 last class we have discussed that if you have a unconstrained optimization problem is there we can solve it by using the KKT condition. And we have discussed the KKT necessary conditions agree. So, next is a sufficient condition whether that point regular point is a minimum value of the function will give you or the maximum value of the function will give you that will be decided by using the sufficient condition. So, let us call that the second order sufficient conditions KKT. So, suppose that extra is the feasible point feasible point means the point which is lies in the what is called feasible set. And at that point that equality constraint and inequality constraint must satisfy. So, that is the feasible point suppose extra is a feasible point for the constraints it must satisfy this constraints is equal to 0. And we have a there are p equality constraint as well as it should satisfy the inequality constraints less than equal to 0 for j is equal to 1 2 dot m this. Now, question is if there is a if there exist this is called there exist vector this symbol is there exist a vector lambda star it is a multi lagrange multiplier what is this one lambda 1 star lambda 2 star dot dot lambda p star. Because, there are p equality constraints are there if there exist vector whose dimension is p cross 1 and mu whose dimension is m cross 1 is equal to mu 1 star mu 2 star dot dot mu m star. If there exist this and this vector so that KKT condition that KKT necessary conditions satisfied condition satisfied what is the necessary condition gradient of Lagrangian function must be equal to 0. So, what is the gradient of Lagrangian function with respect to x if you see that L gradient of L with respect to x is nothing but a gradient of f of x this plus summation of i that is we have shown last class that is equal to lambda i star gradient of h i with respect to x this plus summation of that is j is equal to 1 to m gradient of this x g j x star this must be equal to 0 whose dimension is n cross 1 whose dimension is n cross 1 the dimension is n cross 1. So, and also there are which is called switching function must satisfy mu i mu j g j x star j must be equal to 0 for j is equal to 1 to m and mu j is the Lagrangian multiplier associate with the inequality constraints mu j must be greater than equal to 0 that we have shown last class 1 2 dot dot m. So, this is less than equal to less than my greater than equal to 0 when the constraint will be this type of constraints are less than equal to 0 then this will be greater than equal to 0 non negative mu is non negative that we will see later in your discussions. So, this is the thing are satisfied and for any vector for any you see this is so that the necessary conditions are satisfied this condition the same and for any vector z any non zero vector rather you write it non zero vector any non zero vector z whose dimension is n cross 1 satisfies satisfying the following conditions. What is this following conditions the gradient of g j of x star with respect to x this multiplied by z transpose this must be equal to 0 and mu j star equal to greater than 0 this is greater than 0 for all active inequality constraints for all active for all active inequality constraints constraints. For all active equality constraints means at if x star is a your regular point then at that point that that inequality constraint if it is a at that point g j of x is equal to 0 at x is equal to x star where x star is the regular point then it is called the active constraints. Because I mention it we have a inequality constraint is small m out of small m if at regular point if it is small l is the number of equality constraints satisfied and remaining small m minus l is the satisfy the that inequality conditions i is equal to 1 to j is equal to 1 to dot l and j of x star is less than equal to 0 for j is equal to 1 to dot dot m minus l constraints. These are called active equality constraints and these are called the inactive equality constraints this. So, this condition must be there naturally when this is greater than 0 then g of j is the your active constraints and gradient of that at that point x star gradient of that point with any vector if you multiply by this transpose of this one this must be equal to 0 because that are one is tangent another is perpendicular to that one. So, this is a autonomy condition is must satisfy next is that transpose gradient of x g j of x star is less than equal to 0 when mu j and mu j star