1. Linear Programming Supplement D Copyright ©2013 Pearson Education, Inc. publishing as Prentice HallD- 01

2. What is Linear Programming? Linear Programming A technique that is useful for allocating scarce resources among competing demands. D- 02Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall

3. Basic Concepts Linear programming is an optimization process – Objective Function – Decision variables – Constraints – The feasible region – Parameter or a coefficient – Linearity – Nonnegativity Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall D - 03

4. Formulating a Problem Step 1: Define the decision variables Step 2: Write out the objective function Step 3: Write out the constraints Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall D - 04

5. Example D.1: Stratton Company The Stratton Company produces two basic types of plastic pipe. Three resources are crucial to the output of pipe: extrusion hours, packaging hours, and a special additive to the plastic raw material. The following data represent next week’s situation, with all data being expressed in units of 100 feet of pipe. Product ResourceType 1Type 2 Resource Availability Extrusion4 hr6 hr48 hr Packaging2 hr 18 hr Additive2 lb1 lb16 lb Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall D - 05

6. Example D.1 : Stratton Company Step 1: To define the decision variables that determine product mix, we let: x 1 =amount of type 1 pipe to be produced and sold next week, measured in 100-foot increments (e.g., x 1 = 2 means 200 feet of type 1 pipe) and x 2 =amount of type 2 pipe to be produced and sold next week, measured in 100-foot increments Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall D - 06

7. Example D.1 : Stratton Company Step 2:Next, we define the objective function. The goal is to maximize the total contribution that the two products make to profits and overhead. Each unit of x 1 yields $34, and each unit of x 2 yields $40. For specific values of and x 1 and x 2, we find the total profit by multiplying the number of units of each product produced by the profit per unit and adding them. Thus, our objective function becomes Maximize:$34 x 1 + $40 x 2 = Z Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall D - 07

8. Example D.1 Step 3:The final step is to formulate the constraints. Each unit of x 1 and x 2 produced consumes some of the critical resources. In the extrusion department, a unit of x 1 requires 4 hours and a unit of x 2 requires 6 hours. The total must not exceed the 48 hours of capacity available, so we use the ≤ sign. Thus, the first constraint is: 4x 1 + 6x 2 ≤ 48 Similarly, we can formulate constraints for packaging and raw materials: 2x 1 + 2x 2 ≤ 18 (packaging) 2x 1 + x 2 ≤ 16 (additive mix) Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall D - 08

9. Example D.1 : Stratton Company These three constraints restrict our choice of values for the decision variable because the values we choose for x 1 and x 2 must satisfy all of the constraints. Negative values do not make sense, so we add nonnegativity restrictions to the model: x 1 ≥ 0 and x 2 ≥ 0 (nonnegativity restrictions) D - 09Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall

10. Example D.1 : Stratton Company We can now state the entire model, made complete with the definitions of variables. Maximize: Subject to: 4x 1 + 6x 2 ≤ 48 2x 1 + 2x 2 ≤ 18 2x 1 + x 2 ≤ 16 x 1 ≥ 0 and x 2 ≥ 0 $34x 1 + $40x 2 = Z where x 1 =amount of type 1 pipe to be produced and sold next week, measured in 100-foot increments x 2 =amount of type 2 pipe to be produced and sold next week, measured in 100-foot increments D - 10Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall

11. Application D.1: Crandon Manufacturing The Crandon Manufacturing Company produces two principal product lines, a portable circular saw and a precision table saw. There are two crucial operations: fabrication and assembly. Maximum market demand next year is 3500 saws per month for both products. The average contribution to profits and overhead is $900 for each circular saw and $600 for each table saw. Management wants to determine the best product mix for the next year so as to maximize contribution to profits and overhead. Also, it is interested in the payoff of expanding capacity or increasing market share. Product Resource Circular SawTable SawMaximum Capacity Fabrication 2 hrs/month1 hrs/month4,000 hrs/month Assembly 1 hrs/month2 hrs/month5,000 hrs/month D - 11Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall

