2. Operations Research By: Saeed Yaghoubi 1

3. Graphical Analysis 2

4. 3 Max 8X 1 + 5X 2 (Weekly profit) subject to 2X 1 + 1X 2  1000 (Plastic) 3X 1 + 4X 2  2400 (Production Time) X 1 + X 2  700 (Total production) X 1 - X 2  350 (Mix) X j > = 0, j = 1,2 (Nonnegativity) The Galaxy Linear Programming Model

5. 4 The Graphical Analysis of Linear Programming The set of all points that satisfy all the constraints of the model is called a FEASIBLE REGION

6. 5 Using a graphical presentation we can represent all the constraints, the objective function, and the three types of feasible points.

7. 6 The non-negativity constraints X2X2 X1X1 Graphical Analysis – the Feasible Region

8. 7 1000 500 Feasible X2X2 Infeasible Production Time 3X 1 +4X 2  2400 Total production constraint: X 1 +X 2  700 (redundant) 500 700 The Plastic constraint 2X 1 +X 2  1000 X1X1 700 Graphical Analysis – the Feasible Region

9. 8 1000 500 Feasible X2X2 Infeasible Production Time 3X 1 +4X2  2400 Total production constraint: X 1 +X 2  700 (redundant) 500 700 Production mix constraint: X 1 -X2  350 The Plastic constraint 2X 1 +X 2  1000 X1X1 700 Graphical Analysis – the Feasible Region There are three types of feasible points Interior points. Boundary points.Extreme points.

10. 9 Solving Graphically for an Optimal Solution

11. 10 The search for an optimal solution Start at some arbitrary profit, say profit = $2,000... Then increase the profit, if possible......and continue until it becomes infeasible Profit =$4360 500 700 1000 500 X2X2 X1X1

12. 11 –If a linear programming problem has an optimal solution, an extreme point is optimal. Extreme points and optimal solutions

13. 12 For multiple optimal solutions to exist, the objective function must be parallel to one of the constraints Multiple optimal solutions Any weighted average of optimal solutions is also an optimal solution.

14. Sensitivity Analysis 13

15. 14 The Role of Sensitivity Analysis of the Optimal Solution Is the optimal solution sensitive to changes in input parameters? Possible reasons for asking this question: –Parameter values used were only best estimates. –Dynamic environment may cause changes. –“What-if” analysis may provide economical and operational information.

16. 15 Range of Optimality –The optimal solution will remain unchanged as long as An objective function coefficient lies within its range of optimality There are no changes in any other input parameters. –The value of the objective function will change if the coefficient multiplies a variable whose value is nonzero. Sensitivity Analysis of Objective Function Coefficients.

17. 16 500 1000 500800 X2X2 X1X1 Max 8X 1 + 5X 2 Max 4X 1 + 5X 2 Max 3.75X 1 + 5X 2 Max 2X 1 + 5X 2 Sensitivity Analysis of Objective Function Coefficients.

18. 17 500 1000 400600800 X2X2 X1X1 Max8X 1 + 5X 2 Max 3.75X 1 + 5X 2 Max 10 X 1 + 5X 2 Range of optimality: [3.75, 10] Sensitivity Analysis of Objective Function Coefficients.

19. 18 Reduced cost Assuming there are no other changes to the input parameters, the reduced cost for a variable X j that has a value of “0” at the optimal solution is: –The negative of the objective coefficient increase of the variable X j (-  C j ) necessary for the variable to be positive in the optimal solution –Alternatively, it is the change in the objective value per unit increase of X j. Complementary slackness At the optimal solution, either the value of a variable is zero, or its reduced cost is 0.

20. 19 In sensitivity analysis of right-hand sides of constraints we are interested in the following questions: –Keeping all other factors the same, how much would the optimal value of the objective function (for example, the profit) change if the right-hand side of a constraint changed by one unit? –For how many additional or fewer units will this per unit change be valid? Sensitivity Analysis of Right-Hand Side Values

21. 20 Any change to the right hand side of a binding constraint will change the optimal solution. Any change to the right-hand side of a non- binding constraint that is less than its slack or surplus, will cause no change in the optimal solution. Sensitivity Analysis of Right-Hand Side Values

22. 21 Shadow Prices Assuming there are no other changes to the input parameters, the change to the objective function value per unit increase to a right hand side of a constraint is called the “Shadow Price”

23. 22 1000 500 X2X2 X1X1 2X 1 + 1x 2 <=1000 When more plastic becomes available (the plastic constraint is relaxed), the right hand side of the plastic constraint increases. Production time constraint Maximum profit = $4360 2X 1 + 1x 2 <=1001 Maximum profit = $4363.4 Shadow price = 4363.40 – 4360.00 = 3.40 Shadow Price – graphical demonstration The Plastic constraint

