A clever but ethically corrupted mathematics student used to sell assignment solutions to her lazy fellow students. The student, however, learned that she can make much more money by charging the students fees for doing the assignments with them. This semester the students need to complete three assignments. One in Statistical Models, one in Introduction to Partial Differential Equations and one very difficult one in Optimization and Financial Mathematics. It takes 2 hours to explain the assignment to one student for Statistical Models, 1/2 hours for Introduction to Partial Differential Equations and 5 hours for Optimization and Financial Mathematics. The student can charge 5$, 50 Cents and 10$ per hour for Statistical Models, Introduction to Partial Differential Equations and Optimization and Financial Mathematics, respectively. In Statistical Models there are not more than 10 students wanting her service. In Introduction to Partial Differential Equations there are more than 50 students wanting her service and she has obligations to take at least 50. In Optimization and Financial Mathematics, the course with the most diligent students, there are not more than 2 students wanting her service. How many students from each course will the clever student offer her service to maximize her profit, if she does not want to spend more than 50 hours on the assignments? (a) Formulate this as a linear programming problem (b) Solve the problem using the two-phase simplex algorithm. Matlab could be used for (b) \f\fQ1 Q2 A clever but ethically corrupted mathematics student used to sell assignment solutions to her lazy fellow students. The student, however, learned that she can make much more money by charging the students fees for doing the assignments with them. This semester the students need to complete three assignments. One in Statistical Models, one in Introduction to Partial Differential Equations and one very difficult one in Optimization and Financial Mathematics. It takes 2 hours to explain the assignment to one student for Statistical Models, 1/2 hours for Introduction to Partial Differential Equations and 5 hours for Optimization and Financial Mathematics. The student can charge 5$, 50 Cents and 10$ per hour for Statistical Models, Introduction to Partial Differential Equations and Optimization and Financial Mathematics, respectively. In Statistical Models there are not more than 10 students wanting her service. In Introduction to Partial Differential Equations there are more than 50 students wanting her service and she has obligations to take at least 50. In Optimization and Financial Mathematics, the course with the most diligent students, there are not more than 2 students wanting her service. How many students from each course will the clever student offer her service to maximize her profit, if she does not