12. Application D.1 : Crandon Manufacturing Definition of Decision Variables x 1 = number of circular saws produced and sold per month x 2 = number of table saws produced and sold per month 2x 1 + 1x 2 ≤ 4,000(Fabrication) 1x 1 + 2x 2 ≤ 5,000(Assembly) 1x 1 + 1x 2 ≤ 3,500(Demand) x 1, x 2 ≥ 0(Nonnegativity) Maximize: Subject to: 900x 1 + 600x 2 = Z D - 12Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall

13. Plot the Constraints Disregard the inequality portion of the constraints; plot the equations. Find the axis intercepts by setting one variable equal to zero and solve for the second variable and repeat to get both intercepts. Once both of the axis intercepts are found, draw a line connecting the two points to get the constraint equation. D - 13Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall

14. Graphic Analysis Five basic steps 1.Plot the constraints 2.Identify the feasible region 3.Plot an objective function line 4.Find the visual solution 5.Find the algebraic solution D - 14Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall

15. Plot the Constraints: Crandon Company From Example D.1 4x 1 + 6x 2 = 48 At the x 1 axis intercept, x 2 = 0, so 4x 1 + 6(0) = 48 x 1 = 12 To find the x 2 axis intercept, set x 1 = 0 and solve for x 2 4(0) + 6x 2 = 48 x 2 = 8 We connect points (0, 8) and (12, 0) with a straight line, as shown on the following slide. D - 15Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall

16. Plot the Constraints 4x 1 + 6x 2 ≤ 48 (extrusion) 18 – 16 – 14 – 12 – 10 – 8 – 6 – 4 – 2 – 0 – ||||||||| 24681012141618 x1x1 x2x2 (0, 8) (12, 0) D - 16Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall

17. Example D.2 : Stratton Company For the Stratton Company problem, plot the other constraints: one constraint for packaging and one constraint for the additive mix The equation for the packaging process’s line is 2 x 1 + 2 x 2 = 18. To find the x 1 intercept, set x 2 = 0: 2 x 1 + 2(0) = 18 x 1 = 9 2(0) + 2 x 2 = 18 x 2 = 9 2 x 1 + 0 = 16 x 1 = 8 2(0) + x 2 = 16 x 2 = 16 For the packaging constraint: For the additive constraint: D - 17Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall

18. Example D.2 : Stratton Company 4 x 1 + 6 x 2 ≤ 48 (extrusion) 2 x 1 + 2 x 2 ≤ 18 (packaging) 2 x 1 + x 2 ≤ 16 (additive mix) 18 – 16 – 14 – 12 – 10 – 8 – 6 – 4 – 2 – 0 – ||||||||| 24681012141618 x1x1 x2x2 (0, 9) (9, 0) (0, 16) (8, 0) D - 18Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall

19. Identify the Feasible Region Feasible region – The area on the graph that contains the solutions which satisfy all of the constraints simultaneously, including the nonnegativity restrictions Locate the area that satisfies all of the constraints using three rules : – For the = constraint, only the points on the line are feasible solutions – For the ≤ constraint, the points on the line and the points below and/or to the left are feasible solutions – For the ≥ constraint, the points on the line and the points above and/or to the right are feasible solutions D - 19Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall

20. Identify the Feasible Region When one or more of the parameters on the left-hand side of a constraint are negative, we draw the constraint line and test a point on one side of it 2 x 1 + x 2 ≥ 10 2 x 1 + 3 x 2 ≥ 18 x 1 ≤ 7 x 2 ≤ 5 –6 x 1 + 5 x 2 ≤ 5 x 1, x 2 ≥ 0 12 – 11 – 10 – 9 – 8 – 7 – 6 – 5 – 4 – 3 – 2 – 1 – 0 – |||||||||||| 123456789101112 Feasible region – 6 x 1 + 5 x 2 ≤ 5 2 x 1 + x 2 ≥ 10 2 x 1 + 3 x 2 ≥ 18 x 1 ≤ 7 x 2 ≤ 5 x2x2 x1x1 Test point D - 20Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall

