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Lecture

Lecture. 1. MGMT 650 Management Science and Decision Analysis. Agenda. Operations Management Chapter 1 Management Science & relation to OM Quantitative Analysis and Decision Making Cost, Revenue, and Profit Models Management Science Techniques Introduction to Linear Programming.

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Lecture

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  1. Lecture 1 MGMT 650 Management Science and Decision Analysis

  2. Agenda • Operations Management • Chapter 1 • Management Science & relation to OM • Quantitative Analysis and Decision Making • Cost, Revenue, and Profit Models • Management Science Techniques • Introduction to Linear Programming

  3. Organization Finance Marketing Operations Operations Management The management of systems or processes that create goods and/or provide services

  4. Production of Goods vs. Delivery of Services • Production of goods – tangible output • Delivery of services – an act • Service job categories • Government • Wholesale/retail • Financial services • Healthcare • Personal services • Business services • Education

  5. Scope of Operations Management • Operations Management includes: • Forecasting • Capacity planning • Scheduling • Managing inventories • Assuring quality • Deciding where to locate facilities • And more . . . • The operations function • Consists of all activities directly related to producing goods or providing services

  6. Management Science • The body of knowledge involving quantitative approaches to decision making is referred to as • Management Science • Operations research • Decision science • It had its early roots in World War II and is flourishing in business and industry with the aid of computers

  7. Problem Solving and Decision Making • Steps of Problem Solving (First 5 steps are the process of decision making) • Define the problem. • Identify the set of alternative solutions. • Determine the criteria for evaluating alternatives. • Evaluate the alternatives. • Choose an alternative (make a decision). --------------------------------------------------------------------- • Implement the chosen alternative. • Evaluate the results.

  8. Quantitative Analysis and Decision Making • Potential Reasons for a Quantitative Analysis Approach to Decision Making • The problem is complex • The problem is very important • The problem is new • The problem is repetitive

  9. Physical – – Schematic – Mathematical Models A model is an abstraction of reality. Tradeoffs What are the pros and cons of models?

  10. A Simulation Model

  11. Models Are Beneficial • Easy to use, less expensive • Require users to organize • Systematic approach to problem solving • Increase understanding of the problem • Enable “what if” questions: simulation models • Specific objectives • Power of mathematics • Standardized format

  12. Quantitative Approaches • Linear programming: optimal allocation of resources • Queuing Techniques: analyze waiting lines • Inventory models: management of inventory • Project models: planning, coordinating and controlling large scale projects • Statistical models: forecasting

  13. Analysis of Trade-offs • How many more jeans would Levi need to sell to justify the cost of additional robotic tailors? • Cost of additional robotic tailors vs Inventory Holding Cost

  14. Product Mix Example

  15. A Linear Programming Model • Objective – profit maximization Maximize 60X1 + 50X2 • Subject to Assembly 4X1 + 10X2 <= 100 hours Inspection 2X1 + 1X2 <= 22 hours Storage 3X1 + 3X2 <= 39 cubic feet X1, X2 >= 0

  16. Cost, Revenue and Profit Models

  17. Cost Classification of Owning and Operating a Passenger Car

  18. Cost-Volume Relationship

  19. Cost-Volume Relationship

  20. Amount ($) Total cost = VC + FC Total variable cost (VC) Fixed cost (FC) 0 Q (volume in units) Cost-Volume Relationships Total revenue Amount ($) 0 Q (volume in units)

  21. Cost-Volume Relationships Profit Total revenue Amount ($) Total cost 0 BEP units Q (volume in units)

  22. Example: Ponderosa Development Corp. • Ponderosa Development Corporation (PDC) is a small real estate developer that builds only one style house. • The selling price of the house is $115,000. • Land for each house costs $55,000 and lumber, supplies, and other materials run another $28,000 per house. Total labor costs are approximately $20,000 per house. • Ponderosa leases office space for $2,000 per month. The cost of supplies, utilities, and leased equipment runs another $3,000 per month. • The one salesperson of PDC is paid a commission of $2,000 on the sale of each house. PDC has seven permanent office employees whose monthly salaries are given on the next slide.

