 So, the claim is that does not matter if the polyhedron is bounded or not, if it has at least one extreme point and your linear program has a finite optimal value then it must be that there is at least one extreme point that is a solution, so let us see why this is the case, not necessarily, it depends on how the, so the question here is, is it possible that your feasible region is unbounded, so that means the polyhedron is unbounded and yet you can get a finite optimal value, is that possible and the answer is yes, so for, so it is possible, for example you look at, suppose if I lake a feasible region like this just in R2, I am talking of a feasible region that comprises something like this and the objective function that I am trying to optimize is as this sort of contour which decreases in this direction, so I am looking to minimize this function, this is the direction of decrease of the contour. Now what will happen is this sort of contour the one I have drawn here is cannot be optimal, so but you can move it in this direction to the point, to the extent that eventually it just passes just through this corner point here and that sort of thing will be operable, it is irrelevant for this that the set is unbounded in this direction out here because that is not a direction in which you are getting a better value, so what matters in getting an optimal value finite even in an unbounded feasible region, unbounded polyhedron like that is how the objective function, the direction in which the objective function changes relative to the direction in which the set itself recedes, so there would be extreme rays and extreme direction, there will be rays of this set or the direction in which the set recedes, how is that relative to the direction in which the objective function decreases because we are talking of minimization, so decreases, that is what determines whether you will get a finite value or not, in fact that will be evident as we do this proof also, so let us just go through this proof, so suppose the extreme points say x1 dot xk, these are extreme points of p and suppose the extreme rays of p are say d1 till dm, now the extreme points of p are answer this, so the extreme points of p are they in the set p, are they elements of the set p, yes of course, they are points in the set p, what about extreme rays? Any point plus an extreme ray is a point in the set, any point in the set p plus an extreme ray would be a point in the set B. So, what about the extreme ray itself? The extreme ray is remember just a vector. Is that an element of the set B? Just to give you an example, let us look at this diagram that I drew here, I will just try to draw this with a different color. See, this is for example, an extreme a direction in which the set recedes. You can start from any point in the set and keep going along this direction and you will remain in the set. Now, I can think of this direction as this vector, this little vector here, this vector say alpha, 0 for example, where alpha is some small positive number. Now, this vector, the point alpha, 0, I am imagining that as a direction on which is to be scaled and along which I can keep travelling etc. But the point alpha, 0 is not in the set B, but any point in the set B say x plus alpha, 0, which gives you this point, this sort of point is in the set B. Is this clear? So, these are not necessarily elements of B, d1 to dm, x1 to xk are elements of B, these not necessarily. So, these are suppose the extreme points of B and these are the extreme days of B. Now, what does the Minkowski-Weil theorem tell us? So, by Minkowski-Weil theorem, what do we get? We get that P is equal to x such that x can be written as convex combinations of the x i's, i equals 1 to k, all the lambda i's are greater than equal to 0, the lambda is sum to 1. So, by the Minkowski-Weil theorem, the polyhedron can be written as x such that x equals, x can be written as the sum lambda i x i from 1 to k plus mu i d i, i going from 1 to m. And the lambdas are all non-negative, mu's are also non-negative, but the lambdas sum to 1, this should be k here. Is this clear? So, remember this is different, this is saying that the polyhedron can be written as a convex combination plus a conic combination of these fixed points, you know, these k and points when you are taking the convex combination and these m points that when you are taking, these m points when you are taking the conic combination. Is this clear? This is not saying that we are not allowing in the definition here, we are not allowing the x i's and the d i's to vary, they are being fixed outside the set, outside the set. Is this clear? What is varying here are only the weights, the weights lambdas that sum to 1 and the mu i's which are non, which are just non-negative, yes. So, the extreme ray is a direction, so firstly remember what a ray is, ray is a direction, no, no, no, a ray is a vector such that if I add any, so a ray is a vector, so d is a ray if x plus mu d belongs to, is a ray of p, if x plus mu d, if x belongs to p implies x plus mu d belongs to p for all mu greater than equal to 0. This is what it means for d to be a ray. So, I can take d multiplied by any non-negative scalar, add that to any vector in the set p and the result is going to be a point that lies in the set p. Once if I have a vector like that, then that vector is a ray. So, here for example, as I said alpha comma 0 that is a ray, but