equal to 0. So, this constraint when mu j g j is inactive constraints at that condition this quantity will be less than equal to 0 and next is similarly in equality constraints that with h z transpose gradient of x with h i of x star is equal to 0 for all constraints this is for all constraints h i x star is equal to 0. This is the all constraints for active inequality means when g j of this is 0 this condition this is for inactive constraints this. That means if you have a let us call constraint this is one constraint is there g 1 of x this constraint and another constraint is like this way g 2 of x agree at this point the function value is 0 now what you find out the gradient of at this point because this equality constraints are satisfied at this point you will find out the gradient. That means you draw the tangent corresponding to this curve here and perpendicular to this one is the gradient of g 2 of x this is and here corresponding to this curve draw a tangent and perpendicular to this one is the gradient of g 1 of x star. So, that direction z tangent of this one is z z transpose of this must be this equal to 0 that that one similarly h case also you can show it. So, this is all constraint for i is equal to 1 2 dot dot p. So, if this is the satisfy then this point will that point x star will be a then point x star will be a what is called negate that if the following constraints because we are finding out the sufficient condition if the following constraints is true. That means z transpose z is a any vector is that second partial derivation of derivative of f of x at x is equal to x star this plus summation of i is equal to 1 to p this is nothing but a just like a unconstrained optimization. We have done the second derivative of the function that is hasian matrix must be positive definite for minimum value of the function hasian matrix must be negative definite for maximum value of the function. Similarly, this our now the function is unconstrained function is Lagrangian function the Lagrangian function similarly second derivative of this one with respect to x what we are getting it is that lambda i star gradient of x h of i of x plus summation of j is equal to 1 to m mu j star gradient of x g j of x star into z greater than 0 greater than 0 means it should be a positive definite this indicates this matrix this is the matrix of dimension n cross n. This is a symmetric matrix of dimension n cross n and this quantity will be positive definite provided this matrix is positive definite. So, our condition is this must be in order to x is equal to x star should be a minimum point then this must be a positive definite matrix f of x star plus summation of j i is equal to 1 to p lambda i star gradient of this not gradient second derivative of this hasian matrix of h. So, this is a star this plus summation of j is equal to 1 to m mu j star g j of x star this matrix must be greater than 0 means it should be positive definite then we will call x is equal to x star we will give you the minimum value of the function agree and that point is a isolated point that point is a isolated point optimum point isolated point means near around this point there is no other minimum value of the functions agree. So, this is a you can say isolated point means there is no other around this point will have a minimum value of the function. So, this is the isolated point of this function. So, how to calculate this this one you know this that is if you write more clearly that or isolated point means there is no other local minimum point in the neighborhood of x star agree the isolated point once again isolated point is means that there is no other local minimum point around x star around that x star. So, let us see how to calculate this one x star that is hasian matrix of f of x at x is equal to x star how you calculate this if you recollect this one is nothing but a deleter square of f of x del x 1 del x 2 sorry del x square of f x then del x 1 del x 2 and since we have n variables are there we have n variables are there then we will take del square of f x plus del x 1 del x n. And similarly you can write it that del is del of x square del x 1 del x 2 and del f of x del x 2 square and this way del square of f x del x of 2 del x n and if you continue like this way last term will be del f square del x 1 del x n then del square f and del x 2 del x n and last term is your del f square x del x n square. So, put this values at x is equal to x star. So, you