21. Example D.3 : Stratton Company Identify the feasible region for the Stratton Company problem Because the problem contains only ≤ constraints, and the parameters on the left-hand side of each constraint are not negative, the feasible portions are to the left of and below each constraint. The feasible region, shaded in Figure E.4, satisfies all three constraints simultaneously. 4 x 1 + 6 x 2 ≤ 48 (extrusion) 2 x 1 + 2 x 2 ≤ 18 (packaging) 2 x 1 + x 2 ≤ 16 (additive mix) 18 – 16 – 14 – 12 – 10 – 8 – 6 – 4 – 2 – 0 – ||||||||| 24681012141618 x1x1 x2x2 B D C E A Feasible region D - 21Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall

22. Application D.2: Crandon Manufacturing Plot the constraint equations for Crandon Manufacturing and shade feasible region. 2x 1 + 1x 2 ≤ 4,000(Fabrication) 1x 1 + 2x 2 ≤ 5,000(Assembly) 1x 1 + 1x 2 ≤ 3,500(Demand) x 1, x 2 ≥ 0(Nonnegativity) Point 1Point 2 Constraintx1x1 x2x2 x1x1 x2x2 104,0002,0000 202,5005,0000 303,500 0 D - 22Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall

23. Application D.2: Crandon Manufacturing |||||||||| 500100015002000250030003500400045005000 4000 – 3500 – 3000 – 2500 – 2000 – 1500 – 1000 – 500 – 0 – 2 x 1 + x 2 ≤ 4,000 (Fabrication) x 1 + x 2 ≤ 3,500 (Demand) x 1 + 2 x 2 ≤ 5,000 (Assembly) (2,000, 0)(5,000, 0)(3,500, 0) (0, 4,000) (0, 3,500) (0, 2,500) D - 23Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall

24. Plot an Objective Function Line Limit search for solution to the corner points. A corner point lies at the boundary of the feasible region. Interior points need not be considered. Other points on the boundary of the feasible region may be ignored. If the objective function ( Z ) is profits, each line is called an iso-profit line. If Z measures cost, the line is called an iso-cost line. D - 24Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall

25. From Stratton Company problem, we choose corner point B (0, 8) Plot an Objective Function Line 34x 1 + 40x 2 = Z 34(0) + 40(8) = 320 At corner point E (8, 0) the objective function is 34(8) + 40(0) = 272 Solving for the other axis intercept 34(0) + 40(x 2 ) = 272 x 2 = 6.8 D - 25Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall

26. Plot an Objective Function Line 4 x 1 + 6 x 2 ≤ 48 (extrusion) 2 x 1 + 2 x 2 ≤ 18 (packaging) 2 x 1 + x 2 ≤ 16 (additive mix) 18 – 16 – 14 – 12 – 10 – 8 – 6 – 4 – 2 – 0 – ||||||||| 24681012141618 x1x1 x2x2 B D C E A Optimal solution (3, 6) 34 x 1 + 40 x 2 = 272 D - 26 Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall

27. 4 x 1 + 6 x 2 ≤ 48 (extrusion) 2 x 1 + 2 x 2 ≤ 18 (packaging) 2 x 1 + x 2 ≤ 16 (additive mix) 18 – 16 – 14 – 12 – 10 – 8 – 6 – 4 – 2 – 0 – ||||||||| 24681012141618 x1x1 x2x2 B D C E A Optimal solution (3, 6) 34 x 1 + 40 x 2 = 272 Find the Visual Solution D - 27Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall

28. Application D.3: Crandon Manufacturing Plot iso-profit lines and identify the visual solution for Crandon Manufacturing Let Z = $2,000,000 (arbitrary choice) Plot $900 x 1 + $600 x 2 = $2,000,000 Point 1Point 2 Profitx1x1 x2x2 x1x1 x2x2 $2,000,00003333.332222.220 D - 28Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall

