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Chapter 14 supplement - Linear programming. This chapter includes contents: Model formulation, graphical solution method, linear programming model solution, solving linear programming problems with excel, sensitivity analysis.
OPERATIONS MANAGEMENT: Creating Value Along the Supply Chain, Canadian Edition Robert S Russell, Bernard W Taylor III, Ignacio Castillo, Navneet Vidyarthi CHAPTER 14 SUPPLEMENT Linear Programming Supplement 14-1 Lecture Outline Model Formulation Graphical Solution Method Linear Programming Model Solution Solving Linear Programming Problems with Excel Sensitivity Analysis 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 Supplement 14-3 Types of LP Supplement 14-4 Types of LP Supplement 14-5 Types of LP Supplement 14-6 LP Model Formulation Decision variables symbols representing levels of activity of an operation 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 Constraint linear relationship representing a restriction on decision making Supplement 14-7 LP Model Formulation Max/min z = c1x1 + c2x2 + + cnxn subject to: Constraints a11x1 + a12x2 + + a1nxn (≤, =, ≥) b1 a21x1 + a22x2 + + a2nxn (≤, =, ≥) b2 : an1x1 + an2x2 + + annxn (≤, =, ≥) bn xj = decision variables bi = constraint levels cj = objective function coefficients aij = constraint coefficients Supplement 14-8 Highlands Craft Store Resource Requirements Product Bowl Mug Labor (hr/unit) Clay (lb/unit) Revenue ($/unit) 40 50 There are 40 hours of labor and 120 pounds of clay available each day Decision variables x1 = number of bowls to produce x2 = number of mugs to produce Supplement 14-9 Highlands Craft Store Maximize Z = $40 x1 + 50 x2 Subject to x1 + 4x1 + 2x2 3x2 x1 , x2 Solution is x1 = 24 bowls Revenue = $1,360 40 hr 120 lb (labor constraint) (clay constraint) x2 = mugs Supplement 14-10 Graphical Solution Method x2 50 – 40 – x1 + x2 120 lb Objective function 30 – Area common to both constraints 20 – 10 – 0– x1 + x2 | 10 | 20 | 30 | 40 | 50 | 60 40 hr x1 Supplement 14-12 Computing Optimal Values x2 40 – x1 + x2 = 120 lb 30 – + + + - x1 + x2 = 40 hr x1 + x1 20 – 10 8– 0– x1 4x1 4x1 -4x1 | 10 24 | | 20 30 2x2 3x2 8x2 3x2 5x2 x2 = = = = = = 40 120 160 -120 40 2(8) = = 40 24 | x1 40 Z = $40(24) + $50(8) = $1,360 Supplement 14-13 Extreme Corner Points x1 = bowls x2 = 20 mugs Z = $1,000 x2 40 – 30 – 20 – A 10 – 0– x1 = 224 bowls x2 = mugs x1 = 30 bowls Z = $1,360 x2 = mugs Z = $1,200 B | 10 | 20 C| | 30 40 x1 Supplement 14-14 Objective Function 40 – x2 4x1 + 3x2 = 120 lb 30 – Z = 70x1 + 20x2 20 – 10 – Optimal point: x1 = 30 bowls x2 = mugs Z = $2,100 A 0– B x1 + 2x2 = 40 hr | 10 | 20 | C 30 | 40 x1 Supplement 14-15 Minimization Problem CHEMICAL CONTRIBUTION Brand Gro-plus Crop-fast Nitrogen (lb/bag) Phosphate (lb/bag) Minimize Z = $6x1 + $3x2 subject to 2x1 + 4x2 4x1 + 3x2 x1, x2 16 lb of nitrogen 24 lb of phosphate Supplement 14-16 Graphical Solution x2 14 – x1 = bags of Gro-plus x2 = bags of Crop-fast Z = $24 12 – 10 – 8– A Z = 6x1 + 3x2 6– 4– B 2– 0– | | | | C | 10 | 12 | 14 x1 Supplement 14-17 Simplex Method Mathematical procedure for solving LP problems Follow a set of steps to reach optimal solution Slack variables added to ≤ constraints to represent unused resources x1 + 2x2 + s1 = 40 hours of labor 4x1 + 3x2 + s2 = 120 lb of clay Surplus variables subtracted from ≥ constraints to represent excess above resource requirement 2x1 + 4x2 ≥ 16 is transformed into 2x1 + 4x2 - s1 = 16 Slack/surplus variables have a coefficient in the objective function Z = $40x1 + $50x2 + 0s1 + 0s2 Supplement 14-18 Solution Points With Slack Variables Supplement 14-19 Solution Points With Surplus Variables Supplement 14-20 Solving LP Problems with Excel Click on “Data” to invoke “Solver” Objective function =C6*B10+D6*B11 =E6-F6 =E7-F7 Decision variables bowls (X1) = B10 mugs (x2) = B11 =C7*B10+D7*B11 Supplement 14-21 Solving LP Problems with Excel 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 Supplement 14-22 LP Solution Supplement 14-23 Sensitivity Analysis Sensitivity range for labor; 30 to 80 lbs Shadow prices – marginal values – for labor and clay Sensitivity range for clay; 60 to 160lbs Supplement 14-24 Sensitivity Range for Labor Hours Supplement 14-25 COPYRIGHT Copyright © 2014 John Wiley & Sons Canada, Ltd All rights reserved Reproduction or translation of this work beyond that permitted by Access Copyright (The Canadian Copyright Licensing Agency) is unlawful Requests for further information should be addressed to the Permissions Department, John Wiley & Sons Canada, Ltd The purchaser may make back-up copies for his or her own use only and not for distribution or resale The author and the publisher assume no responsibility for errors, omissions, or damages caused by the use of these programs or from the use of the information contained herein ... Supplement 1 4-1 2 Computing Optimal Values x2 40 – x1 + x2 = 120 lb 30 – + + + - x1 + x2 = 40 hr x1 + x1 20 – 10 8– 0– x1 4x1 4x1 -4 x1 | 10 24 | | 20 30 2x2 3x2 8x2 3x2 5x2 x2 = = = = = = 40 120 160 -1 20... back-up copies for his or her own use only and not for distribution or resale The author and the publisher assume no responsibility for errors, omissions, or damages caused by the use of these... + 4x2 - s1 = 16 Slack/surplus variables have a coefficient in the objective function Z = $40x1 + $50x2 + 0s1 + 0s2 Supplement 1 4-1 8 Solution Points With Slack Variables Supplement 1 4-1 9 Solution