Operations Research

 Operations Research (OR) aims to having the optimization solution for some administrative problems, such as transportation, decision-making, inventory Copyright 2006 John Wiley & Sons, Inc.Supplement 13-2

Operations Research Models  Linear Programming  Markov Chains  Network Optimization  Decision Analysis  Transportation  Inventory Copyright 2006 John Wiley & Sons, Inc.Supplement 13-3

Copyright 2006 John Wiley & Sons, Inc.Supplement 13-4 A model consisting of linear relationships representing a firm’s objective and resource constraints Linear Programming (LP) LP is a mathematical modeling technique used to determine a level of operational activity in order to achieve an objective, subject to restrictions called constraints

Copyright 2006 John Wiley & Sons, Inc.Supplement 13-5 LP Model Formulation (cont.) 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 m1 x1 + a m2 x a mn x n (≤, =, ≥) b m a m1 x1 + a m2 x a mn x n (≤, =, ≥) b m x j = decision variables b i = constraint levels c j = objective function coefficients a ij = constraint coefficients

Copyright 2006 John Wiley & Sons, Inc.Supplement 13-6 LP Formulation: Example Maximize Z = 40 x x 2 Subject to x 1 +2x 2  40 4x 1 +3x 2  120 x 1, x 2  0

x 1 +2x 2  x1 020x2 Copyright 2006 John Wiley & Sons, Inc.Supplement x 1 +3x 2  x1 040x2

Copyright 2006 John Wiley & Sons, Inc.Supplement 13-8 Graphical Solution: Example x x 2  – – – – – 0 0 – | x1x1x1x1 x2x2x2x2

Copyright 2006 John Wiley & Sons, Inc.Supplement 13-9 Graphical Solution: Example 4 x x 2  120 x x 2  – – – – – 0 0 – | x1x1x1x1 x2x2x2x2

Copyright 2006 John Wiley & Sons, Inc.Supplement Graphical Solution: Example 4 x x 2  120 x x 2  40 Area common to both constraints – – – – – 0 0 – | x1x1x1x1 x2x2x2x2

Copyright 2006 John Wiley & Sons, Inc.Supplement Computing Optimal Values 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 – – – – 0 0 – | x1x1x1x1 x2x2x2x2 A B c D

Z = 40 x x 2 (X1, X2) = 0 (0, 0)A = 1200(30, 0)B = 1360(24,8)C = 1000 (0, 20)D Copyright 2006 John Wiley & Sons, Inc.Supplement 13-12

Copyright 2006 John Wiley & Sons, Inc.Supplement Minimization Problem Minimize Z = 6x 1 + 3x 2 subject to 2x 1 +4x 2  16 4x 1 +3x 2  24 x 1, x 2  0

Copyright 2006 John Wiley & Sons, Inc.Supplement – – – 8 8 – 6 6 – 4 4 – 2 2 – 0 0 – |22|222 |44|444 |66|666 |88|888 | x1x1x1x1 x2x2x2x2 A B C Graphical Solution