29. Application D.3: Crandon Manufacturing |||||||||| 500100015002000250030003500400045005000 4000 – 3500 – 3000 – 2500 – 2000 – 1500 – 1000 – 500 – 0 – 2 x 1 + x 2 ≤ 4,000 (Fabrication) x 1 + x 2 ≤ 3,500 (Demand) x 1 + 2 x 2 ≤ 5,000 (Assembly) ( 2,222, 0) (0, 3,333) Visual solution is approximately (1,100, 2,100) D- 29Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall

30. Step 1.Develop an equation with just one unknown by multiplying both sides of one equation by a constant so that the coefficient for one of the two decision variables is identical in both equations. Then subtract one equation from the other and solve the resulting equation for its single unknown variable. Step 2.Insert this decision variable’s value into either one of the original constraints and solve for the other decision variable. Find the Algebraic Solution D - 30Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall

31. Find the optimal solution algebraically for the Stratton Company problem. What is the value of Z when the decision variables have optimal values? Example D.4: Stratton Company Step 1: The optimal corner point lies at the intersection of the extrusion and packaging constraints. Listing the constraints as equalities, we have: 4 x 1 + 4 x 2 = 48 (extrusion) 2 x 1 + 2 x 2 = 18 (packaging) D - 31Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall

32. Multiply each term in the packaging constraint by 2. The packaging constraint now is 4 x 1 + 4 x 2 ≤ 36. Next, subtract the packaging constraint from the extrusion constraint. The result will be an equation from which has x 1 dropped out. (Alternatively, we could multiply the second equation by 3 so that x 2 drops out after the subtraction.) Example D.4 : Stratton Company 4x 1 + 6x 2 = 48 –(4x 1 + 4x 2 = 36) 2x 2 = 12 x 2 = 6 D - 32Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall

33. Step 2: Substitute the value of x 2 into the extrusion equation: Example D.4 : Stratton Company 4x 1 + 6(6)= 48 4x 1 = 12 x 1 = 3 Thus, the optimal point is (3, 6) This solution gives a total profit of: 34(3) + 40(6) = $342 D - 33Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall

34. Application D.4: Crandon Manufacturing Solve Crandon Manufacturing algebraically with two equations and two unknowns From our earlier visual analysis, the optimal solution is at the intersection of the Fabrication and Assembly constraints. 2x 1 + 1x 2 ≤ 4,000(Fabrication) 1x 1 + 2x 2 ≤ 5,000(Assembly) D - 34Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall

35. Application D.4: Crandon Manufacturing Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall D - 35 2x 1 + 1x 2 = 4000 (fabrication constraint as an equality) -2(1x 1 + 2x 2 = 5000) (assembly constraint as an equality) -3x 2 = -6000 (subtracting the 2 constraints) Optimal Z: $900(1,000) + $600(2,000) = $2,100,000

36. Slack and Surplus Variables A binding constraint is a resource which is completely exhausted when the optimal solution is used because it limits the ability to improve the objective function. Insert the optimal solution into a constraint equation and solve it. If the number on the left-hand side and the number on the right-hand side are equal, then the constraint is binding. Relaxing a constraint means increasing the right-hand side for a ≤ constraint and decreasing the right-hand side for a ≥ constraint. Relaxing a binding constraint means a better solution is possible. D - 36Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall

37. Slack and Surplus Variables: Crandon Manufacturing The additive mix constraint, 2x 1 + x 2 ≤ 16, can be rewritten by adding slack variable s 1 : 2x 1 + x 2 + s 1 = 16 We then find the slack at the optimal solution (3, 6): 2(3) + 6 + s 1 = 16 s 1 = 4 For a ≥ constraint we subtract a surplus variable from the left- hand side D - 37Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall

38. Application D.5 : Crandon Manufacturing Find the slack at the optimal solution for Crandon Manufacturing Slack in fabrication at (1000, 2000) 2 x 1 + 1 x 2 ≤ 4,000 2 x 1 + 1 x 2 + s 1 = 4,000 2(1000) + (2000) + s 1 = 4,000 s 1 = 0 D - 38Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall

