David G Luenberger, Yinyu Ye - Linear and Nonlinear Programming International Series Episode 2 Part 6 pps

David G. Luenberger, Yinyu Ye - Linear and Nonlinear Programming International Series Episode 2 Part 6 pps

David G. Luenberger, Yinyu Ye - Linear and Nonlinear Programming International Series Episode 2 Part 6 pps

... 66 45499 10 −375 6 423 20 66  563 77 20 −3759 123 40 66 58443 50 −3 76 5 128 60 66 59191 100 −3771 62 5 80 66 59514 20 0 −3778983 100 66 5 965 6 500 −3787989 120 66 59 825 1000 −37930 12 121 66 59 827 ... (1 /2 of chain) 0 60 .00000 y 1 =−814 8 26 0 10 66 .4 761 0 y 2 =−7 8 26 505 20 66 . 521 80 y 3 =−7 429 208 30 66 .53595 y 4 =− 69 30959 40 66 .54154 y 5 =− 63 1...

Ngày tải lên: 06/08/2014, 15:20

25 325 0
David G. Luenberger, Yinyu Ye - Linear and Nonlinear Programming International Series Episode 2 Part 4 pps

David G. Luenberger, Yinyu Ye - Linear and Nonlinear Programming International Series Episode 2 Part 4 pps

... problem minimize 2x 2 1 +2x 1 x 2 +x 2 2 −10x 1 −10x 2 subject to x 2 1 +x 2 2  5 3x 1 +x 2  6 The first-order necessary conditions, in addition to the constraints, are 4x 1 +2x 2 −10 +2 1 x 1 +3 2 =0 2x 1 +2x 2 −10 +2 1 x 2 + 2 =0  1  ... the problem extremize x 1 +x 2 2 +x 2 x 3 +2x 2 3 subject to 1 2 x 2 1 +x 2 2 +x 2 3  =1 The first-order...

Ngày tải lên: 06/08/2014, 15:20

25 314 0
David G. Luenberger, Yinyu Ye - Linear and Nonlinear Programming International Series Episode 1 Part 3 ppsx

David G. Luenberger, Yinyu Ye - Linear and Nonlinear Programming International Series Episode 1 Part 3 ppsx

... problem x 2 x 3 x 4 x 5 b 01−11 3 12 10 2 c T 23 −11−14 Initial tableau—phase II Transforming the last row appropriately we proceed with: 01−11 3 1 2 −10 2 0 22 0 21 First tableau—phase II −1 /20 −1 /21 2 1 /21 −1 /20 1 10 ... decreasing objective, the algorithm must reach a basis satisfying one of the two terminating conditions. Example 1. Maximize 3x 1 +x 2 +3x 3 subject to 2...

Ngày tải lên: 06/08/2014, 15:20

25 379 0
David G. Luenberger, Yinyu Ye - Linear and Nonlinear Programming International Series Episode 1 Part 6 pptx

David G. Luenberger, Yinyu Ye - Linear and Nonlinear Programming International Series Episode 1 Part 6 pptx

... chapters. Not only have nonlinear methods improved linear programming, but interior- point methods for linear programming have been extended to provide new approaches to nonlinear programming. This chapter ... show how this merger of linear and nonlinear programming produces elegant and effective methods. These ideas take an especially pleasing form when applied to line...

Ngày tải lên: 06/08/2014, 15:20

25 294 0
David G. Luenberger, Yinyu Ye - Linear and Nonlinear Programming International Series Episode 1 Part 7 pps

David G. Luenberger, Yinyu Ye - Linear and Nonlinear Programming International Series Episode 1 Part 7 pps

... found to be 2. Next, u 3 and u 2 are determined, then  3 and  2 , and finally u 1 and  1 . The result is shown below: u 3  468 95 2 2 4 553 22 2 321 3 32 4 2 2  2 −1 120 Cycle of Change In accordance ... positive, indicating that the current solution is optimal. 3  468 95 2 24 553 22 2 321 33 2 4 2 2 2 −1 020 Degeneracy As in all linear programmi...