alpha by 2 comma 0 is also a ray because I can multiply by twice the scalar and I will again get the same property. So, the point here is rays really have only a direction, they do not have a length because I can scale them, the very definition is such that I am supposed to be scaling them by any scalar mu outside. So, we do not really look at, so as a result it does not make sense to talk of a ray being in the set because I can always scale it down and make it out of the set or scale it up and bring it. No, no, no, the ray is simply, the ray is eventually a vector, it is a point in R2. So, the way when I say this is the ray, really what I am doing is actually I am sort of taking this here and moving its origin down to it. I am representing x plus mu d when I am writing that, is this clear? Okay, for clarity make sure that you understand the non-trivality of the Minkowski-Weil theorem which we will use here is that these extreme points and extreme rays are fixed for the set. They are not in, you know, you give me a fixed set of points and I can generate the entire set using the convex combination and conic combination of those fixed points only. I am not changing those points as I get, where as I try to get a different point in the set. Okay, alright. Okay, so if this is the Minkowski, this is my Minkowski-Weil theorem, so now let us come back to our linear program. The linear program says minimizing C transpose x subject to Ax less than equal to b and that is my set, that is your set p. Okay, so by Minkowski-Weil theorem p can be written in this, the p, every point in p can be expressed as convex combination of the Xi's and conic combination of the, plus the conic combination of the d's, of the dI's which means that I can write this optimization in this equivalent form. I just do a change of variable, in place of x, I will substitute this, this form, this expression that I have and then what I will get is an optimization over, so my new variables will now become the lambdas and the mu's. Because as I vary the lambdas and mu's, I get an x and for every x there is a lambda and mu. So it is the standard change of variable that I am doing. So I, in place of x, I am going to put this. Now what are my constraints? Well, my constraints are precisely these, these constraints. So I need, I have to put in the constraints are that lambda i's are all greater than equal to 0, mu i's are all greater than equal to 0 and the lambda i's sum to 1. So let us look, let us, everyone is following this. So now let us take this objective and work with this further. So I, so this is the same as doing minimizing over lambda and mu, the objective this. So what I will do is write this in the following form as summation i from 1 till k lambda i times c transpose x i plus summation mu i lambda to c transpose d i, i sums from 1 to m. I am still lambda greater than equal to 0, mu greater than equal to 0 and lambda sum to 1. So now let us look at this objective a little more closely. So you have the objective comprises of two terms, the, there is here this term which has all the lambdas in it and none of the mu's and there is this term which has only the mu's and no lambdas. Now the lambdas are to be selected so that they are collected effectively satisfy these, these inequality, these constraints. They have to be greater than equal to 0 and they have to sum to 1. What about the mu's? If you look at the mu's, they are all to be selected to be greater than equal to 0 but there is no constraint that connects one mu with the other mu. The mu i's and mu j's are not interconnected with each other through a constraint. So they can be effectively chosen independently. So lambdas they are each greater than equal to 0, your constraints are satisfied. So now look at this objective function. What would happen if, tell me what would happen if one of these is say less than 0. If one of these is less than 0, what would happen? If c transpose di is less than 0 for some one of these i's from 1 to n. So precisely, so if what you can see here is, if one of these c transpose di's is strictly less than 0, they are being multiplied by a mu outside and the mu is for you to choose, right? Say suppose c transpose d1, c transpose d1 was suppose less than 0. If this was less than 0, then what the optimal value of mu that I should choose, mu1 that I should choose. I can just make it as large as I want and I will keep getting a lower and lower value, right? Because since this is less than 0, say suppose this is minus 1 for example, I am allowed to multiply it by a mu1 and my constraints are selling me, I can make my mu1 as large as I want, there is nothing is being violated by doing that. So I can by increasing mu1 and taking it all the way till eventually till plus infinity, I can bring the optimal value down to minus infinity, is this clear? So that then comes in contradiction with what we assumed in the theorem. Remember we said that suppose the optimal value of the LP is finite, right? So if the optimal value of the LP is finite, it cannot happen that this sort of c transpose d1 is negative, strictly negative. Because the optimal value is finite, it ensures that this cannot be the case. So the kind of polyhedron we are dealing with is the one where c transpose d1 is greater than or equal to 0, right? So let me write this down. So suppose c transpose d1 is strictly less than 0, then making mu1 large, so taking it to plus