 last class we have discussed how to solve the LP problems using two phase simplex method. We have taken this example maximize this function subject to this constraint and y 1 is greater than equal to 0 mean non negative number and y 2 is unrestricted sign. So, this y 2 we have to convert into a by introducing two new variables. We have defined y 2 is y 3 and y 4 which is individually that y 3 y 4 is greater than equal to 0, but this problem we have to convert into a standard LP problems that we have seen how to convert the standard LP problems. After converting this one then you see this type of equation when you will get the this type of inequalities greater than equal to equalities then we have to introduce a variable which is called artificial variable. This artificial variable is that x 6 in our case and that artificial variable we consider we have to minimize that means we have considered the artificial variable x 6 as a w 6 a w and this simultaneously you have to optimize that minimize two function objective function one is x 6 which is we denoted by w 6 another is your our original objective function and this first phase what we optimize that x 6 is a w this we have to minimize this one in first phase of the tabular form. After the first phase we got the solution after completing the first phase we got the solution of this that is our artificial objective function value is omega is equal to 0 that should come omega is equal to 0 or x 6 is equal to 0 and corresponding the design variables what we consider x 1, x 2, x 3, x 4, x 5 and x 6 is this value we got it. Then next second that phase 2 since w x of 6 is optimized minimized then we ignored what is called calculation in the table x 6 variables. So, you just ignore x 6 so in second phase it will just find out the cost function of the original systems original objective functions it will find out and during after completing the second phase we got the function value is f is equal to 12 and corresponding the variables value x 1, x 2, x 3, x 4, x 5 we got it this values x this. So, if you look carefully this one we have this is our basic functions of that one if you plot it that one we have seen that this basic function if you plot it graphically if you see this is the our equation one of the constant is this one when you put y 1 is equal to 0 x 2 y 2 is equal to 6. So, this is our 6 when y 2 is 0 x 2 is equal y 1 is 4 that is our 4, 4 0 coordinate this is 0 6 coordinate. Then another constant is 2 y 1 plus 3 y 2 is greater than equal to 6 our original problem if you see putting would just to draw this equation straight line y 1 is 0 y 2 we will get it 2. So, this value is 2 then y 2 value is 0 then where it cuts the x x is this one that is equal to our 3. So, this coordinate 3 0 this coordinate is 0 2. So, it cuts the x x is if you see this cuts x x is y x is this is let us call point b this is point a and this is cross this intersection of two straight line is c. Now, according to our problem if you see y 1 is greater than equal to 0 that means y 1 indicates the write up of the this vertical line y 1 the whole vertical write up of the vertical line and y 2 is unsigned that may y 2 value can be positive and negative that means that may be above this line above this horizontal line or below this horizontal line. So, if you see carefully this one our feasible region is that portion only of our corresponding problems and in this case our objective function if you see the our objective function is nothing but a z maximize z z is equal to y 1 plus twice y 2 and that is we have to maximize and graphically you see when this function value will be 0 a this is the equation of this function will be 0 when this passing through this origin this. So, this is the equation of that line. So, that means we are finding out the what value of y 1 y 2 in the feasible region the function value will be maximum this is the maximum we want maximum value of this function. So, if you put if you just move this line parallel to this that line which is passing through the origin you will see maximum value of this function value will get at this point b point parallel to this one parallel to this objective which is passing through this one. So, that one will give you this one let us see in first phase of our problem what we got it if you see recollect that first phase in first phase first iteration we got the value of if you see x 1 0 x 2 0 x 3 0 x 5 0 and from this one what we can say y 1 y 1 is nothing but x 1 that means x y 1 is 0 y 2 y 2 is what we have defined y 2 is nothing but a x 2 minus x 3. So, x 2 minus x 3 both are 0. So, this point origin point is the in phase phase 1 first iteration we got it this point y, but this point is not in the feasible in feasible and next iteration what we get it in the first phase next iteration if you see we got x 1 is 0 x 2 is 2 and x 3 is 0 in other terms because y 1 is equal to our x 1 that means our y 1 value is 0 if you see y 1 value is 0 and then y 2 value is x 2 minus x 3 means 2 that means we got this point. So, and immediately we can say the function value is increased from previous value previous what at this point function value is 0 at this point one can find out what is the value of this it will be increased physically also you can say this straight line is moving parallel to then cuts y x is some value of this instead of 0 it is cutting this one. So, this and this is the first phase after completion