Q4: Full Simplex Algorithm Consider the following LP. use the full Simplex algorithm to solve it. min6?1+3?2+7?3 s.t. 5?1+?2=8

?1+?2?20

?2+?3?2

?1,?2,?3?0 (a) First, convert the problem to standard form. We will use the Two Phase method to find an initial BFS. (b) Provide the formulation for Phase 1. (c) Using Simplex tableaux, solve the Phase 1 formulation to optimality. Show work. This is a feasible LP, and you should end up with an optimal solution to part (c) that has a value of 0 for the artificial variable(s). Drop the artificial variable(s), and now you have a feasible, initial BFS to the standard form LP presented in part (a). Now we will conduct Phase 2 - solving the LP with this initial BFS. (d) Report the tableau that will start Phase 2. Note: it should not include the artificial variables. (e) Solve this Phase 2 problem to optimality using tableaux. Show work, and briefly note how you know the solution is optimal. (f) Report the optimal solution and optimal value. please show all the iterations done to get to each optimal point of the both phases

R1(new) = R,(old) + 5 R (old) = 8 5 1 0 0 0 0 R1(new) = R, (old) + 5 1 0.2 0 0 0 R2(new) = R2(old) - R, (new) R(old) = 20 1 0 0 R (new) = 1.6 1 0.2 0 0 10 0 R2(new) - R2(old) - R,(new) 18.4 0 0.8 0 1 0 0 R3(new) = R3(old) R3(old) = 2 0 1 - 1 R3(new) = R3(old) 2 0 1 1 0 - 1 1R3(new) = R3(old) R3(old) = 2 0 1 1 0 R3(new) = R3(old) 2 0 1 0 - 1 R1(new) = R1(old) - 0.2R (new) R, (old) = 1.6 0.2 0 0 R,(new) = 2 0 1 1 0 - 1 0.2 x R; (new) = 0.4 0 0.2 0.2 0 -0.2 R, (new) = R,(old) - 0.2R3 (new) 1.2 0 -0.2 0 0.2 R2(new) = R2(old) - 0.8R3(new) R(old) = 18.4 0 0.8 0 0 R (new) = 2 0 0 - 1 0.8 x R;(new) = 1.6 0 0.8 0.8 0 -0.8 R,(new) = R2(old) - 0.8R3(new) 16.8 0 0 - 0.8 0.8R3 (new) = R3(old) R3(old) = 2 0 1 1 0 R3(new) = R3(old) 2 0 1 1 - 1 R1 (new) = R, (old) + 0.2R3 (new) R, (old) = 1.2 0 -0.2 0 0.2 R3(new) = 2 0 1 1 0 0.2 x R, (new) = 0.4 0 0.2 0.2 0 - 0.2 R1(new) = R1(old) + 0.2R,(new) 1.6 1 0.2 0 0 0 R2(new) = R2(old) + 0.8R3(new) R,(old) = 16.8 0 0 -0.8 1 0.8 R3(new) = 2 0 1 1 0 - 1 0.8 x R; (new) = 1.6 0 0.8 0.8 0 -0.8 R2(new) = R2(old) + 0.8R3(new) 18.4 0 0.8 0 1 0R1(new) = R1(old) + 5 R1(old) = 8 5 1 0 0 0 0 R1(new) = R1(old) + 5 1.6 1 0.2 0 0 0 0 R2(new) = R2(old) - R, (new) R,(old) = 20 1 1 0 1 0 0 R (new) = 1.6 1 0.2 0 0 0 0 R2(new) = R2(old) - R1 (new) 18.4 0 0.8 0 1 0 0 R3(new) = R3(old) R3(old) = 2 0 1 1 0 - 1 1 R3 (new) = R3(old) 2 0 1 1 0 - 1 1R3(new) = R3(old) R3(old) = 2 0 1 1 0 R3(new) = R3(old) 2 0 0 - 1 R1(new) = R1(old) - 0.2R3(new) R (old) = 1.6 1 0.2 0 0 0 R3(new) = 2 0 1 0 - 1 0.2 x R3(new) = 0.4 0 0.2 0.2 0 -0.2 R, (new) = R,(old) - 0.23(new) 1.2 1 0 - 0.2 0 0.2 R2(new) = R2(old) - 0.8R3(new) R,(old) = 18.4 0 0.8 0 1 0 R3(new) = 2 0 1 1 0 - 1 0.8 x R3(new) = 1.6 0 0.8 0.8 0 -0.8 R,(new) = R,(old) - 0.8R,(new) 16.8 0 0 -0.8 1 0.8