0:04

Okay, now we are approaching the final question.

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

Â Well, in the previous questions,

Â I have given you the example of optimum power control.

Â 0:37

Of course, power control and schedulings are not all representative

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

Â Congestion control, hand-off control, or whatever.

Â But these are two representative examples of resource management to support

Â a large number of mobile terminals.

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

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

Â 1:14

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

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

Â Meaning that, you are not transmitting some signals consecutively,

Â or continuously.

Â So called, discrete, or discontinuous transmission.

Â So your scheduling is sometimes postponed, or

Â sometimes continuously transmitted, depending on the situations.

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

Â wireless resource management.

Â 1:50

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

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

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

Â But what is missing in the previous slide in the power

Â control section is that maximum power level.

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

Â and the display there's a rectangular region.

Â Meaning that every mobile terminal has their maximum instantaneous transmission

Â power, or like 20 milliwatt or whatever.

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

Â but there is actually a unlimited maximum transmission.

Â It was illustrated for explanation purposes.

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

Â So this area, which is actually intersection of triangular and

Â rectangular, that part is actually the correct power vector or

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

Â So this is little bit about power constraint.

Â Well, with a power constraint,

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

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

Â the other mode is actually high data rate.

Â 3:28

So if you want to have high data rate,

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

Â From 70dB of maybe up to 20dB increasement.

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

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

Â Then how this plotting is going to be?

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

Â each mobile terminal has a two mode each.

Â Like, low data rate, high data rate something.

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

Â 4:05

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

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

Â But actually there is an error type.

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

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

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

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

Â the high data rate.

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

Â you can achieve higher data.

Â 4:45

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

Â Now you have four intersection point.

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

Â mobile terminal is transmitting with a higher transmission power, and

Â then higher data rate.

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

Â And this is the case where both mobile terminal is

Â enjoying the low data rate simultaneously.

Â 5:21

So which is in between?

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

Â Well, now we have two power projects here.

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

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

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

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

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

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

Â 6:03

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

Â then you have only three options.

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

Â then you have two options.

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

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

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

Â where you can minimize your power consumption.

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

Â 6:50

So if you increase your transmission power,

Â you can increase your data rate like this.

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

Â Of course, this is quite complicated

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

Â Well, to understand the combined rate and

Â power control more intuitively, consider the case.

Â We have continuous data rate cases.

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

Â high data rate.

Â 7:40

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

Â Such that in the horizontal line,

Â you have the data rate of mobile terminal number one.

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

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

Â And the curve which is a little bit more steep,

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

Â Where the curve which is a little bit flat,

Â is the data rate region of mobile number two.

Â 8:14

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

Â rate region of two mobile terminal.

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

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

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

Â Red line is the case that the,

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

Â Like this steep red line is the maximum transmission power.

Â A line of the mobile China number one.

Â So they are transmitting the maximum usage of power, and

Â enjoying the highest data rate of mobile number one.

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

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

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

Â within their maximum power whereas this borderline case where the both

Â mobile terminals are transmitting with a maximum power.

Â Like that, so with this intersection area show it,

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

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

Â What would be the point?

Â Well, this very much depend on situation.

Â 10:34

Otherwise if your both mobile terminal transmitted with maximum power,

Â then flippant will be intersection point.

Â That is not the point where the both mobile terminal

Â will be supportive of maximum sum data rate.

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

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

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

Â most mobile terminal is transmitting with the maximum transmission power.

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

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

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

Â Because of too much interference generated by each transmitter if two

Â mobile terminal transmit with their maximum power simultaneously,

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

Â So either one of the mobile terminal should turns off,

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

Â total data rate of a system.

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

Â mobile terminal transmit with the maximize it with power.

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

Â 12:07

Well, that cases so called power-limited cases.

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

Â well our interesting problem or interesting region is actually the lower

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

Â interfaces optimally such that throughput will be maximized, or

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

Â So we are more interested in the interface limited cases.

Â 13:02

However, that may not be the best solution,

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

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

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

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

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

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

Â If only one mobile is transmitting,

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

Â Just to show that.

Â However, if half of the time unit was used for

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

Â 14:15

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

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

Â and mobile number two is shut off at the moment.

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

Â mobile number one is shut off, vice versa.

