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Operations Research.

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Presentation on theme: "Operations Research."— Presentation transcript:

1 Operations Research

2 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.

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

4 Linear Programming (LP)
A model consisting of linear relationships representing a firm’s objective and resource constraints 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.

5 LP Model Formulation (cont.)
Max/min z = c1x1 + c2x cnxn subject to: a11x1 + a12x a1nxn (≤, =, ≥) b1 a21x1 + a22x a2nxn (≤, =, ≥) b2 : am1x1 + am2x amnxn (≤, =, ≥) bm xj = decision variables bi = constraint levels cj = objective function coefficients aij = constraint coefficients Copyright 2006 John Wiley & Sons, Inc.

6 LP Formulation: Example
Maximize Z = 40 x x2 Subject to x1 + 2x2 40 4x1 + 3x2 120 x1 , x2 0 Copyright 2006 John Wiley & Sons, Inc.

7 x1 + 2x2 40 40 x1 20 x2 4x1 + 3x2 120 30 x1 40 x2 Copyright 2006 John Wiley & Sons, Inc.

8 Graphical Solution: Example
x1 + 2 x2 40 50 – 40 – 30 – 20 – 10 – 0 – | 10 60 50 20 30 40 x1 x2 Copyright 2006 John Wiley & Sons, Inc.

9 Graphical Solution: Example
4 x1 + 3 x2 120 x1 + 2 x2 40 50 – 40 – 30 – 20 – 10 – 0 – | 10 60 50 20 30 40 x1 x2 Copyright 2006 John Wiley & Sons, Inc.

10 Graphical Solution: Example
4 x1 + 3 x2 120 x1 + 2 x2 40 Area common to both constraints 50 – 40 – 30 – 20 – 10 – 0 – | 10 60 50 20 30 40 x1 x2 Copyright 2006 John Wiley & Sons, Inc.

11 Computing Optimal Values
x1 + 2x2 = 40 4x1 + 3x2 = 120 4x1 + 8x2 = 160 -4x1 - 3x2 = -120 5x2 = 40 x2 = 8 x1 + 2(8) = 40 x1 = 24 4 x1 + 3 x2 120 lb x1 + 2 x2 40 hr 40 – 30 – 20 – 10 – 0 – | 10 20 30 40 x1 x2 D c A B Copyright 2006 John Wiley & Sons, Inc.

12 Z = 40 x x2 (X1, X2) = 0 (0, 0) A = 1200 (30, 0) B = 1360 (24,8) C = 1000 (0, 20) D Copyright 2006 John Wiley & Sons, Inc.

13 Minimization Problem Minimize Z = 6x1 + 3x2 subject to 2x1 + 4x2  16
Copyright 2006 John Wiley & Sons, Inc.

14 Graphical Solution x2 14 – 12 – 10 – 8 – 6 – 4 – 2 – 0 – A B C | 2 | 4
Copyright 2006 John Wiley & Sons, Inc.


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