of first phase now see after completion of second phase which will minimize the function of that one. In other words our original function will be maximized because we have converted into standard LP problem that is why minimization term we are using. So, in the after the completion of second phase of this one our x 1 value is you see x 1 value is 0 x 2 value is our 6. So, this indicates x 1 0 means x 1 in y 1 is equal to x 1 that means y 1 is 0 y 2 is what x 2 minus x 3 the way we have defined. So, 6 minus 0 6. So, this is this point on this point is I will give the maximum value of the function this time agree. So, graphically also one can represent this one. So, today we will discuss what is called that we can solve the what is called LP problem by using primal and what is called dual problems considering the standard primal and dual problems. So, primal and dual problem. So, what is primal problem first standard primal problem is what we will define standard primal problem. So, first we convert this problem into a standard primal problems. So, what is this what is the standard primal problem maximize z p is the objective function is the way we have defined x 1 d 1 x 1 d 2 x 2 plus dot dot d n x n we have n variables are there. You can write into vector form this d transpose x x is a dimension n cross 1 and d is a vector column vector is your d 1 d 2 dot dot d n and this dimension is n cross 1. So, this is subject to a 1 1 x 1 a 1 2 x 2 plus dot dot a 1 n x n is less than equal to e 1. Similarly, second equation constant equation and constant is of this type less than equal to x 2 1 x 1 a 2 2 x 2 plus dot dot a 2 n x n less than equal to e 2. So, in this way we have a n inequality constant in this way we have a n inequality constant less than equal to type. So, it is a m 1 x 1 plus m 2 x 2 plus dot dot a m n x n dot dot is a less than equal to e m. So, and also x j is greater than equal to 0 for j is equal to 1 2 dot dot n and e j e i i is equal to 1 2 dot dot i we have a m constants are there dot dot m this e i is unrestricted in sign. It can be positive negative 0 whatever may be it is unrestricted in sign. So, no restriction you can just write it no restriction on it sign. So, this is called standard standard there dot dot m this E I is unrestricted in sign it can be positive negative 0 whatever may be it is unrestricted in sign. So, no restriction you can just write it no restriction on it sign. So, this is called standard primal problems. So, this you have to if you are given the LP problem you want to convert into a standard primal problem you have to convert into this structure my mind it from the standard LP problem difference is there the right hand side of this one is not it is not restricted it can be unrestricted it can be any sign positive negative all these things. So, a set of equation I can always write in terms of matrix X A into X less than equal to E that E dimension n cross m cross m cross 1 this dimension is n cross 1 immediately I can write it this dimension is m cross n and E is a vector components of E are E 1 E 2 E 3 dot dot E m this. Suppose in this equation if you get you are given a LP problem which having a equal to sign and we have discussed earlier if you get that equal to sign say like this equality constants if you get it let us call something like 2 x 1 plus 3 x 2 is equal to 5 that equivalently because I have to convert into if you want to convert into standard problem you have to convert into less than equal to. So, this can I can do it immediately I can write by set of 2 equations 3 2 x plus x 2 is less than equal to 5 another I can write 2 x plus 3 x 2 greater than equal to 5 equivalently I can write it this 2 equivalently is same as that one or if you have a equality of this type let us call inequality if you have a inequality constraint of this type then how to convert into a standard primary problems this one. So, suppose you have a equality is like this way 10 x 1 minus 6 x 2 plus 7 x 3 is greater than equal to 12. So, what we will do it both side you multiply by minus. So, if you multiply by minus both side minus that it will be 10 x 1 plus 6 x 2 minus 7 x 3 is less than equal to minus 12. So, this has converted into a standard primal inequality form. So, this keeping this thing in mind any LP problem I can convert into a standard what is called primal problems. Now, what is a dual part of this one that means standard dual problems this problem this standard primary problem also referred as normal maximization problem this standard LP problem also referred as normal maximization problem. Now, next is what is the dual problem of this one next is the dual the dual of this problem dual of primal problems what is the dual of primal problem this what is written maximization I will write minimizations I agree minimize that let us call I have denoted by f d that I have denoted by z p p stands for primal problems means original problem and dual stands f of d is this is a minimization. Then how will you write it see dual problems whatever the inequality constant you have you have a M equality constant and right hand side of the inequality constant E 1 E 2 E dot dot E m that quantity will come to the objective function of the dual problems agree. So, you will write it this E 1 multiplied by some dual some new variable y 1 dual variables you can say E 2 multiplied by y 2 the new variables in this way. So, if you have a M equations are there M