 Myself Satish Halange, Assistant Professor, Department of Civil Engineering, Walsh and History Technology, Solak. In today's session, we are going to see the duality of linear programming problem. At the end of the session, the learner will be able to determine the dual of linear programming problem from the primal one. In short, let's see the linear programming problem and the duality. The linear programming problem is a mathematical modeling technique in which we see the terms included are objective function. It may be of maximized case or minimized case with the n number of decision variables which is non-negative variables and it is subjected to the constraints expressed in the inequality or equality equations. This is an example which is highlighting the particularly objective function that is of maximized case and the subjected to the constraints. There are two constraints with the decision variable that is x1 and x2, which is restricted by the sign. Duality of linear programming problem is what the every primal linear programming problem has an associate with the another program called as a dual and the dual of the dual is a primal one. Here in the table it is shown that the particularly if the primal problem is of maximized case its dual will be of minimized case and similarly if the primal v is of minimized case its dual is of maximized case. Now in short let's see the particular standard forms of LPP required for the duality problem. Example this is a problem which is of maximized case and here we also that there are the particularly four constraints for the particular objective function. Here z is equal to 8x1 plus 10x2 plus 5x3 which is subjected to the four constraints as shown here. Here the number of variables that is n is equal to 3 and number of constraints m is equal to 4 that are four equations are their equation 1, 2, 3 and 4 that are constraints equation. Now before starting to solve the particular primal problem means to convert the primal to the dual one we have to observe that the particular constraints which are here they are in the standard form according to the defined in the earliest slide. Which are the standard required for the solving the particular LPP by the dual methods. Now when we start to see the all the particularly standard forms criteria your first constraint that is 1x1 minus 1x3 is less than or equal to 4 which is according to the objective function case as I said for the max for LPP of maximized case all the constraints should be of LHS less than or equal to right LHS type. So this is as similar to the objective function case. Similarly second a constraint equation 2x1 plus 4x2 is less than or equal to 12. Again this is similar to the objective function case. Now the third and fourth that is 1x1 plus 1x2 plus 1x3 is greater than or equal to 2 which is not according to the objective function case. So it is not in the standard form 3x1 plus 2x2 minus 1x3 equal to 8. Again when to solve the particular problem or you can say the primal to the dual one all the particular constraint equation should be of inequality. So but here in the fourth constraint it is of equality form so it is not in the standard form to convert by the duality method. Now let's start as I have earlier told that the particular step 1 the constraint equation 1 and 2 both are of left hand side is less than or equal to right hand side type according to the maximized case. So there is no need to change so keep as it is. But in steps 2 constraint equation 3 which is of left hand side is greater than or equal to right hand side type which is according to the minimized case. So as our problem is of maximized case we have to convert this particular constraint equation 3 to the objective function case that is of maximized case by multiplying minus 1. Here I have shown you minus 1 multiplying to the constraint equation 3 on the both side that is of left hand side as well as right hand side. Here I am multiplying it we get the a new constraint equation that is minus 1 x 1 minus 1 x 2 1 minus 1 x 3 less than or equal to minus 2. Now when we see this step number 3 constraint equation 4 which is of equality equation here the particular fourth equation was of here when we observe the main problem here it is of equality. But to solve the particular problem by dual method this particular problem should have the inequality equation. So here particularly therefore there is necessary to convert the equality equation to the inequality equation by replacing by a pairs of inequality equations in the opposite direction. So 3 x 1 plus 2 x 2 minus 1 x 3 is equal to 8 this is the equation original equation to convert that here I have in the step 3 by the continuation I have shown you that this particular original constraint equation I have replaced by the 2 new constraints equation. But in the opposite direction that is of inequality