  23. Example: Ponderosa Development Corp. EmployeeMonthly Salary President $10,000 VP, Development 6,000 VP, Marketing 4,500 Project Manager 5,500 Controller 4,000 Office Manager 3,000 Receptionist 2,000

  24. Example: Ponderosa Development Corp. • Identify all costs and denote the marginal cost and marginal revenue for each house. • Write the monthly cost function c (x), revenue function r (x), and profit function p (x). • What is the breakeven point for monthly sales of the houses? • What is the monthly profit if 12 houses per month are built and sold? • Determine the BEP for monthly sale of houses graphically.

  25. Graph of costs and Break-even point

  26. Example: Step Fixed Costs • A manager has the option of purchasing 1, 2 or 3 machines • Fixed costs and potential volumes are as follows: • Variable cost = $10/unit and revenue = $40/unit • If the projected annual demand is between 580 and 630 units, how many machines should the manager purchase?

  27. Break-Even Problem with Step Fixed Costs Total Cost FC + VC = TC Total Revenue BEVs FC + VC = TC 3 machines FC + VC = TC 2 machines 1 machine Quantity Step fixed costs and variable costs.

  28. Assumptions of Cost-Volume Analysis • One product is involved • Everything produced can be sold • Variable cost per unit is the same regardless of volume • Fixed costs do not change with volume • Revenue per unit constant with volume • Revenue per unit exceeds variable cost per unit

  29. Fixed Variable Total City Cost Cost Cost Akron $30,000 $75 $180,000 Bowling Green $60,000 $45 $150,000 Chicago $110,000 $25 $160,000 Cost-Volume & Locational Break-Even Analysis Three locations: Total Cost = Fixed Cost + Variable Cost x Volume

  30. – $180,000 – – $160,000 – $150,000 – – $130,000 – – $110,000 – – – $80,000 – – $60,000 – – – $30,000 – – $10,000 – – Chicago cost curve Annual cost Bowling Green cost curve Akron cost curve Akron lowest cost Chicago lowest cost Bowling Green lowest cost | | | | | | | 0 500 1,000 1,500 2,000 2,500 3,000 Volume Locational Break-Even Analysis Graph of Break-Even Points

  31. Linear Programming – Chapter 2 George Dantzig – 1914 -2005

  32. Linear Programming • Concerned with optimal allocation of limited resources such as • Materials • Budgets • Labor • Machine time • among competitive activities • under a set of constraints

  33. Variables Objective function Constraints Non-negativity Constraints Linear Programming Example Maximize 60X1 + 50X2 Subject to 4X1 + 10X2 <= 100 2X1 + 1X2 <= 22 3X1 + 3X2 <= 39 X1, X2 >= 0 • What is a Linear Program? • A LP is an optimization model that has • continuous variables • a single linear objective function, and • (almost always) several constraints (linear equalities or inequalities)

  34. Linear Programming Model • Decision variables • unknowns, which is what model seeks to determine • for example, amounts of either inputs or outputs • Objective Function • goal, determines value of best (optimum) solution among all feasible (satisfy constraints) values of the variables • either maximization or minimization • Constraints • restrictions, which limit variables of the model • limitations that restrict the available alternatives • Parameters: numerical values (for example, RHS of constraints) • Feasible solution: is one particular set of values of the decision variables that satisfies the constraints • Feasible solution space: the set of all feasible solutions • Optimal solution: is a feasible solution that maximizes or minimizes the objective function • There could be multiple optimal solutions

  35. Another Example of LP: Diet Problem • Energy requirement : 2000 kcal • Protein requirement : 55 g • Calcium requirement : 800 mg

