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In this video, we will consider
a classical application

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of Markov chains, which has to
do with the design of a phone

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system.

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This is a classical problem,
which was posed, analyzed,

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and solved by a Danish
engineer by the name of Erlang.

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It was more than 100
years ago when phones just

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started to exist,
but the technique

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remains relevant today to design
systems of a similar nature.

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As for Erlang, he was
trying to figure out

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how to design the capacity
of a phone system.

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That is, how many
lines should we

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set up for a group of
people, say, in a village,

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to be able to communicate
to the outside world?

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So here is a cartoon
of the problem, where

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these are the
phone lines, and we

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need to decide how many of these
lines to set up, let's say,

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B. How to do that?

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Well, we don't want B to be too
large, much more than needed,

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because too many lines
would be expensive.

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On the other hand, we
want to have enough lines

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so that if a reasonable
number of people place phone

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calls during the
same period, they

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will be able to talk and
not get busy signals.

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So if B is 10 and 15 people
want to talk at the same time,

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then 5 would get a
busy signal, and that's

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probably not what you want as
an acceptable level of service.

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So we would like B to be just
large enough so that there

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is a high probability
that no one is

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going to get a busy signal.

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So how do we go about modeling
a situation like this?

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Well, we need two
pieces of information,

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one describing how
phone calls get

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initiated, and once a phone call
gets started, how long does it

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take until it ends?

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We're going to make some very
simple but somewhat plausible

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assumptions.

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We will assume that phone calls
originate as a Poisson process.

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We will assume that
out of that population,

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there is no coordination.

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At completely random times,
people pick up their phone

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independent of each other's.

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Also, there is nothing special
about the various times,

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and different times
are independent.

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So a Poisson model
is a reasonable way

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of modeling a situation
under these assumptions.

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We also assume that
the rate lambda

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is known or has been estimated.

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Now, it might be the case
that during the night,

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the rate would be different
than during the day.

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In that case, you
would design the system

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to meet the largest
rate of the two.

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For the phone calls
themselves, we

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are going to assume that
the duration of a phone call

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is a random variable that has
an exponential distribution

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with a certain parameter mu.

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So 1/mu is the mean
duration of a phone call.

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Duration of phone calls are
independent between each other.

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So here, again, we assume
that the parameter mu

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has been estimated.

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For example, the mean duration
1/mu could be 3 minutes.

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Now, is the exponential
assumption a good assumption?

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So here is the PDF of an
exponential random variable

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with parameter 1 over three.

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That means that the mean
duration is about three minutes

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here.

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So if you look at
this PDF, it means

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that most phone calls
will be kind of short.

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There is going to be a
fraction of phone calls

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that are going to
be larger, and then

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a very small fraction
of phone calls

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that are going to
be even larger.

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So it sounds reasonable.

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However, it's not exactly
realistic in some situations.

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Typically, phone calls that
last a very short time are not

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that common as opposed to what
an exponential distribution

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would indicate here.

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So some other distribution
might be better, like this one,

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for example, here, where during
a very small period of time

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the wait corresponding to
this very short period of time

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are kind of small as well.

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There are many
distributions of this type.

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I've just provided here
one simple example.

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This one is the Erlang
of parameter 2 and 2/3.

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What it means is that
it is the sum of two

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independent exponential
random variables,

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and each one of them
of parameter 2/3.

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So the mean duration associated
with this distribution

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is also 3 minutes.

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So this might fit better
some practical situation.

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But here we will keep the
simple assumption associated

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with an exponential
distribution.

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All right.

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So let's try now to
come up with the models

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that we can decide how many
lines, B, do we want to set up.

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The Poisson process
run in continuous time.

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And call durations being
exponential random variables

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are also continuous
random variables.

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So it seems that we are in
a continuous time universe.

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Here is a cartoon of the
evolution of the system.

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So here I have in blue when
phone calls get initiated.

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So this is called 1, a second
one, a third, a fourth,

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and a fifth one.

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And also, I have represented
here the duration of the call.

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So call 1 lasted that
long, call 2 lasted long

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until here, 3 up to
here, 4 here, et cetera.

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So when you look at this
kind of system in that way,

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and you run through time
in a continuous manner,

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and here you have 0 line busy.

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You have 1 line used, 0,
1, then 2 becomes busy,

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3, 2, 1, and 0, and
so on and so forth.

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Also note that if I look at
that system at any time t,

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because of our assumptions
of a Poisson process

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and an exponential
duration for phone calls,

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and a memoryless
property associated

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with these processes, it
means that the past really

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has no information
about the future.

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And so, in some sense, the
Markov property is valid.

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So it looks like a
continuous time Markov

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process would be needed here.