inequality equation there in standard primal problems that dual problems you will have a new M variables state of x variables in fiber x n variables in the primal problems in dual problems you will get M variables new variables you have to introduce there. So, I am writing this one E 1 into y 1 E 2 into y 2 agree E 2 into y 2 and this E 1 E 2 is known to us and then in this way E m y m. So, in that standard primal problem inequality constant we have a M inequality constant in dual problem we have to make new M variables y 1 y 2 with an objective function which is minimization of f d is formed with the coefficient you have got in the right hand side of the primal problem agree that coefficient that values you have to introduce in the objective functions. So, this is minimized subject to so this I can write if you write it in a vector form E transpose y and y is what m cross 1 and where is E is at E is a vector of dimension m cross 1 agree or problem is now minimize this one subject to see this one this that a 1 column wise I am reading a 1 multiplied by y 1 a 2 a 2 1 multiplied by y 2 a m and similarly m n multiplied by y m. So, I am writing first constant equation a 1 1 multiplied by y 1 a 2 1 multiplied by y 2 and dot dot a m 1 multiplied by y m and whatever the coefficient was there in the no standard primal problem in the objective function coefficient that coefficient will come in the right hand side of the inequality constant in the dual problem. So, our inequality constant of this type greater than equal to and this is the first constant corresponding to the first one it will come d 1. Similarly, see take the second column second column multiplied by second column is a 1 multiplied by y 1 a 2 2 multiplied by y 2 in this way we have a m 2 multiplied by y m is will be greater than equal to second coefficient associate in the standard primal problem agree in the objective functions. So, this equal to I can write it next a 1 2 y 1 a 2 2 y 2 plus dot dot a m 2 y m is greater than equal to d 2 and if you continue like this way last one will be a 1 m y 1 plus a 2 m 2 m a 2 m here is 2 m a 1 sorry last last will be a 2 m a 1 2 2 1 m 2 1 this is m m 2 y m agree. So, last one will be m this will be m m row m 1 last row you see what will be the this I have written this a 1 m sorry a 1 n I am writing this one last one that how many equations will be there there will be n equations. So, last column is a 1 n into y 1 plus a 2 n see this one 2 n the y 2 plus dot dot a m n m n agree into y m that will be equal to d n. So, you see that how many coefficient in standard primal problems are there in objective function there are n coefficients are there and that and that n coefficient will come in the dual problem inequality constant. That means n inequality constants will be coming here. So, we have written this things if you say a 1 a 1 1 y 1 then a 1 2 a 1 2 this this then this a 1 m y m and similarly this is a second equation we will get it with a d 2 and the n th equation it will be d n. So, this is we will get a set of inequality and inequality constant with the help of standard primal problems objective function coefficient will come to the right hand side of this inequality constants. And if you see this inequality constant this if you express matrix form it is a transpose of y is greater than equal to d d is your dimension n cross 1 agree and this m is m cross 1. So, this we can write and here y i is greater than equal to 0 agree and d 1 d d i i varies from 1 to dot dot n this has no restriction on its sign. So, now, in short what is standard primal problem structure is like this way maximize z p d 1 d 2 d t are the coefficient associated in the objective function and x 1 x 2 dot x n is the design variables and that you convert into a this type of inequality form agree and this e 1 e 2 e n are there is no restriction on its sign, but this design variables are all greater than equal to 0 in turn I can write matrix vector form into this form. This how you convert into dual problem of this is a dual problem of that primal problems that whatever the coefficients are associated in objective function of standard primal problem that will come right hand side of the inequality constant of inequality constants of dual problems. So, it is coming right hand side d 1 d 2 d t and what is the expression we will write it inequality constant this column consider this column a 1 1 into y 1 a 2 1 into y 2 plus a dot dot plus a m 1 into y m will be greater than equal to d 1. Similarly, second column third column in this thing. So, since we have a in primal problem and coefficients are there we will get n constants are there and what are the right hand side in standard problem how many equation inequality equation there are m equations are there. So, that variables will go in the dual problem objective functions it is going and corresponding m new variables are introduced. So, this is called the dual problem of the primal problems and observation if you see this observation of that one it is clearly say that number of dual variables number of dual variables that dual variables is y 1 y 2 y 3 dot dot y m is same as the number of primal the primal constants how many constants are there m constants and dual variables are with a 1 y 1 y 2 y 2. So, number of dual variables is same as number of constants in the primal problems this is the first observation. Second observation number of dual constants is same