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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**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**Organization**Finance Marketing Operations Operations Management The management of systems or processes that create goods and/or provide services**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**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**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**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.**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**Physical**– – Schematic – Mathematical Models A model is an abstraction of reality. Tradeoffs What are the pros and cons of models?**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**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**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**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**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)**Cost-Volume Relationships**Profit Total revenue Amount ($) Total cost 0 BEP units Q (volume in units)**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.**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**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.**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?**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.**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**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**–**$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**Linear Programming – Chapter 2**George Dantzig – 1914 -2005**Linear Programming**• Concerned with optimal allocation of limited resources such as • Materials • Budgets • Labor • Machine time • among competitive activities • under a set of constraints**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)**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**Another Example of LP: Diet Problem**• Energy requirement : 2000 kcal • Protein requirement : 55 g • Calcium requirement : 800 mg**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**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**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 0x14, 0x23, 0x32, 0x48, 0x52, 0x62**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 0x14 0x23 0x32 0x48 0x52 0x62**Optimal Solution – Diet ProblemUsing LINDO 6.1**• Cost of diet = $96.50 per day**Optimal Solution – Diet ProblemUsing Management Scientist**• Cost of diet = $96.50 per day**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**Graphical Solution of LPs**• Consider a Maximization Problem Max 5x1 + 7x2 s.t. x1< 6 2x1 + 3x2< 19 x1 + x2< 8 x1, x2> 0**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**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**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**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**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**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**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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