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

... Proceeding as usual, we obtain the new tableau and new  asfollows.Variable B 1 Values 1 1 10 012 20 1/ 2 0 1 1/2 0 1 0 1/ 2 1 1 1/ 2 1 20 0 0 1 1/2 0s 1 10−2 −2020 1/ 2 0 1 1/230 ... 0 83/2 1 0 1 1/ 21 10 1 11 220 01 110 The optimal solution is x 1 =0, x2 =1, x3=2. The corresponding dual program ismaximize 4 1 +62subject to 2 1 + 2 1 2 1 +22 −4 1 +22 ... introducing slack variables and using the simplex procedure.The appropriate sequence of tableaus is given below without explanation.2 211 0 4 12 20 1 6 1 −4 −3000 1 1 1/ 2 1/ 2 0 2 10 1 11 230 12 ...

... next tableau.a 1 a2a3··b 11 210 3 10 1 11 2 10 12 0−20 01 ··Third tableauOptimizing the new restricted primal we obtain the tableau:a 1 a2a3··b 011 2 11 10 1 11 200 011 00 01 ··Final tableau∗4.7 ... restricted primal by pivoting as indicated we obtaina 1 a2a3··b 11 210 3 10 1 11 2 10 12 0−2 1/ 203/2 ···Now we again calculate the ratios 1 232obtaining 0= 1 2, and add this multiple ofthe ... alsooptimal: x 1 =2, x2 =1, x3=0.∗4.7 REDUCTION OF LINEAR INEQUALITIES Linear programming is in part the study of linear inequalities, and each progressivestage of linear programming theory adds...

... these, giving the equality in the theorem.Then using 1+ xp exp, we havem2m2 1 m 1 /2mm +1 = 1+ 1 m2 1 m 1 /2 1 1 m +1 < exp 1 2m +1 − 1 m +1 =exp− 1 2m ... chapters.Not only have nonlinear methods improved linear programming, but interior-point methods for linear programming have been extended to provide newapproaches to nonlinear programming. This chapter ... show howthis merger of linear and nonlinear programming produces elegant and effectivemethods. These ideas take an especially pleasing form when applied to linear programming. Study of them...

... [T2] and Todd and Ye [T5]. The primal-dual potential reduction algorithm wasdeveloped by Ye [Y1], Freund [F18], Kojima, Mizuno and Yoshise [K7], Goldfarb and Xiao [G1 1] , Gonzaga and Todd [G1 4], ... writing the constraintequations in standard form:x 11 +x 12 +···+x1n=a 1 x 21 +x22+···+x2n=a2xm1+xm2+···+xmn=amx 11 +x 21 xm1=b 1 x 12 + x22+ xm2=b2x1n+ ... McLinden [M3], Megiddo [M4], and Bayer and Lagarias [B3, B4], Gill et al [G5 ].5.5 Path-following algorithms were first developed by Renegar [R1]. A primal barrier or path-following algorithm was...

... 1) (3, 2)(3, 1) (1, 2)(5, 1) 233456222 1 1 1 1 1 1 1 1 1 1 1 (–, ∞)(2, 1) (1, 1) (5, 1) 242 1 1 1 11 12563(–, ∞)24563 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 222354 1 6 (1, 2)(–, ∞)(2, 1) (2, 1) (1, 1) (4, 1) 223456 1 1 1 1 1 1 1 1322(–, ∞)(4, 1) (3, ... every x 1 , x2∈ and every ,0   1, there holdsfx 1 + 1 x2 fx 1  + 1 fx2If, for every ,0<< ;1, and x 1 =x2, there holdsfx 1 + 1 x2 < fx 1  + 1 fx2then...

... −Exk +1 Exk=2k g TkQyk−2k g TkQgkyTkQykUsing g k=Qykwe haveExk −Exk +1 Exk=2 g Tk g k2 g TkQgk− g Tk g k2 g TkQgk g TkQ 1 g k= g Tk g k2 g TkQgk g TkQ 1 g kIn ... analyses:Table 8 .1 Solution to ExampleStep kfxk00 1 −2 .15 636252 −2 .17 440623 −2 .17 464404 −2 .17 465855 −2 .17 46 595 6 −2 .17 46 595 Solution point x∗= 1 53 496 5 0 12 20 097  1 97 515 6 1 412 95 48.7 ... in Section 7 .1. A solution tothis equation will satisfyyk +1 − 1 yk→0ThuslogMk +1 − 1 logMk→0 or logMk +1 Mk 1 →0 and hencek +1  1 k→M 1 1 Having derived the...