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. Now actually in this question of the example, I will also give you an example of how the packet scheduling or transmission scheduling will increase the capacity. Of course, power control and schedulings are not all representative resource management techniques, 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. But anyway, the scheduling is very fundamental technique to support a large number of mobile terminal simultaneously. 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. It's 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. 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 the maximum power level. 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 an 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 the intersection of triangular and rectangular, that part is actually the correct power vector or power value so that you can support both mobile terminals within the power range. So this is a little bit of 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. 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 increment. 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 is this plotting going to be? For example, so assume the case you actually have two mobile terminals, and each mobile terminal has two modes each. Like, low data rate, high data rate something. Then this plotting will be generalized like in the next slide. If you look at this here, first, x-axis and y-axis are 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. Now in the previous slide, we have just one intersection point. Now you have four intersection point. From the outer point, the most outer 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. 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 this 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. 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, both mobile terminals are also supported with the lower data range. Well, this is actually how the rate and power controls are combined. 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 a 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 each mobile has only two options, like lower data rate and high data rate. But for better understanding, to better understand actually, assume the case where the data rate is actually the continuous function of your transmission power. That is actually the topic of your next slide. 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 can ctually see a two curves here, like this. And the curve which is a little bit more steep, it 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. And then the intersection area of both curves is the feasible data rate region of two mobile terminals. So this little bit bright intersection area, is feasible data rate region of the both mobile and terminal and then now, a quite interesting case is that actually,is thatit will be this red line itself. Red line is the case that, 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 actually shows maximum power level of mobile number two. So with a maximum power level that they can actually enjoy higher data rate. So inside this bright region, if both mobile terminals are transmitting within their maximum power whereas this borderline case where both mobile terminals are transmitting with a maximum power. Like that, so this intersection area shows, 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 that point be ? Well, this very much depend on the situation. Well this is, for example, sometimes the intersection area looks a little bit like a convex curve type shown in the roll part, or sometimes concave type of curvature shown in the blue curve in the next plotting. So, it's shown it in this red line curve. In order to maximize some of R1 and R2, then I mean optimal point will be either R1 or either R2 or either one mobile terminal transmitted with maximum power. Otherwise if your both mobile terminal transmitted with maximum power, then flippant will be the intersection point. That is not the point where 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, that will maximize R1 plus R2, that is the point most mobile terminals are transmitting with the maximum transmission power. So we have two cases of our feasible data region which are 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 an interference limited system. Because of too much interference generated by each transmitter if two mobile terminals transmit with their maximum power simultaneously, you will lose your data rate because of interference with each other. So either one of the mobile terminals should turn 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 terminals transmit to maximize it with power. That is the point where the critical width, the R1 to R2 point is maximized. Well, that cases are 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. Well, let's go further on interface remedy system using next slide. Well, this is once again interface limited system. And to maximize the throughput of the system. Either user number one or Mobile Terminal, MT one or mobile two should transmit alone. However, that may not be the best solution, because if only one mobile terminal will be transmitting, others should shut off. So if they would, there will be some vanished issues. And there is other perspective 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. Half, divide by two, amount of data rate will be available. That means, for example, what is possible to divide one unit time into two. And one fraction of time is devoted to the transmitting of mobile number one. 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 are 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 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 some of the real numbers, then you have a tilted line data rate region, and if you look at this data rate region by tilted line which is actually larger than just the blue 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 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 do not transmit. But depending on the time fraction, you have this dotted line capacity region. Well, this capacity region is actually called average capacity region. Whereas this low-lying curve or bright region is called this instantaneous data rate region. Well, average capacity has been introduced and you see the average capacity is much larger than instantaneous data read. So by introducing this scheduling, we have the concept of average data rate. 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? 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. 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. 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. 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 terminals simultaneously, perhaps total throughput is not that much high. And then there is something which is between what is called proportional fairness. In the standard review system, there is a scheduler called PF Scheduler. In the PF scheduler, PF means proportional fairness. Now this is time, what is the proportional fairness? 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. 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. 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. 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. Now, we can think of the example of total fair. For example, then, if you have total fairness, like maximum fairness, the solution will be 3, 3, 3. So everybody will be equalized. However, user 1 uses more resources. 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. 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. Well, total throughput is now 10 which is in between 9 and 12. However, there is a fairness issue little bit resolved. Because we have given two amount of resources to the first circuit. How did we get this solution? Well, objective function is we maximize product of X(1) X(2) X(3). So by having such X(1), X(2), X(3) product maximization, we should exclude the solution where one of these becomes 0. So, by having this proportional fairness objective function, we could achieve what is called, we could both balance between Fairness and the effectiveness. Well, product maximization x1, x2, x3 is equivalent to, mathematically, is a log sum maximization. So, I have introduced some criteria which all the time we 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. 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 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 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. 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.