Operations Management Lecture 7 Linear Programming PowerPoint presentation to accompany Heizer/Render Principles of Operations Management, 7e Operations Management, 9e

Recap Implications of Quality tools and Problem solving Check Sheets Scatter Diagrams Cause-and-Effect Diagrams Pareto Charts Flowcharts Histograms Statistical Process Control (SPC)

Outline Linear Programming Requirements of a Linear Programming Problem Maximization Problems Minimization Problems Linear Programming Methods Graphical Method Simplex Method (will do later)

Outline – Continued Formulating Linear Programming Problems Shader Electronics Example Graphical Solution to a Linear Programming Problem Graphical Representation of Constraints Iso-Profit Line Solution Method Corner-Point Solution Method

Outline – Continued Linear Programming Applications Production-Mix Example Diet Problem Example Labor Scheduling Example

Learning Objectives When you complete this module you should be able to: Formulate linear programming models, including an objective function and constraints Graphically solve an LP problem with the iso-profit line method Graphically solve an LP problem with the corner-point method

Linear Programming A mathematical technique to allocate limited resources to achieve an objective linear programming (LP) is a technique for optimization of a linear objective function, subject to linear equality and linear inequality constraints. This slide provides some reasons that capacity is an issue. The following slides guide a discussion of capacity.

Linear Programming Linear programming determines the way to achieve the best outcome (such as maximum profit or minimum cost) in a given mathematical model and given some list of requirements represented as linear equations. It is one of more powerful technique for managerial decisions This slide provides some reasons that capacity is an issue. The following slides guide a discussion of capacity.

Linear Programming A Linear Programming model seeks to maximize or minimize a linear function, subject to a set of linear constraints. The linear model consists of the following components: A set of decision variables. An objective function. A set of constraints This slide provides some reasons that capacity is an issue. The following slides guide a discussion of capacity.

Linear Programming Will find the minimum or maximum value of the objective Guarantees the optimal solution to the model formulated This slide provides some reasons that capacity is an issue. The following slides guide a discussion of capacity.

LP Applications Scheduling school buses to minimize total distance traveled Allocating police patrol units to high crime areas in order to minimize response time Scheduling tellers at banks so that needs are met during each hour of the day while minimizing the total cost of labor This slide can be used to frame a discussion of capacity. Points to be made might include: - capacity definition and measurement is necessary if we are to develop a production schedule - while a process may have “maximum” capacity, many factors prevent us from achieving that capacity on a continuous basis. Students should be asked to suggest factors which might prevent one from achieving maximum capacity.

LP Applications Selecting the product mix in a factory to make best use of machine- and labor-hours available while maximizing the firm’s profit Picking blends of raw materials in feed mills to produce finished feed combinations at minimum costs Determining the distribution system that will minimize total shipping cost This slide can be used to frame a discussion of capacity. Points to be made might include: - capacity definition and measurement is necessary if we are to develop a production schedule - while a process may have “maximum” capacity, many factors prevent us from achieving that capacity on a continuous basis. Students should be asked to suggest factors which might prevent one from achieving maximum capacity.

LP Applications Developing a production schedule that will satisfy future demands for a firm’s product and at the same time minimize total production and inventory costs This slide can be used to frame a discussion of capacity. Points to be made might include: - capacity definition and measurement is necessary if we are to develop a production schedule - while a process may have “maximum” capacity, many factors prevent us from achieving that capacity on a continuous basis. Students should be asked to suggest factors which might prevent one from achieving maximum capacity.

Requirements of an LP Problem There should be an objective function to maximize or minimize some quantity (usually profit or cost) The restrictions or constraints are present which limit the ability to achieve objective This slide can be used to frame a discussion of capacity. Points to be made might include: - capacity definition and measurement is necessary if we are to develop a production schedule - while a process may have “maximum” capacity, many factors prevent us from achieving that capacity on a continuous basis. Students should be asked to suggest factors which might prevent one from achieving maximum capacity.

Requirements of an LP Problem There must be alternative courses of action from which to choose The objective and constraints in linear programming problems must be expressible in terms of linear equations or inequalities This slide can be used to frame a discussion of capacity. Points to be made might include: - capacity definition and measurement is necessary if we are to develop a production schedule - while a process may have “maximum” capacity, many factors prevent us from achieving that capacity on a continuous basis. Students should be asked to suggest factors which might prevent one from achieving maximum capacity.

Mathematical formulation of Linear Programming model Step 1 - Study the given situation - Find the key decision to be made Identify the decision variables of the problem Step 2 Formulate the objective function to be optimized Step 3 Formulate the constraints of the problem Step 4 - Add non-negativity restrictions or constraints The objective function , the set of constraints and the non-negativity restrictions together form an LP model.

