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PROFESSOR: OK.
So let's get started.
We want to first review Wald's equality a little bit.
Wald's equality is very tricky thing.
If you think you understand it, you will go along.
And at some point, you will be using it and you will say, I
don't understand what this says.
And that will happen for a long time.
It still happens to me occasionally.
What happens is you work with it for
longer and longer times.
The periods when it becomes confusing become rarer.
And the expected time to straighten
it out becomes smaller.
It is a strange kind of result.
So we started out with a stopping trial definition.
J is a stopping trial for a sequence of random variables.
If it's a random variable and it has the property that for
each n greater than or equal to 1, the indicator random
variable indicator of J equals n is a function
of X1 to X sub n.
In other words, the decision of whether to stop at time n
is a function of 1 up to n.
A possibly effective stopping trial is the same, except that
might be a defective random variable.
And the reason you want to have possibly defective random
variables is that before you start analyzing something that
might be a stopping rule, you generally have no way of
knowing whether that actually is a stopping rule or whether
it's a detective stopping rule.
So you might as well just say it's a defective stopping rule
to start with and show that it's not.
Then from that, we went on to Wald's equality which added
the condition that X sub n that this is based on is a
sequence of IID random variables.
To be a stopping trial, you don't
need IID random variables.
You don't need any restriction at all, other than the fact
that you can make a decision on when to stop based on what
you've already seen.
That's the only condition there.
Wald's equality is based on this extra condition that the
random variables are IID.
Each of them, with some mean, X bar.
If J is a stopping trial and as the expected value of J is
less than infinity, then the sum S sub J at the stopping
trial, J, satisfies this relationship here.
And remember we proved that last time.
And the key to proving it, the hard part of it--
well, it's not hard if you see it.
But what's difficult is when you start looking at how you
find the expected value of S sub J that's equal to a sum
over n of X sub n times the indicator function of J being
greater than or equal to n.
Now the condition here is that this indicator at J equals n
is a function of these quantities here.
When you add the IID quantity down here, what you find then
is the indicator function for J greater than or equal to n.
It's also the indicator function for J 1 minus the
indicator function for J less than n.
It's independent of X1 up to X sub blah, blah, blah.
That indicator random variable is then a function only of X1
up to X sub n minus 1, because it's 1 minus the indicator of
J less than or equal to n minus 1.
J less than n, which means that X1 up to X n minus 1 is
independent of Xn.
And X1 up to X n minus 1 is what determines this indicator
function J less than n.
It's the J less than n which is important.
So we got that Wald's equality.
What I want to do today to start off with is to do the
elementary renewal theorem, which is a strange result.
Wald's equality, you can use it to determine the expected
value of N of t.
Now why do we want to determine the expected
value of N of t?
We already have shown in great generality that there's a very
nice, with probability, one type limit theorem associated
with N of t over t.
We know that N of t over t approaches t
with probability 1.
Namely, all the sample functions, except a set of
probability 0, all approach this same
constant, 1 over X bar.
So it seems that we know everything we want to know
about the expected value of N of t, about N of t, and
everything else.
But no.
And there are two reasons why you want to do
something more than that.
One of them is that very often you want to know how N of t
over t approaches the limit of 1 over X bar.
The other is that sometimes you really are interested in,
what is N of t at some finite value of t?
N of t is a random variable, some finite value of t.
So you can't evaluate it.
But at least you would like to know what its
expected value is.
You might want to know what it's variance is too.
But people who study renewal theory have spent an enormous
amount of time on trying to find the expected
value of N of t.
It's the basic problem that people work on all the time.
The elementary renewal theorem is something which says a
little more for finite times.
And it actually says the expected value of N of t over
t and the limit is equal to 1 over X bar.
Sounds like much less than what we've done before.
And perhaps this is because this was a computational thing
that people could do without computers, before computers
came along.
And all the work on renewal theory went on basically from
1900 until about 1970.
People didn't have any computers to compute things.
And all the mathematicians who were interested in this field
really loved to compute ugly things.
And they computed ugly things all the time.
And the expected value of N of t was one of the ugly things
that they could really do a lot with.
So it probably has more importance in this field than
it deserves.
But it does have some importance.
Anyway, what we want to do is to get some idea of why this
elementary renewal theorem is true.
Later on we will study how to actually find the expected
value of N of t.
N of t is the number of arrivals that have occurred up
to and including t.
So N of t plus 1 is the number of the first
arrival after t since--
that should be a t there--
the expected value of N of t is finite, the expected value
J is also finite.
In other words, we're defining J to be N of t plus 1.
We want to show the J is actually a stopping trial.
And therefore we can use Wald's equality on it.
Let's stop and think for a minute.
Why don't we just define a stopping trial to be the
number of arrivals that have occurred up until time t?
What's the matter with that?
If I let you observe a sequence of these IID random
variables, the first inter-arrival interval, the
second inter-arrival interval, the third inter-arrival
interval, and so forth, and I say stop when you have the
last dozen that's less than t.