have computed and similarly if you see this one del x that it hasian matrix of h 1 it varies from i is equal to p h 1 you will again you will get a matrix you will get a matrix of dimension say with respect to a dimension n cross n. So, this dimension is n cross n. So, similarly I am not writing this one del square f h i of x you can get it in similar only you can say f will be replaced by h i this f will be replaced by h i function. Then you can write it this is also n cross n and you have to put the value x is equal to x star. So, I have just write it first for you that h i of x del x 1 square h del i of x then del x 1 del x 2 and this way del square i of x del x 1 del x n. In this way last is del square i of x del x 1 del x n and if you do this one last one is del square i of x del x n square. So, this is the second part of this sufficient conditions we have written all i. Now here we to write it if you see this one here you have to write it i varies from 1 2 dot dot p similarly you can write it delta x that g j of x. So, I will not repeat only in place of f you replace by g j and that is also n by n matrix put the value x is equal to x star and it is also symmetric matrix. Once you do this one next is you have to check this matrices you know add all this together you will result and you will get another matrix. So, you will see whether that matrix is positive definite or not and you know how to test the positive definite matrix or not by silver blaster equality constraints, equality conditions or you can find out if a matrix is positive definite matrix. Then you find out it's Eigen values of the matrix if all the Eigen values are positive. Then it is a positive definite matrix if the all the Eigen values are negative then it is a negative definite matrix. Similarly, positive definite matrix. You that matrix you find out the Eigen values of this matrix if some of the Eigen are positive, some are what is 0, then it is a positive semi definite matrix. Similarly, positive negative semi definite matrix case also you will get some of the Eigen values are negative and some of the Eigen values are 0, at least 1 will be 0. So, once you know all this one, then I can find out the second order sufficient conditions. Again if this matrix is positive definite, we will call it is a what is called the function will give you the minimum value of the function and x star is the isolated point. Isolated point means this that the function does not have any other local minimum point around that x star. So, let us now solve a problem, then how to work out this problems using our technique that what is called constant optimization problem, how to find out. So, our problem is minimize example minimize f of x is equal to x 1 square plus x 2 square minus 4 x 1 minus 6 x 2, subject to constraints x 1 plus x 2 is equal to 2, 2 x 1 plus 3 x 2 is less than equal to 12. And we are assuming that our side constraint x is greater than 0 and x 2 is greater than 0 this. So, this problem you have to solve it by using KKT conditions that is our problem. So, what is our job, this is our equality of constraints is there and inequality constraints are both are inequality constraints again this are inequality constraints this are. So, we do not have any equality constraints. So, in that case h i of x is equal to 0 that lambda is associated with the equality constraints. So, lagrange multiplier lambda i will not be there in our solution. So, let us say first you write it the solution of the problems. So, let us call first we write it the necessary conditions again. So, first we write the lagrangian functions that function f of x plus summation of there is no h equality constraints. So, that term will not come into the picture that is a inequality constraint j is equal to 1 to m in our case m is equal to 2 there are 2 inequality constraints if you see here this and this. So, 2 then mu j then our inequality constraints we have converted into a equality constraint that is if you see g j of x plus as j square this one. So, now I use the necessary conditions for this one. So, I will find out del x 1 is what is equal to 0 because. So, what is this one f of x you see f of x is that one. So, you differentiate with respect to x 1. So, it will be 2 x 1 minus 4. So, it will be a 2 x 1 minus 4 and we have a also if you see this one or I write it then it will be clear instead of calling this one. What is f of x it is a x 1 square plus x 2 square minus 4 x 1 minus 6 x 2 