39. Application D.5 : Crandon Manufacturing Slack in assembly at (1000, 2000) x 1 + 2x 2 ≤ 5,000 x 1 + 2x 2 + s 2 = 5,000 (1000) + 2(2000) + s 2 = 5,000 s 2 = 0 Slack in demand at (1000, 2000) x 1 + x 2 ≤ 3,500 x 1 + x 2 + s 3 = 3,500 1000 + 2000 + s 3 = 3,500 s 3 = 500 Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall D - 39

40. Sensitivity Analysis Parameters in the objective function and constraints are not always known with certainty. – Usually parameters are just estimates which don’t reflect uncertainties. – After solving the problem using these estimated values, the analysts can determine how much the optimal values of the decision variables and the objective function value Z would be affected if certain parameters had different values. This type of post solution analysis for answering “what-if” questions is called sensitivity analysis. D - 40Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall

41. Sensitivity Analysis SENSITIVITY ANALYSIS INFORMATION PROVIDED BY LINEAR PROGRAMMING TermDefinition Reduced Cost How much the objective function coefficient of a decision variable must improve (increase for maximum or decrease for minimization) before the optimal solution changes and the decision variable “enters” the solution with some positive number Shadow price The marginal improvement in Z (increase for maximization and decrease for minimization) caused by relaxing the constraint by one unit Range of optimality The interval (lower and upper bounds) of an objective function coefficient over which the optimal values of the decision variables remain unchanged Range of feasibility The interval (lower and upper bounds) over which the right-hand-side parameter can vary while its shadow price remains valid D - 41Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall

42. Simplex method – An iterative algebraic procedure – The initial feasible solution starts at a corner point – Subsequent iterations result in improved intermediate solutions – In general, a corner point has no more than m variables greater than 0, where m is the number of constraints. – When no further improvement is possible, the optimal solution has been found and the algorithm stops. Computer Solution D - 42Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall

43. 43 LP-3 Symplex Method

44. 44  When decision variables are more than 2, it is always advisable to use Simplex Method to avoid lengthy graphical procedure.  The simplex method is not used to examine all the feasible solutions.  It deals only with a small and unique set of feasible solutions, the set of vertex points (i.e., extreme points) of the convex feasible space that contains the optimal solution. The Simplex Method

45. 45  Steps involved: 1.Locate an extreme point of the feasible region. 2.Examine each boundary edge intersecting at this point to see whether movement along any edge increases the value of the objective function. 3.If the value of the objective function increases along any edge, move along this edge to the adjacent extreme point. If several edges indicate improvement, the edge providing the greatest rate of increase is selected. 4.Repeat steps 2 and 3 until movement along any edge no longer increases the value of the objective function. The Simplex Method (2)

46. 46 The Simplex Method (3) Example: Product Mix Problem The N. Dustrious Company produces two products: I and II. The raw material requirements, space needed for storage, production rates, and selling prices for these products are given below: The total amount of raw material available per day for both products is 15751b. The total storage space for all products is 1500 ft 2, and a maximum of 7 hours per day can be used for production. The company wants to determine how many units of each product to produce per day to maximize its total income.

47. 47 The Simplex Method (4) Solution  Step 1: Convert all the inequality constraints into equalities by the use of slack variables. Let: As already developed, the LP model is: …..Eq (4)

48. 48 The Simplex Method (5)  Introducing these slack variables into the inequality constraints and rewriting the objective function such that all variables are on the left-hand side of the equation. Equation 4 can be expressed as: …..Eq (5)

49. 49 The Simplex Method (6)  Since the coefficients of x 1 and x 2 in Eq. (A1) are both negative, the value of Z can be increased by giving either x 1 or x 2 some positive value in the solution.  In Eq. (B1), if x 2 = S 1 = 0, then x 1 = 1500/4 = 375. That is, there is only sufficient storage space to produce 375 units at product I.  From Eq. (C1), there is only sufficient raw materials to produce 1575/5 = 315 units of product I.  From Eq. (D1), there is only sufficient time to produce 420/1 = 420 units of product I.  Therefore, considering all three constraints, there is sufficient resource to produce only 315 units of x 1. Thus the maximum value of x 1 is limited by Eq. (C1).