The question is, is that a stopping trial?
You have to be able to determine that just from what
you've seen up until the present.
But when you see that last arrival, there's no way you
know that it's the last arrival less than t unless you
see the arrival after that, also to see whether that's
greater than t.
So the only way you can define a stopping trial of this form
is to define the stopping trial not at N of t, but of N
of t plus 1.
So now we want to apply Wald's equality to the expected value
of the time at which N of t plus 1 arises.
Do you see now why we want to call this a stopping trial
instead of a stopping time?
If you call that a stopping time, you want to know what
you were doing at this point.
Because the stopping trial says you're looking at the
first arrival that occurs after time t.
Then you're looking at the time at which
that arrival occurs.
Wald's equality says the expected value of that time is
equal to the expected value of the inter-arrival time times
the expected value of J. J is a stopping trial.
The expected value of J is just--
J is N of t plus 1.
So that's expected value of N of t plus 1 times X bar.
That's simple.
It's simple until you try to recreate that argument.
I would suggest to all of you that sometime later today you
go back and try to recreate this argument.
And if you can do it, then you understand it.
If you can't, think about it a little more.
Because it is deceptively simple.
So that was a relationship we had.
Wald's equality then is relating two unknown
quantities.
It's relating the expected value of N of t with the
expected value of the first arrival time after t.
That's one of the troubles with Wald's equality.
It relates two things that, neither of which you know in
many cases.
There are many situations in which you do
know one or the other.
We talked about one last time.
But in most of them, you don't know either of them.
So it doesn't help you immediately.
You have to find something else.
So what we're going to do here is say, well we can't find
expected value of N of t from this.
But we can at least bound it.
And we can bound it by saying that S sub N of t plus 1.
This is the time of the first arrival after t.
The first arrival after t has to be at a time greater than
or equal to t.
So expected value of S sub N of t plus 1 also is greater
than or equal to t.
So expected value of N of t, solving this equation is
expected value of S sub N of t plus 1 over X bar minus 1.
And then you lower bound expected value of this first
arrival after t by t.
So the expected value of N of t has to be greater than t
over X bar minus 1.
Well this seems to be useful, because if what you're
interested in is the expected value of N of t over t.
And that's what we're interested in for the
elementary renewal theorem.
You're saying that that's greater than 1 over X bar
minus 1 over t.
And when t gets large, that 1 over t term gets very small.
What that means is you have a lower bound on expected value
of N of t, which is going to 1 over X bar.
The elementary renewal theorem then says, let X bar be the
mean inter-renewal.
Then the limit of t approaches infinity of the expected value
of N of t over t is equal to 1 over X bar.
Very weak result.
At least it seems to be a very weak result.
But we need an upper bound on the expected value of N of t.
We don't have any upper bound on this quantity here.
Because you remember these--
Actually S sub N of t plus 1 minus t is this residual
life at time t.
And we found that the residual life at time t can be pretty
badly behaved.
So we're in trouble.
So what do we do?
Well the argument is done carefully in the notes.
It's very discouraging to me, because this should be a
simple result.
And here this argument is incredibly tricky.
So we're working very hard to do something that's not--
Well, it seems like it ought to be elementary.
It is an elementary result in terms of its usefulness.
It's a very hard result in terms of trying to prove it.
So the way you prove this is first you truncate the random
variable X. Namely instead of having a random variable with
this arbitrary, non-negative probability distribution, what
you do is you look at another random variable, which has the
same distribution up until some large value B. And then
you truncate it at that point.
So the distribution function looks like--
Here's f of x, however it might look.
This is 1 here.
This is supposed to be going up towards 1.
And what you do is you truncate it at some point B.
So this is a truncated version which stops at B. And this
here is the actual Fx of x.
So first you truncate the random variable.
Then you prove the elementary renewal theorem for that
truncated random variable, which is easy.
Because at that point, the expected value of S sub N of t
plus 1 is less than or equal to t plus B.
And then the next thing you have to do is monkey around
with the difference between X bar and X tilde bar.
And you have to go through a bunch of strange inequalities.
You have to then let B approach infinity as t is
approaching infinity in just the right way.
And when you get all of done doing this, bingo.
You get the elementary renewal theorem.
You see why I don't want to go through that in class.
And you see, I hope, why probably most of you will not
want to go through it in detail either.
Because if you want to learn how to do truncation
arguments, this is probably not the best one to use as a
way of figuring out how to do that.
If you want to learn how to use truncation arguments, do
the Weak law of large numbers where you
don't have a variance.
It's done back in chapter one.
It's a nice relatively simple argument where you truncate a
random variable.
This one is just peculiar.
We do all of that with stopping trials as we've
defined them.
Very often, and particularly when you get to queuing
theory, you want to define stopping trials in a more
general way.
And here's a more general definition.
