... x 1 , x 2 ∈ and ,0 1. Setting x =x 1 + 1 x 2 and alternatively y =x 1 or y =x 2 , we have fx 1 fx +fxx 1 −x (10 ) fx 2 fx +fxx 2 −x (11 ) Multiplying (10 ) by and ... by and (11 ) by (1 ) and adding, we obtain fx 1 + 1 fx 2 fx +fxx 1 + 1 x 2 −x But substituting x =x 1 + 1 x 2 , we obtain fx 1 + 1...
Ngày tải lên: 06/08/2014, 15:20
... 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 operations ar...
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 First table...
Ngày tải lên: 06/08/2014, 15:20
David G. Luenberger, Yinyu Ye - Linear and Nonlinear Programming International Series Episode 1 Part 1 doc
... Conditions 333 11 .6. Eigenvalues in Tangent Subspace 335 11 .7. Sensitivity 3 39 11 .8. Inequality Constraints 3 41 11. 9. Zero-Order Conditions and Lagrange Multipliers 346 11 .10 . Summary 353 11 .11 . Exercises ... Newton Methods 4 79 15 .6. Descent Properties 4 81 15.7. Rate of Convergence 485 15 .8. Interior Point Methods 487 15 .9. Semidefinite Programming 4 91 15 .1...
Ngày tải lên: 06/08/2014, 15:20
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 1...
Ngày tải lên: 06/08/2014, 15:20
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 2 obta...
Ngày tải lên: 06/08/2014, 15:20
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...
Ngày tải lên: 06/08/2014, 15:20
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], ... writing the constraint equations in standard form: x 11 +x 12 +···+x 1n =a 1 x 21 +x 22 +···+x 2n =a 2 x m1 +x m2 +···+x mn =a m x 11 +x 21 x m1 =b 1 x 12 +...
Ngày tải lên: 06/08/2014, 15:20
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
David G. Luenberger, Yinyu Ye - Linear and Nonlinear Programming International Series Episode 1 Part 10 pot
... −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 ... analyses: Table 8 .1 Solution to Example Step kfx k 00 1 −2 .15 63625 2 −2 .17 44062 3 −2 .17 46440 4 −2...
Ngày tải lên: 06/08/2014, 15:20