plus mu 1 g 1 of x g 1 of x is your what you see g 1 of x is equal to x 1 plus g 1 of x is equal to you see if you call this is g 1 of x x 1 minus x. So, I can write it this if you see this one I can write it this g 1 of x g 1 of x is equal to x 1 plus x 2 minus 2 less than equal to 0. So, this is our standard form we have written similarly, this is also you can write g 2 of x you can write it this one g 2 of x is equal to g 2 of x you can write it is equal to 2 x 1 plus 3 x 2 minus 12 is less than equal. So, you have written our standard form type less than equal to form. So, our g 1 of x if you see x 1 plus x 2 minus 2 and we are adding with a some positive quantity. So, that it will turns out to be a equality constant plus mu 2 then 2 x 1 that g 2 g 2 is your 2 x 1 plus 3 x 2 minus 12 plus s 2 square. Now, I am differentiating this with respect to x 1. So, it will be a 2 x 1 4 x then here is x 1 is there mu 1 and no other x is here here again is there there is 2 mu 2. So, this equal to 0 this is one equation again you have to differentiate with respect to x 2. So, if you differentiate with x 2 this is 2 x 2 minus 6 then it is a plus mu 1 then plus 3 mu 2 because I am differentiate with respect to x 2 mu 2 then this is 0 with respect to this with respect to x 2. So, this is 0 then this is the 1 equation let us call this is equation number 1 this is equation number 1 and this is equation number 2 and we have a in fact we have to differentiate L with respect to there is no h. So, we need not to differentiate with respect to lambda. So, we have to differentiate with respect to our mu and another thing L we have to differentiate with respect to s we have shown it that these two conditions can be combined together and finally, we can write it that condition is mu i g j of x is equal to 0 and our case j is equal to 1 and 2. So, what is this condition we will write it mu 1 g 1 of x must be 0 another condition is there mu 2 g 2 is 0. So, we have to solve this two equation we have a how many unknowns are there x 1 x 2 mu 1 mu 2 and we have a 4 unknowns two equation and from there we have to see how to solve it. Look at this one that there is this has a two possibilities to satisfy this condition one condition is mu 1 if it is a mu 1 is 0 and mu what is called g 1 is not equal to 0 another choice is g 1 is 0, but mu 1 is not equal to 0. When mu 1 is not is equal to 0 g 1 is not equal to mu 1 is equal to 0 case mu 1 is equal to 0 g 1 is not equal to 0 that means, it is a inactive conditions when mu g 1 is 0, but mu 1 is not equal to 0, but this condition is satisfied in that situation it is called active condition is satisfied. Similarly, you have this in general if you have a m small m number of inequality constant we have a 2 to the power of m possibilities are there to satisfy this equation. Again 2 to the power of m possibilities are there when switching we can switching 2 to the power of m switching is possible to satisfy this equation when that we have a j is a j is equal to 1 to m that means, m constants are there. So, in short if you have a m constants are the inequality constants are there then we have a such type of equation we have m equations are there which in turn we have a 2 to the power of m switching conditions are there to satisfy this equations. So, let us solve for this case. So, our first case we are considering that case 1 mu 2 is let us call 0, if mu 2 is 0 then g 2 cannot be 0 again so g 2 is our condition is g 2 is less than equal to 0. So, equal to cannot be so it is this. So, it is a inactive condition the condition is more relaxed when it will g 2 is a is equal to 0 it is a active condition. That means, it is a more tightened conditions that means it should be on the curve or on the line if it is g x is a straight line then it should be on the line only when g x is a active condition 0. So, we have a this one condition mu is 0 then we can get that g 1 is 0 g 1 of x g 1 is equal to 0, but if g 1 is 0 mu 1 is greater than 0 both cannot be 0 mu 1 g 1 mu 1 is equal to 0 if this is 0 it is enough to satisfy that condition. So, in this case that means mu is 0 g 1 is 0 in this case what is this situation we will see from 1 and 2. So, from 1 and 2 equation I will put our if you see mu 2 is 0 and there is a constraints are there g 1 is 0. So, we will put mu 2 is 0 from 1 and 2 then we will get it 2 x 1 minus 4 plus mu 1 is equal to 0. Another equation I will get 2 x 2 minus 6 plus mu 1 is equal to 0 this is and we