A generalized stopping trial for a sequence of pairs of
random variables, X1, V1, X2, V2, X3, V3, is a positive
integer random variable such as for each n greater than or
equal to 1, the indicator function, which tells you
whether to stop or not, is a function of X1, V1, X2,
V2, up to Xn, Vn.
In other words, you don't have to decide whether to stop or
not in terms of these random variables X1 to Xn that you're
interested in.
You have this other set of random variables too.
Typically in queuing situations, these other random
variables that you're interested
in are service times.
And the X's are arrival times.
And these things are going on in parallel with each other.
And still, you can do the same arguments about stopping
trials by looking at just the past history of the input
random variables and these other random variables,
whatever they happen to be.
Wald's equality then says that the expected value of S sub n
is equal to X bar times the expected value J, where Sn is
the sum of the X's.
And Wald's equality holds by exactly the
same proof as before.
There's nothing new.
You just look at the proof and you say, OK.
All these V's are involved there also.
You can replace each of Vi in this definition with a whole
vector of random variable.
So you can make this as general as you
want to make it.
In fact, a lot of people prove Wald's equality by skipping
all of this stuff about stopping trials.
And what they say is that the rule for stopping at time J is
independent of all of the future arrivals.
And that lets you prove the theorem immediately.
It doesn't give you any clue at all as to when that holds,
so that in fact you're taking Wald's equality and making it
essentially a truism.
And you're avoiding the real problem, which is to know when
you have a stopping rule and when you don't
have a stopping rule.
But anyway, we can generalize it in this way.
And now we want to use this to look at a fairly general
queueing situation.
We talked about the G/G/M queue before.
Let's just talk about the G/G/1 queue at this point.
This is a queue with general independent IID and her
arrival times, IID service times.
So what you wind up with is at time 0, you assume that
there's an arrival at time 0.
You don't count it as part of the arrival renewal process.
So this arrival is here.
That arrival goes into service.
There's some service time V sub 0.
At some inter-arrival time, X1, another arrival comes in.
In this particular sample function here, the second
arrival comes in before this 0 arrival that we didn't account
finishes service.
And I'm sorry.
I should have--
It might make better sense to call this X1 and call this V1.
It would make all the notation simpler.
Or call this X0 and call this V0.
But unfortunately that's not consistent with any of the
notation that anybody uses for talking
about renewal processes.
Because you want these X's to be the elements
of a renewal process.
You want the V's to be IID random variables.
And this, unfortunately, is the way it has to be.
So these arrivals come in.
This is the inter-arrival time before--
This is S1 here, which is the time of the first
arrival after 0.
This is S2, which is the time of the first
arrival after time 0.
And when you count the arrival at time 0 in, this function
here is N of t plus 1.
So this is N of t plus 1 going along here.
This is the arrival process plus 1, where now we've
counted this arrival at time 0.
You have these services going on.
V0 for this particular sample function is the time of the
service of this 0 arrival.
V1 is the time of the service of this X1--
well, it's the service time of the arrival that has come in
at this point.
This is the service time of the arrival at time 0.
This is the service time of the arrival at this time.
This as the service time of the arrival at
this time and so forth.
I want to make sure that you understand this model here
before we go on.
Because if you don't understand this, you won't
understand anything we do after this.
If you look at this now, you see a departure process which
is going on also.
These departure times are determined partly as a sum of
the number of customers to have departed by time t.
And this example here is just the sum of V0 plus V1 plus V2.
So it's almost the same as this renewal process here,
except here you're talking about a renewal process of
dealing with the departure times.
This changes any time that the queue empties out.
Because when the queue empties out, the server isn't doing
anything for a while.
So suddenly, the number of departures slows up.
And in some sense, you reset things here.
What I want to do in this argument today is to explain
much more carefully why you can actually restart things
when this new arrival occurs.
So I want to say that, in fact, this is a renewal time
for this entire queuing process.
These are renewal times for the arrivals
to the queuing process.
This thing here, the next arrival to an empty server and
so forth, are all, in fact, renewals of the
entire queuing system.
We want to understand why that is.
We want to use this idea of stopping trials
to understand this.
So look at the first arrival to start a new busy period.
Well, think of that as a generalized stopping trial.
Why is it generalized stopping trial?
Well the sequence of paired random variables we want to
look at is X1, V0, X2, V1, X3, V2, and so forth.
And stopping at J equals 3 is actually a function of this
first pair, the second pair, and the third pair, which is
the restriction we need for a stopping trial.
So the stopping trial stop when the first arrival comes
to an empty system is in fact a function of the arrivals and
as the departure intervals up until that time.
Wald's equality holds because, in fact, every time you get a
new arrival, it's independent of all the old arrivals.
It's independent of all the old service times.
It's independent of all the new service
times also, sort of.
But you have to be more careful with that.
But in fact, the service times are all IID.
So that works too.
So Wald's equality holds here.
But that's not what we're interested in here.
What we're interested in is trying to show that you do get
renewals at these times here.
So let's try to argue why that is.
Did I--
AUDIENCE: Question?