want to spend more than 50 hours on the assignments? (a) Formulate this as a linear programming problem (b) Solve the problem using the two-phase simplex algorithm. Matlab could be used for (b) Q3 Q4 \f\fThe problem is converted to canonical form by adding slack, surplus and artificial variables as appropriate As the constraint 1 is of type '' we should add the slack variable X3. As the constraint 2 is of type '' we should add the slack variable X4. MAZIMIZE: 2 X1 + 1 X2 MAXIMIZE: 2 X1 + 1 X2 + 0 X3 + 0 X4 1 X1 -1 X2 10 2 X1 -1 X2 40 1 X1 -1 X2 + 1 X3 = 10 2 X1 -1 X2 + 1 X4 = 40 X1, X2 0 X1, X2, X3, X4 0 We'll build the first tableau of the Simplex method. Tableau 1 2 1 0 0 P2 P3 P4 Base Cb P0 P P3 0 10 1 -1 1 0 P4 0 40 2 -1 0 1 0 -2 -1 0 0 Z 1 The leaving variable is P3 and entering variable is P1. Tableau 2 2 1 0 0 P2 P3 P4 Base Cb P0 P P1 2 10 1 -1 1 0 P4 0 20 0 1 -2 1 20 0 -3 2 0 Z 1 The leaving variable is P4 and entering variable is P2. Tableau 3 2 1 0 0 P2 P3 P4 Base Cb P0 P P1 2 30 1 0 -1 1 P2 1 20 0 1 -2 1 80 0 0 -4 3 Z 1 Solution is unbounded. MAZIMIZE: 2 X1 + 1 X2 1 X1 -1 X2 10 2 X1 -1 X2 40 X1, X2 0 The problem is unbounded. In green the points where the solution is located. In red (below table) the points that not belong to the feasible region. Point X coordinate (X1) Y coordinate (X2) Value of the objective function (Z) O 0 0 0 A 10 0 20 B 30 20 80 C 20 0 40 Now we change the problem into minimization problem. The problem is converted to canonical form by adding slack, surplus and artificial variables as appropriate As the constraint 1 is of type '' we should add the slack variable X3. As the constraint 2 is of type '' we should add the slack variable X4. MINIMIZE: 2 X1 + 1 X2 MAXIMIZE: -2 X1 -1 X2 + 0 X3 + 0 X4 1 X1 -1 X2 10 2 X1 -1 X2 40 1 X1 -1 X2 + 1 X3 = 10 2 X1 -1 X2 + 1 X4 = 40 X1, X2 0 X1, X2, X3, X4 0 We'll build the first tableau of the Simplex method. Tableau 1 -2 -1 0 0 P2 P3 P4 Base Cb P0 P P3 0 10 1 -1 1 0 P4 0 40 2 -1 0 1 0 2 1 0 0 Z 1 The optimal solution value is Z = 0 X1 = 0 X2 = 0 MINIMIZE: 2 X1 + 1 X2 1 X1 -1 X2 10 2 X1 -1 X2 40 X1, X2 0 The problem is unbounded but since it is a minimizing problem can find a solution. In green the points where the solution is located. In red the points that not belong to the feasible region. Point X coordinate (X1) Y coordinate (X2) Value of the objective function (Z) O 0 0 0 A 10 0 20 B 30 20 80 C 20 0 40 (a) The problem is converted to canonical form by adding slack, surplus and artificial variables as appropriate As the constraint 1 is of type '', and the independent term is negative or zero (the constraint is multiplied by -1), we should add the slack variable X 3. As the constraint 2 is of type '', and the independent term is negative or zero (the constraint is multiplied by -1), we should add the slack variable X 4. MINIMIZE: 2 X1 -1 X2 -1 