Presentation is loading. Please wait.

Presentation is loading. Please wait.

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

Similar presentations


Presentation on theme: "Linear Programming Chapter 14 Supplement. Lecture Outline Model Formulation Graphical Solution Method Linear Programming Model Solution Solving Linear."— Presentation transcript:

1 Linear Programming Chapter 14 Supplement

2 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

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

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

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

7 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

8 LP Model Formulation Max/min z = c 1 x 1 + c 2 x 2 +... + c n x n subject to: a 11 x 1 + a 12 x 2 +... + a 1n x n (≤, =, ≥) b 1 a 21 x 1 + a 22 x 2 +... + a 2n x n (≤, =, ≥) b 2 : a n1 x1 + a n2 x 2 +... + 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

9 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

10 Highlands Craft Store Copyright 2011 John Wiley & Sons, Inc.Supplement 14-10 Maximize Z = $40 x 1 + 50 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

11 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

12 Graphical Solution Method Copyright 2011 John Wiley & Sons, Inc.Supplement 14-12 4 x 1 + 3 x 2  120 lb x 1 + 2 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

13 Computing Optimal Values Copyright 2011 John Wiley & Sons, Inc.Supplement 14-13 40 – 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 1 + 3 x 2  120 lb x 1 + 2 x 2  40 hr | 10 | 20 | 30 | 40 x1x1 x2x2 Z = $40(24) + $50(8) = $1,360 24 8

14 Extreme Corner Points Copyright 2011 John Wiley & Sons, Inc.Supplement 14-14 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 –

15 Objective Function Copyright 2011 John Wiley & Sons, Inc.Supplement 14-15 40 – 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 1 + 20x 2 Optimal point: x 1 = 30 bowls x 2 =  0 mugs Z = $2,100

16 Minimization Problem Copyright 2011 John Wiley & Sons, Inc.Supplement 14-16 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

17 Graphical Solution Copyright 2011 John Wiley & Sons, Inc.Supplement 14-17 14 – 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

18 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

19 Copyright 2011 John Wiley & Sons, Inc.Supplement 14-19 Solution Points With Slack Variables

20 Copyright 2011 John Wiley & Sons, Inc.Supplement 14-20 Solution Points With Surplus Variables

21 Solving LP Problems with Excel Copyright 2011 John Wiley & Sons, Inc.Supplement 14-21 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”

22 Solving LP Problems with Excel Copyright 2011 John Wiley & Sons, Inc.Supplement 14-22 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

23 LP Solution Copyright 2011 John Wiley & Sons, Inc.Supplement 14-23

24 Sensitivity Analysis Copyright 2011 John Wiley & Sons, Inc.Supplement 14-24 Sensitivity range for labor; 30 to 80 lbs. Sensitivity range for clay; 60 to 160lbs. Shadow prices – marginal values – for labor and clay.

25 Copyright 2011 John Wiley & Sons, Inc.Supplement 14-25 Sensitivity Range for Labor Hours

26 Copyright 2011 John Wiley & Sons, Inc.Supplement 14-26 Sensitivity Range for Profit for Bowls

27 Copyright 2011 John Wiley & Sons, Inc.Supplement 14-27 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.


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

Similar presentations


Ads by Google