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QMB 4701 MANAGERIAL OPERATIONS ANALYSIS

QMB 4701 MANAGERIAL OPERATIONS ANALYSIS. Chapter 3 Linear Programming: Graphical Solution Procedure & Spreadsheet Solution Procedure. Jay Marshall Teets Decision and Information Sciences University of Florida. Agenda. Graphical linear programming Example: Par, Inc. Problem

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QMB 4701 MANAGERIAL OPERATIONS ANALYSIS

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  1. QMB 4701 MANAGERIAL OPERATIONS ANALYSIS Chapter 3 Linear Programming: Graphical Solution Procedure & Spreadsheet Solution Procedure Jay Marshall TeetsDecision and Information Sciences University of Florida

  2. Agenda • Graphical linear programming • Example: Par, Inc. Problem • Graphical solution procedure • Example: Dorian problem • Additional considerations • Degeneracy • Inconsistent problem • Multiple optimal solutions • Unbounded problem • Spreadsheet solution basics

  3. Cutting & Dyeing Sewing Finishing Inspection & Packaging Example: Par, Inc. Problem Par, Inc: Manufacturer of golf supplies New products: Standard golf bags and Deluxe golf bags Manufacturing operations: Objective: Maximize total profit

  4. Par Inc. Model Formulation 1.What is the objective in words? 2. What are the constraints in words? Maximize the total profit Cutting & Dyeing Cutting & Dyeing Hours Allocated  Hours Available Sewing Hours Allocated  Sewing Hours Available Inspection & Packaging Inspection & Packaging Hours Allocated  Hours Available Finishing Hours Allocated  Finishing Hours Available

  5. Par Inc. Model Formulation 3. What are the decision variables? 4. Formulate the objective function: 5. Formulate the constraints: 6. Do we need non-negative constraints? x1 = Number of Standards Produced x2 = Number of Deluxes Produced Profit Contribution Maximize 10 x1 + 9 x2 Subject to: Cutting & Dyeing: 7/10 x1 + x2  630 Sewing 1/2 x1 + 5/6 x2  600 Finishing x1 + 2/3 x2  708 Inspection & Packaging 1/10 x1 + 1/4 x2  135 Logic x1  0 , x2  0

  6. Profit Contribution Maximize 10 x1 + 9 x2 x1 = 600, x2 = 100? Profit = 6900 x1 = 700, x2 = 100? Profit =7900 Subject to: Cutting & Dyeing: 7/10 x1 + x2  630 Sewing 1/2 x1 + 5/6 x2  600 Finishing x1 + 2/3 x2  708 Inspection & Packaging 1/10 x1 + 1/4 x2  135 520 383.33 666.67 85 590 433.33 766.67 95 Logic x1  0 , x2  0 How to Solve It? Trial and Error? Feasible Solution Infeasible Solution

  7. A. Determine feasible region (draw constraint lines) Feasible solution: solution that satisfies all constraints Feasible solutions B. Construct isoprofit lines C. Locate the optimal solution Optimal solutions Graphical Approach Solution: any production plan Solutions Optimal solution: feasible solution that achieves the maximal objective value

  8. Constraints Determine Feasible Region Approach: 1. Draw lines representing the constraints solved as equalities 2. Show the direction of feasibility 3. The feasible region is the set of values which simultaneously satisfies all the constraints Cutting & Dyeing 7/10 x1 + x2 630 Sewing 1/2 x1 + 5/6 x2 600 Finishing x1 + 2/3 x2 708 Inspection & Packaging 1/10 x1 + 1/4 x2 135 x1 0, x2 0,

  9. A. Constraints Determine Feasible Region

  10. Redundant constraint A. Constraints Determine Feasible Region

  11. Construct Isoprofit Lines Approach: 1. Select a constant C 2. Draw the “profit line” 10 x1 + 9 x2 = C 3. Construct parallel profit lines Slope of the objective function: x2= -10/9 x1+C/9

  12. Optimal Solution Locate The Optimal Solution - Observations Observations: 1. The optimal solution cannot fall in the interior of the feasible region 2. The optimal solution must occur at a “corner point” of the feasible region 3. In searching for the optimal solution we only have to examine the corner points

  13. Locate The Optimal Solution Find extreme points (corner points) and calculate objective values Extreme Points: the vertices or “corner points” of the feasible region. With two variables, extreme points are determined by the intersection of the constraint lines. 5 I &P 4 C&D 3 F 1 2

  14. Locate The Optimal Solution - Find Extreme Points Example: Find extreme point #3: Intersection of the cutting & dyeing constraint and the finishing constraint: Cutting & Dyeing: 7/10 x1 + x2= 630 Finishing x1 + 2/3 x2= 708 Solve the two equations for x1 and x2: 7/10 x1 + x2= 630 7/10 x1 = 630 x2 x1 = 900  10/7 x2 x1+ 2/3 x2= 708  (900  10/7 x2) +2/3 x2 = 708  900 - 30/21x2 + 14/21 x2 = 708  16/21 x2 =  192 x2= 21/16  192 = 252 x1 = 900 - 10/7 x2  x1 = 900  10/7 (252)  x1 = 900  360 = 540

  15. Slack constraint I &P Binding constraints C&D F Slack and Binding Constraints At optimal solution (540, 252) Hrs used Hrs available C&D:630 = 630 S: 480 < 600 F:708 = 708 I&P: 117 < 135 Binding Slack = 600  480 = 120 Binding Slack = 135  117 = 18

  16. Summary of Graphical Approach A. Graph Feasible Region • Draw each constraint line • For each constraint, plot points on the x1 axis (x2=0) and x2axis (x2=0) and connect the points with a line. • Indicate the feasible area with arrow • Check if (0,0) is include, if so, draw arrow towards (0,0). If not, draw arrow away from (0,0). If (0,0) is on the constraint line, you must choose another point to use as a check • Hatch the area in common for all constraints and their respective arrows B. Find Extreme Points • Set the intersecting constraints equal to one another - basic algebra

  17. Summary of Graphical Approach C. Find optimal solution Option #1 • List all the extreme points and calculate objective function values for each extreme point. • Choose the point with the highest (profit) or lowest (cost) value Option #2 • Find the slope of the objective function • Choose which extreme point is last touched by the isoprofit/isocost line as it is increased/decreased. This point is the optimal point • Calculate the objective value for the optimal point.