X1 + 1 X2 -2 -1 X1 -1 X2 -6 X1, X2 0 So the standard form of the given LPP is MAXIMIZE: -2 X1 + 1 X2 + 0 X3 + 0 X4 1 X1 -1 X2 + 1 X3 = 2 1 X1 + 1 X2 + 1 X4 = 6 X1, X2, X3, X4 0 (b) MAXIMIZE: -2 X1 + 1 X2 + 0 X3 + 0 X4 1 X1 -1 X2 + 1 X3 = 2 1 X1 + 1 X2 + 1 X4 = 6 X1, X2, X3, X4 0 Let, X1 be the number of students for Statistical Models, X2 be the number of students for Introduction to Partial Differential Equation and X3 be the number of students for Optimization and Financial Mathematics. So the formulated LPP is MAZIMIZE: 5 X1 + .5 X2 + 10 X3 Subject to, 2 X1 + 0.5 X2 + 5 X3 50, X1 10, X2 50, X3 2, X1 , X2 ,X3 0 The problem is converted to canonical form by adding slack, surplus and artificial variables as appropriate As the constraint 1 is of type '' we should add the slack variable X 4. As the constraint 2 is of type '' we should add the slack variable X 5. As the constraint 3 is of type '' we should add the surplus variable X 6 and the artificial variable X8. As the constraint 4 is of type '' we should add the slack variable X 7. MAZIMIZE: 5 X1 + .5 X2 + 10 X3 2 X1 + 0.5 X2 + 5 X3 50 1 X1 + 0 X2 + 0 X3 10 0 X1 + 1 X2 + 0 X3 50 0 X1 + 0 X2 + 1 X3 2 X1, X2, X3 0 MAZIMIZE: 5 X1 + 0.5 X2 + 10 X3 + 0 X4 + 0 X5 + 0 X6 + 0 X7 + 0 X8 2 X1 + 0.5 X2 + 5 X3 + 1 X4 = 50 1 X1 + 1 X5 = 10 0 X1 + 1 X2 -1 X6 + 1 X8 = 50 0 X1 + 1 X3 + 1 X7 = 2 X1, X2, X3, X4, X5, X6, X7, X8 0 We'll build the first tableau of Phase I from Two Phase Simplex method. Tableau 1 0 0 0 0 0 0 0 -1 P2 P P4 P5 P P7 P8 Base Cb P0 P P4 0 50 2 0.5 5 1 0 0 0 0 P5 0 10 1 0 0 0 1 0 0 0 P8 -1 50 0 1 0 0 0 -1 0 1 P7 0 2 0 0 1 0 0 0 1 0 -50 0 -1 0 0 0 1 0 0 Z 1 3 6 The leaving variable is P8 and entering variable is P2. Tableau 2 0 0 0 0 0 0 0 -1 P5 P6 P7 P8 Base Cb P0 P1 P2 P3 P P4 0 2 5 2 0 5 1 0 0. 5 0 -0.5 P5 0 1 0 1 0 0 0 1 0 0 0 P2 0 5 0 0 1 0 0 0 -1 0 1 P7 0 2 0 0 1 0 0 0 1 0 4 Z 0 0 0 0 0 0 0 0 1 There is any possible solution for the problem, so we can continue to Phase II to calculate it. Tableau 1 5 0. 5 10 0 0 0 0 Base Cb P0 P1 P2 P3 P4 P5 P6 P7 P4 0 25 2 0 5 1 0 0.5 0 P5 0 10 1 0 0 0 1 0 0 P2 0. 5 50 0 1 0 0 0 -1 0 P7 0 2 0 0 1 0 0 0 1 25 -5 0 -10 0 0 0.5 0 Z The leaving variable is P7 and entering variable is P3. Tableau 2 5 0.5 1 0 0 0 0 0 P6 P7 Base Cb P0 P1 P2 P3 P4 P P4 0 1 5 2 0 0 1 0 0.5 -5 P5 0 1 0 1 0 0 0 1 0 0 P2 0. 5 5 0 0 1 0 0 0 -1 0 P3 10 2 0 0 1 0 0 0 1 4 5 -5 0 0 0 0 -0.5 10 Z 5 The leaving variable is P4 and entering variable is P1. Tableau 3 5 0. 5 10 0 0 0 0 P6 P7 Base Cb P0 P1 P2 P3 P4 P P1 5 7.5 1 0 0 0.5 0 0.25 -2.5 P5 0 2.5 0 0 0 -0.5 1 0.25 2.5 P2 0.5 50 0 1 0 0 0 -1 0 P3 10 2 0 0 1 0 0 0 1 82. 