have a g 1 is 0 g 1 is what x 1 plus x 2 minus 2 is equal to 0. So, we have a 3 unknowns are there we have to solve this 3 equations by solving this equation we get we obtain x 1 is equal to half or is equal to 0.5 x 2 is equal to 3 by 2 is equal to 1.5 and mu 1 we got it 3 and mu 1 must be I told you when it is a g 1 is active constraint mu 1 must be positive non negative number. If you get it by solving this one negative that means it does not give any solution for to become a what is called optimal value of the function. So, this must be this is 3 means greater than 0 and mu 2 is equal to 0 this we got it. Now, we have to see whatever this point is it must satisfy all feasible condition I mean all equality condition. And equality conditions if it is satisfy then we will call the point is in the inside the feasible space or feasible set or feasible region. So, let us see feasibility check. So, our g 2 we have to check it now g 2 of x if you see g 2 of x is what twice x 1 plus 3 x 2 minus 12 if you see the g 2 of x ever expression. And this we have to show it is our condition must satisfy our condition is less than equal to 0. So, this condition if you put the value of x 1 is equal to half value of x 2 is equal to 3 by 2 then what will get it that one will get what is called this is half means this is 1. And this is 3 by 2 3 by 2 means this is 9 by 2 means 4.5 minus 12 and that is less than 0 that 5.5 minus 12 is less than 0. So, this satisfy this condition. So, this point may be one of the what is called point which will give may give the your minimum value of the functions. So, that we have to check it next. So, how will you check it that one. So, this is this satisfy all what is called constant equation at this experiment that x 1 is half and 3 by 2 it satisfy all constant. So, we can see this is the solution that means x 1 is x 1 star is equal to half x 2 star is equal to 3 by 2 is the solution. Now, here also you can check it whether it is the minimum value of the function or maximum value of function without checking the sufficient efficiency condition. And that condition will call that how to check that one let us see. So, let us call a function at x is equal to x 1 is equal to x 1 star and x 2 is equal to x 2 star. This value is I know x 1 plus x 2 minus 4 x 1 minus 6 x 2 minus 4 x 1 minus put the value x 1 is equal to half and x 2 is equal to what is x 2 star is equal to 3 by 2. If you put this value of this one you will get this value is 8.5. So, this you got it. Now, how you say this is the you got the value of the function at that point, but how you ensure that this will give you the minimum value of the function. One can check this thing by using the monotonicity analysis that is monotonicity analysis. What is this? I also we have discussed you earlier that at this point you give a small perturbation. If it is a minimum value and you give a perturbation around this point the function value if it is a increase then it indicates we have reached the minimum value. So, this is called the monotonic analysis. So, what is this? Let us call we put the value of the function x is equal to this at x 1 is equal to let us call we have found this value at x 1 is 1.495. Instead of instead of what is called that finding out the x 1 is not this is x 2 is 1.1 instead of 1.5 I am writing 1.495 x 2. That means what is the change I made it 0.005 and x 1 I made it let us call 0.505, 0.005 added and from 0.005 is part of it from this one and if you put this value at this one you will get that value is 4.8.495. That means function value is increased from minus 8.5 it is a minus 4.5 it increased. If you see in the dimension x 1 and x 2 dimension x 1 and x 2 dimension if it is 8.5. This is corresponding to 8.5 this is 8.5 then this is 8.495 that means this function value is increased from here to here. So, this indicates this is what is called give a minimum value of the function also we will check through what is called necessary sufficient condition of KKD condition. So, case 2 may be case 1 you say we have made it case 1 mu 2 is 0 I will make it mu 2 is 0. Now, there is another possibility is there that mu 2 mu 1 also 0. So, our case keeping mu 2 0 same that means g 2 of x is less than 0 and we are meeting making that mu 1 is 0 that means g 1 of x is less than 0. So, again from 1 and 2 again from 1 and 2 we can write it twice x I will put the value of mu 1 and mu 2 in that equation 1 and 2 equation. So, I will get twice x is equal