Q3. Solve the given LP problem on the right by using...

Q3. Solve the given LP problem on the right by using graphical solution method.\ a- Find the optimal solution.\ b- Find the optimality range for the changes in the objective coefficient $c_2$.\ c- Find the feasibility range for the changes in the RHS of one of the binding constraints.\ (LP): Max $Z = 3X_1 + 2X_2$\ st.\ $2X_1 + 2X_2 \le 5$\ $2X_1 + X_2 \le 4$\ $X_1 + 2X_2 \ge 4$\ $X_1, X_2 \ge 0$

Transcript

00:01 Hello everyone in this problem we are given with the objective function maximum of z to be equal to x1 plus 3 x2 and we are given with the subject to constraints x1 plus x2 less than or equal to 2 and minus x1 plus x2 less than are equal to 4 also given that x1 is unrestricted and x2 is greater than are equal to 0 since x1 is unrestricted so x1 has both the positive and negative values.

00:36 Now in the sub part a we need to find the all basic feasible solution so let us add the stack variables x3 and x4 in equation the constraints first and second so we have x1 plus x2 plus x3 to be equal to two let us take this to be equation 1 and minus x1 plus x2 plus x4 which is equal to 4.

01:06 Let us take this to be equation number.

01:10 So in any basic feasible solution, at least n minus m.

01:14 So here, n minus m to be equal to 4 minus 2, which is equal to 2.

01:20 So there are two variables must be 0.

01:25 These two variables are non -basic solutions.

01:28 So let us put x1 and x2 is equal to 0 in equations 1.

01:40 And 2 we have x2 sorry x3 which is equal to 3 and x4 which is equal to 4 so we have the values 0 0 3.

01:57 So this is a basic feasible solution.

02:06 Similarly for the different values of we have so here for different values of x to be 0 we have.

02:20 Corresponding values.

02:25 So from this the possible basic feasible solutions are 0 .024, 0202, 2006...