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And this is indeed an
option, but we have not

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studied those in this class.

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So we will discretize
time instead

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and work with a Markov chain.

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We are discretizing time in the
familiar way, the way we did it

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when we studied the
Poisson process.

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We are going to
take the time axis

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and split it into
little discrete time

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slots, each of duration delta.

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And delta is supposed to be
a very, very small number.

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So now under this
discretization,

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by the definition of
the Poisson process

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the probability that we'll
see 1 arrival during any time

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slots of duration delta
will be lambda times delta.

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Also, if at any time, like here
we have 1 simple call active,

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the probability that this call
will end during any future time

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slot of duration delta
is mu delta, like here.

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Indeed, as we have
seen in Unit 9,

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an exponential
random variable can

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be thought of as
representing the time

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until the first arrival of a
Poisson process with rate mu.

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What if you have i busy
calls at the same time?

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Then the probability
of having 1 call ending

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in a time slot of duration
delta will be i mu delta.

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Like, for example here, this one
could correspond to something

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as 2 mu delta.

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Indeed, each of the Poisson
processes associated

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with these calls
with rate mu can

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be combined into a merged
Poisson process of rate i times

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mu.

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And a call completion will
correspond to the time

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until the first arrival of
this merged Poisson process.

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For example, if I go back here
in my situation here at time t,

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there were still 3
phone calls active.

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And I represent here the
call number 2, call number 3,

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and call number 4 and
their remaining duration.

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And if you look at these
and you combine these 3

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associated Poisson
processes into 1,

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you get a merged
Poisson process.

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And if you look now at the time
arrival of the first event,

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which would
correspond to here, it

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would be an exponential
random variable.

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The duration here
would correspond

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to an exponential random
variable of parameter 3 mu.

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So in that case, if you go back
to that representation here,

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the probability of
a departure would

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be 3 times mu times delta.

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OK?

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So let us continue
with our discrete time

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approximation of our system.

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Again, we have the village,
and the lines, the B

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that we would like to decide.

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We have discretized
the time steps.

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We have made some approximation.

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And we know that during any
of these time slots here,

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the probability that
you would get a new call

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is about lambda times delta.

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Lambda is the rate of
the Poisson process.

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And given that you have
i calls, the probability

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that one of these calls ends
will be i times mu times delta.

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OK.

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If we want to propose a Markov
chain model for this system,

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we need to specify the
states and the transition

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probabilities.

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What are the states
of the system?

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If you look at the system
at any particular time,

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the minimum relevant
information to collect

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would be the number
of busy lines,

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something like these 2 lines are
busy, or all of them are busy,

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or none of them are used.

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Now, because of our assumptions,
again, about the Poisson

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process arrivals and
exponential duration

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of calls and their
memoryless property,

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that information is
enough to fully describe

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the state of our
system in such a way

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that we get a Markov chain.

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So the states are numbers
from 0 to B. 0 corresponds

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to a state in which all
the phone lines are free.

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No one is talking.

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B corresponds to a case where
all the phone lines are busy.

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And then you've got
states in between.

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Now, let us look at the
transition probabilities.

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Suppose that right now,
you are in that state.

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What can happen next?

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Well, a new phone call
gets placed, in which case

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the state moves up by 1.

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Or an existing call
terminates, in which case

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the state goes down by 1.

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Or none of the two
happens, in which case

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00:11:00,930 --> 00:11:03,090
the state stays the same.

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00:11:03,090 --> 00:11:06,990
Well, it is also possible that
a phone call gets terminated,

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00:11:06,990 --> 00:11:10,510
and a new phone call gets
placed in the same time period.

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00:11:10,510 --> 00:11:12,430
But when the duration
of the time slots

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are very, very small,
the delta here,

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this event is going to have
a negligible probability,

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00:11:19,180 --> 00:11:21,800
order of delta squared.

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00:11:21,800 --> 00:11:26,500
So we ignore it, as
we ignore the fact

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00:11:26,500 --> 00:11:29,160
that more than one
new call can happen,

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or more than one call can be
terminated during a given slot.

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So what is the probability
of an upward transition?

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00:11:37,470 --> 00:11:40,610
That's the probability that the
Poisson process has an arrival

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00:11:40,610 --> 00:11:42,900
during the slots
of duration delta.

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00:11:42,900 --> 00:11:46,850
And as we have seen, this
is lambda times delta.

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So each one of these
upward transitions

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has the same probability
of lambda times delta.

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How about phone
call terminations?

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00:11:58,520 --> 00:12:02,000
If we have i phone calls
that are currently active,

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00:12:02,000 --> 00:12:07,480
the probability that one of them
terminates becomes i mu delta.