equation form 3 x 1 plus 2 x 2 minus 1 x 3 greater than or equal to 8 3 x 1 plus 2 x 2 minus 1 x 3 less than or equal to 8 these are the 2 inequality equations. But here the obtained fifth constraint equation is not according to the objective function case. So it is necessary to convert it into the standard form by multiplying by minus 1 as the constraint equation 1 is not in the standard form I will multiply it by minus 1 and convert into the standard form according to the objective function case that is of maximize case. This after multiplying on the both side by minus 1 LHS as well as on the right hand side minus 3 x 1 minus 2 x 2 plus 1 x 3 less than or equal to minus 8 this is a newly obtained constraint equation of 5. Now once again I will write all the particularly converted constraint equations in the slide that is a standard primal problem earlier I have only said that is a primal problem but now I will say it is standard primal problem because this is the problem is totally converted according to the objective function case. So the newly obtained primal problem is the maximized case that is equal to 8 x 1 plus 10 x 2 plus 5 x 3 and subjected to all this newly obtained constraints. Now this when we observe all the constraints here we see observe that left hand side is less than or equal to right hand side which is according to the objective function case. Now let's start to convert this particular standard primal problem to the dual one. Now once I start to convert into the dual form here we see that as per as there are 5 constraints equation there will be 5 new non-negative variables y 1, y 2, y 3 and y 4 and y 5. Now earlier the objective function was of maximized case now it will convert into the minimized case as earlier told. Now z is equal to 4 y 1 plus 12 y 2 minus 2 y 3 minus 8 y 4 plus 8 y 5 which is the values from the constraint equations right hand sides. Now the first x 1 this is a marking a box marking which I shown you this will be the newly obtained constraint equation number 1 in the dual one that is 1 y 1 plus 2 y 2 minus 1 y 3 minus 3 y 4 plus 3 y 5 greater than or equal to 8. Similarly for x 2 and x 3 that is 4 y 2 minus 1 y 3 minus 2 y 4 plus 2 y 5 greater than or equal to 10. Similarly for the third that is x 3 minus 1 y 1 minus 1 y 3 plus 1 y 4 minus 1 y 5 greater than or equal to 5. This is the dual which is obtained from the standard primal problem. But this obtained dual problem is again not in the standard form here let's see the number of variables in the original primal problem is 3 and the number of constraints are 4. But when we observe the dual of the particular problem here the number of variables n is equal to 5 and number of constraint m is equal to 3. As I said that when from the original primal problem the number of constraints should be equal to the number of variables in the dual. But when we observe this that m is equal to 4 but the number of variable n is equal to 5. So this particular dual is not in the standard form. So to convert this particular dual in the standard form again I can replace this particular dual variables how first of all we have to observe that in the dual problem which variables are having same coefficient that is when we observe here that y 4 and y 5 have the same coefficient. So I can replace this y 4 and y 5 by the new variable which will be undistricted by the sign to convert the particular dual into the standard dual form. So to obtain the dual problem can be converted into the standard dual problem by replacing difference of y 4 and y 5 by the new variables that is y 6 which will be undistricted by sign. This is a representation of a separation of the coefficients of both the y 4 and y 5 and I will replace this by the y 6 why I have selected y 4 and y 5 as I earlier told both have the same coefficients. So after replacing I get the new particular standard dual problem that is g is equal to 4 y 1 plus 12 y 2 minus 2 y 3 minus 8 y 6. Similarly I will replace the in the constraints also these are the constraints I will replace here also this is again 1 y 1 plus 2 y 2 minus 1 y 3 minus 3 y 6 greater than or equal to 8 similar to the other 2 also. But finally I will mention that y 1 and y 2 and y 3 are restricted by sign but y 6 is undistricted by the sign. But here we can we observe that the number of variables in the original primal problem that is and when we observe the standard dual problem the number of constraints are 3 and the number of constraints in the original primal problem m is equal to 4 and the number of variables in the particular standard dual problem n is equal to 4 which is satisfying the standard requirement. Now this is the particular standard dual problem of the original primal problem. Let's select the correct answers for this m secues. Hope so you have selected this one. These are the references for these situations. Thank you.