24. 23 Range of Feasibility Assuming there are no other changes to the input parameters, the range of feasibility is –The range of values for a right hand side of a constraint, in which the shadow prices for the constraints remain unchanged. –In the range of feasibility the objective function value changes as follows: Change in objective value = [Shadow price][Change in the right hand side value]

25. 24 Range of Feasibility 1000 500 X2X2 X1X1 2X 1 + 1x 2 <=1000 Increasing the amount of plastic is only effective until a new constraint becomes active. The Plastic constraint This is an infeasible solution Production time constraint Production mix constraint X 1 + X 2  700 A new active constraint

26. 25 Range of Feasibility 1000 500 X2X2 X1X1 The Plastic constraint Production time constraint Note how the profit increases as the amount of plastic increases. 2X 1 + 1x 2  1000

27. 26 Range of Feasibility 1000 500 X2X2 X1X1 2X 1 + 1X 2  1100 Less plastic becomes available (the plastic constraint is more restrictive). The profit decreases A new active constraint Infeasible solution

28. 27 –Sunk costs: The shadow price is the value of an extra unit of the resource, since the cost of the resource is not included in the calculation of the objective function coefficient. –Included costs: The shadow price is the premium value above the existing unit value for the resource, since the cost of the resource is included in the calculation of the objective function coefficient. The correct interpretation of shadow prices

29. 28 Other Post - Optimality Changes Addition of a constraint. Deletion of a constraint. Addition of a variable. Deletion of a variable. Changes in the left - hand side coefficients.

30. 29 Infeasibility: Occurs when a model has no feasible point. Unboundness: Occurs when the objective can become infinitely large (max), or infinitely small (min). Alternate solution: Occurs when more than one point optimizes the objective function Models Without Unique Optimal Solutions

31. 30 1 No point, simultaneously, lies both above line and below lines and. 1 23 2 3 Infeasible Model

32. 31 Unbounded solution The feasible region Maximize the Objective Function

33. Coordinate Axes Graphical Solution of Maximization Model (1 of 12) Coordinates for Graphical Analysis Maximize Z = $40x 1 + $50x 2 subject to:1x 1 + 2x 2  40 4x 1 + 3x 2  120 x 1, x 2  0 X 1 is bowls X 2 is mugs

34. Labor Constraint Graphical Solution of Maximization Model (2 of 12) Graph of Labor Constraint Maximize Z = $40x 1 + $50x 2 subject to:1x 1 + 2x 2  40 4x 1 + 3x 2  120 x 1, x 2  0

35. Labor Constraint Area Graphical Solution of Maximization Model (3 of 12) Labor Constraint Area Maximize Z = $40x 1 + $50x 2 subject to:1x 1 + 2x 2  40 4x 1 + 3x 2  120 x 1, x 2  0

36. Clay Constraint Area Graphical Solution of Maximization Model (4 of 12) Clay Constraint Area Maximize Z = $40x 1 + $50x 2 subject to:1x 1 + 2x 2  40 4x 1 + 3x 2  120 x 1, x 2  0

37. Both Constraints Graphical Solution of Maximization Model (5 of 12) Graph of Both Model Constraints Maximize Z = $40x 1 + $50x 2 subject to:1x 1 + 2x 2  40 4x 1 + 3x 2  120 x 1, x 2  0

38. Feasible Solution Area Graphical Solution of Maximization Model (6 of 12) Figure 2.7 Feasible Solution Area Maximize Z = $40x 1 + $50x 2 subject to:1x 1 + 2x 2  40 4x 1 + 3x 2  120 x 1, x 2  0

39. Objective Function Solution = $800 Graphical Solution of Maximization Model (7 of 12) Objection Function Line for Z = $800 Maximize Z = $40x 1 + $50x 2 subject to:1x 1 + 2x 2  40 4x 1 + 3x 2  120 x 1, x 2  0

40. Alternative Objective Function Solution Lines Graphical Solution of Maximization Model (8 of 12) Alternative Objective Function Lines Maximize Z = $40x 1 + $50x 2 subject to:1x 1 + 2x 2  40 4x 1 + 3x 2  120 x 1, x 2  0

41. Optimal Solution Graphical Solution of Maximization Model (9 of 12) Identification of Optimal Solution Point Maximize Z = $40x 1 + $50x 2 subject to:1x 1 + 2x 2  40 4x 1 + 3x 2  120 x 1, x 2  0

42. Optimal Solution Coordinates Graphical Solution of Maximization Model (10 of 12) Optimal Solution Coordinates Maximize Z = $40x 1 + $50x 2 subject to:1x 1 + 2x 2  40 4x 1 + 3x 2  120 x 1, x 2  0

43. Extreme (Corner) Point Solutions Graphical Solution of Maximization Model (11 of 12) Solutions at All Corner Points Maximize Z = $40x 1 + $50x 2 subject to:1x 1 + 2x 2  40 4x 1 + 3x 2  120 x 1, x 2  0

