#### Transcript Linear Programming

Linear Programming Chapter 13 Supplement Pottery Example Beaver Creek Pottery Company is located on a Native American reservation. Each day, the company has available 40 hours of labor and 120 pounds of clay. The firm makes two products, bowls and mugs. A bowl requires 1 hour of labor and 4 pounds of clay. A mug requires 2 hours of labor and 3 pounds of clay. The firm's profit is $40 per bowl and $50 per mug. The company wants to maximize profit. Pottery Example Business Objective: Determine • Number of bowls to make • Number of mugs to make to maximize profit. Decision variables: x = number of bowls to make y = number of mugs to make Objective Function • Profit • $40 per bowl (x) • $50 per mug (y) • Maximize Z = 40x + 50y Constraint Table Constrained Bowls Quantity x Labor 1 Clay 4 Mugs y Max. or Type Minimum 2 < 40 3 < 120 Constraints: x + 2y < 40 4x + 3y < 120 Nonnegativity Constraints • We cannot make a negative amount of either product • Add nonnegativity constraints x>0 y>0 Problem Statement Maximize Z = 40x + 50y subject to x + 2y < 40 4x + 3y < 120 x>0 y>0 Linear Programming Terminology • Feasible solution: Any solution which satisfies all constraints. • Feasible region: Set of points which satisfy all constraints. • Optimal solution(s): A point which • Satisfies all constraints • Maximizes or minimizes the value of the objective function. Solving a Linear Programming Problem by the Graphical Method (1) Set up the problem: • Decision variables and their definitions. • Write objective function and state whether it should be maximized or minimized. • Write the constraints as mathematical inequalities or equalities. (2) Draw the feasible region. Solving a Linear Programming Problem by the Graphical Method(2) (3) Determine which point(s) in the feasible region give an optimal solution • Point(s) which maximize or minimize the objective function. • Note: To use the graphical method, the problem must have only 2 variables. To Draw the Feasible Region • Convert each constraint into an equation. • Draw the corresponding line. • The set of points bounded by these lines is the feasible region. Problem Statement Maximize Z = 40x + 50y subject to x + 2y < 40 4x + 3y < 120 x>0 y>0 Feasible Region for This Problem This region is bounded by the lines x + 2y = 40 (Labor) 4x + 3y = 120 (Clay) x=0 y=0 Plot x + 2y = 40 • x-intercept: Set y = 0. x + 0 =40 ==> x = 40. Point is (40,0) • y-intercept: Set x=0. (0)+2y=40 ==> y = 20. Point is (0,20) Mugs (y) 40 Clay Points in feasible region satisfy all constraints. 30 20 10 Feasible Region 10 20 Bowls (x) Labor 30 40 Iso-Profit Lines A set of points on which the objective function is constant Example: The set of points satisfying 40x + 50y = 800 x-intercept: (20,0) y-intercept: (0,16) Iso-profit Line Mugs 40 Clay 30 20 10 Labor 10 20 Bowls 30 40 Mugs 40 Clay Iso-profit Lines for profits of $800 and $1,000 30 20 10 Labor 10 20 Bowls 30 40 Iso-profit Lines Mugs 40 Clay 30 20 Optimal Point=(24,8) 10 Labor 10 20 Bowls 30 40 Optimal Solution • Satisfies x + 2y = 40 and 4x + 3y = 120 • Solution is (24,8): 24 bowls and 8 mugs • Verification: 24 + 2(8) = 40 4(24) + 3(8) = 120 Maximum Profit • Optimal solution is (24,8) • Objective function is 40x + 50y $40(24) + $50(8) = $1,360 • Maximum profit is $1,360 Fundamental Theorem of Linear Programming In a linear programming problem, the optimal solution occurs at an extreme point (vertex or corner point) of the feasible region. Objective of Linear Programming • Maximize the use of resources (or minimize costs) to achieve competitive priorities. • Optimum solution is a base case which must be adjusted to reflect business realities. Linear Programming Model • Decision variables are mathematical symbols representing activity levels. • Objective function is a mathematical, linear function which represents the organization’s objectives. • Used to compare alternative courses of action. © 1998 by Prentice-Hall Inc Russell/Taylor Oper Mgt 2/e Ch 11 Supp - 2 Linear Programming Model (2) • Constraints are mathematical, linear relationships representing restrictions on decision making • Resource constraints. • Policy or legal constraints. • Sales constraints. • Constraints may be <, =, or >. © 1998 by Prentice-Hall Inc Russell/Taylor Oper Mgt 2/e Ch 11 Supp - 2 Linear programming maximizes or minimizes the objective function subject to constraints. Uses of Linear Programming • Production Scheduling • Maximize profit • Minimize cost Different objectives yield different schedules. • Determine product or service mix • Maximize profit • Maximize revenue • Minimize costs Uses of Linear Programming (2) • Scheduling labor in services • Minimize cost • Production-location problem • Allocate products and customers to plants to minimize total cost of production and distribution Uses of Linear Programming (3) • Distribution • Minimize cost • Facility location • Minimize transportation cost. • Emergency response systems • Minimize average response time. Conditions to Use Linear Programming • Objective function must be linear • Constraints must be linear inequalities or linear equations Methods for Solving LP Problems • Graphical method: limited to 2 variables • Any linear programming problem • Simplex method • Karmarkar’s algorithm • Transportation method: • Facility location problems • Production-location problem: minimize total cost of production and shipment to customers Slack Variables • Slack = unused amount of any resource or constrained quantity • S1 = Amount of unused labor = 40 - (x + 2y) = 40 - x - 2y = 40 - 24 - 8(2) = 0. Optimum solution uses all labor. • S2 = Amount of unused clay = 120 - (4x + 3y) = 120 - 4x - 3y = 120 - 4(24) - 3(8) = 0. Optimum solution uses all clay. Sensitivity Analysis (Ranging) • Dual value or shadow price • Incremental increase or decrease in the objective function if one more unit of a resource is added. • The amount we would pay to get one more unit of a resource. • Valid only for resources which have no slack. Shadow Price for Labor • Dual value = shadow price = $16. • If labor hours increase from 40 to 41, profit increases from $1360 to $1376. • Amount to buy = upper bound - original value = 80 hours - 40 hours = 40 hours Mugs Optimal point = (0,40). Profit = $2,000 40 30 Clay New Labor 20 10 Labor 10 20 Bowls 30 40 80