0:04

Okay, now we are approaching the final question.

How does scheduling increase the capacity of wireless network or average capacity?

Well, in the previous questions,

I have given you the example of optimum power control.

0:37

Of course, power control and schedulings are not all representative

resource management technique, but there will be some other techniques as well.

Congestion control, hand-off control, or whatever.

But these are two representative examples of resource management to support

a large number of mobile terminals.

So if you are interested in some other techniques as well,

perhaps actually you could take the other more detailed lectures as well.

1:14

When you say schedule, scheduling in the way the resource management,

the question is, when to transmit, when not to transmit?

Meaning that, you are not transmitting some signals consecutively,

or continuously.

So called, discrete, or discontinuous transmission.

So your scheduling is sometimes postponed, or

sometimes continuously transmitted, depending on the situations.

That's the proper mobile transmission scheduling from the perspective of

wireless resource management.

1:50

Well, in my previous slide, in question, the PowerPoint slide,

you may remember this plotting here in the two-dimensional spaces.

We have a power control level that you can support to mobile terminal simultaneously.

But what is missing in the previous slide in the power

control section is that maximum power level.

Here actually the maximum power level is actually shown here in both terminal,

and the display there's a rectangular region.

Meaning that every mobile terminal has their maximum instantaneous transmission

power, or like 20 milliwatt or whatever.

When previously we have shown that there is no maximum transmission,

but there is actually a unlimited maximum transmission.

It was illustrated for explanation purposes.

But however, a more practical constraint is that you have a power range like this.

So this area, which is actually intersection of triangular and

rectangular, that part is actually the correct power vector or

power value that you can support both mobile terminal within the power range.

So this is little bit about power constraint.

Well, with a power constraint,

now let me think a little bit about more on multi-rate systems.

For example, your mobile terminal has two modes, one is low data rate,

the other mode is actually high data rate.

3:28

So if you want to have high data rate,

perhaps you should have a higher target then, like 70 dB.

From 70dB of maybe up to 20dB increasement.

So, you have a high target area with high data rate, or

lower target value, like ten dB or something, your lower data rate.

Then how this plotting is going to be?

For example, so assume the case actually you have two mobile terminal, and

each mobile terminal has a two mode each.

Like, low data rate, high data rate something.

Then this plotting will be generalized like in the next slide.

4:05

If you look at this here, first, x-axis and y-axis actually the same format.

There's a p1, there's actually a p3 here.

But actually there is an error type.

p3 should be the p2 replaced by p2, so you may understand.

Where there's a p1 and p2 in a y-axis, now you have four lines here.

First two lines, I mean each terminal has two lines each.

Actually, the inner curve is for the low data rate, and outer curve is for

the high data rate.

So if you increase your transmitter power, you can achieve high target, and

you can achieve higher data.

4:45

Now in the previous slide, we have just one intersection point.

Now you have four intersection point.

From the outer point, the outer most point is actually the case where both

mobile terminal is transmitting with a higher transmission power, and

then higher data rate.

Now most inner point, which is actually close point of both red lines here.

And this is the case where both mobile terminal is

enjoying the low data rate simultaneously.

5:21

So which is in between?

One is higher data rate, the other one is lower data rate, like this.

Well, now we have two power projects here.

One is green line here, and you have a larger relaxed power budget where you

have a like dotted line more, more rigid, or more restricted power budget here.

For the case, if you have a green line power budget, like relaxed power budget.

Perhaps if you want to maximize data rate of both mobile terminal,

then the point where the blue line meet with each other will be the optimal

solution where you can maximize the data rate of both mobile terminal.

6:03

However, if your power budget is very much limited like in a dotted line,

then you have only three options.

Well, if your goal is to maximize the sum of rate of both mobile terminal,

then you have two options.

Either, the one actually in upper, and the one in the slightly in the lower.

Well, depending on your criteria, perhaps you can choose one of these.

Perhaps most avoidable, the solution would be to list one,

where you can minimize your power consumption.

But however, the both mobile terminal is also supported with the lower data range.

6:50

So if you increase your transmission power,

you can increase your data rate like this.

So with this figure, you can have combined control of data rate and power control.

Of course, this is quite complicated

thing because you have a discrete number of data rate cases.

Well, to understand the combined rate and

power control more intuitively, consider the case.

We have continuous data rate cases.

Here we have each mobile has only two options, like lower data rate and

high data rate.

7:40

So with some mathematics, we now have this transformation of the curves.

Such that in the horizontal line,

you have the data rate of mobile terminal number one.

In a vertical line, you have a data rate of the mobile number two.

And then you see actually two curves here, like this.

And the curve which is a little bit more steep,

is actually the data range region of the mobile number one.

Where the curve which is a little bit flat,

is the data rate region of mobile number two.

8:14

And then the intersection area of both curves are the feasible data

rate region of two mobile terminal.

So this little bit bright intersection area, is feasible data rate

region of the both mobile and terminal and then now, quite interesting

the case is that actually, what it will be the this, red line itself.

Red line is the case that the,

each mobile terminal is all transmitting with their maximum transmission power.