Ngày tải lên: 06/08/2014, 15:20

25 399 0
David G. Luenberger, Yinyu Ye - Linear and Nonlinear Programming International Series Episode 1 Part 9 ppsx

David G. Luenberger, Yinyu Ye - Linear and Nonlinear Programming International Series Episode 1 Part 9 ppsx

... algorithms for solving nonlinear programming problems are not globally convergent in their purest form and thus occasionally generate sequences that either do not converge at all or converge ... GLOBAL CONVERGENCE OF DESCENT ALGORITHMS A good portion of the remainder of this book is devoted to presentation and analysis of various algorithms designed to solve nonlinear programming...

Ngày tải lên: 06/08/2014, 15:20

25 263 0
David G. Luenberger, Yinyu Ye - Linear and Nonlinear Programming International Series Episode 2 Part 1 pot

David G. Luenberger, Yinyu Ye - Linear and Nonlinear Programming International Series Episode 2 Part 1 pot

... 2 1745 82 2 174048 12 2 17 465 5 2 17 464 3 2 174054 13 2 17 465 8 2 17 465 6 2 17 460 8 14 2 17 465 9 2 17 465 6 2 17 460 8 15 2 17 465 9 2 17 465 8 2 174 62 2 16 2 17 465 9 2 17 465 5 17 2 17 465 9 2 17 465 6 18 ... 2 14 969 0 2 06 023 4 6 2 17 027 2 2 14 969 3 2 06 023 7 7 2 1 727 86 2 167 983 2 165 641 8 2 17 427 9 2 173 169 2 165 704 9 2 174583 2 1743 92 2 168 440 10 2 17 4...

Ngày tải lên: 06/08/2014, 15:20

25 310 0
David G. Luenberger, Yinyu Ye - Linear and Nonlinear Programming International Series Episode 2 Part 2 ppt

David G. Luenberger, Yinyu Ye - Linear and Nonlinear Programming International Series Episode 2 Part 2 ppt

... hypothesis both g k and Qd k belong to g 0  Qg 0 Q k+1 g 0 , the first by (a) and the second by (b). Thus g k+1 ∈ g 0  Qg 0 Q k+1 g 0 . Furthermore g k+1  g 0  Qg 0 Q k g 0  =d 0  ... g 0  g 1  g k  = g 0  Qg 0 Q k g 0  b) d 0  d 1 d k  = g 0  Qg 0 Q k g 0  c) d T k Qd i =0 for i  k −1 d)  k =g T k g k /d T...

Ngày tải lên: 06/08/2014, 15:20

25 283 0
David G. Luenberger, Yinyu Ye - Linear and Nonlinear Programming International Series Episode 2 Part 3 pot

David G. Luenberger, Yinyu Ye - Linear and Nonlinear Programming International Series Episode 2 Part 3 pot

... Self-scaling 1 20 0.333 20 0.333 20 0.333 20 0.333 2 2.7 327 89 93 .65 457 93 .65 457 2. 811 061 338 368 99×10 2 56. 929 99 56. 929 99 35 62 7 69 ×10 2 46 3 764 61×10 −4 1. 62 0 68 8 1. 62 0 68 8 4 20 060 0 ×10 −4 51 21 9515×10 −5 5 25 1115×10 −1 5 25 1115×10 −1 4 7 26 918×10 6 62 457944 ... restart) Self-scaling 1 96. 3 066 9 96. 3 066 9 96. 3 066 9 96. 3 066 9 2 1. 56...

Ngày tải lên: 06/08/2014, 15:20

25 353 0
David G. Luenberger, Yinyu Ye - Linear and Nonlinear Programming International Series Episode 2 Part 5 potx

David G. Luenberger, Yinyu Ye - Linear and Nonlinear Programming International Series Episode 2 Part 5 potx

... problem minimize x 2 1 +x 2 2 +x 2 3 +x 2 4 −2x 1 −3x 4 subject to 2x 1 +x 2 +x 3 +4x 4 =7 (20 ) x 1 +x 2 +2x 3 +x 4 =6 x i  0i=1 2 3 4 Suppose that given the feasible point x = 2 2 1 0 we ... and therefore g 2 =0 is adjoined to the set of working constraints. g 1 = 0 ∇f T g 2 = 0 x Feasible region g 1 T Fig. 12. 4 Constraint to be dropped 11.9 Zero-Ord...

Ngày tải lên: 06/08/2014, 15:20

25 474 0
w