Do V's have to be independent somehow?
PROFESSOR: What?
AUDIENCE: Do the V's have to be independent somehow?
PROFESSOR: Yes.
Yes.
AUDIENCE: So they have to be independent from one another?
PROFESSOR: Yes.
AUDIENCE: It was the case here, you said.
PROFESSOR: Yes.
That's what you mean by a G/G/1 queue.
A G/G/1 queue is--
AUDIENCE: You have a service time and the arrival.
PROFESSOR: Service time.
AUDIENCE: They're both independent.
PROFESSOR: They're both independent, yes.
AUDIENCE: And that makes the departure?
PROFESSOR: Right.
And in fact, the first G says that the arrivals are IID,
have an arbitrary distribution.
The second G says that the service times are IID, but
have an arbitrary distribution.
And then the 1, our the M, or whatever, says how many
servers there are.
AUDIENCE: And that makes the departures independent?
PROFESSOR: Yeah.
So you can really have much more general queuing
situations than this.
And you often do.
Yes?
AUDIENCE: With the departure process, it's not a renewal?
PROFESSOR: What?
AUDIENCE: The departures.
PROFESSOR: The departure process is not a renewal
process, because every once in while, the
server stops doing anything.
What?
AUDIENCE: That's how I got confused actually.
Because every once in a while, the server somehow--
PROFESSOR: Every once in a while, the
server stops working.
But V1, V2, V3 are defined as the service time
of the first customer.
Well actually V0, the service time of the zeroth customer.
V1 is the service time of the first customer.
Namely the time from when the customer enters service to
when the customer's finished with service.
And the reason why these things--
If you looked at the sequence V0, V1, V2, V3, you could call
that a renewal process.
But it wouldn't be a renewal process of any interest to
you, because it would be a renewal process where the
renewals occur at V0, at V0 plus V1, at V0
plus V1 plus V2.
And those might not be the time when these customers are
finishing service.
Because there are these idle times to take
into account also.
If you didn't have the ideal times to take into account,
you wouldn't have to think about queuing theory at all.
Because all you'd have is two separate and independent
renewal processes.
So this is where all of the gimmickry in all of queuing
theory comes from.
But what I wanted to come back and say that I didn't say very
well before.
In the notes, the assumption is that these
pairs here are IID.
What you really would like is something a little more
general than that, which says that each Xi is independent of
all of the earlier pairs.
So that that's giving you a little bit of flexibility for
how these other random
variables impact the situation.
And in fact, they can be anything at all, just so long
as you have this condition that each Xi is independent of
all these past things here.
So we said that Wald's equality holds.
But the other thing, which is what we really want, is the
new arrivals.
Namely the arrival after S3, which comes at S3 plus X3.
That's X sub J plus 1, X sub J plus 2, and so forth.
And the service times, in this case V sub J, is the service
time that occurs after this.
V sub J plus 1, and so forth, are all independent of the old
arrivals and the old service times.
So in fact, this is actually saying that everything that
happens after this J's arrival is independent of everything
that happens before.
And when I say everything, you have to be careful about that,
because we're talking about all the inter-arrival times
that occur after this.
And we're talking about all of the service times that occur
after this.
We're not talking about things like S sub n plus 1, which is
the time of the next arrival, because that's this arrival
time plus that next inter-arrival time.
So each of these intervals are in fact
independent of each other.
And what we've done here is use this idea of a stopping
trial not to define when we stop this whole process, but
to define when a renewal occurs.
In other words, we're using it to define when the old
inter-renewal period ends and a new one begins.
This can be called a stopping trial, if in fact you stop the
whole process at that point and then
don't do anything further.
So that's what's nice about this idea of stopping trials.
They let you not only do Wald's equality.
But they also let you actually see why it is that these
arrivals that start after this point are independent of the
arrivals and departures before that.
So let me just try to say that again.
The stopping rule is the index of the first arrival in a new
busy period.
The arrivals and departures in the new busy period are
independent and identically distributed to
those in the old.
Thus the intervals between new busy periods
form a renewal process.
And that's the thing that we're going to use in
everything else we do here.
So we have one renewal process, which is embedded in
another renewable process.
The outside renewal process of the arrivals are what you
start out with.
These embedded renewals are, in fact, functions of both the
arrival process and the service distribution.
So that's a more complicated thing.
But at least you now know that that's an arrival process.
It says that you can analyze any one of these queuing
systems or any considerably broader type of queueing
system, which has the same property of having renewals
every once in a while, by looking at what happens within
a renewal period.
Then using what happens within a renewal period and applying
one of these limit theorems on the embedded renewals to get
some time average overall time.
So that's what we're really up to here.
This same analysis applies to G/G/m.
Applies to lots of queueing systems.
You have to look at a queuing system and
see whether it applies.
To get some idea of how you can see that without going
through this whole analysis, suppose that you decided to
try to call this point here, at which the system first
became empty the end of a renewal period.