50. 50 The Simplex Method (7)  Step 2: From Equation CI, which limits the maximum value of x 1. …..Eq (6) Substituting this equation into Eq. (5) yields the following new formulation of the model. …..Eq (7)

51. 51 The Simplex Method (8)  It is now obvious from these equations that the new feasible solution is: x 1 = 315, x 2 = 0, S 1 = 240, S 2 = 0, S 3 = 105, and Z = 4095  It is also obvious from Eq.(A2) that it is also not the optimum solution. The coefficient of x 1 in the objective function represented by A2 is negative ( -16/5), which means that the value of Z can be further increased by giving x 2 some positive value.

52. 52 The Simplex Method (9)  Following the same analysis procedure used in step 1, it is clear that:  In Eq. (B2), if S 1 = S 1 = 0, then x 2 = (5/13)(240) = 92.3.  From Eq. (C2), x 2 can take on the value (5/3 )(315) = 525 if x 1 = S 2 = 0  From Eq. (D2), x 2 can take on the value (5/7)(105) = 75 if S 2 = S 3 = 0  Therefore, constraint D 2 limits the maximum value of x 2 to 75. Thus a new feasible solution includes x 2 = 75, S 2 = S 3 = 0.

53. 53 The Simplex Method (10)  Step 3: From Equation D2: …..Eq (8) Substituting this equation into Eq. (7) yield: …..Eq (9) From these equations, the new feasible solution is readily found to be: x 1 = 270, x 2 = 75, S 1 = 45, S 2 = 0, S 3 = 0, Z = 4335.  Because the coefficients in the objective function represented by Eq. (A3) are all positive, this new solution is also the optimum solution.

54. 54 Simplex Tableau for Maximization(1)  Step I: Set up the initial tableau using Eq. (5). In any iteration, a variable that has a nonzero value in the solution is called a basic variable.

55. 55 Simplex Tableau for Maximization(2)  Step II:. Identify the variable that will be assigned a nonzero value in the next iteration so as to increase the value of the objective function. This variable is called the entering variable.  It is that nonbasic variable which is associated with the smallest negative coefficient in the objective function.  If two or more nonbasic variables are tied with the smallest coefficients, select one of these arbitrarily and continue.  Step III: Identify the variable, called the leaving variable, which will be changed from a nonzero to a zero value in the next solution.

56. 56 Simplex Tableau for Maximization(3)  Step IV:. Enter the basic variables for the second tableau. The row sequence of the previous tableau should be maintained, with the leaving variable being replaced by the entering variable.

57. 57 Simplex Tableau for Maximization(4)  Step V: Compute the coefficients for the second tableau. A sequence of operations will be performed so that at the end the x 1 column in the second tableau will have the following coefficients: The second tableau yields the following feasible solution: x 1 = 315, x 2 = 0, SI = 240, S2 = 0, S3 = 105, and Z = 4095

58. 58 Simplex Tableau for Maximization(5)  The row operations proceed as follows:  The coefficients in row C2 are obtained by dividing the corresponding coefficients in row C1 by 5.  The coefficients in row A2 are obtained by multiplying the coefficients of row C2 by 13 and adding the products to the corresponding coefficients in row Al.  The coefficients in row B2 are obtained by multiplying the coefficients of row C2 by -4 and adding the products to the corresponding coefficients in row Bl.  The coefficients in row D2 are obtained by multiplying the coefficients of row C2 by -1 and adding the products to the corresponding coefficients in row Dl.