0 0 0 2.5 0 0.75 -2.5 Z 5 5 The leaving variable is P5 and entering variable is P7. Tableau 4 5 0.5 1 0 0 0 0 0 P2 P3 P4 P5 P6 P7 Base Cb P0 P P1 5 10 1 0 0 0 1 0 0 P7 0 1 0 0 0 -0.2 0.4 -0.1 1 P2 0.5 50 0 1 0 0 0 -1 0 P3 10 1 0 0 1 0.2 0.4 0.1 0 85 0 0 0 2 1 0.5 0 Z 1 The optimal solution value is Z = 85 X1 = 10 X2 = 50 X3 = 1 \f\fThe problem is converted to canonical form by adding slack, surplus and artificial variables as appropriate As the constraint 1 is of type '' we should add the slack variable X3. As the constraint 2 is of type '' we should add the slack variable X4. MAZIMIZE: 2 X1 + 1 X2 MAXIMIZE: 2 X1 + 1 X2 + 0 X3 + 0 X4 1 X1 -1 X2 10 2 X1 -1 X2 40 1 X1 -1 X2 + 1 X3 = 10 2 X1 -1 X2 + 1 X4 = 40 X1, X2 0 X1, X2, X3, X4 0 We'll build the first tableau of the Simplex method. Tableau 1 2 1 0 0 P2 P3 P4 Base Cb P0 P P3 0 10 1 -1 1 0 P4 0 40 2 -1 0 1 0 -2 -1 0 0 Z 1 The leaving variable is P3 and entering variable is P1. Tableau 2 2 1 0 0 P2 P3 P4 Base Cb P0 P P1 2 10 1 -1 1 0 P4 0 20 0 1 -2 1 20 0 -3 2 0 Z 1 The leaving variable is P4 and entering variable is P2. Tableau 3 2 1 0 0 P2 P3 P4 Base Cb P0 P P1 2 30 1 0 -1 1 P2 1 20 0 1 -2 1 80 0 0 -4 3 Z 1 Solution is unbounded. MAZIMIZE: 2 X1 + 1 X2 1 X1 -1 X2 10 2 X1 -1 X2 40 X1, X2 0 The problem is unbounded. In green the points where the solution is located. In red (below table) the points that not belong to the feasible region. Point X coordinate (X1) Y coordinate (X2) Value of the objective function (Z) O 0 0 0 A 10 0 20 B 30 20 80 C 20 0 40 Now we change the problem into minimization problem. The problem is converted to canonical form by adding slack, surplus and artificial variables as appropriate As the constraint 1 is of type '' we should add the slack variable X3. As the constraint 2 is of type '' we should add the slack variable X4. MINIMIZE: 2 X1 + 1 X2 MAXIMIZE: -2 X1 -1 X2 + 0 X3 + 0 X4 1 X1 -1 X2 10 2 X1 -1 X2 40 1 X1 -1 X2 + 1 X3 = 10 2 X1 -1 X2 + 1 X4 = 40 X1, X2 0 X1, X2, X3, X4 0 We'll build the first tableau of the Simplex method. Tableau 1 -2 -1 0 0 P2 P3 P4 Base Cb P0 P P3 0 10 1 -1 1 0 P4 0 40 2 -1 0 1 0 2 1 0 0 Z 1 The optimal solution value is Z = 0 X1 = 0 X2 = 0 MINIMIZE: 2 X1 + 1 X2 1 X1 -1 X2 10 2 X1 -1 X2 40 X1, X2 0 The problem is unbounded but since it is a minimizing problem can find a solution. In green the points where the solution is located. In red the points that not belong to the feasible region. Point X coordinate (X1) Y coordinate (X2) Value of the objective function (Z) O 0 0 0 A 10 0 20 B 30 20 80 C 20 0 40 (a) The problem is converted to canonical form by adding slack, surplus and artificial variables as appropriate As the constraint 1 is of type '', and the independent term is negative or zero (the constraint is multiplied by -1), we should add the slack variable X 3. As the constraint 2 is of type '', and the independent term is negative or zero (the constraint is multiplied by -1), we should add the slack variable X 4. MINIMIZE: 2 X1 -1 X2 -1 X1 + 1 X2 -2 -1 X1 -1 X2 -6 X1, X2 0 So the standard form of the given LPP is MAXIMIZE: -2 X1 + 1 