Linear Programming Chapter 14 Supplement Lecture Outline Model Formulation Graphical Solution Method Linear Programming Model Solution Solving Linear.

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Linear Programming Chapter 14 Supplement

Lecture Outline Model Formulation Graphical Solution Method Linear Programming Model Solution Solving Linear Programming Problems with Excel Sensitivity Analysis Copyright 2011 John Wiley & Sons, Inc.Supplement 14-2

Linear Programming (LP) A model consisting of linear relationships representing a firm’s objective and resource constraints A mathematical modeling technique which determines a level of operational activity in order to achieve an objective, subject to restrictions called constraints Copyright 2011 John Wiley & Sons, Inc.Supplement 14-3

4 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. Introduction to Linear Programming

Types of LP Models Copyright 2011 John Wiley & Sons, Inc.Supplement 14-5

Types of LP Models, Cont’d Copyright 2011 John Wiley & Sons, Inc.Supplement 14-6

Types of LP Models, Cont’d Copyright 2011 John Wiley & Sons, Inc.Supplement 14-7

LP Model Formulation Define Decision variables symbols representing levels of activity of an operation Define Objective function linear relationship for the objective of an operation most frequent business objective is to maximize profit most frequent objective of individual operational units (such as a production or packaging department) is to minimize cost Define Constraints linear relationships representing a restriction on decision making Copyright 2011 John Wiley & Sons, Inc.Supplement 14-8

General LP Model Formulation Max/min z = c 1 x 1 + c 2 x c n x n subject to: a 11 x 1 + a 12 x a 1n x n (≤, =, ≥) b 1 a 21 x 1 + a 22 x a 2n x n (≤, =, ≥) b 2 : a n1 x1 + a n2 x a nn x n (≤, =, ≥) b n x j = decision variables b i = constraint levels c j = objective function coefficients a ij = constraint coefficients Copyright 2011 John Wiley & Sons, Inc.Supplement 14-9 Constraints

Highlands Craft Store Copyright 2011 John Wiley & Sons, Inc.Supplement LaborClayRevenue Product(hr/unit)(lb/unit)($/unit) Bowl1440 Mug2350 There are 40 hours of labor and 120 pounds of clay available each day. Management wants to max Revenue Decision variables x 1 = number of bowls to produce x 2 = number of mugs to produce Resource Requirements

Highlands Craft Store Copyright 2011 John Wiley & Sons, Inc.Supplement Maximize Z = $40 x x 2 Subject to x 1 +2x 2  40 hr(labor constraint) 4x 1 +3x 2  120 lb(clay constraint) x 1, x 2  0 Solution is x 1 = 24 bowls x 2 = 8 mugs Revenue = $1,360

Graphical Solution Method 1.Plot model constraint on a set of coordinates in a plane 2.Identify the feasible solution space on the graph where all constraints are satisfied simultaneously 3.Plot objective function to find the point on boundary of this space that maximizes (or minimizes) value of objective function Copyright 2011 John Wiley & Sons, Inc.Supplement 14-12

Graphical Solution Method Copyright 2011 John Wiley & Sons, Inc.Supplement x x 2  120 lb x x 2  40 hr Area common to both constraints 50 – 40 – 30 – 20 – 10 – 0 – | 10 | 60 | 50 | 20 | 30 | 40 x1x1 x2x2 Objective function

Computing Optimal Values Copyright 2011 John Wiley & Sons, Inc.Supplement – 30 – 20 – 10 – 0 – x 1 +2x 2 =40 4x 1 +3x 2 =120 4x 1 +8x 2 =160 -4x 1 -3x 2 =-120 5x 2 =40 x 2 =8 x 1 +2(8)=40 x 1 =24 4 x x 2  120 lb x x 2  40 hr | 10 | 20 | 30 | 40 x1x1 x2x2 Z = $40(24) + $50(8) = $1,

Extreme Corner Points Copyright 2011 John Wiley & Sons, Inc.Supplement x 1 = 24 bowls x 2 =  8 mugs Z = $1,360 x 1 = 30 bowls x 2 =  0 mugs Z = $1,200 x 1 = 0 bowls x 2 =  20 mugs Z = $1,000 A B C | 20 | 30 | 40 | 10 x1x1 x2x2 40 – 30 – 20 – 10 – 0 –

Objective Function Copyright 2011 John Wiley & Sons, Inc.Supplement – 30 – 20 – 10 – 0 – 4x 1 + 3x 2  120 lb x 1 + 2x 2  40 hr B | 10 | 20 | 30 | 40 x1x1 x2x2 C A Z = 70x x 2 Optimal point: x 1 = 30 bowls x 2 =  0 mugs Z = $2,100

Minimization Problem Copyright 2011 John Wiley & Sons, Inc.Supplement CHEMICAL CONTRIBUTION BrandNitrogen (lb/bag)Phosphate (lb/bag) Gro-plus24 Crop-fast43 Minimize Z = $6x 1 + $3x 2 subject to 2x 1 +4x 2  16 lb of nitrogen 4x 1 +3x 2  24 lb of phosphate x 1, x 2  0

