David G Luenberger, Yinyu Ye - Linear and Nonlinear Programming International Series Episode 2 Part 8 pot

... 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 ... problem minimize x 2 1 +2x 2 2 subject to x 1 +x 2 ≥1 /2 x 1 x 2 nonnegative integers It is clear that the solution is x 1 = 1x 2 = 0, with objective value f ∗ = 1....

Ngày tải lên: 06/08/2014, 15:20

... −1446460 4 2 130796 2 0 529 49 2 0 529 49 5 2 163 586 2 149690 2 06 023 4 6 2 17 027 2 2 149693 2 06 023 7 7 2 1 727 86 2 167 983 2 165641 8 2 17 427 9 2 173169 2 165704 9 2 174 583 2 1743 92 2 1 684 40 10 ... 2 1 684 40 10 2 1746 38 2 174397 2 173 981 11 2 174651 2 1745 82 2 1740 48 12 2 174655 2 174643 2 174054 13 2 1746 58 2 174656 2 1746 08 14 2 174659 2 174656 2 1746 08 15...

Ngày tải lên: 06/08/2014, 15:20

... 97.33665 2 1. 58 625 1 1. 621 9 08 1. 621 9 08 0.7 024 8 72 32 989 875×10 2 8 26 88 93×10 −1 8 26 88 93×10 −1 4090350 ×10 −3 459 081 01×10 −4 43 029 43×10 −1 43 029 43×10 −1 1779 424 ×10 −5 51194144×10 −5 444 98 52 ×10 −3 444 98 52 ... Self-scaling 1 20 0.333 20 0.333 20 0.333 20 0.333 2 2.7 327 89 93.65457 93.65457 2. 81 1061 33 83 689 9×10 2 56. 929 99 5...

Ngày tải lên: 06/08/2014, 15:20

... 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 we find d =−Pg = 1 11  8 24  8 0 12. 4 The Gradient ... 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 fea...

Ngày tải lên: 06/08/2014, 15:20

... extension of linear programming. In linear programming, the variables form a vector which is required to be component- wise nonnegative, while in semidefinite programming the variables are compo- nents ...  higher than the minimal objective cost. Example 2 (Linear Programming) . To see that the problem (SDP) (that is, (56)) generalizes linear programing define C = diagc 1 c...

Ngày tải lên: 06/08/2014, 15:20

Ngày tải lên: 06/08/2014, 15:20

... −1. 6 .8 Maximal Flow 171 (a) (b) (c) (d) (e) (4, 1) (3, 1) 2 1 1 1 3 1 2 2 2 3 5 4 1 6 (1, 2) (–, ∞) (2, 1) (2, 1)(1, 1) (4, 1) 2 2 3 4 5 6 1 1 1 1 1 1 1 1 3 2 2 (–, ∞) (4, 1) (3, 2) (3, 1) (1, 2) (5, ... following requirements: a) a =1015 7 8 b = 8 6 9 12 5 b) a = 2 3 4 56 b =6 5 4 3 2 c) a = 2 4 3 15 2 b = 6 4 2 3 2 2. Transform the following t...

Ngày tải lên: 06/08/2014, 15:20

... form x k+1 =x k −  g T k g k g T k Qg k  g k  ( 32) where g k =Qx k −b. 21 8 Chapter 8 Basic Descent Methods N = 2 N = 3 N = 4 N = 5 4 4 5 3 3 3 2 2 2 2 1 1 1 1 1 8 1 5 1 3 2 3 1 2 2 5 3 5 2 8 3 8 5 8 Fig. 8 .2 ... −Ex k+1  Ex k  = 2 g T k g k  2 g T k Qg k  − g T k g k  2 g T k Qg k  g T k Q −1 g k = g T k g k  2 g...

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  ... Method 28 7 Proof. We have by direct substitution E  x k  −E  x k+1  E  x k  =  g T k S k g k  2  g T k S k QS k g k  g T k Q −1 g k   Letting T k =...

Ngày tải lên: 06/08/2014, 15:20

... 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