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00:12:07,480 --> 00:12:13,140
So here it would be mu
delta, and here B mu delta.

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Now, let us analyze this chain.

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It has the birth and
death form that we

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discussed in the
previous lecture.

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00:12:19,710 --> 00:12:22,560
So instead of writing
down the balance equation

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00:12:22,560 --> 00:12:24,850
in a general form,
we think in terms

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00:12:24,850 --> 00:12:27,980
of frequency of transitions
across some particular cut

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00:12:27,980 --> 00:12:30,760
in this diagram, so
for example here.

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00:12:33,320 --> 00:12:36,970
The frequency with which
transition of this kind

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00:12:36,970 --> 00:12:40,190
happen or are observed
has to be the same

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00:12:40,190 --> 00:12:43,740
as the frequency of
transition of this kind.

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00:12:43,740 --> 00:12:46,290
The frequency of
transition of this type

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00:12:46,290 --> 00:12:52,060
will be, if you look at pi i
here and pi of i minus 1 here,

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00:12:52,060 --> 00:13:01,910
this transition here will happen
with pi i times i mu delta.

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00:13:01,910 --> 00:13:06,510
And the transition
of this type here

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00:13:06,510 --> 00:13:13,620
will be pi i minus 1
times lambda times delta.

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00:13:13,620 --> 00:13:16,230
And the frequency
of these transitions

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00:13:16,230 --> 00:13:19,400
have to be the same as the
frequency of these transitions,

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00:13:19,400 --> 00:13:21,520
so we have that equals that.

247
00:13:21,520 --> 00:13:25,770
And then we can cancel
the delta in both,

248
00:13:25,770 --> 00:13:28,230
and we are left with
this equation here.

249
00:13:31,980 --> 00:13:38,340
So this equation expresses pi of
i in terms of pi of i minus 1.

250
00:13:38,340 --> 00:13:43,900
So if we knew pi of 0, then
we can calculate pi of 1,

251
00:13:43,900 --> 00:13:48,960
and then in turn calculate pi
of 2, and so on and so forth.

252
00:13:48,960 --> 00:13:51,330
And the general formula
that comes out of this,

253
00:13:51,330 --> 00:13:55,020
after some algebra, is
given by this expression,

254
00:13:55,020 --> 00:13:57,200
which involves pi of 0.

255
00:14:00,260 --> 00:14:01,860
Now, what is pi 0?

256
00:14:01,860 --> 00:14:06,715
Well, we can find it by using
the normalization equation,

257
00:14:06,715 --> 00:14:10,620
the summation of pi i equals 1.

258
00:14:10,620 --> 00:14:13,680
You use this normalization,
replace each pi

259
00:14:13,680 --> 00:14:17,790
i by their quantities
as a function of pi 0,

260
00:14:17,790 --> 00:14:23,000
and then we obtain
this equation for pi 0.

261
00:14:23,000 --> 00:14:25,450
So here, again, we use
that normalization.

262
00:14:25,450 --> 00:14:28,090
We replaced pi i by their value.

263
00:14:28,090 --> 00:14:31,610
We sum to 1, and
we obtain pi of 0.

264
00:14:31,610 --> 00:14:34,070
And then in turn,
from this pi of 0,

265
00:14:34,070 --> 00:14:37,430
you can replace the
pi of 0 in pi of i,

266
00:14:37,430 --> 00:14:45,230
and you obtain a pi of i as a
function of B, lambda, and mu.

267
00:14:45,230 --> 00:14:49,180
So if we know B
and lambda and mu,

268
00:14:49,180 --> 00:14:54,660
we can set up this Markov chain,
and we can calculate pi 0,

269
00:14:54,660 --> 00:14:58,410
and then pi of i for all i's.

270
00:14:58,410 --> 00:15:00,775
We can then answer a
question like this.

271
00:15:00,775 --> 00:15:04,120
After the chain has
run for a long time,

272
00:15:04,120 --> 00:15:07,380
how likely is it that at
any given random time,

273
00:15:07,380 --> 00:15:10,740
you will find the system
with i busy lines?

274
00:15:10,740 --> 00:15:13,950
Well, it will be pi of i.

275
00:15:13,950 --> 00:15:17,010
And also, we can interpret
the steady-state probabilities

276
00:15:17,010 --> 00:15:18,360
as frequencies.

277
00:15:18,360 --> 00:15:21,520
So once I found pi of
i, it also tells me

278
00:15:21,520 --> 00:15:25,770
what fraction of the time
I will have i busy lines.

279
00:15:25,770 --> 00:15:29,710
And you can answer that
question for every possible i.