of same as number of number of variables in the objective function primal variables. Number of constants in the dual problem is same as number of primal variables in the primal problems this is the second observation we got it and coefficient associated with the standard primal problem coefficient associated with x is a matrix what you get coefficient matrix in the primal problem. If you take that transpose that will get you the coefficient matrix of dual problems this is the third observation we have seen then we can see this one that inequality constant are reverse in reverse in direction the primal problem inequality is less than equal to type and dual problem is reverse in direction greater than equal to this is the third fourth one we will get it and minimization or maximization problem of primal problem becomes a minimization problem of dual problems. So, you see this is the maximization problems now become a minimization problem of dual problems this. So, these are the observation we got it this one and you have seen this coefficient associated with the primal in objective function coefficient associated with the variables that comes in the what is called objective that comes in the dual problem inequality constant right hand side and vice versa which is coming in the standard problem inequality constants the right hand side what is coming e 1 e 2 e dot dot e m that will come as a coefficient in the dual problem objective functions. And this primal and the dual variables are non negative that means x i is greater than equal to 0 and y i is greater than equal to 0 this. So, keeping this thing in mind we further see how to convert that a dual problem into primal problems and primal problem into dual problem we can do it. So, let us take in one example and we will see how to solve it. So, let us call the example the primal and dual problems using simplex method. So, let us I will just give I will give you the primal problem convert into dual problem agree. So, primal problem primal so what is the primal problem maximize maximize z p is given 5 x 1 minus 2 x 2 which I can write it d transpose general structure into the x this is our 2 cross 1. If you see our d where d is equal to 5 minus 2 which dimension is 2 cross 1 subject to 2 x 1 plus x 2 is less than equal to 9 x 1 minus 2 x 2 minus 2 x 2 minus 2 x 2 minus is less than equal to 2 minus 3 x 1 plus 2 x 2 is less than equal to 3 and x i is greater than equal to 0 for i is equal to 1 to this. Now, you see it is already in standard what is called primal problem if it is not there in standard primal problem you convert into standard primal problems agree. So, this is the standard primal problems first your job is let us call I given the problem you convert into a standard dual problems and now you see so this if I ask you let us call first I ask you this is the primal problem solve this problem by using simplex method. Then you can solve it because you have to convert this thing into a standard LP problems this is one way and if I tell you that this is the primal problem convert it dual problem and then solve it by using simplex method that is why you have to convert into dual problem. Sometimes when you convert from what is called dual you have a standard problem to dual problem the computational advantage you can get the gather you can get it while you will solving the multi objective optimization problems linear optimization problems that is may be the one of the advantage you can get it from this one. So, let us see first I our question is solve convert this thing into standard LP problem and then solve it standard LP problem then solve it, but I will not solve this by standard LP problem because I here because I already discussed how to solve a standard LP problem using simplex method means in tabular form or it is a algebraic approach or matrix form you can solve it. So, solution suppose you are asked to solve the problem is primal dual problem using simplex method. So, this solution convert standard LP problem and then solve and then solve using simplex method. So, I will just tell you how to convert the this how to convert the standard LP what is the standard LP this you have to convert equal to sign and not only that and right hand side of this equation must be positive right hand side of this equation must be must be positive right hand side of this equation must be positive this is already in positive terms. So, we have to introduce in this equation one slack variable here one slack variable here one slack variable. Suppose you have a this is the primal problems you have a suppose in the. So, then after that you can solve it. So, let us call how to convert this thing into a standard what is called convert I just convert into a standard LP problem. So, our standard LP minimize that is Z p f of x is equal to minus Z p which is equal to see this one minus mean minus 5 x minus it is minus there. So, it will be plus 2 x and this you can convert by adding one slack variables because this is less than this you have to add some slack variable and slack variable value greater than equal to 0. So, that slack variable I introduce as a x 3. So, 2 x 1 plus x 2 plus x 3 plus x 2 plus is equal to 9 this is the slack variable. Then we have to find out that next equation you see similarly second and third I can write it. So, second equation is x 1 minus 2 x 2 plus x 4. So, this is also slack variable is equal to your 2 then third