X2 + 0 X3 + 0 X4 1 X1 -1 X2 + 1 X3 = 2 1 X1 + 1 X2 + 1 X4 = 6 X1, X2, X3, X4 0 (b) MAXIMIZE: -2 X1 + 1 X2 + 0 X3 + 0 X4 1 X1 -1 X2 + 1 X3 = 2 1 X1 + 1 X2 + 1 X4 = 6 X1, X2, X3, X4 0 Let, X1 be the number of students for Statistical Models, X2 be the number of students for Introduction to Partial Differential Equation and X3 be the number of students for Optimization and Financial Mathematics. So the formulated LPP is MAZIMIZE: 5 X1 + .5 X2 + 10 X3 Subject to, 2 X1 + 0.5 X2 + 5 X3 50, X1 10, X2 50, X3 2, X1 , X2 ,X3 0 The problem is converted to canonical form by adding slack, surplus and artificial variables as appropriate As the constraint 1 is of type '' we should add the slack variable X 4. As the constraint 2 is of type '' we should add the slack variable X 5. As the constraint 3 is of type '' we should add the surplus variable X 6 and the artificial variable X8. As the constraint 4 is of type '' we should add the slack variable X 7. MAZIMIZE: 5 X1 + .5 X2 + 10 X3 2 X1 + 0.5 X2 + 5 X3 50 1 X1 + 0 X2 + 0 X3 10 0 X1 + 1 X2 + 0 X3 50 0 X1 + 0 X2 + 1 X3 2 X1, X2, X3 0 MAZIMIZE: 5 X1 + 0.5 X2 + 10 X3 + 0 X4 + 0 X5 + 0 X6 + 0 X7 + 0 X8 2 X1 + 0.5 X2 + 5 X3 + 1 X4 = 50 1 X1 + 1 X5 = 10 0 X1 + 1 X2 -1 X6 + 1 X8 = 50 0 X1 + 1 X3 + 1 X7 = 2 X1, X2, X3, X4, X5, X6, X7, X8 0 We'll build the first tableau of Phase I from Two Phase Simplex method. Tableau 1 0 0 0 0 0 0 0 -1 P2 P P4 P5 P P7 P8 Base Cb P0 P P4 0 50 2 0.5 5 1 0 0 0 0 P5 0 10 1 0 0 0 1 0 0 0 P8 -1 50 0 1 0 0 0 -1 0 1 P7 0 2 0 0 1 0 0 0 1 0 -50 0 -1 0 0 0 1 0 0 Z 1 3 6 The leaving variable is P8 and entering variable is P2. Tableau 2 0 0 0 0 0 0 0 -1 P5 P6 P7 P8 Base Cb P0 P1 P2 P3 P P4 0 2 5 2 0 5 1 0 0. 5 0 -0.5 P5 0 1 0 1 0 0 0 1 0 0 0 P2 0 5 0 0 1 0 0 0 -1 0 1 P7 0 2 0 0 1 0 0 0 1 0 4 Z 0 0 0 0 0 0 0 0 1 There is any possible solution for the problem, so we can continue to Phase II to calculate it. Tableau 1 5 0. 5 10 0 0 0 0 Base Cb P0 P1 P2 P3 P4 P5 P6 P7 P4 0 25 2 0 5 1 0 0.5 0 P5 0 10 1 0 0 0 1 0 0 P2 0. 5 50 0 1 0 0 0 -1 0 P7 0 2 0 0 1 0 0 0 1 25 -5 0 -10 0 0 0.5 0 Z The leaving variable is P7 and entering variable is P3. Tableau 2 5 0.5 1 0 0 0 0 0 P6 P7 Base Cb P0 P1 P2 P3 P4 P P4 0 1 5 2 0 0 1 0 0.5 -5 P5 0 1 0 1 0 0 0 1 0 0 P2 0. 5 5 0 0 1 0 0 0 -1 0 P3 10 2 0 0 1 0 0 0 1 4 5 -5 0 0 0 0 -0.5 10 Z 5 The leaving variable is P4 and entering variable is P1. Tableau 3 5 0. 5 10 0 0 0 0 P6 P7 Base Cb P0 P1 P2 P3 P4 P P1 5 7.5 1 0 0 0.5 0 0.25 -2.5 P5 0 2.5 0 0 0 -0.5 1 0.25 2.5 P2 0.5 50 0 1 0 0 0 -1 0 P3 10 2 0 0 1 0 0 0 1 82. 0 0 0 2.5 0 0.75 -2.5 Z 5 5 The leaving variable is P5 and entering variable is P7. Tableau 4 5 0.5 1 0 0 0 0 0 P2 P3 P4 P5 P6 P7 Base Cb P0 P P1 5 10 1 0 0 0 1 0 0 P7 0 1 0 0 0 -0.2 0.4 -0.1 1 P2 0.5 50 0 1 0 0 0 -1 0 P3 10 1 0 0 1 0.2 0.4 0.1 0 85 0 0 0 2 1 0.5 0 Z 1 The optimal solution value is Z = 85 X1 = 10 X2 = 50 X3 = 1