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

... 2 14 9690 2 06 023 4 6 2 17 027 2 2 14 9693 2 06 023 7 7 2 1 727 86 2 16 7983 2 16 56 41 8 2 17 427 9 2 17 316 9 2 16 5704 9 2 17 4583 2 17 43 92 2 16 8440 10 2 17 4638 2 17 4397 2 17 39 81 11 2 17 46 51 2 17 45 82 ... 2 17 45 82 2 17 4048 12 2 17 4655 2 17 4643 2 17 4054 13 2 17 4658 2 17 4656 2 17 4608 14 2 17 4659 2 17 4656 2 17 4608 15 2 17 4659 2 17 4658 2 17 4 622 16 2 1...

Ngày tải lên: 06/08/2014, 15:20

... 10 −4 51 21 9 515 10 −5 5 25 11 15 10 1 5 25 11 15 10 1 4 726 918 10 −6 62 457944 10 −7 3 323 745 10 1 3 323 745 10 1 76 15 0890 10 −3 8 1 027 00 10 −3 83 025 393 10 −3 2 973 0 21 10 −3 93 025 476 10 −5 1 95 01 52 10 −3 10 ... 97.33665 2 1. 58 625 1 1. 6 21 908 1. 6 21 908 0.7 024 8 72 32 989875 10 2 8 26 8893 10 1 8 26 8893 10 1 4090350...

Ngày tải lên: 06/08/2014, 15:20

... find A q A T q  1 = 1 11 ⎡ ⎣ 6 −5 19 −5 614 19 14 73 ⎤ ⎦ and finally P = 1 11 ⎡ ⎢ ⎢ ⎣ 1 − 310 −39−30 1 − 310 0000 ⎤ ⎥ ⎥ ⎦  (22 ) The gradient at the point (2, 2, 1, 0) is g = 2 4 2 −3 and hence ... solution is x 1 = x 2 = 1 and the Lagrange multiplier is  = 1. The Lagrangian problem is minimize 1 2 x 2 1 +x 2 2  1 x 1 +x 2 2 subject t...

Ngày tải lên: 06/08/2014, 15:20

... absolute-value penalty function. We minimize the function 2x 2 +2xy +y 2 −2y +cx (66) We rewrite (66) as 2x 2 +2xy +y 2 −2y +cx =2x 2 +2xy +cx+y 1 2 1 =2x 2 +2x +cx+y 1 2 +2xy 1 1 =x 2 +2x ... region, and let 0 ≤ ≤ 1. Then  1 + 1  2  = inf fx + 1 + 1  2  T g xx ∈  ≥inf fx 1  + T 1 g x 1 x 1 ∈ +inf  1 −f...

Ngày tải lên: 06/08/2014, 15:20

... expressed in Theorem 2, Section 5.6. 15 .9 SEMIDEFINITE PROGRAMMING Semidefinite programming (SDP) is a natural extension of linear programming. In linear programming, the variables form a vector which ... satisfying the first- order conditions for problems when fx and g i x are not generally convex functions. Quadratic Programming Let fx = 1 /2 x T Qx +c T x and g...

Ngày tải lên: 06/08/2014, 15:20

Ngày tải lên: 06/08/2014, 15:20

... 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  ... 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...

Ngày tải lên: 06/08/2014, 15:20

... problem minimize 2x 2 1 +2x 1 x 2 +x 2 2 10 x 1 10 x 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  ... yields the equations 4x 1 +2x 2 10 +2 1 x 1 =0 2x 1 +2x 2 10 +2 1 x 2 =0 x 2 1 +x...

Ngày tải lên: 06/08/2014, 15:20

... −36660 61 10 −6645499 10 −3756 423 20 −6656377 20 −3759 12 3 40 −6658443 50 −3765 12 8 60 −665 919 1 10 0 −377 1 625 80 −6659 514 20 0 −3778983 10 0 −6659656 500 −3787989 12 0 −6659 825 10 00 −37930 12 12 1 ... −37930 12 12 1 −6659 827 15 00 −3794994 12 2 −6659 827 20 00 −3795965 25 00 −3796489 y 1 = 410 9 519 y 1 =988 622 3 376 Chapter 12 Prima...

Ngày tải lên: 06/08/2014, 15:20

... identity  A 1 A 2 A T 2 A 3  1 =  A 1 −A 2 A 1 3 A T 2  1 −A 1 −A 2 A 1 3 A T 2 A 2 A 1 3 −A 1 3 A T 2 A 1 −A 2 A 1 3 A T 2  1 A 3 −A T 2 A 1 1 A 2  1   (29 ) Using the fact that A 1 3 c → 0 gives the result. To ... 388565 3 8 24 388563 5 3 15 388563 7 3 21 388563 c = 20 0 ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ 1 23 0 ∗ 23 0 488607 3...

Ngày tải lên: 06/08/2014, 15:20