Like this steep red line is the maximum transmission power.

A line of the mobile China number one.

So they are transmitting the maximum usage of power, and

enjoying the highest data rate of mobile number one.

Whereas this flat line shows actually maximum power level of mobile number two.

So with a maximum power level that they actually can enjoy higher data rate.

So inside this bright region, if both mobile terminals are transmitting

within their maximum power whereas this borderline case where the both

mobile terminals are transmitting with a maximum power.

Like that, so with this intersection area show it,

with colored by this bright region, you are quite interested in such edge point.

For example, you are interested in maximizing some of R1 and R2.

What would be the point?

Well, this very much depend on situation.

10:34

Otherwise if your both mobile terminal transmitted with maximum power,

then flippant will be intersection point.

That is not the point where the both mobile terminal

will be supportive of maximum sum data rate.

Whereas in the blue line curves, if you want to maximize sum of R1 and

R2 then close of a point which is just in the middle is the point,

point that will maximize R1 plus R2, that is the point

most mobile terminal is transmitting with the maximum transmission power.

So we have two cases of our feasible data region which is actually the red

line top of the region and the other one is actually the blue line type of region.

In the red line type of region we call this a interference limited system.

Because of too much interference generated by each transmitter if two

mobile terminal transmit with their maximum power simultaneously,

you will lose your data rate because of interference with each other.

So either one of the mobile terminal should turns off,

and the other terminal should transmit it alone to maximize the total throughput of

total data rate of a system.

Whereas in the blue curve, to maximize the throughput in both

mobile terminal transmit with the maximize it with power.

That is the point where the critical width, the R1 to R2 point is maximized.

12:07

Well, that cases so called power-limited cases.

Because throughput is very much dependent on the maximum transmission power,

well our interesting problem or interesting region is actually the lower

curved area because in the radio reserves management our goal is to control

interfaces optimally such that throughput will be maximized, or

data rate will be maximized or system capacity will be maximized.

So we are more interested in the interface limited cases.

13:02

However, that may not be the best solution,

because if only one mobile terminal will be transmitting, others should shut off.

So did they would, they will be some vanished issues.

And there is other perspective so called the scheduling or time multiplex issues.

That means they can also still opportunity to increase the capacity,

of the system, for example, just one time unit, like one second or

one ever whatever, assuming you have a one time unit.

If only one mobile is transmitting,

then you could achieve a wrong amount of data, right?

Just to show that.

However, if half of the time unit was used for

transmitting of mobile terminal number one you could have like R1 divide by two.

14:15

And the remaining fraction of time is devoted to the mobile number two.

Like in the first time fraction, mobile number one is transmitting,

and mobile number two is shut off at the moment.

In the second fraction of time, mobile number two is transmitting and

mobile number one is shut off, vice versa.

So, if two mobile stations actually be transmitting

half unit of time in the round robin way, you have a tripper which is

noted in this slide like R1 divided by two, R2 divided by two.

Means you draw a line, which is dotted line between R1 and R2 and

14:58

have the meaning that if your fraction is changing, like most of the fraction is,

all of the fraction is devoted to the transmission of mobile number one,

then you have one extreme.

And the other extreme is actually the R2 data rate.

If fraction of time is determined with the some of real number,

then you have a tilted line data rate region, and

if you look at this data rate region by tilted line which is actually

the larger than just the blue the intersection area of yellow regions.

That means by introducing scheduling which we record as time multiplexing.

So, multiple access of two mobile terminal is done through the time

15:50

by time way, or time division way or something.

For this we record as time multiplexing,

all cellular radio system we called this as a scheduling.

Sometimes you transmit and sometimes you not to transmit.

16:04

But depending on the time fraction,

you have this dotted line capacity region.

Well, this capacity region is actually so-called average capacity region.

Whereas this low-lying curve or

bright region is so-called this instantaneous data rate region.

16:46

How much data you have received for a given time unit on average?

That's actually the concept of every data rate.

And every data has been achieved like that.

So this actually gain of scheduling.

So you have this plotting point here.

My next question is, so out of this plotting point,

a fraction of the plotting point, which point should be chosen?

17:13

Like, how much fraction of time is devoted to R1 and

how much fraction of time is devoted to R2?

Well, this is very much dependent on the designer's perspective.

Now, we have two issues here.

One is efficiency.

The other one is actually the fairness.

So for example, if you are interested in the efficiency,

that means maximize the R1 + R2.

Then, in our previous slide,

either user number one or user number two should transmit alone.

Either R1 or R2 should be achieved alone.

We can not transmit simultaneously a workflow by terminal for

having this efficiency.

18:07

But either, R1 is maximized where R2 is 0 or R2 is maximized, R1 is 0.

We have some issues so-called fairness because those mobile terminal,

who receives any data rate will be very unhappy.

This is so-called issues of fairness.

18:25

Well, when we have the fairness issues, we have such formula in a the bottom.

Your goal is to maximize of minimum R1 and R2.

So, your goal is to maximize the minimum data rate of each user.

Where this is to secure, fairness among the two transmitters.