to 4. So, x 1 is equal to 2. So, twice x is equal to 6 x 2 is equal to 3. So, I have to check this value I got it. So, I have to check the feasibility conditions and mu 1 mu 2 is 0 feasibility conditions I have to check the feasibility condition. What is the feasibility condition g 2 of x I have to find out g 2 of x I have to find out at x 1 is equal to 2 x 2 is equal to 3. Let us see this one what is g 2 of x if you see this one x 2 minus x 2 plus 3 x sorry x 2 x 1 plus 3 x 2 minus 12. So, this is our g 2 of x put this value x 1 is equal to 2 x 2 is equal to 3 and you will get this value g is 13 minus 12 which is not equal to which is less than is not less than 0 this is not less than 0 this is means this is equal to 1. So, it does not satisfy this one that means we have a point 2 and 3 this is outside the feasible region feasible space. Similarly, one is enough to check that means this is not the acceptable solution here. So, you can also check g 1 of x at x 1 is equal to 2 x 2 is equal to 3 and in this way g 1 of x you know x 1 plus x 2 minus 2 put the value of x 1 is 2 x 2 is equal to 3. So, x 1 is a 5. So, it is a 5 minus 2 which is not less than 0 which is not less than 0 agree. So, this 2 condition g 1 d. So, our conclusion is that g 1 and g 2 g 1 of x and g 2 of x at x 1 is equal to 2 x 2 is equal to 3 violates the conditions. So, our conclusion then this is not the our or condition that this indicates this implies the x 1 is 2 x 2 is 3 is not does not belong to feasible space. This does not belong to a feasible space or design space agree. So, this case 2 will not be acceptable that way case 3 on other words is not feasible space there is no solution at that point. So, case 3 now I made mu 2 is 0 agree then this another possibility is you make it mu 1 is 0 is mu 1 g 1 of x is less than 0. Then you have a g 2 of x g 2 of x is equal to 0. That means mu 2 greater than 0 agree. So, this greater than equal to 0 this. So, let us see this one mu 1 is 0. So, from equation 1 and 2 from 1 and 2 I will put mu 1 is 0 in this. So, I will get it once again if you see I will get twice x 1 minus 4 plus mu 1 is 0. So, twice mu 2 is equal to 0 from equation 1 I will get this one putting mu 1 is 0 from equation 2 2 x 2 minus 6 plus 3 mu 2 is equal to 0. I will get it and we have a g 2 is 0 g 2 is what if you see g 2 over 2 x 2 sorry 2 x 1 plus 3 x 2 minus 12 agree g 2 is 0 this one. So, solving we have a 1 2 3 unknowns are the three equations are solving this solving we get x 1 is equal to x 1 is equal to 1.846 x 2 is equal to 2.769 and mu 1 you have considered is 0 and mu 2 is and mu 2 we are getting is 0.31 which is greater than 0 agree. So, this satisfy all the conditions agree. So, this may be a 1 of the possible solution to get the minimum value of the function, but first still you have to check it either what is called monotonicity analysis or using the what is called by KKT second order conditions. So, let us check the feasibility conditions our feasibility condition is this condition must be satisfied g 1 is less than 0 this must be satisfied. So, our feasibility conditions if you see this one our feasible condition g 1 of g 1 of x what is g 1 x 1 plus x 2 minus g 1 x is minus 2 I think. So, put the value x 1 is equal to that we got it x 1 value is how much we got it x 1 we got it if you see this x 1 is 1.846 x 2 is equal to 2.6 9. So, if you put this one g 1 is you see does not satisfy. So, it is a 1.846 plus 2.769 minus 2 which is greater than 0 agree which is greater than 0. So, it does not satisfy the our monotonicity analysis you know what is does not satisfy our constraints this must be less than 0. So, this cannot be a once again this cannot be a g 1 of x violates the condition. So, the point x 1 is equal to 1.846 and x 2 is equal to 2.769 cannot be a solution of our optimization problem because it does not belongs to in feasible region feasible space this point is not belongs to feasible space. So, reject that solution agree. So, you see this one our case 3 what we made it your case 3 case 3 is that one we made it that g 1 is 0 natural g 1 into g 2 must be 0. So, this will be less than 0. So, this equation g is coming from equation number 2 and 3 putting mu 1 is 0 this then I have g 2 is 0 this. So, solving this equation I got this. So, it must satisfy this equation and when we are putting this equation we got it that one is it does not satisfy this one. Now, case 4 what is the choice is left now that g 1 is 0 mu 1 I