Duality 2 - Dual of Maximization LPP and Minimization LPP

hello welcome now let us say take very simple case of writing or constructing duel the first linear programming problem which was is maximize the objective function is Z equals 240 X 1 plus 35 X 2 subject to the constraints 2x1 + 3 x2 less than or equal to 60 4x1 + 3 x2 less than or equal to 96 X 1 and X 2 both have no negative first of all checking what is the objective maximize if the objective is maximization the signs of all the constraints in the primal must be less than or equal to 0 yes they are less than or equal to 0 in both Dawkins he saw both the constraints the same are less than or equal to 0 which are in matching with the objective so we can directly write the dual of this primer so this problem becomes primal for us let us give title primal and now we are going to write a skewer now let us remember the recap the rules what we should do first of all the objective of primary's maximization so the objective of the duel will be minimization size are less than or equal to in case of primal so science will be greater than or equal to in case of primer that is very simple thing but another two things there are two constraints in the primal so there will be two variables interview and there are two variables in the dual there will be two constraints in the primal now how to write the dual of a brain a very simple thing is to be done we will have to variables this is a variables in the dual because we have two constraints in the Bible similarly we have two this is a variance in the primal we will have two constraints now what the right hand side of all the constraints will be now the part of our objective function in the viewer now we can easily write the dual of this frame objective will be minimized because the objective of the primal is maximized in objective function we are going to use sine Z star or Z a stirring or you can write Z Y because this is objective function of a problem which very high decision variables X so this is called Z s this can be called Genoa it is now upon you what will happen these two will be now the coefficients of the decision variables in the object u function y1 into 60 so it will be 60 y1 this is positive or negative so plus 96 Y to see carefully the right hand side values of the constraints become the coefficients of the decision variables in the objective function in the viewer now the first constraint of the DL will be on the basis of the first decision variable of the primal on the basis of x1 we are going to write the first constraint subject to 2 to y 1 the coefficient of x1 in the first constraint becomes the of why Monica first and to y1 this is four wins positive 4 plus 4 y 2 plus 4 y 2 sign will be matching with objective now the objective is minimization sign will be greater than or equal to or remember the moon if sign is less than or equal to in primal sign will be greater than or equal to e do and now the coefficient of first of the decision variable in the objective function of a primal will become the right hand side value of the first constraint for T similarly the second constraint will be written on the basis of the second decision variable of the primal 3 into y1 3 y 1 plus 3 plus 3 y 2 again sign greater than or equal to and coefficient of x 2 will be now right hand side value of the second constraint 35 and y1 y2 both are non-negative so this is the dual of this primer now let us take an interesting thing what will happen if we write the doer of this viewer what will happen if we add the dual of this dual what will be there see now if we take this as a primal so tip you off primarily is minimization the objective of jewelry the maximization there are two constraints so there will be two decision variables in case of its fewer there are two decision variables in this primal there will be two constraints in its doer okay let us write you of this dual minimize objective will be now maximize Z or the X will be X 1 into 40 40 X 1 plus 35 into X 2 subject to the constraints the first constraint will be 2 into X 1 plus 3 into X 2 sine will be now less than or equal to opposite to greater than or equal to and this 60 will be now right hand side we similarly on the basis of y2 we can write the second constraint for x1 plus 3x2 again less than or equal to 96 and X and X 1 and X 2 both are see this is given of the primer that means our original linear programming problem this is doer of the doer this is fuel now this one is doer of this dual what is this exactly it is the primal itself so the broad conclusion is dual of a dual is the primal itself dual of the dual is the primal itself now let us take another case which is minimization case minimize Z equals 240 X 1 + 24 X 2 subject to the constraint 20 X 1 plus 50 X 2 greater than or equal to 4000 radiant red 80 X 1 plus 50 X 2 greater than or equal to 7200 X 1 and X 2 both are non-negative objective is minimization signs are greater than or equal to no unusual thing so now we have two constraints so there will be two variables in the dual we have two two decision variables in the primal we will have two constraints in the dual let us write its fuel objective of dual will be maximize Z star or zero I will be now y1 into 4800 so 4800 y1 plus 7200 into I 2 subject to now the first constraint will be written on the basis of column of x1 in the primal 20 y1 plus 80 by 2 less than or equal to 40 20 x1 plus 80 y 2 less than or equal to 40 now objective is maximization sine will be less than or equal to R sine will be exactly opposite to the sign in the primal center the strain will be written on the basis of the second decision variable in the primal that is X 250 y1 plus 50 y 2 less than or equal to 24 50 y1 3 plus 50 by 2 less than or equal to 24 the coefficient of the decision variable in the objective function in the primal becomes the right hand side value of the constraint in the viewer and here in this case y1 and y2 both are greater than or equal to 0 that is non-negative again let us slide the dual of the dual what is going to happen we have two constraints so we have now two variables the right hand side values of the constraints will be now the coefficients in their function or objective function so against maximized object we will be minimize Z or there X will be 14 2 X 1 plus 24 into X 2 subject to 20 X 1 plus 50 X 2 now against less than or equal to it will be greater than or equal to and 4800 from the Z function the second will be 80 X 1 plus 50 X 2 greater than or equal to 7500 X 1 and X 2 both are non-negative I mean see what is this this is the triangle which was our original linear programming problem but let us take let us let us see what is happening with the check points object you minimize subject you maximize sign greater than or equal to sign less than or equal to number of constraints in the primal equals to number of variables in the viewer number of variables in the dual equals to number of constraints in the primal all the check points are satisfied that means this is await you but the important thing we arrive at the end of both the solutions is dual of dual is primal itself that's it for now in the next lecture we are going to cover some complicated problems thank you

Channel: PUAAR Academy

Views: 181,698

Keywords: OR, Operations RKesearch, Linear Programming, Duality, Dual, MBA, MCA, CA, CS, CWA, CPA, CFA, CMA, BBA, BCom, MCom, CAIIB, FIII

Id: 4AL-xv_oOl4

Length: 13min 12sec (792 seconds)

Published: Sun May 29 2016