59. 59 Simplex Tableau for Maximization(6)  Step VI: Check for optimality. The second feasible solution is also not optimal, because the objective function (row A2) contains a negative coefficient. Another iteration beginning with step 2 is necessary.  In the third tableau (next slide), all the coefficients in the objective function (row A3) are positive. Thus an optimal solution has been reached and it is as follows: x 1 = 270, x 2 = 75, SI = 45, S2 = 0, S3 = 0, and Z = 4335

60. 60

61. 61 Marginal Values of Additional Resources (1)  The simplex solution yields the optimum production program for N. Dustrious Company.  The company can maximize its sale income to $4335 by producing 270 units of product I and 75 units of product II.  There will be no surplus of raw materials or production time.  But there will be 45 units of unused storage space.  The managers are interested to know if it is worthwhile to increase its production by purchasing additional units of raw materials and by either expanding its production facilities or working overtime.

62. 62 Marginal Values of Additional Resources (2)  The critical questions are:  What is the income value (or marginal value) of each additional unit of each type of resources?  What is the maximum cost ( or marginal cost) that they should be willing to pay for each additional unit of resources?  Answers to these questions can be obtained from the objective function in the last tableau of the simplex solution:

63. 63 Marginal Values of Additional Resources (3)  Because S 1, S 2 and S 3 represent surplus resources, the negatives of these variables (i.e., -S 1, -S 2, -S 3 ) represent additional units of these resources that can be made available.  The income values (or marginal values of additional units of these resources can be obtained by taking the partial derivatives of Z with respect to -S 1, -S 2 and -S 3.  Therefore, the marginal value of one additional unit of:

64. 64 Marginal Values of Additional Resources (4)  Thus, the marginal values of additional units of resources can be obtained directly from the coefficients of the objective function in the last tableau of a simplex solution.  The N. Dustrious Company should be willing to pay up to $15/7 for an additional unit of raw materials and $16/7 for an additional unit of production time.  If the actual cost of an additional unit (i.e., marginal cost) of these resources are smaller than the marginal value, the company should be able to increase its income by increasing production.  The marginal values above are valid, however, only as long as there is surplus storage space available.

65. 65 Sensitivity Analysis  Sensitivity analysis helps to test the sensitivity of the optimum solution with respect to changes of the coefficients in the objective function, coefficients in the constraints inequalities, or the constant terms in the constraints.  For Example in the case study discussed:  The actual selling prices (or market values) of the two products may vary from time to time. Over what ranges can these prices change without affecting the optimality of the present solution?  Will the present solution remain the optimum solution if the amount of raw materials, production time, or storage space is suddenly changed because of shortages, machine failures, or other events?  The amount of each type of resources needed to produce one unit of each type of product can be either increased or decreased slightly. Will such changes affect the optimal solution ?

66. 66 Complications in Simplex Method (1)  An objective function to be minimized instead of maximized.  Greater-than-or-equal-to constraints.  Equalities instead of inequalities for constraints.  Decision variables unrestricted in signs.  Zero constants on the right-hand side of one or more constraints.  Some or all decision variables must be integers.  Non-positive constants on the right-hand side of the constraints.  More than one optimal solution, that is, multiple solutions such that there is no unique optimal solution.

67. 67 Complications in Simplex Method (2)  The constraints are such that no feasible solution exists.  The constraints are such that one or more of the variables can increase without limit and never violate a constraint (i.e., the solution is unbounded).  Some or all of the coefficients and right-hand-side terms are given by a probability distribution rather than a single value.

68. 68 Complications in Simplex Method (3)  This objective function can be converted to the standard form of maximization. Let Z’ = -Z, so: Minimization Problem – Solution 1  Since maximum Z’ = minimum (Z), the objective function becomes:  After the Z’ value is found, replace Z = -Z’

69. 69 Complications in Simplex Method (4)  In the case of a minimization problem, an optimum solution is reached when:  All the nonbasic variables have nonpositive coefficients in row 1 of the simplex tableau  The entering variable will be one which has the largest positive coefficient in row I.  All the other operations in the simplex method remain unchanged. Minimization Problem – Solution 2

70. 70 Complications in Simplex Method (5)  In the standard form of the linear programming model, the constraints are all expressed as less than or equal to (  ) a certain amount, that is, Greater- Than-Or-Equal- To Constraints  In many occasions, the constraints must specify the lower bounds rather than the upper bounds such as: which involves the inequality "greater than or equal to"