280
00:15:29,710 --> 00:15:32,770
Now, we were initially
interested in the probability

281
00:15:32,770 --> 00:15:36,160
that the entire system is
busy at any point in time,

282
00:15:36,160 --> 00:15:39,350
in other words, in
that state here.

283
00:15:39,350 --> 00:15:42,160
So if a new phone
call gets placed,

284
00:15:42,160 --> 00:15:45,390
it is going to find the
system in a random state.

285
00:15:45,390 --> 00:15:47,830
That random state is
described in steady-state

286
00:15:47,830 --> 00:15:50,270
by the probability pi's.

287
00:15:50,270 --> 00:15:54,020
And the probability that
the entire system is busy

288
00:15:54,020 --> 00:15:59,270
is going to be given by pi of
B, and this is the probability

289
00:15:59,270 --> 00:16:03,740
that we would like to be small
in a well-engineered system.

290
00:16:03,740 --> 00:16:08,900
So again, given lambda,
mu, the design question

291
00:16:08,900 --> 00:16:14,040
is to find B so that this
probability is small.

292
00:16:14,040 --> 00:16:17,190
Could we figure out a
good value for B by doing

293
00:16:17,190 --> 00:16:19,760
a back-of-the-envelope
calculation?

294
00:16:19,760 --> 00:16:27,630
Well, let's suppose that
lambda is 30 calls per minute.

295
00:16:27,630 --> 00:16:31,180
And let's assume
that mu is 1/3 so

296
00:16:31,180 --> 00:16:34,350
that the mean
duration is 3 minutes.

297
00:16:34,350 --> 00:16:38,870
So on average, a call
lasts for 3 minutes,

298
00:16:38,870 --> 00:16:42,440
and you get 30 calls
on average per minute.

299
00:16:42,440 --> 00:16:45,390
Then how many calls would
be active on the average?

300
00:16:45,390 --> 00:16:48,590
If a call lasted exactly
1 minute, then at any time

301
00:16:48,590 --> 00:16:51,500
you would have 30
calls being active.

302
00:16:51,500 --> 00:16:54,270
Now, a call lasts, on the
average, for 3 minutes.

303
00:16:54,270 --> 00:16:56,562
So by thinking in
terms of averages,

304
00:16:56,562 --> 00:16:58,020
you would expect
that, at any time,

305
00:16:58,020 --> 00:17:05,490
there would be about 90 calls
that are active, 3 times 30.

306
00:17:05,490 --> 00:17:08,180
And if 90 calls are
active on the average,

307
00:17:08,180 --> 00:17:14,720
you could say, OK, I'm going
to set up my B to be 90.

308
00:17:14,720 --> 00:17:17,420
But that's not very good,
because if the average number

309
00:17:17,420 --> 00:17:21,480
of phone calls that want to
happen is, on the average, 90,

310
00:17:21,480 --> 00:17:23,740
sometimes you are
going to have 85,

311
00:17:23,740 --> 00:17:26,319
and sometimes you'll get 95.

312
00:17:26,319 --> 00:17:28,830
And to be sure that the
phone calls will go through,

313
00:17:28,830 --> 00:17:30,630
you probably want
to choose your B

314
00:17:30,630 --> 00:17:34,100
to be a number a
little larger than 90.

315
00:17:34,100 --> 00:17:36,030
How much larger than 90?

316
00:17:36,030 --> 00:17:40,820
Well, this is a question that
you can answer numerically.

317
00:17:40,820 --> 00:17:43,550
By looking at these
formulas, if you

318
00:17:43,550 --> 00:17:48,210
decide that your acceptable
level of service, pi of B,

319
00:17:48,210 --> 00:17:52,190
has to be less than
1%, then you will

320
00:17:52,190 --> 00:17:58,820
find that the B that you need
to design is to be at least 106.

321
00:17:58,820 --> 00:18:02,720
So you actually need some margin
to protect against a situation

322
00:18:02,720 --> 00:18:04,990
if suddenly, by
chance, more people

323
00:18:04,990 --> 00:18:06,910
want to talk than
on an average day.

324
00:18:06,910 --> 00:18:08,950
And if you want to
have a good guarantee

325
00:18:08,950 --> 00:18:11,830
that an incoming person will
have a very small probability

326
00:18:11,830 --> 00:18:14,690
of finding a busy
system, here 1%,

327
00:18:14,690 --> 00:18:18,840
then you will need
about 106 phone lines.

328
00:18:18,840 --> 00:18:20,950
So that's the calculation
and the argument

329
00:18:20,950 --> 00:18:23,730
that Erlang went
through a long time ago.

330
00:18:23,730 --> 00:18:26,600
It's actually interesting that
Erlang did this calculation

331
00:18:26,600 --> 00:18:28,994
before Markov chains
were invented.