:29,039
put one summation here and this become summation
over y and this is over, this is over y, this
259
00:37:29,039 --> 00:37:47,289
is over x. So this summation now can be written
as a+1-an-1 same I can do for this and I can
260
00:37:47,289 --> 00:38:09,630
write a+1-an-1.
So I can remove this one this turns have to
261
00:38:09,630 --> 00:38:15,069
be nothing but unity this is also turns out
to be unity so this is gone and these what
262
00:38:15,069 --> 00:38:32,410
we are left with okay. Now solving it further
out of these now writing it 1/1-a2 this k!
263
00:38:32,410 --> 00:38:41,369
is
independent of x and y, so this can come out
264
00:38:41,369 --> 00:38:56,750
I can write it here rest everything I will
kept inside n-1!2 can also come out okay.
265
00:38:56,750 --> 00:39:10,130
So but rest everything has to be kept inside.
So I will have ax+y yeah this x! and y! we
266
00:39:10,130 --> 00:39:44,460
cancel it from here so these will not be here
these have gone out and that is what we are
267
00:39:44,460 --> 00:39:55,619
left with okay. So we will now actually split
into two separate parts, so what I will do
268
00:39:55,619 --> 00:40:12,579
is I will take X+y-k which actually means
k s k has to come somewhere here I will put
269
00:40:12,579 --> 00:40:36,259
1-a ,n-1 ľy here so I need to take somewhere
here 1-a, n-1-x this
270
00:40:36,259 --> 00:40:46,380
is because 1-a2 has to come up so these -1
have setting in so I have to remove this okay,
271
00:40:46,380 --> 00:41:09,900
we will also add a factorial here so that
we can solve it later on n-1+x-k so ill put
272
00:41:09,900 --> 00:41:29,960
the same thing in denominator here
and I will keep this n-x so i will take this
273
00:41:29,960 --> 00:41:37,239
part now you can see there in no the x is
something which is the in this summation i
274
00:41:37,239 --> 00:41:44,789
have to solve this thing first with respect
to y okay.
275
00:41:44,789 --> 00:41:50,480
This is the complete binomial you can actually
add these two a and 1-a this terms out to
276
00:41:50,480 --> 00:42:07,230
be n-1-y will cancel with this n-1 +x ľk
okay x+ y-k is here n-1-y is here so this
277
00:42:07,230 --> 00:42:13,819
forms a binomial actually but whatĺs the
range of y in this case so thatĺs a one of
278
00:42:13,819 --> 00:42:21,400
the interesting question so range will be
when x+ y ľk because this is something which
279
00:42:21,400 --> 00:42:25,930
has to be certainly observed otherwise you
know itĺs a binomial, I can simply solve
280
00:42:25,930 --> 00:42:39,220
it by adding a+1-a s something okay.
So when the x +y ľk 0 thatĺs one possibility
281
00:42:39,220 --> 00:42:51,410
okay secondly this can go from 0 to the largest
value which is this one ,so second possibility
282
00:42:51,410 --> 00:43:06,249
is that x+ y-k will become equal to n-1 +
x-k, in this case that k will cancelled x
283
00:43:06,249 --> 00:43:16,230
will cancel y will be equal to n-1 is the
range here so y will be ,so y will range now
284
00:43:16,230 --> 00:43:35,829
n-1 coming from here n y = k-x y ,so infect
by you should note here if y is not greater
285
00:43:35,829 --> 00:43:40,559
then k-x switch cannot be in blocking states
which can only be in blocking states it is
286
00:43:40,559 --> 00:43:47,019
greater then, k-x remember with this is we
did while estimating the Jacobian approximation
287
00:43:47,019 --> 00:43:50,880
.
So this is the range in which y will be valid
288
00:43:50,880 --> 00:43:58,289
so this range has to come here in the summation
and this forms a complete binomial so complete
289
00:43:58,289 --> 00:44:05,529
binomial means this plus this a+1-as power
whole term so I can represent. So I can actually
290
00:44:05,529 --> 00:44:27,190
re write it, a+1-a,n-1+x-k so
this is one 1s power anything going to be
291
00:44:27,190 --> 00:44:34,579
1 so this is a material so this is go away
.only thing is remains its summation with
292
00:44:34,579 --> 00:44:58,019
respective to s and I can rewrite is this
in this case I can actually sum these two
293
00:44:58,019 --> 00:45:06,559
things up and make a add a factorial here
multiplied by that I should now remove this
294
00:45:06,559 --> 00:45:12,849
one.
This will be two end x will cancelled 2n-2-k
295
00:45:12,849 --> 00:45:21,250
thatĺs what it will be , so I have to divide
it also so but this is independent of x I
296
00:45:21,250 --> 00:45:32,859
can write it here now here x will be actually
also going taking up a range so x can take
297
00:45:32,859 --> 00:45:50,809
,now ak going to come out which is a second
term here 1-a this so 1-a this so 1something
298
00:45:50,809 --> 00:46:05,119
is the second term so that second term is
1 2n-2-k this actually it will be n-1+x-k
299
00:46:05,119 --> 00:46:12,339
okay.
So these forms complete binomial so whatĺs
300
00:46:12,339 --> 00:46:25,930
a range of x here so when n-1-x goes to 0
so x will be equal to n-1, thatĺs one possibility
301
00:46:25,930 --> 00:46:48,960
secondly when this term will go to 0 which
is n-1 + x ľk so k=n-1 +x so thatĺs equivalent
302
00:46:48,960 --> 00:46:52,400
to the range in which is going to operate
.
303
00:46:52,400 --> 00:47:03,720
So this also become a full binomial I can
replace
304
00:47:03,720 --> 00:47:47,440
it by 1+1-a 2n-2-k,so which is in case be
the expression for t (coral loss )
305
00:47:47,440 --> 00:47:56,619
and remember this you can only give you a
value if you are going to have k<2n-1,okay
306
00:47:56,619 --> 00:48:22,720
if k> that then it will make since otherwise
it wonĺt and of course one important thing
307
00:48:22,720 --> 00:48:27,609
this is the call loss probability there has
been a another approach by which a computation
308
00:48:27,609 --> 00:48:32,589
can be done for the switch being in the blocked
state okay .
309
00:48:32,589 --> 00:48:42,650
So I have not done that here and if you do
that I will end up in a expression which will
310
00:48:42,650 --> 00:49:06,720
be PB which will turn out to be ak and an
interesting expression it will be n! 2 2-a
311
00:49:06,720 --> 00:49:16,450
2n-k/k!2n-k and you can actually clear the
relationship is still holds the relationship
312
00:49:16,450 --> 00:49:40,170
which we had PBn-1 ,this relationship which
we have derived earlier also holds to here
313
00:49:40,170 --> 00:50:22,089
okay. So that what is the method of estimating
the blocking probability using Carlos approach.