infinity would give an optimal value minus infinity, which is a contradiction to the assumptions. And what is true for c transpose d1, I just said c transpose d1 is to be non-negative, so this means c transpose, implies c transpose d1 is greater than equal to 0 and what is true for c transpose d1 is also true for c transpose d2 and c transpose d3 and so on. So in short all the c transpose dis have to be greater than equal to 0. So your c has to be aligned in such a way that it makes an acute angle with all the dis, all the extreme rays. Now this answers also the kind of the question you had asked earlier, is it possible that your objective function, that your optimal value is finite even when this polyhedron itself is unbounded, well it is possible, it is possible for that to happen, but for that it is necessary condition is that for that to happen is that c, these the extreme rays and your vector c make a acute angle, acute angle because we are talking of minimization, is this clear? So this is absolutely necessary for that. So this has to be greater, these have to be greater than equal to 0. So now let us come back here, so if this is, so the question then is well if these are all greater than equal to 0, then what is the right value of mu to choose for this, 0 because I am looking to minimize, I want to get as small a value as possible, I can choose any value for mu which is greater than equal to 0, all of them can be chosen independently so long as any of them they are all greater than equal to 0, I do not care, the optimal value for me to choose is therefore 0 because if any of these is, if they are 0, if any of these c transpose Di's are actually equal to 0 themselves, then they drop out of the equation, the ones that are positive I can scale them down to 0 by taking mu equal to 0. So in short this gives us for free the optimal value of mu 1 to mu m is equal to 0. So by just inspection of the linear program and by working with the assumptions that we have which is that the LP must have a finite optimal solution, optimal value, we have got that the mu's are 0. So what has that done? That has reduced our LP to this problem, now we have minimizing, you are minimizing summation lambda i c transpose xi, the mu's are all been said to be equal to 0, so my variables are only lambda or only my lambdas, let me write them explicitly here, lambda 1 to lambda, they are constrained to be greater than equal to 0 and they have to be, they have to sum to 1. So now what can we say about this optimization? So what is the objective here? The objective, if you look at the objective function, it is a weighted linear combination of the lambdas with what weights? With weights c transpose xi, I am taking a weighted linear combination of the lambdas with weights c transpose xi. So if I want this weighted combination to be the largest possible value, what should I take my lambda as, sorry I want it to be the least possible value, what should I take my lambdas as? So this is simply an average weighted average of these c transpose xi with the weights as lambda and the lambdas sum to 1. So this being a weighted average will be greater than equal to the minimum of the individual terms. So this is always, see look at this term c transpose xi times lambda i summed over 1 to k, this is always greater than equal to the minimum of the c transpose xi, the smallest value of c transpose xi as i ranges from 1 to k. And moreover, if I take, if you take lambda so suppose this is the minimum of this is suppose say c transpose xi star, suppose i star is the index that gives you the least possible value, then if I take lambda i star as 1 and lambda i equal to 0 for i not equal to i star, then actually I can get this equal to. So what this means is in this linear program if I take all I have to do is look at the i that look at these coefficients here c transpose xi and look for the i star which is for where the c transpose xi is least, i equal to i star for which the c transpose xi is least put all my weight of lambdas on to that one and put everything else equal to 0. That is that sort of point is feasible because it satisfies the constraint and it achieves this low least possible value you can take, is this clear? So what this means is the optimal solution. So an optimal solution is lambda i star equal to 1 and lambda i equal to 0 for all i not equal to i star, this is an optimal solution. Now what is this, what does this mean then? What is the, what is your x? Let us we did a change of variable and got the problem in terms of mu and lambda, now let us go back and recover the x. Remember we said we will put x, express x in terms of mu and lambda in this sort of way, right? So now we found our mu's and we found our lambdas, what is the corresponding x now? It is x i star, right? Because all the mu's are 0, lambdas are also 0 except for i star, i equal to i star. So my x which means the optimal, an optimal solution of the original problem is x equal to x i star. Now is x i star equal to x i star, okay? Now what is x i star? x i star go back was one of these points, it was one of these ways, right? And what were these? These were the extreme points, these were the extreme points of the polyhedron, right? So what have we got for free? We have got that there exists, right? So these were your extreme points. So what have we got? We got therefore that the LP as a solution is optimal solution at an extreme point.