Graphical Solution Copyright 2011 John Wiley & Sons, Inc.Supplement – 12 – 10 – 8 – 6 – 4 – 2 – 0 – |2|2 |4|4 |6|6 |8|8 | 10 | 12 | 14 x1x1 x2x2 A B C x 1 = 0 bags of Gro-plus x 2 = 8 bags of Crop-fast Z = $24 Z = 6x 1 + 3x 2

Solving LP Problems with Excel Copyright 2011 John Wiley & Sons, Inc.Supplement Objective function =C6*B10+D6*B11 =E6-F6 =E7-F7 =C7*B10+D7*B11 Decision variables bowls (X 1 ) = B10 mugs (x 2 ) = B11 Click on “Data” to invoke “Solver”

Solving LP Problems with Excel Copyright 2011 John Wiley & Sons, Inc.Supplement After all parameters and constraints have been input, click on “Solve” Objective function Decision variables C6*B10+D6*B11≤40 and C7*B10+D7*B11≤120 Click on “Add” to insert constraints Click on “Options” to add non-negativity and linear conditions

LP Solution Copyright 2011 John Wiley & Sons, Inc.Supplement 14-21

Sensitivity Analysis Copyright 2011 John Wiley & Sons, Inc.Supplement Sensitivity range for labor; 30 to 80 lbs. Sensitivity range for clay; 60 to 160lbs. Shadow prices – marginal values – for labor and clay.

Copyright 2011 John Wiley & Sons, Inc.Supplement Sensitivity Range for Labor Hours

Copyright 2011 John Wiley & Sons, Inc.Supplement Sensitivity Range for Profit for Bowls

25 THE GALAXY INDUSTRY PRODUCTION PROBLEM - A Prototype Example Galaxy manufactures two toy models: Space Ray. Zapper. Purpose: to maximize profits How: By choice of product mix How many Space Rays? How many Zappers? A RESOURCE ALLOCATION PROBLEM

26 Galaxy Resource Allocation Resources are limited to 1200 pounds of special plastic available per week 40 hours of production time per week. All LP Models have to be formulated in the context of a production period In this case, a week

27 Marketing requirement Total production cannot exceed 800 dozens. Number of dozens of Space Rays cannot exceed number of dozens of Zappers by more than 450. Technological input Space Rays require 2 pounds of plastic and 3 minutes of labor per dozen. Zappers require 1 pound of plastic and 4 minutes of labor per dozen.

28 Current production plan calls for: Producing as much as possible of the more profitable product, Space Ray ($8 profit per dozen). Use resources left over to produce Zappers ($5 profit per dozen). The current production plan consists of: Space Rays = 550 dozens Zapper = 100 dozens Profit = 4900 dollars per week

29 MODEL FORMULATION :Decisions variables: X1 = Production level of Space Rays (in dozens per week). X2 = Production level of Zappers (in dozens per week). Objective Function: Weekly profit, to be maximized

30 The Objective Function Each dozen Space Rays realizes $8 in profit. Total profit from Space Rays is 8*X1. Each dozen Zappers realizes $5 in profit. Total profit from Zappers is 5*X2. The total profit contributions of both is 8X1 + 5X2 8X1 + 5X2 (The profit contributions are additive because of the linearity assumption)

31 we have a plastics resource constraint, a production time constraint, and two marketing constraints. PLASTIC: each dozen units of Space Rays requires 2 lbs of plastic; each dozen units of Zapper requires 1 lb of plastic and within any given week, our plastic supplier can provide 1200 lbs.

32 The Linear Programming Model Max 8X1 + 5X2 (Weekly profit) subject to 2X1 + 1X2 < = 1200 (Plastic) 3X1 + 4X2 < = 2400 (Production Time) X1 + X2 < = 800 (Total production) X1 - X2 < = 450 (Mix) X j > = 0, j = 1,2 (Nonnegativity)

33 The Set of Feasible Solutions for Linear Programs The set of all points that satisfy all the constraints of the model is called a FEASIBLE REGION

34 Using a graphical presentation we can represent all the constraints, the objective function, and the three types of feasible points.

The Plastic constraint Feasible The plastic constraint: 2X1+X2<=1200 X2 Infeasible Production Time 3X1+4X2<=2400 Total production constraint: X1+X2<= Production mix constraint: X1-X2<=450 There are three types of feasible points Interior points. Boundary points. Extreme points. X1

36 Recall the feasible Region X2 X1 We now demonstrate the search for an optimal solution Start at some arbitrary profit, say profit = $2, Profit = $ 000 2, Then increase the profit, if possible... 3, 4,...and continue until it becomes infeasible Profit =$5040

X2 X1 Let’s take a closer look at the optimal point Feasible region Feasible region Infeasible

38 Summary of the optimal solution Space Rays = 480 dozens Zappers = 240 dozens Profit = $5040 This solution utilizes all the plastic and all the production hours. Total production is only 720 (not 800). Space Rays production exceeds Zapper by only 240 dozens (not 450).