equation you see minus 3 x 1 plus 2 x 2 and I have to add another slack variable with this one. So, that it will be equal to 3. So, this is also slack variable and right hand side of this equation all are positive. So, our x I is greater than equal to 0 I is very to 1 2 dot dot 5. So, this is converted into standard LP problem. Then you know in tabular form how to solve this problem agree by simple method that you solve it. If you solve this problem then you will get the solution of this one that our x 1 optimally of this one you will get 4 x 2 star you will get 1 x 3 is equal to x 4 this star it will get it 0 then x 5 star you will get it 13. So, this is the solution and in if you complete this solution in tabular form that our basic variables are x 1 x 2 and our x 5. These are the 3 this and this is the basic variables basic variables and remaining are non basic variables this is non basic variables. So, immediately what is our minimum value of the function if you just put it here that x 1 is 4 means minus 20 minus 20 plus 2 f d or f is you will get it f star is minus 18. Our problem is if you see our original problem is maximization problems agree our maximization of problem of that one. So, that will be a our maximum z will be if you get plus 18 so, this is the solution you have a given LP problem if you are asked to ask to convert into a dual what is called primal problem and then solve it by using simplex method you have to do like this way. Now, next is I told you I mentioned you if you convert the primal problem that that is why called if you convert the primal problem into dual problem that maximization problem you convert into what is called minimization problems. And the way it is converting into dual problem just now I explain it if you convert into dual problem and then solve it by what is called simplex method agree, but that variable dual problem variables are new things that is no way linked with the I mean no way relation direct relation with the our primal variables x 1 x 2. So, there must be some link between the dual variables and our primal variables because our objective is to find out the primal variables of that one. There are two objectives in mind if you convert the primal problem into dual problem we can get the what is called the computational advantage while we will solve the multi objective optimization problems agree. And when you are going in that domain what is called dual problem solution our variables are y 1 y 2 y 3 dot dot y m that variables we have to make a direct connections with the x how to convert that values of that dual variables values how to convert into a that standard variables values that is next issue. So, let us first let us now we convert our problem is standard l p problem if you see this standard l p problems this standard l p problem now we will convert into a dual problems how will you do now you remember first you have to minimize minimize that function how will you objective function of that dual problem you will be considered whatever the coefficients are there that coefficient will come in the objective function of dual problems and that dual problem will be minimization of the objective function. So, if you do this one dual problem now keeping that in mind minimize f of d is say 9 2 x there are 3 variables 3 what is called in standard problem 3 inequality constraints are there 9 into y 1 9 into y 1 new variables then it is a 2 into y 2 that 2 that 2 is y 2 then 3 into y 3. So, this is our objective function dual problem objective function is subject to now see this one 2 into y 1 this coefficient is 1 plus y 2 then minus 3 into minus 3 into y 3. So, I will write it this first equation I will write it 2 y 1 plus y 2 minus 3 y 3 is greater than equal to your 5. So, this is our objective function 2 y 1 plus y 2 minus 3 y 3 is greater than equal to 5 that all coefficient of standard L P problem of standard primal problems standard primal problems will come right inside of the inequality constraint of dual problems agree this. Now, this I can write it second equation that you see this coefficient is 1. So, 1 into y 1 plus that is minus sign minus twice y 2 plus twice y 3 is greater than equal to what minus 2. So, I will write it y 1 minus twice y 2 plus twice y 3 is greater than equal to minus 2. So, this one agree and we have a now new variables a dual problem variables are y 1 y 1 greater than equal to 0 y 2 greater than equal to 0 y 3 greater than equal to 0 agree. So, as I mentioned is here when you convert into dual problems the right hand side of this constraint does not have any restriction on a sign. So, we have converted into same this form. So, now if you see this one then what is this this is as converted into a that our standard dual problem. Now, if you want to solve this this standard dual problem you have converted then you want to solve it by using simplex method. That means, you have to convert next is convert into convert the dual problem into standard LP problem and then solve and then solve by simplex method. So, first you convert into standard LP problems this is minimization part is already there we need not to but this part is not in standard form. So, you have to make it in equality form now see this one when you will convert into standard LP problem right hand side must be sign must be positive. So, this one I can write it that you see this is greater than equal to this that means minus some variable