made it 0 g 1 of is g 1 is equal to 0 which means mu 1 is greater than 0 and next choice is left g 2 is 0 that means mu 2 is greater than 0. So, let us say using this equation in from 1 and 2 from 1 and 2 what will get it 2 x 1 minus 4 plus mu 1 plus twice mu 2 is equal to 0 this I am getting from equation 1 and then from equation 2 I am getting is 2 x 2 minus 6 plus mu 1 plus 3 mu 2 is equal to 0. So, this is you are getting from equation 2 this one. Now, so this we have a how many variables are there x 1 x 2 then your x 1 x 2 mu 1 mu 2. So, you have a 4 variables are there agree. So, now use this conditions because this is the we consider g 1 is 0 g 1 of this 0 means g 1 is what x 1 plus x 2 minus 2 is equal to 0 agree. So, this and g 2 is 0 means that is 2 x 1 plus 3 x 2 minus 12 is equal to 0. So, you have a 4 equations are there 4 unknowns are there mu 1 mu 2 mu 3 mu 4 x 1 x 2 mu 1 mu 2. So, you solving this equation solving the above equation we get agree solving the above equation what will get or you see this one either this first you solve this and this is a variable of 2 variables g 1 x 2 x this you have a 2 variables 2 equation. So, I can solve this or you can write it solving solving solving we get solving this equations this equations we get x 1 is equal to minus 6 and x 2 is equal to 8 agree. And it does not satisfy our what is call our side constants we have put it x 1 is greater than equal to 0 x 2 is greater than 0. So, let us call and solving the other 2 equation other solving solving the other 2 equation other other 2 equations again solving other 2 equations or you just write it solving other 2 equations we get mu 1 is equal to 68 and mu 2 is equal to minus 26 which is not greater than 0. So, it also violates our what is called lagrange multiplier conditions there are side constants are violated and our what is called the lagrange multiplier for inequality constant that should be positive, but it is become negative. So, mu 2 violates the conditions. So, no solution at this point point what is this point x 1 is equal to we got it x 1 is equal to minus 6 x 2 is equal to 8 not only this you can stop it here because this does not satisfy our side constants. So, now if you see out of the 4 cases now in short if you have a n if sorry if you have a m small m inequality constant you will you have to solve the 2 to the power of m cases for considering the switching possibilities cases. Out of this you have to search on which one will give you the feasible design space or feasible solution and at that point whether it is a minimum or maximum one can check by monotonicity analysis or you can check by considering the second order KKT conditions condition. So, if you see that second order conditions. So, our acceptable our acceptable things is check our acceptable x 1 is equal to 0.5 and x 2 is equal to 1.5 star agree this. So, you have to check the KKT conditions second KKT conditions what is the second order KKT conditions f of x this f of x plus summation of j is equal to 1 2 2 mu 2 mu 2 mu 2 mu 2 mu 2 mu j mu j del f del x g j of x star at this star agree. So, this you have to because we do not have any equality constant. So, this 2 things you have to find out. So, let us call what is f of x is our x 1 square plus x 2 square minus 4 x 1 minus 6 x 2. And what is g 1 of x is equal to x 1 plus x 2 minus 2 g 2 of x is equal to 2 x 1 plus 3 x 2 minus 6. So, find out the this this of this at x 1 is equal to x 1 is equal to 0.5 x 2 is equal to 1.5 agree and this will come if you do this one it will come 0 0 to 0. And this second derivative of g 1 of g 1 of x star which is equal to x you will get a null matrix agree. And similarly g x of g 2 of x you will get a null matrix. So, at this matrix at this matrix and ultimately you will get this one. So, this is positive definite matrix. So, 0 0 2 is a this matrix is a positive definite matrix by using single step. So, our conclusion is we will get the optimum value means minimum value of the function at this point. And corresponding mu is what if you see there is corresponding mu of that one is that mu you just see that mu you got it positive quantity. If you recollect this one mu 1 is equal to what and mu 2 is equal to what you will see is a positive quantity. So, we will stop it here next class we will discuss some of the issues related with the what is called parameter variation in the objective functions effect of parameter variation in the objective function. Thank you.