Now how do we think about this capacity allocation problem here? First of all, in the rest of this lecture, when we say a flow, we mean a session, we mean the source. Because we equate a session with a source. As each session traverse a single path. That is already given to us. And a session has one source node and one destination node. So this is a unicastic session over a single path. We're going to use these three words interchangeably, flow, session, outsource. We're going to index each of them by say, I. And routing is already fixed, so you cannot change what path will a session take. Okay. You can only change the transmission rate of the source of that session or flow, not the path that it takes. Now the interaction between routing and congestion control which we briefly mentioned last lecture, in network architecture, you can think about that, and maybe in the homework problem you can encounter one of those. But for the lecture, fixed routing, you only adjust the degree of freedom for congestion control, namely, the source rates. Now this capacity allocation problem. We first have to make sure that the solution we propose is a feasible one. Meaning that, it can satisfy the link capacity constraints on all the links in the network. Otherwise the vector is infeasible. It might be great to have but you cannot accommodate that. There might be a single bottleneck link somewhere. And then, there will be likely uncountable number of feasible allocation vectors. The question then is which one you should pick as the optimal. Optimal with respect to what? What you want, efficiency and you want fairness in this allocation. You don't want to waste link capacity. For example, assigning zero to all the sources as the rates met as a feasible solution, but clearly is very inefficient. You also want it to be fair. Let's take a look at an example. Okay. Here is example with four links, kay, and four nodes in a graph and there are three sessions. So, four links. Right by one, two, three, four, and three sessions labeled by a, b, and c. Session a traverses all three lengths in this tandem, okay? So from source node to destination node. Section, session b goes from this source to this destination, sharing one link with session a, and then go through another link by itself. Session c also share one link, session a in fact, that is the only link to session c goes through. Okay. Now the question is. If I have a unit capacity on each of these links then what should be the, the allocation of the capacity? So first of all you need to be feasible. Okay, for example I want to give all three sessions one megabit per second. So I want to give the vector 111 to sessions A, B, and C. Now that's not feasible. Right? Cuz then link one cannot accommodate two megabit per second. It can only have the capacity of one megabit per second. Same thing for link four. So, you have to pick a feasible solution. And you start to think that, hey maybe links one and four are kind of the bottleneck links, right? If they are saturated, then even if link three has leftover capacity, I cannot add more rates to session a. Because the bottleneck might be on links one and four, competing with the other two sessions. So it is not just about the number of links that you traverse. Its not like c goes to one link, b goes to two link, a goes to three link and therefore you know, a should fewer rate than b, b should get fewer rate than c is really about what kind of links are you tra, traversing. If you're using a lot of bottlenecks links then may be you should not receive too many rates, too much rates. And in that sense, we can see that A uses two bottleneck links, one and four, B uses one bottleneck link and C uses one bottleneck link. So, maybe B and C even though B traverse one more link it should be treated in similar ways because the additional link B traverses is really not a bottleneck link. Okay? In the configuration of equal one unit. One mega per second capacity for all four links. Of course, if you change that configuration, the, the implications can be differe nt. Then we want it to be efficient. Now, it is actually impossible to fully utilize four megabits per second across all four links. Okay, but you can try to use as much as possible. For example, if I give session one, session A, one megabit per second and the other two sessions zero. Okay. That is a particular allocation. Right? It actually would use three out of four megabits available in the network. Hm, here's another way, for example I can give half, half, half negative per second through the three sessions. That actually also uses three out of four negative per second available in the network. Right, cuz this link and this link will be fully utilized. That's two megabit per second, this link got half, this link got half for sessions A B respectively. So, if you look at these two vectors, they are both efficient. Okay. But clearly they look very different. In particular, this allocation actually starves, sessions B and C, giving them zero rates. So you probably think, this is unfair. Later, in lecture twelve, the last lecture of the course, we'll devote a whole lecture to quantifying notions of fairness. But you may remember alpha fairness, okay? And this one looks more fair than that one. But you may think, gee, session A is actually taking up a huge amount of resources. So why should we treat it equally as if this is same as sessions B and C, right? It goes through two bottleneck lengths. Not one. And indeed, if you go through the derivation of a proportionally fair. You may remember if we do logarithmic utility maximization, we get proportional fairness. Then the answer is that you could give session A one-third of a unit, so one-third of a megabit per second. And sessions B and C two-thirds Again the efficiency is that you are using four, three out of four megabit per second in the network. But allocation is. One unit of share for A. Two units for B and C. And that confor, conforms to our intuition that B and C should be treated about the same, because they use one single bottleneck link. And A should not be the same cuz it uses two. And indeed, you see a sense of proportionality. If you use twice as many bottleneck links you will be given, half of the capacity, as the other sessions. And if you impose that condition you can quickly see that this one-third two-thirds, two-thirds will be the most efficient solution. So If we want feasible solution and then strike a trade-off between efficiency in utilizing the link capacities, and the fairness of allocating them among competing sessions. So now the question is how do we formulate this problem in general. Let's first look at the constraints of this optimization problem which is the link capacity constraint. We have to find a representation of the routing decisions