71. 71 Complications in Simplex Method (6) Example: Greater- Than-Or-Equal- To Constraints  To start the solution, slack variables must first be assigned to convert all in-equalities to equalities. Let S 1 and S 2 be slack variables.  Re- arrange the objective function so that all the variables are on the left-hand side of the equation. …….Eq. (1)

72. 72 Complications in Simplex Method (7)  The negative signs for S 1 and S 2 make it no longer feasible to set all the decision variables (i.e., y 1, y 2, y 3 ) equal to zero as the initial solution.  To assure a starting feasible solution, artificial variables can be added to the greater-than-or-equal-to constraints. Let W 1 and W 2 be two artificial variables. Hence the Eq. (2) becomes: …….Eq. (2) …….Eq. (3)

73. 73 Complications in Simplex Method (8)  A starting feasible solution can be easily derived from Eq. (3) as follows:  The objective function in Eq. (3) then becomes:  From Eq. (3): …….Eq. (4)

74. 74 Complications in Simplex Method (9)  Substituting these expressions in Eq. (4) yields the following new expression for the objective function:  The objective function may now be combined with Eq. (3) to express the problem model as follows:  The coefficients and constants in Eq. (5) can now be arranged in the Tableau format as shown in next slide. …….Eq. (5)

75. 75 Complications in Simplex Method (10)

76. 76 Complications in Simplex Method (11)  An equality constraint has the following general form: Equality Constraint  An artificial variable must be assigned to each equality constraint to start the simplex solution. Otherwise, the constraint would be violated when all the decision variables are assumed to be zero.  Sometimes a decision variable may take on either negative or positive values. If x 1 is unrestricted in sign, replace it throughout the model by the difference of two new nonnegative variables: x j = x j ’ – x j ’’ where x j ’  0, x j ’’  0

77. 77 Complications in Simplex Method (12)  Because x j ’ and x j ’’ can have any nonnegative values their difference (x j ’ –x j ’’) can have any value (positive or negative). After substitution, the simplex method can proceed with just nonnegative variables. Equality Constraint

78. 78 Complications in Simplex Method (13)  If the number of basic variables is fewer than the number of constraints in a solution, the solution is said to be degenerate.  A zero constant term for one or more basic variables in any iteration of the simplex solution would be a clear indication of a degenerate solution.  The normal simplex procedure cannot solve a degenerate problem.  Advanced methods are available to solve degenerate problems. Degenerate Solution

79. 79 Complications in Simplex Method (14)  A linear programming problem in which all the decision variables must have integer values is called an integer programming problem.  A problem in which only some of the decision variables must have integer values is called a mixed-integer programming problem.  Sometimes, some (or all) of the decision variables must have the value of either 0 or 1. Such problems are then called zero-one mixed-integer programming problems.  Simplex method cannot be used to such problems. Advanced methods are available for this purpose. Integer and Mixed-Integer Problems

80. Most real-world linear programming problems are solved on a computer, which can dramatically reduce the amount of time required to solve linear programming problems POM for Windows in MyOMLab can handle small- to medium-sized linear programming problems Microsoft’s Excel Solver offers a second option for similar problem sizes Computer Output D - 80Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall

81. Computer Output: Stratton Company D- 81 Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall Data Entry Screen

82. Computer Output : Stratton Company D- 82Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall Data Entry Screen (Using POM for Windows)

83. Results and Ranging Screens D- 83 Results Screen Ranging Screen Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall

84. Tips on Interpreting Output Reduced cost – The sensitivity number is relevant only for a decision variable that is 0 in the optimal solution – It reports how much the objective function coefficient must improve before it would enter the optimal solution at some positive level Shadow prices – The number is relevant only for binding constraints – The shadow price as either positive or negative The number of variables in the optimal solution > 0 never exceeds the number of constraints Degeneracy occurs when the number of variables ≠ 0 in the optimal solution is less than the number of constraints D - 84Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall