Download presentation

Presentation is loading. Please wait.

Published byGian Jobes Modified over 3 years ago

1
1. Set up the phase 1 dictionary for this problem and make the first pivot: Maximize X 1 + X 2 + 2 X 3 + X 4 subject to -X 1 + X 3 + 2 X 4 ≤ -3 -X 1 + X 2 ≤ -7 -X 1 + 2 X 2 + X 4 ≤ -5 X 1, X 2, X 3, X 4 ≥ 0 1

2
The equations: -X 1 + X 3 + 2 X 4 –x 0 ≤ -3 -X 1 + X 2 -x 0 ≤ -7 -X 1 + 2 X 2 + X 4 –x 0 ≤ -5 Phase 1: Before pivoting to make feasible. X5 = -3 + X1 - X3- 2 X4 + X0 X6 = -7 + X1 - X2 + X0 X7 = -5 + X1 -2 X2 - X4 + X0 ------------------------------------- z = 0 - X0 2

3
Taking the first pivot: X0 enters. X6 leaves. z = -0.000000 Taking the first pivot: X0 enters. X6 leaves. z = -0.000000 The initial dictionary: X5 = 4+ 0 X1 + 1 X2 - 1 X3 - 2 X4 + 1 X6 X0 = 7- 1 X1 + 1 X2 + 0 X3 + 0 X4 + 1 X6 X7 = 2+ 0 X1 - 1 X2 + 0 X3 - 1 X4 + 1 X6 ----------------------------------------- z = -7+ 1 X1 - 1 X2 + 0 X3 + 0 X4 - 1 X6 3

4
The optimal solution to the phase 1 problem is: After 1 pivot: X5= 4 +1 X2 -1 X3 -2 X4 +1 X6 + 0 X0 X1= 7 +1 X2 +0 X3 +0 X4 +1 X6 - 1 X0 X7= 2 -1 X2 +0 X3 -1 X4 +1 X6 + 0 X0 ------------------------------------- z = 0 +0 X2 +0 X3 +0 X4 +0 X6 - 1 X0 2. Set up the initial dictionary for the Phase 2 problem. Recall the objective function is: Maximize X 1 + X 2 + 2 X 3 + X 4 4

5
Answer: X5 = 4 +1 X2 -1 X3 -2 X4 +1 X6 X1 = 7 +1 X2 +0 X3 +0 X4 +1 X6 X7 = 2 -1 X2 +0 X3 -1 X4 +1 X6 ------------------------------- z = 7 +2 X2 +2 X3 +1 X4 +1 X6 5

6
The final dictionary is: X5 = 6 -1 X3 - 3 X4 + 2 X6 - 1 X7 X1 = 9 +0 X3 - 1 X4 + 2 X6 - 1 X7 X2 = 2 +0 X3 - 1 X4 + 1 X6 - 1 X7 ----------------------------------- z = 11 +2 X3 - 1 X4 + 3 X6 - 2 X7 3. What can you conclude about this problem? 6

7
A problem that could cause grief for a computer implementation (small example found by Fay Ye): Maximize X 1 + X 2 subject to 1 X 1 + 1 X 2 ≤ 1.5 1 X 1 – 1 X 2 ≤ -1.5 X 1, X 2 ≥ 0 7

8
Phase 1 Problem: Maximize –x 0 subject to 1 x 1 + 1 x 2 - x 0 ≤ 1.5 1 x 1 – 1 x 2 - x 0 ≤ -1.5 x 0, x 1, x 2 ≥ 0 Initial dictionary: x 3 = 1.5 -1 x 1 - 1 x 2 + x 0 x 4 = -1.5 -1 x 1 + 1 x 2 + x 0 -------------------------- z= 0.0 - x 0 8

9
Initial dictionary: x 3 = 1.5 -1 x 1 - 1 x 2 + x 0 x 4 = -1.5 -1 x 1 + 1 x 2 + x 0 (*) -------------------------- z= 0.0 - x 0 Pivot to make feasible: X5 enters. X4 leaves. z = 0.0 X3 = 3.0+ 0.0 X1 - 2.0 X2 + 1.0 X4 X0 = 1.5+ 1.0 X1 - 1.0 X2 + 1.0 X4 ------------------------------------ z = -1.5- 1.0 X1 + 1.0 X2 - 1.0 X4 9

10
X3 = 3.0+ 0.0 X1 - 2.0 X2 + 1.0 X4 X0 = 1.5+ 1.0 X1 - 1.0 X2 + 1.0 X4 ------------------------------------ z = -1.5- 1.0 X1 + 1.0 X2 - 1.0 X4 X2 enters. X3 leaves. z = -1.5 After 1 pivot: X2 = 1.5 + 0.0 X1 - 0.5 X3 + 0.5 X4 X0 = 0.0 + 1.0 X1 + 0.5 X3 + 0.5 X4 ----------------------------------- z = 0.0 - 1.0 X1 - 0.5 X3 - 0.5 X4 The optimal solution: 0 X1= 0, X2= 1.5, X3= 0, X4= 0 10

11
Problem: X0 is still in the basis! X2 = 1.5 + 0.0 X1 - 0.5 X3 + 0.5 X4 X0 = 0.0 + 1.0 X1 + 0.5 X3 + 0.5 X4 ----------------------------------- z = 0.0 - 1.0 X1 - 0.5 X3 - 0.5 X4 What happens if you leave z there? Use any variable with non-zero coefficient in the X0 row to pivot it out. 11

12
Problem: X0 is still in the basis! X2 = 1.5 + 0.0 X1 - 0.5 X3 + 0.5 X4 X0 = 0.0 + 1.0 X1 + 0.5 X3 + 0.5 X4 ----------------------------------- z = 0.0 - 1.0 X1 - 0.5 X3 - 0.5 X4 Use any variable with non-zero coefficient in the X0 row to pivot it out. X1 enters. X5 leaves. X2 = 1.5 - 0.5 X3 + 0.5 X4 + 0.0 X0 X1 = -0.0 - 0.5 X3 - 0.5 X4 + 1.0 X0 -------------------------------------- z = 0.0 + 0.0 X3 + 0.0 X4 - 1.0 X0 12

13
X2 = 1.5 - 0.5 X3 + 0.5 X4 + 0.0 X5 X1 = 0.0 - 0.5 X3 - 0.5 X4 + 1.0 X5 -------------------------------------- z = 0.0 + 0.0 X3 + 0.0 X4 - 1.0 X5 Recall: z= x 1 + x 2 The new objective function row is: X1 + X2 = 1.5 - 0.5 X3 + 0.5 X4 + 0.0 X5 0.0 - 0.5 X3 - 0.5 X4 + 1.0 X5 ================================ 1.5 – 1.0 X3 + 0 X4 13

14
Finishing the problem (Phase 2): The initial dictionary: X2 = 1.5 - 0.5 X3 + 0.5 X4 X1 = 0.0 - 0.5 X3 - 0.5 X4 ------------------------------ z = 1.5 - 1.0 X3 + 0.0 X4 The optimal solution: 1.5 X1= 0.0, X2= 1.5, X3=0.0, X4=0.0 14

Similar presentations

Presentation is loading. Please wait....

OK

Introduction to Operations Research

Introduction to Operations Research

© 2018 SlidePlayer.com Inc.

All rights reserved.

To ensure the functioning of the site, we use **cookies**. We share information about your activities on the site with our partners and Google partners: social networks and companies engaged in advertising and web analytics. For more information, see the Privacy Policy and Google Privacy & Terms.
Your consent to our cookies if you continue to use this website.

Ads by Google

Ppt on uk economy Free download ppt on surface chemistry Ppt on production planning File type ppt on cybercrime training Ppt on sigma and pi bonds Ppt on critical thinking Ppt on swami vivekananda death Ppt on self development quotes Ppt on microsoft excel formulas Ppt on single phase and three phase dual converter kit