which is again given to us already. We don't get to change that. But we have to write it down cuz that clearly changes the configuration of what session uses what path, what links. There are a few different ways to do that is to say that, if you look at each source. Okay. Is that a source that uses link l? For each link l, you can write out a subset of nodes which are the sources that use this link l. Denote that set by S(l) and you can ask yourself is node i in that set. Conversely you can say is link l being used by the source given the routing you know for each source i there is a set of links a concatenation of which provides the end to end path. Is this link l in that path? Is it an element in the set l of i? There is yet another way to represent this in a matrix. Think about a number R which is a binary number 01 sub li. It is one if node i uses link l and zero otherwise. So if link l out is all the path used by source i then Rli is one otherwise it's zero. And soon you'll see an example illustrating this using the same topology that you just saw. So now I can write down the link capacity constraint by saying that multiply Rli Xi which is my source rate as source i summing over all the I's. What is this? Well, we can give it a symbol. This is y sub l because we sub over all the i's fixing a particular l. This is the load on link l. So the loa d on link l. Okay? Or equivalent, we can write it as the summation of all the Xi's but summing only over those Is in the set of sources using link l. We can either represent it in this index way, or represent it in this binary, matrix way. And what we want is to say this load on link l should be no bigger than the capacity on link l, denoted as C sub l. So each link indexed by L, got a fixed known number capacity C. Now you see that the routing matrix collectively called R and the link capacity vector, which we can put this number into a vector C, are both given constants. Our vec, our matrix, and C vector our constants. We do not get to change the link capacities or the routing decisions. We get to change the X vector. Hopefully distributively. That is the source rates. Okay, y is just a shorthand notation of a sum of certain X with respect to a particular link, you know?'K? So we want a linear function of X called y to be less than the capacity C. During the transient the, this inequality will be valid as you try to search for the right loading, on the link l. But at equilibrium you want this inequality to be satisfied for all links l. Okay. Now we can of course write this as shorthand matrix notation R, matrix multiply X vector should be less than equal to the C vector. So this is yet another way, a shorthand notation to represent this link capacity constraint. Of course this inequality is comparing two vectors, the Y vector and the C vector. This inequality here means component wise. the first component's less than the first component, the second component's less than the second component, and so on. Okay? Now what about the objective function? Well the objective function here as we just mentioned can be alpha fair utility function. They capture both efficiency and a notion of fairness. Promethized by this alpha. So again, reminding ourselves from lecture eleven that the utility function is a function of the convex right at the source promethized by alpha equals x to the power of one minus alpha over one minus a lpha, if alpha is not equal to one. If it is equal to one then, in the limit, this becomes log of x Okay. So different sources I may have different alpha. So you can say, this source have one alpha. Another source has a different shape. Or we can say they all share the same Alpha. In which case we are just maximizing the summation of U times, a U(Xi) subs, alpha sharing the same shape, summing over all of these sources i. And this will be the objective function: maximize the sum U(Xi). Subject to the constraint of RX less than equal to C and therefore we have our formulation of Network Utility Maximization, N U M the NUM formulation. Maximize summation U(Xi) summing over i such that RX is less than or equal to C where the variables is the X vector, collecting all the source transmission rates Xi for a given constant. The routing matrix given, and the capacity given. And say the shape of the utility function Alpha, given. If you look at this optimization problem, see that it is linearly constrained and objective function assuming smooth increasing and concave functions of the utility function which is the case Alpha fear utility function, this is a concave maximization which as you may recall from lecture four. is equivalent to minimizing a convex function. Same as minimizing minus the sum of utilities. Okay. So instead of maximizing this you are talking about minimizing this. Subject to linear constrains. And this is a nice problem. It is a convex optimization. Because you are minimizing convex function or equivalently maximizing concave functions. Subject to convex constrains that which turns out to be a set of linear constrains in equalities. So it is a nice optimization. Even though we did not spend time talking about the exact algorithms that will belong to an operations research course. but you can believe me that they are very efficient centralized algorithm to solve it. Now better still, and as needed in this case, we need more than just efficient centralized computation. We need distributed algorithm to solve this efficiently. And. That indeed turns out to be the case because this problem is so called decomposable. Clearly the objective functions is a sum of different sources utility and therefore it is already decomposed the problem, the trouble resides in this coupling constraint, Okay? Each links shared by many sessions and each session traverses through many links just like what we saw already in this example. Okay. So we cannot ignore those coupling represented by this RX less than equal to C. But fortunately we can do what we call dual decomposition. Is called dual decomposition because it's actually solving the so called La Grange dual problem. Give me an optimization problem, I can always derive, so called dual problem. And there are many strong, nice properties between the primal and the dual problem. Okay, one of which allows us to look for cracks in the original problem and solve it through solving the dual problem. And you get a distributed solution. This is a very interesting topic. And it's quite a, a powerful mathematical tool in designing networks of various kinds. But we will have to defer this to advanced material part of the lecture. Okay. So suffice it to say at this point that, indeed this is a convex and decompose for problem and we'll now present the distributed algorithm. If you want to see the derivation, please wait until the advanced material part.