39 Extreme points and optimal solutions If a linear programming problem has an optimal solution, it will occur at an extreme point. Multiple optimal solutions For multiple optimal solutions to exist, the objective function must be parallel to a constraint that defines the boundary of the feasible region. Any weighted average of optimal solutions is also an optimal solution.

40 The Role of Sensitivity Analysis of the Optimal Solution Is the optimal solution sensitive to changes in input parameters? Possible reasons for asking this question: Parameter values used were only best estimates. Dynamic environment may cause changes. “What-if” analysis may provide economical and operational information.

41 Range of Optimality The optimal solution will remain unchanged as long as An objective function coefficient lies within its range of optimality There are no changes in any other input parameters. The value of the objective function will change if the coefficient multiplies a variable whose value is nonzero. Sensitivity Analysis of Objective Function Coefficients.

X2 X1 The effects of changes in an objective function coefficient on the optimal solution Max 8x1 + 5x2 Max 4x1 + 5x2 Max 3.75x1 + 5x2 Max 2x1 + 5x2

X2 X1 The effects of changes in an objective function coefficient on the optimal solution Max8x1 + 5x2 Max 3.75x1 + 5x2 Max8x1 + 5x2 Max 3.75 x1 + 5x2 Max 10 x1 + 5x Range of optimality

44 Multiple changes The range of optimality is valid only when a single objective function coefficient changes. When more than one variable changes we turn to the 100% rule. This is beyond the scope of this course

45 Reduced costs The reduced cost for a variable at its lower bound (usually zero) yields: The amount the profit coefficient must change before the variable can take on a value above its lower bound. Complementary slackness At the optimal solution, either a variable is at its lower bound or the reduced cost is 0.

46 Sensitivity Analysis of Right-Hand Side Values Any change in a right hand side of a binding constraint will change the optimal solution. Any change in a right-hand side of a non-binding constraint that is less than its slack or surplus, will cause no change in the optimal solution.

47 In sensitivity analysis of right-hand sides of constraints we are interested in the following questions: Keeping all other factors the same, how much would the optimal value of the objective function (for example, the profit) change if the right-hand side of a constraint changed by one unit? For how many additional UNITS is this per unit change valid? For how many fewer UNITS is this per unit change valid?

X2 The Plastic constraint Feasible X Production time constraint Maximum profit = x1 + 1x2 <=1200 The new Plastic constraint 2x1 + 1x2 <=1350 Production mix constraint Infeasible extreme points

49 Range of feasibility The set of right - hand side values for which the same set of constraints determines the optimal extreme point. The range over-which the same variables remain in solution (which is another way of saying that the same extreme point is the optimal extreme point) Within the range of feasibility, shadow prices remain constant; however, the optimal objective function value and decision variable values will change if the corresponding constraint is binding

50 Models Without Optimal Solutions Infeasibility: Occurs when a model has no feasible point. Unboundedness: Occurs when the objective can become infinitely large.

51 Infeasibility No point, simultaneously, lies both above line and below lines and

52 Unbounded solution The feasible region Maximize the Objective Function

53 Navy Sea Ration A cost minimization diet problem – Mix two sea ration products: Texfoods, Calration. – Minimize the total cost of the mix. – Meet the minimum requirements of Vitamin A, Vitamin D, and Iron.

54 Decision variables X1 (X2) -- The number of two-ounce portions of Texfoods (Calration) product used in a serving. The Model Minimize 0.60X X2 Subject to 20X1 + 50X2 100 Vitamin A 25X1 + 25X2 100 Vitamin D 50X1 + 10X2 100 Iron X1, X2 0 Cost per 2 oz. % Vitamin A provided per 2 oz. % required

55 The Graphical solution Feasible Region Vitamin “D” constraint Vitamin “A” constraint The Iron constraint

56 Summary of the optimal solution Texfood product = 1.5 portions (= 3 ounces) Calration product = 2.5 portions (= 5 ounces) Cost =$ 2.15 per serving. The minimum requirements for Vitamin D and iron are met with no surplus. The mixture provides 155% of the requirement for Vitamin A.

57 Linear programming software packages solve large linear models. Most of the software packages use the algebraic technique called the Simplex algorithm. The input to any package includes: The objective function criterion (Max or Min). The type of each constraint:. The actual coefficients for the problem. Computer Solution of Linear Programs With Any Number of Decision Variables

58 The typical output generated from linear programming software includes: Optimal value of the objective function. Optimal values of the decision variables. Reduced cost for each objective function coefficient. Ranges of optimality for objective function coefficients. The amount of slack or surplus in each constraint. Shadow (or dual) prices for the constraints. Ranges of feasibility for right-hand side values.

Copyright 2011 John Wiley & Sons, Inc.Supplement Copyright 2011 John Wiley & Sons, Inc. All rights reserved. Reproduction or translation of this work beyond that permitted in section 117 of the 1976 United States Copyright Act without express permission of the copyright owner is unlawful. Request for further information should be addressed to the Permission Department, John Wiley & Sons, Inc. The purchaser may make back-up copies for his/her own use only and not for distribution or resale. The Publisher assumes no responsibility for errors, omissions, or damages caused by the use of these programs or from the use of the information herein.