44. Optimal Solution for New Objective Function Graphical Solution of Maximization Model (12 of 12) Maximize Z = $70x 1 + $20x 2 subject to:1x 1 + 2x 2  40 4x 1 + 3x 2  120 x 1, x 2  0 Optimal Solution with Z = 70x 1 + 20x 2

45. Standard form requires that all constraints be in the form of equations (equalities). A slack variable is added to a  constraint (weak inequality) to convert it to an equation (=). A slack variable typically represents an unused resource. A slack variable contributes nothing to the objective function value. Slack Variables

46. Linear Programming Model: Standard Form Max Z = 40x 1 + 50x 2 + 0s 1 + 0s 2 subject to:1x 1 + 2x 2 + s 1 = 40 4x 1 + 3x 2 + s 2 = 120 x 1, x 2, s 1, s 2  0 Where: x 1 = number of bowls x 2 = number of mugs s 1, s 2 are slack variables Solution Points A, B, and C with Slack

47. Minimize Z = $6x 1 + $3x 2 subject to:2x 1 + 4x 2  16 4x 1 + 3x 2  24 x 1, x 2  0 Graph of Both Model Constraints Constraint Graph – Minimization

48. Feasible Solution Area Feasible Region– Minimization Minimize Z = $6x 1 + $3x 2 subject to:2x 1 + 4x 2  16 4x 1 + 3x 2  24 x 1, x 2  0

49. Optimum Solution Point Optimal Solution Point – Minimization Minimize Z = $6x 1 + $3x 2 subject to:2x 1 + 4x 2  16 4x 1 + 3x 2  24 x 1, x 2  0

50. A surplus variable is subtracted from a  constraint to convert it to an equation (=). A surplus variable represents an excess above a constraint requirement level. A surplus variable contributes nothing to the calculated value of the objective function. Subtracting surplus variables in the farmer problem constraints: 2x 1 + 4x 2 - s 1 = 16 (nitrogen) 4x 1 + 3x 2 - s 2 = 24 (phosphate) Surplus Variables – Minimization

51. Graph of Fertilizer Example Graphical Solutions – Minimization Minimize Z = $6x 1 + $3x 2 + 0s 1 + 0s 2 subject to:2x 1 + 4x 2 – s 1 = 16 4x 1 + 3x 2 – s 2 = 24 x 1, x 2, s 1, s 2  0

52. For some linear programming models, the general rules do not apply. Special types of problems include those with:  Multiple optimal solutions  Infeasible solutions  Unbounded solutions Irregular Types of Linear Programming Problems

53. Example with Multiple Optimal Solutions Multiple Optimal Solutions Beaver Creek Pottery The objective function is parallel to a constraint line. Maximize Z=$40x 1 + 30x 2 subject to:1x 1 + 2x 2  40 4x 1 + 3x 2  120 x 1, x 2  0 Where: x 1 = number of bowls x 2 = number of mugs

54. An Infeasible Problem Graph of an Infeasible Problem Every possible solution violates at least one constraint: Maximize Z = 5x 1 + 3x 2 subject to:4x 1 + 2x 2  8 x 1  4 x 2  6 x 1, x 2  0

55. An Unbounded Problem Graph of an Unbounded Problem Value of the objective function increases indefinitely: Maximize Z = 4x 1 + 2x 2 subject to: x 1  4 x 2  2 x 1, x 2  0

56. The Complete LP Model MAX: 350X 1 + 300X 2 S.T.: 1X 1 + 1X 2 <= 200 9X 1 + 6X 2 <= 1566 12X 1 + 16X 2 <= 2880 X i >= 0 i=1, 2 The general form of an LP model: MAX (or MIN): c 1 X 1 + c 2 X 2 + … + c n X n Subject to: a 11 X 1 + a 12 X 2 + … + a 1n X n <= b 1 : a k1 X 1 + a k2 X 2 + … + a kn X n >= b k : a m1 X 1 + a m2 X 2 + … + a mn X n = b m X i >= 0 i=1,n

57. Graphical solution approach X2X2 X1X1 250 200 150 100 50 0 0 100 150 200 250 240 Feasible Region 12X 1 + 16X 2 = 2880 X 1 + X 2 = 200 9X 1 + 6X 2 = 1566 261 174 180

58. Enumerating the corner points X2X2 X1X1 250 200 150 100 50 0 0 100 150 200250 o.f.v. = $54,000 (0, 180) o.f.v. = $64,000 (80, 120) o.f.v. = $66,100 (122, 78) o.f.v. = $60,900 (174, 0) o.f.v. = $0 (0, 0) $15,000

59. Linear Programming When we attempt to solve an LP, one of the following will occur: Unique optimal solution Multiple (alternate) optimal solutions  Unbounded solutions  Infeasible solution