Why couldn't you do that?
Why is this not a renewal period?
Well, is it a stopping trial?
It certainly isn't a stopping trial for the X's.
Because at this time when this last departure has left, you
have no idea how long this new inter-arrival period that's
going on here is going to be.
That has some arbitrary distribution.
All we know at this point is that it's longer than this.
And if you had a busy period end at some point close to the
beginning of this inter-arrival period, you have
a different distribution from when it occurs close to the
end of an inter-arrival period.
Well, because these inter-arrivals are not
memoryless.
They have memory which says their conditional distribution
depends on how long they've been running.
So this is the only sensible place you can define a renewal
process here.
So let's go on.
There is something called Little's theorem, which was
invented by John Little not all that long ago.
And Little's theorem is curious, because you can view
it as being either trivial or non-trivial.
I will try to convince you that it's trivial and also
convince you that it's non-trivial.
And what I mean by that is that it's trivial in terms of
trying to use it someplace.
It's then nontrivial to try to justify that you
can actually do that.
But you get the idea of doing it in many, many places.
We're going to assume an arrival at time 0, like we've
been doing before.
Service process can be almost anything.
But let's assume a G/G/1 queue to be specific.
The system empties out eventually with probability 1.
And assume that it restarts on the next arrival.
We've seen that intervals between restarting form a
renewal process for the G/G/1 queue.
And we've argued that it forms a renewal process for an even
broader class of queues.
That's normally.
Who can tell me when you don't get a renewal process when you
have a G/G/1 queue?
We made an assumption here without talking
at all about it.
And what's that assumption?
What?
AUDIENCE: Starting with an empty queue.
PROFESSOR: Yes.
Every once in a while, you get an empty queue.
If you have a system where service times are greater than
inter-arrival times, what's going to happen is that
arrivals keep coming in, eventually
building up more and more.
The queue gets longer and longer.
And you never have any renewals occurring.
So one thing we didn't say here is we're talking about
G/G/1 queues and more general queues, and the more important
case where the queue sometimes empty out.
We want to use this to analyze when they
actually do empty out.
So here's the picture.
The same kind of picture as before of an arrival process.
And every time we're talking about these kinds of things,
these queuing situations, we will distinguish the arrival
process, which is a renewal process.
But when we talk about the renewal process, we'll talk
about this renewal process which starts on arrivals to an
empty system.
So here's a renewal.
Here's a renewal, and so forth.
Suppose we look at the difference between A
of t and D of t.
A of t is the number of arrivals that have come in up
until time t, counting this arrival at time 0.
D of t is the number of departures that have occurred.
What is the difference?
The difference, L of t, is the number of customers in the
system at time t.
And one of the things you'd like to be able to analyze in
a queuing system is, what's the distribution?
What do you know about how busy the queue is?
If you build storage for 100 customers, you would like to
know what's the likelihood that that
storage will fill up.
And you have to drop customers on the floor or do whatever
you do to customers when the queue is full.
So L of t is clearly something we want to look at.
It's the number of customers in the system at any time.
Not the number of customers waiting in queue, but the
number of customers both to waiting in queue and being
served currently.
We can view this as a generalized renewal reward
function at this point, because this number in the
system at any given time t, is a function what happens within
this inter-renewal period here.
It's a function of these arrivals within this
inter-renewal period and these departures within this
inter-renewal period.
So we can use renewal reward theory.
The total reward with an inner renewal period is the integral
of L of tau over that period.
This is the way we found--
Every time we looked at renewal reward theory, we
looked at the reward within one inter-renewal period.
And then we tried to look at what happened when we started
to look at many, many inter-renewal periods.
So we want to look at L of tau integrated
over one renewal period.
And then we want to go on to see how we apply renewal
theory over that.
So in each inter-renewal period, the aggregate number
and queue number in the system integrated over the duration
of that inter-renewal period is this integral of L of tau.
Here I've stated Little's theorem really.
Let's go back and see where that comes from.
If I want to look at the integral of L of t over this
period of time, what is it?
I can view it as W1 plus W2 plus W3.
That's the easy way to integrate this function.
You start out here.
You're integrating up to here.
1 times D tau.
And you have 2 times D tau.
Then 1, then 2, and then 1 again.
But the easy way to integrate it is just say the integral is
W1 plus W2 times this height, plus W3.
For those of you who have studied Lebesgue integration,
it's the idea of a Lebesgue integral as opposed to a
Riemann integral.
If you haven't studied it, don't worry about it.
It's just a trivial idea that you can integrate this way as
well as integrating this way.
That's the crux of Little's theorem.
The crux of Little's theorem is this equality between the
integral of L of t and the sum of the waiting
times of each customer.
W1 is the waiting time of customer--
Well it's the waiting time of customer 0 in this case.
W2 is the waiting time of customer 1, from the time it
comes in, it's waiting in queue until this time.
Then it starts service.
Next customer comes in here.
Waits in queue.
Finally starts service.
