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.

Business Objective: Determine 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 Maximize Z = 40x + 50y $40 per bowl (x) $50 per mug (y) Maximize Z = 40x + 50y

Constraint Table Constraints: x + 2y < 40 4x + 3y < 120 Constrained Quantity Max. or Minimum Bowls x Mugs y Type Labor 1 4 2 3 < 40 120 Clay 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

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

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)

Points in feasible region satisfy all constraints. 40 Clay Mugs (y) 30 20 10 Feasible Region Labor 10 20 30 40 Bowls (x)

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 40 Clay 30 Mugs 20 10 Labor 10 20 30 40 Bowls

40 Clay 30 Mugs 20 10 Labor 10 20 30 40 Bowls Iso-profit Lines for profits of $800 and $1,000 40 Clay 30 Mugs 20 10 Labor 10 20 30 40 Bowls

Iso-profit Lines 40 Clay 30 Mugs 20 10 Labor 10 20 30 40 Bowls Optimal Point=(24,8) 10 Labor 10 20 30 40 Bowls

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 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 20 New Labor 10 Labor 10 20 30 40 80 Bowls