 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 dual a linear programming problem from the primal one. Now let us see what is linear programming problem and the duality in short. What is linear programming problem? It is a mathematical modeling technique in which we see the terms like objective function that is maximized case or minimize case and the n number of decision variables are present which are non-negative variables. This particular objective function is subject to the set of constraints expressed by the equations of an inequality equations. Here this is an example of one of the LPP which is saying that particularly objective function is of maximized case and it is subjected to the two constraints that is of two equation inequality equations we can say and there are two decision variables that is x1 and x2 which is finally restricted by this sign. Now what is duality of LPP program? Now every primal linear programming problem has an associate with the another linear programming problem called as dual. Here the dual of the dual is a primal one. Now table is showing that the primal LPP if it is of maximized case, your dual will be of minimized case. Similarly, if the primal is of minimized case, your dual will be of maximized case. Here it is vice versa. Now let us see in short in earlier session we have seen in detail what are the standard format of the LPP required for the duality and the making of changes if there is necessity. Now in short see what are the points which are to be observed in the LPP to solve the particular LPP by duality form duality. Here the first point is all the variables that are the decision variables it should be of nonnegative type as shown in this particular bracket x1 and x2 are these two non these are nonnegative decision variables which are restricted by sign that is greater than or equal to 0. And for LPP of maximized case all constraints should be of left hand side should be less than or equal to right hand side type. And for LPP of minimized case all the constraints should be of left hand side should be greater than or equal to right hand side type. Here the fourth point is if the variable LPP is unrestricted by the sign means what in the first point I have already mentioned that here the if the particular problem have two variables x1 and x2 it is restricted by the sign that is greater than or equal to 0. And means this particular variables either it will be equal to 0 positive. But sometime the sign the particular variables are unrestricted by sign that are the which can be replaced by the difference of two variables that are the two new decision variables. When all the constraints are in right direction as per the objective function there is no any problem. But sometime we observe that all the constraints may be not according to the objective function case means what if it is of maximized case all the constraints should be of LHS should be less than or equal to right hand side that is RHS. But sometime we observe that it is not in the right direction. At that time to convert that particular constraint according to the objective function it should be multiplied by minus 1. If the constraints involve an inequality equation then to convert that particular equality equation in the inequality equation it is replaced by pair of inequality equations in the opposite direction. And if the primal LPP contained n non-negative variables and m set of constraints then the DL LPP should have the m non-negative variables and n set of constraints. Now duality problem this is a particular maximized case here we observing that the particular primal problem is of maximized case here Z is equal to 40x1 plus 35x2. Subjective to the constraint 2x1 plus 3x2 less than or equal to 15 this is first constraint equation second constraint equation is 4x1 plus 3x2 less than or equal to 10 this second constraint equation. In this particular problem we are observing that this particular x1 and x2 are finally restricted by the sign that is x1 comma x2 is greater than or equal to 0. Now we have to observe that particular problem is in standard form or not. First of all it is of maximized case yes for the maximized case all the constraints should be of left hand side should be of less than or equal to right hand side type. So observe here the constraint the particular left hand side is 2x1 plus 3x2 which is less than or equal to 15 similarly second constraint 4x1 plus 3x2 is less than or equal to 10 this particular constraints are as per the requirement of the objective function. And finally these two constraints are restricted by the sign means this is a primal problem which is in the standard form. Now let us convert this primal problem to the dual one. Here one more point we have to observe that the number of variable in this particular primal problem is n is equal to 2 that is x1 and x2 and the number of constraints m is equal to 2 that is equation 1 and equation 2 constraint equation 1 and constraint equation 2. As per the earlier slide it is mentioned that if the primal LPP contain n non-negative variables and m set of constraints then the dual LPP should have m non-negative variables and n set of constraints. So here particularly in the dual problem we should have the number of ray variables equal to the number of constraints present in the primal one. Now the number of constraints m is equal to 2 in the primal one. So for so I will take 2 new variables y1 and y2 as the non-negative variables in the dual. Now objective function will be of minimize case because in the primal one the particular objective function was of maximize case but now as we are converting the primal to the dual one its case changes and it will become minimize case. Now here z is equal to particularly 15 y1 plus 10 y2 here how it is obtained when we observe the primal problem here the particular there are 2 constraints and both the left hand side and the right hand side we have to observe the right hand side values are 15 and 10. This values will become the coefficient of a particular variables in the dual problem here. Now there are 2 variables y1 and y2 and here the y1 will be now its coefficient will be 15 y1 plus 10 y2 in the dual problem. This particular problem is subjected to the constraint again here we have to observe in the primal problem here I have made a particular box for the x1 variable as well as for the x2 variable when we see this particular first column I will say the box in the column forms that is why I am saying in the column here 40 x1 is there 2 x1 is there 4 x1 is there. This particular column will convert into horizontal form in the dual with the 2 new variables that is y1 and y2 here this 2 x1 will become 2 y1 plus 4 x1 will become 4 y2 now as the particular problem is changing from maximize case to minimize case its sign changes it will become greater than or equal to 40 here this is a coefficient of particular variable in the objective function. So it will become now the right hand side of the constraint equation 2 y1 plus 4 y2 greater than or equal to 40 similarly the second constraint which is obtained from the second column that is from the primal that is of x2 variable here this 3 x2 will become 3 y1 plus 3 x2 will become 3 y2 greater than or equal to 35 here this is a 35 coefficient which is present in the objective function case of the primal one that will become the right hand side in the particular dual in the constraint equation of second. Now the obtained objective function and the subject to the constraints I have to give the restriction to this particular newly variables that is y1 from y2 will be greater than or equal to 0 when we observe this particular according to the last point the number of variables n is equal to 2 and the number of constraints m is equal to means it is equal again second number of constraint in the primal is 2 and the number of variables in the dual is 2 now this particular problem is in a standard form here we are getting both the as per the requirement now let us select the correct answers for this particular mcqs hope so you have selected the correct answer for this one as shown in the slide these are the references for today's session thank you.