you have to make it to make it is equal to I mentioned you that when it is this type of equation is there if you just put a new variable y 4 minus y 4 is equal to 5. Then you will see when you will solve this problem the y 4 value will not be restricted with a non negative number it may be negative number to avoid that problem we introduce what is called the artificial variables. So, this thing I can immediately I can write it twice y 1 plus y 2 minus 3 y 3 minus y 4 plus y 2 minus y 3 minus y 5 and this our y 5 is your our artificial variable and this is we called if you remember this is a surplus variable select variable surplus variable again artificial variable. This is this equation is converted into a now this equation you can convert into a plus sign because I have to make it plus sign and equality sign this. So, if you most side you multiply by minus this equation if you both side multiplied by minus then it will be y 1 minus y 1 plus twice y 2 minus y 3 then it will be less than equal to 2. That means we have to add another artificial variable let us call this is y 6 that y 6 is equal to 2 and this is a slack variable. Now, tell me how you solve this standard LP problem minimize this one subject to this two equality constraints and out of this we have a this is an artificial variable and this is the what is called surplus variable this is our slack variables. So, this we have to solve for y i is greater than equal to 0 for i is equal to 1 plus 2 6 this is our problems. So, if you recollect this thing if you have a artificial variable we have to solve this problem in two stages first stage it will minimize the artificial variable that is y 5. If you have a more than this type of inequality is there in your equation then it will come for each equation it will come one artificial variable again and resultant cost function for the artificial variables artificial variable cost function will be the sum of algebraic sum of all artificial variables will be the objective function. So, in this case only one artificial variable y 5 I have to minimize then y 5 expression I can write it keeping all this variable in right hand side then y 5 is equal to I can write this expression. So, our phase one if you see our phase one problem is like this way minimize w is equal to y 5 and what is y 5 if you see this one 5 minus twice y minus twice twice y 1 minus y 1 plus 3 y 2 if you take this side plus y 4. So, I am writing 5 minus twice y 1 minus y 2 plus 3 y 3 plus y 4. So, this is our objective problem subject to what this constraints subject to 2 y 1 plus y 2 minus 3 y 3 minus y 4 minus plus y 5 is equal to 5 and minus y 1 plus y 2 y 2 2 this 2 equation what we have written same equation I am writing here minus 2 y 3 plus y 6 is equal to 2. We have our original objective function is what if you see our objective original objective function is minimize if you see this one minimize f d is equal to 9 y 1 again plus 2 y 2 plus 3 y 3. So, in phase 1 in tabular form we have to minimize first this one then in phase 2 we have to minimize that one. So, when you will minimize in first one tabular form both the function you have to write it simultaneously agree. So, this you have to write simultaneously this one. So, I leave it and this gives an exercise to work out and you will get it after phase 1 after completion of phase 1 the solution of this one you will get it that after completion of phase 1 you will get the value of this that phase 1 end of phase 1 end of phase 1 you will get it the value of w is equal to 0 then 1 in phase 1 you have a 2 iteration that I am telling you. So, after the end of the phase 1 you will get w 1 is 0 means w is 0 means y 5 y 5 is equal to 6 this 0 then you will get it y 1 value 2.5 y 2 y 3 y 4 y 5 these values are 0s and y 6 values is 4.5. These are the basic variables you will get it and these are non basic variables non basic variables. So, this is the end of phase 1 that means we have minimized the artificial variable which we are expecting that y 5 value should be 0 it is coming 0 then at the similarly next you ignore that y 5 in the tabular form remove it from the tabular form this y 5 and then proceed for phase 2. If you see the phase 2 at the end of phase 2 in the phase 2 you are minimizing the objective function the change objective function during this process when you did it phase 1 in this process the objective function is changed because we have done lot of elementary row operations there. So, it is changed it. So, phase 2 the end of phase 2 you will get basic variables y 1 star is equal to 1.6 y 2 star you will get 1.8 then y 3 y 4 y 4 y 5 then y 5 then y 6 no that y 5 is not there yes y 3 y 4 then y 6 these values will be 0. Now, you see what is the objective function value y 1 is 1.6 y 2 value is your that 1.6 y 3 value is 0. So, 9 point if you see the f of d this is 9 into 1.6 plus 2 into 1.8 plus 2 into 1.8 if you do this one this will be a 3.6 14. So, it will be a this will be a 18. So, f d value is will be a 18 then what is our z value z value how you find out z value we do not know the value of x. So, there must be a some transformation from what is called dual variables to a primal variables that we will discuss next class, but we can say this value objective function value in the primary must be also same agree that 18. We will discuss next class in details how to convert the dual variables into a primal variables design variables we will stop it