Then the system is empty.
And you integrate L of t from here up to here.
You sum the W's.
And that's what I did in the next slide.
And I can integrate over as many inter-renewal periods as
you want to integrate over.
And what's going to happen?
I'm adding up these individual inter-renewal periods.
I'm adding up this integral over each of them.
I'm also adding up the sum of the waiting times over that
larger interval.
And this equality still occurs.
Any time I integrate over an integer number of
inter-renewal periods, I have this equality between the
integral of L of t and the sum of W sub i's.
One of the things that we observe from renewal theory is
when you start taking limits between 0 and infinity, one
inter-renewal period doesn't make any difference.
When we go on through the mathematics there, we've done
it carefully.
We've upper bound it.
And we've lower bound it.
And we've then shown that it doesn't make any difference.
But at this point, you know that it doesn't make any
difference.
So that you know that when you take the time average, the
time average number in the system is equal to the limit
as t goes to infinity of 1 over t times i equals 1 up to
the number of arrivals up until time t times W sub i.
When you take that limit, it's W sub i over A of t times the
limit of A of t over t.
Both of these go to a limit with probability 1, because
this is just a strong law of large numbers again.
This is just the renewal theorem again.
And that limit here is the time average of W. It's a
little awkward calling this a time average.
Because what this really is is a sample path average.
That's a sample path average which exists
with probability 1.
This is a sample path every over any old sample path you
want to look at of the time average over all of the
customers that come in.
You add up all the customers that come in and you divide by
the number of customers.
And that gives you the time average delay that a customer
experiences.
So what Little's theorem says is that the time average of
the expected number of customers in the system at a
given time is equal to lambda, which is the arrival rate--
A of t over t, is arrival rate--
times the expected delay that each customer experiences.
That's a very useful relationship.
Like all of the things that come from the Wald equality,
it only gives you a relationship between two
things that you don't know.
But one relationship between two things you don't know is
better than no relationships between two
things you don't know.
Here we have this one useful relationship.
We can understand intuitively what's going on here if we
look this diagram again.
If you look at the diagram again, if you expand this
whole diagram in t, if you measure it in hours instead of
measuring it in milliseconds, for example.
This whole thing spreads out enormously.
And what happens?
The W's all get multiplied by a very large amount.
A of tau stays the same.
Lambda, the expected arrival rate,
becomes very much larger.
Arrivals per hour rather than arrivals per millisecond.
So you see that at least as far as scaling is scaling,
there's no way of relating this to this without having
something there, which is something per unit time.
This is time.
This is a time average.
So we have Little's theorem.
Useful in many, many situations.
It's useful as an approximation in many places
where it doesn't hold.
But the right way to look at this, this is really an
accounting identity.
There was this first thing we brought out, that this
integral of L of t over 1 inter-renewal period was equal
to the sum of the waiting times over
that one renewal period.
All of the mathematics and all of the stuff we've been
struggling with have all been concerned with showing that
when you go to the limit as t goes to infinity, all of these
things make sense.
Question is not whether L equals lambda W, but whether
these quantities exist as sensible time averages or is
limiting ensemble averages.
What we're dealing with now is time averages.
But you can do all of these same things with limiting
ensemble averages.
And hopefully, you get the same answer when you do it.
And you have to go through a lot of analysis in order to
show that these are the same.
But the idea that L is equal to lambda W is, in some sense,
much more simple and fundamental than that.
Last thing I want to do is talk about the
Pollaczek-Khinchin formula for M/G/1 queues.
The thing that you get here is you improve on Little by one
step, because now if you have arrivals that are coming in a
Poisson way--
Yes?
AUDIENCE: I'm sorry.
I just have a question.
Is there an easy way to tell when we have the condition we
need with probability 1 that restarts?
That the system empties out with probability--
is there some kind of condition like the expectation
of the probable [INAUDIBLE]?
PROFESSOR: Well certainly one thing that you want is that
the expected time between arrivals has to be less than
the expected service time if you only have one server.
And if you have n servers, then you want the similar
situation that the expected inter-arrival arrival time of
arriving customers is less than or equal to the expected
service time divided by n.
Because you have n servers.
Now the expected arrival time is less than or equal to n
times the expected service time for each server, because
you have n servers working at a time.
You have to be careful on that one.
Because you can dream up situations where you can have
half the servers busy all the time.
The system never empties out.
And you always keep oscillating between half of
the servers full and all of the servers full.
You can even control systems and you sometimes want to
control systems so that you have about half the servers
operating at all time.
If you have half the servers operating at all times, then
you have a fair amount of leeway before
the queue fills up.
You also have a fair amount of leeway before the queue
empties out.
And you have all these servers that you're paying $100 an
hour to and they're not doing anything.
So you'd like to keep, somehow,
sitting in the middle.
You would like to keep servers busy doing other things when
they're not busy serving customers, to
put it another way.
AUDIENCE: So there is no way to obviously tell when the
[INAUDIBLE]?
