... −Ex k +1 Ex k = 2 k g T k Qy k − 2 k g T k Qg k y T k Qy k Using g k =Qy k we have Ex k −Ex k +1 Ex 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 T k Qg k g T k Q 1 g k In ... false position (11 ) converges to x ∗ with order 1 1 618 . x k x k + 1 x k – 1 x g Fig. 8.6 False po...
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 fx and g i x are not generally convex functions. Quadratic Programming Let fx = 1/ 2x T Qx +c T x and g...
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David G. Luenberger, Yinyu Ye - Linear and Nonlinear Programming International Series Episode 1 Part 1 doc
... in Tangent Subspace 335 11 .7. Sensitivity 339 11 .8. Inequality Constraints 3 41 11. 9. Zero-Order Conditions and Lagrange Multipliers 346 11 .10 . Summary 353 11 .11 . Exercises 354 Chapter 12 . Primal ... Conditions 3 21 11. 1. Constraints 3 21 11. 2. Tangent Plane 323 11 .3. First-Order Necessary Conditions (Equality Constraints) 326 11 .4. Examples 327 11 .5. Second-Order C...
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David G. Luenberger, Yinyu Ye - Linear and Nonlinear Programming International Series Episode 1 Part 2 ppsx
... then obtain x 1 x 2 x 3 x 4 x 5 x 6 3/ 51/ 5 010 −2/ 518 /5 2/5 1/ 50 01 3/57/5 1/ 53/ 510 0 1/ 5 1/ 5 Continuing, there results x 1 x 2 x 3 x 4 x 5 x 6 1 1 − 210 0 4 1 −2 −3 010 2 1 −3 −50 01 1 From this ... tableau e 1 e 2 e 3 a 1 a 2 a 3 b 10 0 1 1 15 010 2− 313 0 01 12 1 1 and replace e 1 by a 1 e 2 by a 2 , and e 3 by a 3 . The required operati...
Ngày tải lên: 06/08/2014, 15:20
David G. Luenberger, Yinyu Ye - Linear and Nonlinear Programming International Series Episode 1 Part 3 ppsx
... Method x 2 x 3 x 4 x 5 x 6 x 7 b 1 10 1 10 1 12 10 11 2 1 − 210 20−2 Second tableau—phase I 01 11 01 3 12 10 11 2 00 00 11 0 Final tableau—phase I Now we go back to the equivalent reduced problem x 2 x 3 x 4 x 5 b 01 11 3 12 10 ... 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 −220− 21 Fir...
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David G. Luenberger, Yinyu Ye - Linear and Nonlinear Programming International Series Episode 1 Part 4 doc
... Proceeding as usual, we obtain the new tableau and new as follows. Variable B 1 Value s 1 1 10 012 2 0 1/ 2 0 1 1/2 0 1 0 1/ 2 1 1 1/ 2 1 2 0 0 0 1 1/2 0 s 1 10 2 −20 2 0 1/ 2 0 1 1/2 3 0 ... 0 8 3/2 1 0 1 1/ 21 10 1 11 2 20 01 110 The optimal solution is x 1 =0, x 2 =1, x 3 =2. The corresponding dual program is maximize 4 1 +6 2 subject to 2...
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David G. Luenberger, Yinyu Ye - Linear and Nonlinear Programming International Series Episode 1 Part 5 docx
... next tableau. a 1 a 2 a 3 ··b 11 210 3 10 1 11 2 10 12 0−2 0 01 ·· Third tableau Optimizing the new restricted primal we obtain the tableau: a 1 a 2 a 3 ··b 011 2 11 10 1 11 2 00 011 0 0 01 ·· Final tableau ∗ 4.7 ... restricted primal by pivoting as indicated we obtain a 1 a 2 a 3 ··b 11 210 3 10 1 11 2 10 12 0−2 1/ 203/2 ··· Now we again calculate the ratios 1 2 3...
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David G. Luenberger, Yinyu Ye - Linear and Nonlinear Programming International Series Episode 1 Part 6 pptx
... these, giving the equality in the theorem. Then using 1+ x p e xp , we have m 2 m 2 1 m 1 /2 m m +1 = 1+ 1 m 2 1 m 1 /2 1 1 m +1 < exp 1 2m +1 − 1 m +1 =exp − 1 2m ... chapters. Not only have nonlinear methods improved linear programming, but interior- point methods for linear programming have been extended to provide new approaches to...
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David G. Luenberger, Yinyu Ye - Linear and Nonlinear Programming International Series Episode 1 Part 7 pps
... [T2] and Todd and Ye [T5]. The primal-dual potential reduction algorithm was developed by Ye [Y1], Freund [F18], Kojima, Mizuno and Yoshise [K7], Goldfarb and Xiao [G1 1] , Gonzaga and Todd [G1 4], ... following example. Example 3. A completed solution array is shown below: 10 0 10 20 − 10 + 30 20 + 10 0 30 − 60 10 0 10 10 − + 40 0 50 40 10 30 40 40 In thi...
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David G. Luenberger, Yinyu Ye - Linear and Nonlinear Programming International Series Episode 1 Part 8 pdf
... 1) (3, 2) (3, 1) (1, 2) (5, 1) 2 3 3 4 5 6 2 2 2 1 1 1 1 1 1 1 1 1 1 1 (–, ∞) (2, 1) (1, 1) (5, 1) 2 4 2 1 1 1 11 1 2 5 6 3 (–, ∞) 24 5 6 3 1 Fig. 6.6 Example of maximal flow problem 17 8 Chapter ... value 0, +1, or 1. 6.8 Maximal Flow 17 1 (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,...
Ngày tải lên: 06/08/2014, 15:20