Â So, if two mobile stations actually be transmitting

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

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

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

Â 14:58

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

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

Â then you have one extreme.

Â And the other extreme is actually the R2 data rate.

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

Â then you have a tilted line data rate region, and

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

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

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

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

Â 15:50

by time way, or time division way or something.

Â For this we record as time multiplexing,

Â all cellular radio system we called this as a scheduling.

Â Sometimes you transmit and sometimes you not to transmit.

Â 16:04

But depending on the time fraction,

Â you have this dotted line capacity region.

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

Â Whereas this low-lying curve or

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

Â 16:46

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

Â That's actually the concept of every data rate.

Â And every data has been achieved like that.

Â So this actually gain of scheduling.

Â So you have this plotting point here.

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

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

Â 17:13

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

Â how much fraction of time is devoted to R2?

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

Â Now, we have two issues here.

Â One is efficiency.

Â The other one is actually the fairness.

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

Â that means maximize the R1 + R2.

Â Then, in our previous slide,

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

Â Either R1 or R2 should be achieved alone.

Â We can not transmit simultaneously a workflow by terminal for

Â having this efficiency.

Â 18:07

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

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

Â who receives any data rate will be very unhappy.

Â This is so-called issues of fairness.

Â 18:25

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

Â Your goal is to maximize of minimum R1 and R2.

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

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

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

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

Â That means R1, R2 should be the same level.

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

Â 19:03

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

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

Â mobile number two.

Â As I introduce there are cases R1 plus R2,

Â where only one mobile terminal will be transmitted.

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

Â Even though we have a support both mobile terminal simultaneously,

Â perhaps total throughput not be that much high.

Â 19:39

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

Â In the PF scheduler, PF means proportional fairness.

Â Now this is time, what is the proportional fairness?

Â 19:53

Well, to introduce proportional fairness more correctly.

Â Let me take an example of this wire line communications or

Â any communication network system.

Â You see, there are three circuits.

Â One is an upper and the two circuits down.

Â 20:12

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

Â that uses common resources.

Â In the middle there is actually the green part,

Â which is actually the router for example.

Â So, when there is traffic circulation

Â of the first circle upper, they used two routers.

Â One if left, the other one right, however,

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

Â They are using one router respectively.

Â 20:47

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

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

Â the first circuit in upper.

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

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

Â example, if I want to maximize efficiency,

Â our goal would be x(1), x(2), x(3).

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

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

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

Â 21:32

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

Â The total throughput will be 12.

Â 6 plus 6, 12.

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

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

Â So, if you want to maximize the total throughput,

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

Â Then you have a solution 0,6,6.

Â So, from the first circuit x(1)

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

Â 22:32

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

Â So by comparison between throughput maximum fairness.

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

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

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

Â actually significantly lower than the max throughput cases.

Â 23:05

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

Â And that is actually the topic of the Proportional Fairness.

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

Â 24:23

So, I have introduced some criteria

Â which we are all the time actually should very carefully

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

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

Â And this efficiency and fairness issue is also applied to

Â the case when we design scheduling systems.

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

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

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

Â So there is actually the unfairness in issues here.

Â And on the other hand if you only consider fairness,

Â you have to equalize data rate of every user.

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

Â every user should be equalized.

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

Â of the efficiency maximization case, this is significantly lower.

Â So we have to compromise between efficiency and

Â fairness because there is a fundamental trade off for that.

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

Â that actually the proportion in our fairness cases.

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

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

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

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

Â 26:06

And then depending on your scheduling parameter,

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

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

Â So our goal is to consider both

Â 26:23

effectiveness and fairness, for that we introduce proportional fairness.

Â So proportional fairness point is the point where the R1 and

Â R2 multiplication should be maximized, that is actually the cross

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

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

Â This is so called proportional fair scheduling, so

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

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

Â 27:01

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

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

Â 4G system as of today.

Â And then I think that this will be also used for

Â the scheduler for the next generation system.

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

Â the very fundamental question and the last two question is for

Â the some examples of the radio resource management technique,

Â for example power controlling and scheduling.

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

Â other level of resource management technique like congestion control or

Â admision control and hand of control, whatever, but

Â that will be covered in the other lectures as well.

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

Â insight into the issues so called wireless resource management.

Â Thank you for participation.

Â Thank you.

Â