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

Types of LP Copyright 2011 John Wiley & Sons, Inc.Supplement 14-4

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

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

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 14-8 Constraints

Highlands Craft Store Copyright 2011 John Wiley & Sons, Inc.Supplement 14-9 LaborClayRevenue Product(hr/unit)(lb/unit)($/unit) Bowl1440 Mug2350 There are 40 hours of labor and 120 pounds of clay available each day 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 14-11

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

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 x 1 + 2x 2 + s 1 = 40 hours of labor 4x 1 + 3x 2 + s 2 = 120 lb of clay Surplus variables subtracted from ≥ constraints to represent excess above resource requirement. 2x 1 + 4x 2 ≥ 16 is transformed into 2x 1 + 4x 2 - s 1 = 16 Slack/surplus variables have a 0 coefficient in the objective function Z = $40x 1 + $50x 2 + 0s 1 + 0s 2 Copyright 2011 John Wiley & Sons, Inc.Supplement 14-18

Copyright 2011 John Wiley & Sons, Inc.Supplement Solution Points With Slack Variables

Copyright 2011 John Wiley & Sons, Inc.Supplement Solution Points With Surplus Variables

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 14-23

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

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