Formulating LP Problems The product-mix problem at Shader Electronics Two products Shader X-pod, a portable music player Shader BlueBerry, an internet-connected color telephone Determine the mix of products that will produce the maximum profit

Formulating LP Problems X-pods BlueBerrys Available Hours Department (X1) (X2) This Week Hours Required to Produce 1 Unit Electronic 4 3 240 Assembly 2 1 100 Profit per unit $7 $5 Table B.1 Decision Variables: X1 = number of X-pods to be produced X2 = number of BlueBerrys to be produced

Formulating LP Problems Objective Function: Maximize Profit = $7X1 + $5X2 There are three types of constraints Upper limits where the amount used is ≤ the amount of a resource Lower limits where the amount used is ≥ the amount of the resource Equalities where the amount used is = the amount of the resource

Formulating LP Problems First Constraint: Electronic time available time used is ≤ 4X1 + 3X2 ≤ 240 (hours of electronic time) Second Constraint: Assembly time available time used is ≤ 2X1 + 1X2 ≤ 100 (hours of assembly time)

Graphical Solution Can be used when there are two decision variables Plot the constraint equations at their limits by converting each equation to an equality Identify the feasible solution space Create an iso-profit line based on the objective function Move this line outwards until the optimal point is identified

4X1 + 3X2 = 240-------(I) 2X1 + 1X2 = 100-----(II) If X2 = 0, X1 = 60 Point (60, 0) If X2 = 0, X1 = 50 Point (50, 0) If X1 = 0, X2 = 80 Point (0, 80) If X1 = 0, X2 = 100 Point (0, 100)

Graphical Solution X2 II(0, 100) I(0, 80) Assembly (constraint B) – 80 – 60 – 40 – 20 – | | | | | | | | | | | 0 20 40 60 80 100 Number of BlueBerrys Number of X-pods X1 X2 II(0, 100) I(0, 80) Assembly (constraint B) Electronics (constraint A) Feasible region I (60, 0) II(50, 0) Figure B.3

Graphical Solution Iso-Profit Line Solution Method $210 = 7X1 + 5X2 – 80 – 60 – 40 – 20 – | | | | | | | | | | | 0 20 40 60 80 100 Number of Watch TVs Number of X-pods X1 X2 Assembly (constraint B) Electronics (constraint A) Feasible region Figure B.3 Choose a possible value for the objective function $210 = 7X1 + 5X2 Solve for the axis intercepts of the function and plot the line X2 = 42 X1 = 30

Graphical Solution $210 = $7X1 + $5X2 X2 Number of BlueBerrys (0, 42) – 80 – 60 – 40 – 20 – | | | | | | | | | | | 0 20 40 60 80 100 Number of BlueBerrys Number of X-pods X1 X2 Figure B.4 $210 = $7X1 + $5X2 (0, 42) (30, 0)

Graphical Solution $350 = $7X1 + $5X2 $280 = $7X1 + $5X2 – 80 – 60 – 40 – 20 – | | | | | | | | | | | 0 20 40 60 80 100 Number of BlueBeryys Number of X-pods X1 X2 $350 = $7X1 + $5X2 $280 = $7X1 + $5X2 $210 = $7X1 + $5X2 $420 = $7X1 + $5X2 Figure B.5

Optimal solution point Graphical Solution – 80 – 60 – 40 – 20 – | | | | | | | | | | | 0 20 40 60 80 100 Number of BlueBerrys Number of X-pods X1 X2 Maximum profit line Optimal solution point (X1 = 30, X2 = 40) $410 = $7X1 + $5X2 Figure B.6

Corner-Point Method 2 3 1 4 X2 Number of BlueBerrys X1 – 80 – 60 – 40 – 20 – | | | | | | | | | | | 0 20 40 60 80 100 Number of BlueBerrys Number of X-pods X1 X2 2 3 1 Figure B.7 4

Corner-Point Method The optimal value will always be at a corner point Find the objective function value at each corner point and choose the one with the highest profit Point 1 1 : (X1 = 0, X2 = 0) Profit $7(0) + $5(0) = $0 Point 2 : (X1 = 0, X2 = 80) Profit $7(0) + $5(80) = $400 Point 4 : (X1 = 50, X2 = 0) Profit $7(50) + $5(0) = $350

Corner-Point Method The optimal value will always be at a corner point Find the objective function value at each corner point and choose the one with the highest profit Solve for the intersection of two constraints 2X1 + 1X2 ≤ 100 (assembly time) 4X1 + 3X2 ≤ 240 (electronics time) 4X1 + 3X2 = 240 - 4X1 - 2X2 = -200 + 1X2 = 40 4X1 + 3(40) = 240 4X1 + 120 = 240 X1 = 30 Point 1 : (X1 = 0, X2 = 0) Profit $7(0) + $5(0) = $0 Point 2 : (X1 = 0, X2 = 80) Profit $7(0) + $5(80) = $400 Point 4 : (X1 = 50, X2 = 0) Profit $7(50) + $5(0) = $350

Corner-Point Method The optimal value will always be at a corner point Find the objective function value at each corner point and choose the one with the highest profit Point 1 : (X1 = 0, X2 = 0) Profit $7(0) + $5(0) = $0 Point 2 : (X1 = 0, X2 = 80) Profit $7(0) + $5(80) = $400 Point 4 : (X1 = 50, X2 = 0) Profit $7(50) + $5(0) = $350 Point 3 : (X1 = 30, X2 = 40) Profit $7(30) + $5(40) = $410