  18. Note: Simplex Method 1. Start with a feasible corner point solution 2. Check to see if a feasible neighboring corner point is better 3. If not, stop; otherwise move to that better neighbor and return to step 2.

  19. x x 2 2 16 16 14 14 0 , 14 12 12 10 10 HIW 8 8 6 6 4 4 HIM 2 2 3.6 , 1.4 12 , 0 x x 1 1 2 2 4 4 6 6 8 8 10 10 12 12 14 14 16 16 Graphing Example: Minimization Problem Dorian Problem MIN 50,000 x1 + 100,000 x2 ST HIW: 7x1 + 2 x2  28 HIM: 2x1 + 12 x2  24 x1 0, x2  0

  20. Additional Consideration: Degeneracy Degeneracy: Assume Inspection & Packaging time is 117 hours (instead of 135) Profit Contribution Maximize 10 x1 + 9 x2 ST Cutting & Dyeing: 7/10 x1 + x2 630 Sewing 1/2 x1 + 5/6 x2 600 Finishing x1 + 2/3 x2 708 Inspection & Packaging 1/10 x1 + 1/4 x2117 Logic x1  0 , x2  0

  21. Additional Consideration: Inconsistent Problem Assume there is an additional constraint: need to produce at least 725 standard bags, i.e., x1  725 Profit Contribution Maximize 10 x1 + 9 x2 ST Cutting & Dyeing: 7/10 x1 + x2 630 Sewing 1/2 x1 + 5/6 x2 600 Finishing x1 + 2/3 x2 708 Inspection & Packaging 1/10 x1 + 1/4 x2 135 x1  725 Logic x1  0 , x2  0

  22. Additional Consideration: Multiple Optimal Solutions Assume the objective function is 12 x1 + 8 x2 (instead of 10 x1 + 9 x2) Profit Contribution Maximize12 x1 + 8 x2 ST Cutting & Dyeing: 7/10 x1 + x2 630 Sewing 1/2 x1 + 5/6 x2 600 Finishing x1 + 2/3 x2 708 Inspection & Packaging 1/10 x1 + 1/4 x2 135 Logic x1  0 , x2  0 Slope of the new objective function: x2 =-12/8 x1 +C/8 = =-3/2 x1 +C/8

  23. Additional Consideration: Unbounded Problem Assume we have the following problem: Profit Contribution Maximize 10 x1 + 9 x2 ST Demand: x1 + x2  1000 x1  725 Logic x1  0 , x2  0

  24. Excel Solver Solution Procedure • Excel can do the following: • Solve linear programs Note: Activate the “Assume Linear Model” option if your problem is a LINEAR program, otherwise the Solver will treat it as a NONLINEAR program. • Solve nonlinear programs • Solve integer programs • Perform sensitivity analysis

  25. Implementing an LP Model in a Spreadsheet • Organize the data for the model on the spreadsheet including: • Objective Coefficients • Constraint Coefficients • Right-Hand-Side (RHS) values • clearly label data and information • visualize a logical layout • row and column structures of the data are a good start

  26. Implementing an LP Model in a Spreadsheet • Reserve separate cells to represent each decision variable • arrange them such that the structure parallels the input data • keep them together in the same place • Right-Hand-Side (RHS) values • Create a formula in a cell that corresponds to the Objective Function • For each constraint, create a formula that corresponds to the left-hand-side (LHS) of the constraint • NOTE: Once you’ve formulated a spreadsheet model for a certain type of application, it’s useful to save it as a template for future use!

  27. Solver Terminology • Target cell • Represents the objective function • Must indicate max or min • Changing cells • Represents the decision variables • Constraint cells • the cells in the spreadsheet that represent the LHS formulas of the constraints in the model • the cells in the spreadsheet that represent the RHS values of the constraints in the model • TIP: If you scale the constraints such that they are all "" or all "", then it's easier to input the constraints as a block. Then, you avoid having to enter each constraint as a separate line.

  28. Solver Example: Par, Inc. Model

  29. Par, Inc. Model - Answer Report Solver Parameters: Objective: MAX F9 Variables: B11:C11 Constraints: D3:D6 <= F3:F6 Options: Assume Linear Model Assume Non-Negative Optimal objective value Optimal solution Constraints Binding & Slack Information

  30. Summary of Excel Solver Procedure

  31. Solver Example: Minimization Example Dorian Problem MIN 50,000 x1 + 100,000 x2 ST 7x1 + 2 x2  28 2x1 + 12 x2  24 x1 0, x2 0

  32. Solver Example: Minimization Example

  33. Summary of Additional Considerations

  34. Summary • Graphical solution procedure • Feasible region • Extreme points • Optimal solution • Binding and Slack Constraints • Additional considerations • Degeneracy • Inconsistent problem • Multiple optimal solutions • Unbounded problem • Spreadsheet solution basics

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