PROFESSOR: There is no way in general.
For particular simple kinds of queueing
systems, yes you can tell.
We've been going through some of that.
For the M/G/1 queue, it's very easy to tell, because there's
a nice formula.
And you'll see from the simple formula that it's not as
simple as it looks.
Because first I'll tell you what the formula is.
Suppose that X1, X2, as usual, are IID exponential arrivals
at rate lambda.
So the arrival process is Poisson.
That's what the M there means, memoryless arrivals.
V1, V2, and so forth are going to be IID service times.
We're going to assume they have first and second moments.
First moment of the service time, we'll call it V bar.
The second moment of the service time, we'll call it V
squared bar.
What the Pollaczek-Khinchin formula says is that the
expected waiting time in queue, namely before you get
into service, expected waiting time in queue is lambda times
V squared bar divided by 2 times 1 minus rho, where rho
is the service factor, which is the arrival rate times the
expected service time.
This quantity here, as the simplest
answer to your question.
If the service factor, which is the arrival rate times the
expected service time, if that's bigger than 1, you're
in trouble.
If that's less than 1, normally
you're not in trouble.
This formula says it's something worse than that.
It says that as the service factor, if the duty factor is
less than 1, this quantity here is positive.
But if the second moment of the service time is infinite,
you're still going to spend an infinite amount of time
waiting in the queue.
And that's unfortunate.
You might expect this by now after talking about residual
life on all of these things.
But it's still is unpleasant to see it.
When you go from the expected number in the queue to the
expected number in the system, what is it?
Well, the expected number in the system over time is going
to be expected number in queue plus the
expected service time.
Yes.
A customer comes into the system.
He waits in the queue for a while.
Then he gets served.
This is his expected service time.
This his expected time waiting for service.
The expected number in this system, how do you get that?
We get that from this by using Little's relationship.
So the expected number in the system is lambda squared times
V squared bar over 2 times 1 minus rho.
You multiply this times lambda.
So you get lambda squared from this lambda.
From the V bar, you get lambda V bar,
which is the duty factor.
The expected number sitting in the queue is just going to be
this quantity.
I won't derive that for you.
You can work it out.
If you look at some examples of this, before we do try to
derive it, expected number in a queue for an M/D/1 system,
namely Poisson arrivals, deterministic service time.
When you have deterministic service time, the second
moment of the service time is just the first moment squared.
Every arrival takes you exactly the same amount of
time to be served so that V bar is that service time.
V bar squared is a second moment of the service time.
So we just get the formula we had before.
I hope it's the formula we had before.
Lambda times V bar squared.
Oh yes, rho is lambda times V bar.
So we have lambda times V bar squared over 2
times 1 minus rho.
You look at exponential inter-arrivals, suddenly you
get twice as much wait in queue.
It is worse to have exponential service times than
it is to have Fx deterministic service times.
If you look at this strange kind of distribution we talked
about last time, a very, very heavy tailed distribution in a
sense, binary distribution where V is equal to epsilon
with probability 1 minus epsilon.
And V is equal to 1 over epsilon
with probability epsilon.
Then if you work it out, the expected W sub q is equal to
rho over epsilon times 1 minus rho.
This is the same kind of behavior we had before.
You have an enormous residual life.
And therefore, the time that you wait for a customer in
service to get finished is very, very large.
It's what you notice all the time if you're waiting in some
kind of system where many of the customers
get very rapid service.
And every once in while, there's a disaster.
A good example of this is at an airline, when you've missed
your plane or something.
And you go up to a booth there.
And many customers, it's just a matter of
printing them out a ticket.
They're done in 20 or 30 seconds.
Every once in while, there's a customer who
has enormous problems.
And it takes an hour for them to be served.
That's exactly this situation here.
This says that the expected waiting time in q, before you
even get up to get your own service has this factor of 1
over epsilon in it.
This is a nice kind formula, because, in fact, it separates
two different kinds of things.
It has this one factor of 1 over 1 minus rho, which is a
typical factor that you get.
Because if you have a server which is overwhelmed by too
many customers, if the average service time is very, very
close to the average inter-arrival time, then
you're going to have relatively large
build ups of queues.
Every once in while, it'll empty out.
But it will be rare.
And you're going to have a large amount of delay.
This term here is a separate kind of term.
And this has to do simply with a service time distribution
that says that bad service time distributions affect you
in exactly the same way, no matter what
the duty factor is.
So there's this nice separation between these two
things, which is just the way it is.
Why does the wait in q go up with--
that should be V squared bar.
I'm sorry.
Look at the time average weight, the expected R of t,
for the customer in service to finish service.
And picture of that is down here.
If you look at this over time, what you see is at time 0,
there's a customer that just came in.
The residual time until it gets finished, residual life,
is this triangle that we're used to.
Then at this point, the next customer starts to be served.
That time is there.
And the third customer starts to be served.
The difference between this and a renewal process we
looked at before, is that when we're looking at this residual
life in a queueing system, every once in while the queue
empties out.
And then there's a period when nothing happens.
And then suddenly again we start these triangles here.
So there's a question as to how you
evaluate this time average.
Very difficult question, isn't it?
If you had to do this in a quiz, how would you do it?
Would you throw up your arms?
Or would you say, well let me just figure out
what this has to be.
I have these triangles here.
I have this idle time here.
How much idle time is there?
How much triangle time is there?
The triangles are these one half V sub i's that we're used
to from before.
And then this empty time here is a 1 minus rho term.
It's 1 minus the duty factor of the system.
And that's the amount of time over a very long period of
time, the fraction of time, that the system is empty.
That fraction has to be 1 minus rho, because the system
is busy with fraction rho.
It's empty with fraction 1 minus rho.
So you can write down the answer to this without really
having much of a justification for it.
So now we want to figure out why this is what it is.
This is all right.
I want to take the limit as tau goes to infinity of this
integral here over time.
And what that is going to be is 1 over tau.
And what this is is an analytic way of finding that
factor of 1 minus rho that I really
didn't talk about before.
So this integral here, if I look at it out to A of tau, is
going to be 1 over tau times the sum of all of these
triangles here up to the number of customers which have
arrived up until time tau.
So I have all of these out to time tau, which is going to be
out here now.
So I sum all of these up.
And now I ask the question, what is that sum going to be
and the limit as tau goes to infinity?
Well, we know how to evaluate that kind of thing, because
we've done it four or five times already.
The limit as tau goes to infinity of 1 over tau times
the summation from i equals 1 to A of tau Vi
squared divided by 2.
You can write this as limit of A of tau over tau times 1 over
A of tau times the sum from i equals 1
to A of tau Vi squared.
Now as tau increases, this A of tau is going to increase.
These are IID random variables.
So what we're doing is increasing the sample average
of these IID random variables.
So this quantity goes to what?
It goes to the expected value of V squared by the strong law
of large numbers.
And what does this quantity go to?
A of tau over tau, that's the number of
arrivals per unit time.
And that goes to lambda.
Is that what I got there?
I forgot this 2 here.
So that goes 1/2 expected value of V squared.
So this is the careful way of doing it, where you bring all
of this enormous power to bear for something that you could
intuit without using any of the power.
And what is W sub q then?
What is the average time we spend in the queue?
Well you have two components of time you
spend in the queue.
If you arrive when the server is busy, you're going to have
to wait this residual lifetime, which is the
expected value of r, which is residual life.
The other thing you're going to have to wait for is for
every customer in the queue to be served.
Now how long does that take?
Well here you have to be quite careful, because you have a
number of customers that have been built up in the queue.
Each customer that's in the queue has a service time
associated with it.
Think of a customer.
When it comes in, it's given a little packet with its service
time in it.
So you look at the sum of all those terms.
The question is, is the expected number in the queue
independent of the service time of the
customers in the queue?
And how do you reason that out?
And at this point, you really don't want to reason it out in
terms of writing a lot of equations.
You want to just think about it.
How does a queueing system work, if you have a first come
first served queuing system?
Customers come in.
Each of them has a certain amount of time to be served,
which it's holding in a little bag at its side, but which
nobody else sees.
And these are independent, random variables.
But when it comes in, there are a certain number of
customers in the queue in front of it.
So the number of customers that are in the queue before
it each have their own service times sitting in their own
little pockets.
So the amount of time that it takes for me to be served is
this sum of all of the previous times.
And all these previous times are
independent, random variables.
And each of them are independent of how many
customers are in the queue at the time they arrive.
So what I'm trying to argue is that the number of customers
in the queue at time t are independent of the service
time of each of those customers, because the service
times are not apparent to the system until a person getting
into service takes it out of his pocket and looks at what
his service time is.
And that's unknown to the system
until the service starts.
So when you take N q bar independent of V bar0--
Incidentally, if you read the notes, this is explained in a
slightly more mathematical than customers having little
pockets which they read.
But it doesn't help a whole lot.
But anyway, when you go through that argument, you
then get this expression.
And then you add a little Little.
Namely a little Little's theorem, which says that N sub
q is equal to lambda times W sub q.
So you can take this over on the other side.
And this gives you lambda V squared over 2,
which is this term.
And at the same time, you get 1 over--
Excuse me.
Lambda V squared over 2 is this term.
And W q minus N q V bar is 1 minus is W q times 1--
I'll do it at the board.
Probably I'll see it.
But W sub q is equal to lambda V squared over 2.
That's the expected value of r plus lambda W sub Q bar.
So we take this over on this side.
And then we get W q bar equals lambda V squared over 2 times
1 over 1 minus lambda.
1 minus lambda V bar.
Where did the V bar come from?
Beats me.
Oh, came from this quantity there.
Let's patch up our lambda V bar times W sub q bar.
So now we get lambda V bar.
That is all I had to say today.
And so have any questions?