  36. Example of LP : Diet Problem • oatmeal: at most 4 servings/day • chicken: at most 3 servings/day • eggs: at most 2 servings/day • milk: at most 8 servings/day • pie: at most 2 servings/day • pork: at most 2 servings/day Design an optimal diet plan which minimizes the cost per day

  37. Step 1: define decision variables • x1 = # of oatmeal servings • x2 = # of chicken servings • x3 = # of eggs servings • x4 = # of milk servings • x5 = # of pie servings • x6 = # of pork servings Step 2: formulate objective function • In this case, minimize total cost • minimize z = 3x1 + 24x2 + 13x3 + 9x4 + 24x5 + 13x6

  38. Step 3: Constraints • Meet energy requirement 110x1 + 205x2 + 160x3 + 160x4 + 420x5 + 260x6 2000 • Meet protein requirement 4x1 + 32x2 + 13x3 + 8x4 + 4x5 + 14x6  55 • Meet calcium requirement 2x1 + 12x2 + 54x3 + 285x4 + 22x5 + 80x6  800 • Restriction on number of servings 0x14, 0x23, 0x32, 0x48, 0x52, 0x62

  39. So, how does a LP look like? minimize 3x1 + 24x2 + 13x3 + 9x4 + 24x5 + 13x6 subject to 110x1 + 205x2 + 160x3 + 160x4 + 420x5 + 260x6 2000 4x1 + 32x2 + 13x3 + 8x4 + 4x5 + 14x6  55 2x1 + 12x2 + 54x3 + 285x4 + 22x5 + 80x6  800 0x14 0x23 0x32 0x48 0x52 0x62

  40. Optimal Solution – Diet ProblemUsing LINDO 6.1 • Cost of diet = $96.50 per day

  41. Optimal Solution – Diet ProblemUsing Management Scientist • Cost of diet = $96.50 per day

  42. Guidelines for Model Formulation • Understand the problem thoroughly. • Describe the objective. • Describe each constraint. • Define the decision variables. • Write the objective in terms of the decision variables. • Write the constraints in terms of the decision variables • Do not forget non-negativity constraints

  43. Graphical Solution of LPs • Consider a Maximization Problem Max 5x1 + 7x2 s.t. x1< 6 2x1 + 3x2< 19 x1 + x2< 8 x1, x2> 0

  44. Graphical Solution Example • Constraint #1 Graphed x2 8 7 6 5 4 3 2 1 1 2 3 4 5 6 7 8 9 10 x1< 6 (6, 0) x1

  45. Graphical Solution Example • Constraint #2 Graphed x2 8 7 6 5 4 3 2 1 1 2 3 4 5 6 7 8 9 10 (0, 6 1/3) 2x1 + 3x2< 19 (9 1/2, 0) x1

  46. Graphical Solution Example • Constraint #3 Graphed x2 (0, 8) 8 7 6 5 4 3 2 1 1 2 3 4 5 6 7 8 9 10 x1 + x2< 8 (8, 0) x1

  47. Graphical Solution Example • Combined-Constraint Graph x2 x1 + x2< 8 8 7 6 5 4 3 2 1 1 2 3 4 5 6 7 8 9 10 x1< 6 2x1 + 3x2< 19 x1

  48. Graphical Solution Example • Feasible Solution Region x2 8 7 6 5 4 3 2 1 1 2 3 4 5 6 7 8 9 10 Feasible Region x1

  49. Graphical Solution Example • Objective Function Line x2 8 7 6 5 4 3 2 1 1 2 3 4 5 6 7 8 9 10 (0, 5) Objective Function 5x1 + 7x2 = 35 (7, 0) x1

  50. Graphical Solution Example • Optimal Solution x2 Objective Function 5x1 + 7x2 = 46 8 7 6 5 4 3 2 1 1 2 3 4 5 6 7 8 9 10 Optimal Solution (x1 = 5, x2 = 3) x1

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