To achieve such fairness, one solution is that you balance R1 to R2.

That means R1 and R2 should be equalized in the end.

That means R1, R2 should be the same level.

Well, when you have fairness issues, you lose efficiency.

19:03

So, whether there is actually the some other objective function,

when you determining scheduling fraction of time for mobile number one and

mobile number two.

As I introduce there are cases R1 plus R2,

where only one mobile terminal will be transmitted.

Where maximum fairness, where you equalize data rate for both mobile terminal.

Even though we have a support both mobile terminal simultaneously,

perhaps total throughput not be that much high.

19:39

In the standard review system, there is a scheduler so-called PF Scheduler.

In the PF scheduler, PF means proportional fairness.

Now this is time, what is the proportional fairness?

19:53

Well, to introduce proportional fairness more correctly.

Let me take an example of this wire line communications or

any communication network system.

You see, there are three circuits.

One is an upper and the two circuits down.

20:12

For example, we can think of this as a communication network

that uses common resources.

In the middle there is actually the green part,

which is actually the router for example.

So, when there is traffic circulation

of the first circle upper, they used two routers.

One if left, the other one right, however,

in the bottom part there are two circuits circulating their traffic each other.

They are using one router respectively.

20:47

And each router has a capacity of handling six amount of capacity maximum.

So, when we say that x(1) is the throughput capacity of

the first circuit in upper.

And x(2) is the second circuit, and x(3) is the third circuit.

And then our goal is, for example, the max throughput case is for

example, if I want to maximize efficiency,

our goal would be x(1), x(2), x(3).

Like, sum of the total throughput circulating by each circuit should be

maximized, with the constraint that the x(1) plus x(2) is less than 6,

and x(1) plus x(3) is also less than 6.

21:32

An optimal solution of this one is actually 0, 6, 6.

The total throughput will be 12.

6 plus 6, 12.

The Y at the X one becomes zero, because X one is

a little bit inefficient in terms of using two routers simultaneously.

So, if you want to maximize the total throughput,

some of the throughput of each circulation data rate that is circulating circuit...

Then you have a solution 0,6,6.

So, from the first circuit x(1)

perspective this is quite unfair because they do receive any data rate.

22:32

By circulating one amount of traffic, it uses two routers simultaneously.

So by comparison between throughput maximum fairness.

Where throughput is 0,6,6,12, in this case you has 3,3,3,9.

So, there actually the some differences in amount in total throughput.

So, this is very fair in cases, where the total throughput is,

actually significantly lower than the max throughput cases.

23:05

Well, there is anything in which we'll consider both fairness and throughput.

And that is actually the topic of the Proportional Fairness.

If you look at proportional fairness allocation, we have 2,4,4.

24:23

So, I have introduced some criteria

which we are all the time actually should very carefully

take into consideration when we're designing the resource allocation,

which is, one is efficiency and the other one is fairness.

And this efficiency and fairness issue is also applied to

the case when we design scheduling systems.

When we only consider efficiency, of course, we could have a bigger pie.

In our previous case we have total of 12 capacity but, however,

there is actually the user who do not receive any data rate.

So there is actually the unfairness in issues here.

And on the other hand if you only consider fairness,

you have to equalize data rate of every user.

That means you have to put your resources such that the data rate of

every user should be equalized.

In our previous case it was three, three, three, nine, compared to three twelfths

of the efficiency maximization case, this is significantly lower.

So we have to compromise between efficiency and

fairness because there is a fundamental trade off for that.

Actually, achieving to slice bigger pie considering the little bit of fairness,

that actually the proportion in our fairness cases.

So coming back to our scheduling example, we have this plot here.

Originally, we have two user's data rates region there and

we have a simultaneous or instantaneous data region colored by yellow curves here.

Now we have a dotted line region, which is the average data rate region.

26:06

And then depending on your scheduling parameter,

like which fraction of a time should be devoted to R1,

and which fraction of time should be devoted to by R2.

So our goal is to consider both

26:23

effectiveness and fairness, for that we introduce proportional fairness.

So proportional fairness point is the point where the R1 and

R2 multiplication should be maximized, that is actually the cross

over point where the dotted line is meeting the sum convex curvilinear.

Which can be achieved by mathematically by maximizing log R1 + log R2.

This is so called proportional fair scheduling, so

that you could devote some amount of time for R1 and some amount of

time R2 depending on the position of that crossover point just in the middle.

27:01

Well, in practice, we design proportional fair scheduler based on this concept.

And PF scheduler is very well used in practical system like 3G and

4G system as of today.

And then I think that this will be also used for

the scheduler for the next generation system.

So by concluding, I said actually we have touched upon six questions starting with

the very fundamental question and the last two question is for

the some examples of the radio resource management technique,

for example power controlling and scheduling.

But as I said in the beginning of this lecture, there are same more detailed

other level of resource management technique like congestion control or

admision control and hand of control, whatever, but

that will be covered in the other lectures as well.

But, however, these six questions will give you some fundamental idea or

insight into the issues so called wireless resource management.

Thank you for participation.

Thank you.