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. 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 - 43Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall

44. Computer Output: Stratton Company D- 44 Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall Data Entry Screen

45. Computer Output : Stratton Company D- 45Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall Data Entry Screen (Using POM for Windows)

46. Results and Ranging Screens D- 46 Results Screen Ranging Screen Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall

47. 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 - 47Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall

48. Example D.5: Stratton Company The Stratton Company needs answers to three important questions: Would increasing capacities in the extrusion or packaging area pay if it cost an extra $8 per hour over and above the normal costs already reflected in the objective function coefficients? Would increasing packaging capacity pay if it cost an additional $6 per hour? Would buying more raw materials pay? Expanding extrusion capacity would cost a premium of $8 per hour, but the shadow price for that capacity is only $3 per hour. Expanding packaging hours would cost only $6 per hour more than the price reflected in the objective function, and the shadow price is $11 per hour. Buying more raw materials would not pay because a surplus of 4 pounds already exists; the shadow price is $0 for that resource. D - 48 Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall

49. The Transportation Method A special case of linear programming – Represented as a standard table, sometimes called a tableau – Rows of the table are linear constraints that impose capacity limitations – Columns are linear constraints that require a certain demand level to be met – Each cell in the tableau is a decision variable, and a per-unit cost is shown in each cell – The focus in this section is on the setup and interpretation of the problem D - 49Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall

50. Transportation Method for Production Planning Making sure that demand and supply are in balance is central to Sales and Operations Planning (SOP) Transportation method for production planning is based on the assumptions – Demand forecast and workforce adjustment plan is available for each period – Capacity limits on overtime and the use of subcontractors also are required – All costs are linearly related to the amount of goods produced D - 50Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall

51. 1.Obtain the demand forecasts for each period to be covered by the SOP, identify initial inventory levels 2.Select a candidate workforce adjustment plan and specify capacity limits of each production alternative for each period 3.Estimate the cost of holding inventory and the cost of possible production alternatives and any cost of undertime 4.Input the information gathered in steps 1-3 into a computer routine that solves the transportation problem and use the output to calculate the anticipation inventory levels and identify high-cost elements 5.Repeat the process with other plans until you find the solution that best balances cost and qualitative considerations Developing an SOP D - 51Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall

52. Example D.6 The Tru-Rainbow Company produces a variety of paint products. The demand for paint is highly seasonal. Initial inventory is 250,000 gallons, and ending inventory should be 300,000 gallons. Manufacturing manager wants to determine the best production plan. Regular-time cost is $1.00 per unit, overtime cost is $1.50 per unit, subcontracting cost is $1.90 per unit, and inventory holding cost is $0.30 per unit per quarter. Undertime is paid and the cost is $0.50 per unit. D - 52Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall

53. Example D.6 The following constraints apply: a.The maximum allowable overtime in any quarter is 20 percent of the regular-time capacity in that quarter. b.The subcontractor can supply a maximum of 200,000 gallons in any quarter. Production can be subcontracted in one period and the excess held in inventory for a future period to avoid a stockout. c.No backorders or stockouts are permitted. D - 53Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall

54. Example D.6 Demand Regular-time Capacity Overtime Capacity Subcontracting Capacity Quarter 130045090200 Quarter 285045090200 Quarter 31,500750150200 Quarter 435045090200 Totals3,0002,100420800 D - 54Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall

55. Example D.6 D - 55 POM for Windows Data Table Screen Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall

56. Example D.6 From POM for Windows Screen: – The demand for quarter 4 is shown to be 650,000 gallons rather than the demand forecast of only 350,000. – The larger number reflects the desire of the manager to have an ending inventory in quarter 4 of 300,000 gallons. D - 56Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall

57. Example D.6 1.There is a row for each supply alternative on a quarter- by-quarter basis. The last column in each row indicates the maximum amount that can be used to meet demand. 2.A column indicates each future quarter of demand and the last row gives its demand forecast. The Excess Capacity column shows the cost of unused capacity. 3.The numbers in the other cells (excluding the cells in the last row or last column) show the cost of producing a unit in one period and, in some cases, carrying the unit in inventory for sale in a future period. D - 57Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall

58. Example D.6 4.The cells in the bottom left portion of the tableau with a cost of $9,999 are associated with backorders (or producing in a period to satisfy demand in a period after it was needed). If backorder costs are so large, the transportation method will try to avoid backorders because it seeks a solution that minimizes total cost. 5.The least expensive alternatives are those in which the output is produced and sold in the same period. We may not always be able to avoid alternatives that create inventory because of capacity restrictions. 6.The per-unit holding cost for the beginning inventory in period 1 is 0 because it is a function of previous production- planning decisions. D - 58Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall

59. Example D.6 D - 59 Transportation Method (Production Planning) Results Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall

60. Example D.6 Quarter Regular-time Production Overtime ProductionSubcontracting Total Supply Anticipation Inventory 14509020560250 + 560 – 300 = 510 245090200740510 + 740 – 850 = 400 37501502001,100400 + 1,100 – 1,500 = 0 4450901106500 + 650 – 350 = 300 Totals2,1004205303,050 Note: Anticipation inventory is the amount at the end of each quarter, where Beginning inventory + Total production – Actual Demand = Ending inventory D - 60Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall

61. Example D.6 Computing the cost column by column (it can also be down on a row-by-row basis) yields a total cost of $4,010,000, or $4,010 x 1,000. Cost Calculations by Column Quarter 1230($0) + 50($1.00) + 20($1.90)= $88 Quarter 2400($1.30) + 450($1.00)= $970 Quarter 3 20($0.60) + 90($2.10) + 90($1.80) + 200($2.20) + 750($1.00)+ 150($1.50) + 200($1.90) = $2,158 Quarter 4450($1.00) + 90($1.50) + 110($1.90)= $794 Total= $4,010 D - 61Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall

62. Application D.6: Bull Grin The Bull Grin Company makes an animal-feed supplement. Sales are seasonal, but Bull Grin's customers refuse to stockpile the supplement and won’t accept backorders. The reactive alternatives that they use, in addition to work-force variation, are regular time, overtime, subcontracting, and anticipation inventory. Backorders are not allowed. Complete the tableau below by entering the cost per pound produced with each production alternative to meet demand in each period. Bull Grin employs workers who produce 1,000 pounds of supplement for $830 on regular time and $910 on over-time. Holding 1000 pounds of feed supplement in inventory per quarter costs $100. There is no cost for unused regular-time, overtime or subcontracting capacity. D - 62 Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall

63. Application D.6: Bull Grin Now enter data for the capacity column of the tableau as thousands of pounds. The work-force plan being investigating now would provide regular-time capacities of 390 in quarter 1, 400 in quarter 2, 460 in quarter 3, and 380 in quarter 4. Overtime is limited to production of a total of 20,000 pounds per quarter, and subcontractor capacity to only 30,000 pounds per quarter. The current inventory level is 40,000 pounds. Next enter the data for the demand row. The demand forecasts are 130 in quarter 1, 400 in quarter 2, 800 in quarter 3, and 470 in quarter 4. Management wants 40,000 pounds available at the end the year. D - 63Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall

64. QuarterUnusedTotal Alternatives1234Capacity Beginning$0$100$200$300 Inventory400 Regular$830$930$1,030$1,130 Time 90 220 - 80 - 1Overtime$910$1,010$1,110$1,210 - - 20 - - Subcontract$1,000$1,100$1,200$1,300 - - - - 30 Regular$99,999 Time - 2Overtime$99,999 - Subcontract$99,999 - Regular$99,999 Time - 3Overtime$99,999 - Subcontract$99,999 - Regular$99,999 Time -380 4Overtime$99,999 -20 Subcontract$99,999 -30 Demand130301,870 180 220 20 30 460 20 30 380 20 30 Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall D - 64

65. Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall QuarterUnusedTotal Alternatives1234Capacity Beginning$0$100$200$300 Inventory400 Regular$830$930$1,030$1,130 Time 90 220 - 80 - 1Overtime$910$1,010$1,110$1,210 - - 20 - - Subcontract$1,000$1,100$1,200$1,300 - - - - 30 Regular$99,999 Time - 2Overtime$99,999 - Subcontract$99,999 - Regular$99,999 Time - 3Overtime$99,999 - Subcontract$99,999 - Regular$99,999 Time -380 4Overtime$99,999 -20 Subcontract$99,999 -30 Demand130301,870 40 390 20 30 400 20 30 460 20 30 400800470 180 220 20 30 460 20 30 380 20 30 $830$930$1,030 $910$1,010$1,110 $1,000$1,100$1,200 $99,999$830$930 $99,999$910$1,010 $99,999$1,000$1,100 $99,999 $830 $99,999 $910 $99,999 $1,000 D - 65Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall

66. Application D.6 : Bull Grin Quarter 1Quarter 2Quarter 3Quarter 4Totals Production Regular-time1,630 Overtime80 Subcontract90 Total Supply1,800 Shipments1,800 Anticipation Inventory810 390400460380 20 030 410450570430 130400800470 3203708040 D - 66Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall What production levels, shipments, and anticipation inventories are called for by the optimal solution in the tableau above?

67. Application D.6 : Bull Grin Quarter 1:= $ 74,700 Quarter 2:= $ 354,000 Quarter 3:= $ 710,000 Quarter 4:= $ 454,000 Total= $1,592,700 What is the total cost of the optimal solution, except for the cost of hiring and layoffs? D - 67Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall

68. Application D.6 : Bull Grin Quarter 1:= $ 74,700 Quarter 2:= $ 354,000 Quarter 3:= $ 710,000 Quarter 4: 80($1,300) + 380($830) + 20($910) + 30($1,000) = $ 454,000 Total= $1,592,700 What is the total cost of the optimal solution, except for the cost of hiring and layoffs? D - 68Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall 40($0) + 90($830) 220($930) + 180($830) 20($1,110) + 220($930) + 20($1,010) + 30($1,110) + 460($830) + 20($910) + 30($1,000)

69. Other Applications Examples in operations management and other functional areas – Constraint management – Shipping assignments – Inventory control – Shift scheduling D - 69Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall

70. Solved Problem 1 O’Connel Airlines is considering air service from its hub of operations in Cicely, Alaska, to Rome, Wisconsin, and Seattle, Washington. O’Connel has one gate at the Cicely Airport, which operates 12 hours per day. Each flight requires 1 hour of gate time. Each flight to Rome consumes 15 hours of pilot crew time and is expected to produce a profit of $2,500. Serving Seattle uses 10 hours of pilot crew time per flight and will result in a profit of $2,000 per flight. Pilot crew labor is limited to 150 hours per day. The market for service to Rome is limited to nine flights per day. a.Use the graphic method to maximize profits. b.Identify slack and surplus constraints, if any. D - 70Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall

71. Solved Problem 1 a.The objective function is to maximize profits, Z: Maximize: $2,500x 1 + $2,000x 2 = Z where x 1 =number of flights per day to Rome, Wisconsin x 2 =number of flights per day to Seattle, Washington The constraints are x 1 + x 2 ≤ 12(gate capacity) 15x 1 + 10x 2 ≤ 150(labor) x 1 ≤ 9(market) x 1, x 2 ≥ 0 D - 71Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall

72. Solved Problem 1 A careful drawing of iso-profit lines will indicate that point D is the optimal solution. It is at the intersection of the labor and gate capacity constraints. Solving algebraically: The maximum profit results from making six flights to Rome and six flights to Seattle: $2,500(6)+ $2,000(6) = $27,000 15x 1 + 10x 2 = 150(labor) –10x 1 – 10x 2 = –120(gate  –10) 5x 1 + 0x 2 = 30 x 1 = 6 6 + x 2 = 12(gate) x 2 = 6 D - 72Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall

73. Solved Problem 1 ||| 51015 15 – 10 – 5 – 0 – x1x1 x2x2 2,500 x 1 + 2,000 x 2 = $20,000 (iso-profit line) x 1 + x 2 ≤ 12 (gate ) x 1 ≤ 9 (market) 15 x 1 + 10 x 2 ≤ 150 (labor) B C D E A Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall D - 73

74. Solved Problem 1 b.The market constraint has three units of slack, so the demand for flights to Rome is not fully met: x 1 ≤ 9 x 1 + s 3 = 9 6 + s 3 = 9 s 3 = 3 D - 74Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall

75. Solved Problem 2 The Arctic Air Company produces residential air conditioners. The manufacturing manager wants to develop a sales and operations plan for the next year based on the following demand and capacity data (in hundreds of product units): Demand Regular-time Capacity Overtime Capacity Subcontractor Capacity Jan-Feb (1)50651310 Mar-Apr (2)60651310 May-Jun (3)90651310 Jul-Aug (4)120801610 Sep-Oct (5)70801610 Nov-Dec (6)40651310 Totals4304208460 D - 75Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall

76. Solved Problem 2 Undertime is unpaid, and no cost is associated with unused overtime or subcontractor capacity. Producing one air conditioning unit on regular time costs $1,000, including $300 for labor. Producing a unit on overtime costs $1,150. A subcontractor can produce a unit to Arctic Air specifications for $1,250. Holding an air conditioner in stock costs $60 for each 2-month period, and 200 air conditioners are currently in stock. The plan calls for 400 units to be in stock at the end of period 6. No backorders are allowed. Use the transportation method to develop a plan that minimizes costs. D - 76Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall

77. Solved Problem 2 The following tables identify the optimal production and inventory plans. An arbitrarily large cost ($99,999 per period) was used for backorders, which effectively ruled them out. All production quantities are in hundreds of units. Note that demand in period 6 is 4,400. That amount is the period 6 demand plus the desired ending inventory of 400. The anticipation inventory is measured as the amount at the end of each period. Cost calculations are based on the assumption that workers are not paid for undertime or are productively put to work elsewhere in the organization whenever they are not needed for this work. D - 77Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall

78. D - 78Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall

79. Solved Problem 2 Production Plan Period Regular-time Production Overtime ProductionSubcontractingTotal 16,500—— 2 400—6,900 36,5001,300—7,800 48,0001,6001,00010,600 57,000—— 64,400—— D - 79Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall

80. Solved Problem 2 Anticipation Inventory Period Beginning Inventory Plus Total Production Minus Demand Anticipation (Ending) Inventory 1200 + 6,500 – 5,0001,700 21,700 + 6,900 – 6,0002,600 32,600 + 7,800 – 9,0001,400 41,400 + 10,600 – 12,0000 50 + 7,000 – 7,0000 60 + 4,400 – 4,000400 D - 80Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall

81. D- 81 All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior written permission of the publisher. Printed in the United States of America.