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

David G. Luenberger, Yinyu Ye - Linear and Nonlinear Programming International Series Episode 1 Part 1 doc
... 5 .1 ∗ 5.2 ∗ 5.3 5.4 5.5 5.6 5.7 5.8 5.9 93 98 10 3 11 1 PART II Chapter 11 2 11 4 11 5 11 8 12 1 12 6 13 4 13 9 14 0 Transportation and Network Flow Problems 14 5 6 .1 6.2 6.3 6.4 6.5 6.6 6.7 6.8 6.9 6 .10 ... Subspace 11 .7 Sensitivity 11 .8 Inequality Constraints 11 .9 Zero-Order Conditions and Lagrange Multipliers 11 .10 Summary 11 .11 Exercises Chapter 12 Primal Methods 12 .1 Advantage of Primal Methods 12 .2 ... Chapter 11 Constrained Minimization Conditions 11 .1 Constraints 11 .2 Tangent Plane 11 .3 First-Order Necessary Conditions (Equality Constraints) 11 .4 Examples 11 .5 Second-Order Conditions 11 .6 Eigenvalues...
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David G. Luenberger, Yinyu Ye - Linear and Nonlinear Programming International Series Episode 1 Part 2 ppsx

David G. Luenberger, Yinyu Ye - Linear and Nonlinear Programming International Series Episode 1 Part 2 ppsx
... x1 + x2 − x3 = 2x1 − 3x2 + x3 = −x1 + 2x2 − x3 = 1 To obtain an original basis, we form the augmented tableau e1 0 e2 e3 0 a1 a2 a3 b 1 1 −3 1 1 1 and replace e1 by a1 e2 by a2 , and e3 ... 2. 2 Examples of Linear Programming Problems 15 c x + c2 x2 + · · · + c n x n subject to the nutritional constraints a 11 x1 + a 12 x2 + · · · + a1n xn a 21 x1 + a 22 x2 + · · · + a2n xn · · · am1 ... canonical form: x1 x2 xm + y1 m +1 xm +1 + y1 m +2 xm +2 + · · · + y1 n xn = y10 + y2 m +1 xm +1 + y2 m +2 xm +2 + · · · + y2 n xn = y20 · · · · · · + ym m +1 xm +1 + ··· + ym n xn = ym0 (4) 3 .1 Pivots 35 Corresponding...
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David G. Luenberger, Yinyu Ye - Linear and Nonlinear Programming International Series Episode 1 Part 3 ppsx

David G. Luenberger, Yinyu Ye - Linear and Nonlinear Programming International Series Episode 1 Part 3 ppsx
... Method x2 1 1 x3 1 −2 x4 1 x5 0 x6 1 x7 b −2 1 Second tableau—phase I 1 1 1 0 1 Final tableau—phase I Now we go back to the equivalent reduced problem cT x3 x2 x4 1 1 1 x5 1 b 14 Initial ... 1 1 xB 0 1 Now T = 1 3 B 1 = −2 and r1 = −7 r4 = 3 r5 = ∗ 3. 8 The Simplex Method and LU Decomposition 59 We select a1 to enter the basis We have the tableau B 1 Variable −2 −4 1 1 xB 0 y1 ... with: 1 −2 1 1 0 − 21 0 19 First tableau—phase II 1/ 2 1/ 2 1 1/ 2 1/ 2 Final tableau—phase II The solution x3 = 1, x5 = can be inserted in the expression for x1 giving x1 = −7 + · + · = 1 thus...
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David G. Luenberger, Yinyu Ye - Linear and Nonlinear Programming International Series Episode 1 Part 4 doc

David G. Luenberger, Yinyu Ye - Linear and Nonlinear Programming International Series Episode 1 Part 4 doc
... usual, we obtain the new tableau and new follows B 1 Variable Value 0 2 s1 3 T 1 1/2 1/ 2 0 0 1 1 1/ 2 1/ 2 1/ 2 0 s1 1/ 2 1/ 2 −2 −2 1 1 1/ 2 1/ 2 = 4 −2 −7 B 1 = 1 −2 −5 as 70 Chapter The Simplex ... explanation 1 2 4 1 0 3/2 1 2 −3 0 0 0 1/ 2 1/ 2 1 1 2 1/ 2 1 10 1 The optimal solution is x1 = 0, x2 = 1, x3 = The corresponding dual program is maximize subject to 2 +6 1+ 1+ 2 1+ 2 2 2 1 4 −3 ... x4 = x1 + 3x2 − x3 + 2x4 = x + x2 + x =5 x2 x3 x4 74 Chapter The Simplex Method 22 Find a basic feasible solution to x1 + 2x2 − x3 + x4 = 2x1 + 4x2 + x3 + 2x4 = 12 x1 + 4x2 + 2x3 + x4 = x1 i=1...
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David G. Luenberger, Yinyu Ye - Linear and Nonlinear Programming International Series Episode 1 Part 5 docx

David G. Luenberger, Yinyu Ye - Linear and Nonlinear Programming International Series Episode 1 Part 5 docx
... /y2j and select the minimum positive ratio This yields the pivot indicated Continuing, the remaining tableaus are 0 − 1 1 0 5/ 2 1/ 2 7/2 1/ 2 1/ 2 3/2 −2 Second tableau 5/ 2 1 1/2 −2 1 1 11 0 ... problem, and the corresponding dual solution is ˆ = − r ii) Show that this scheme fully allocates H 12 Solve the linear inequalities −2x1 + 2x2 2x1 − x2 − 4x2 15 x1 − 12 x2 12 x1 + 20x2 1 −2 1 Note ... extremal 25 Eliminate the null variables in the system 2x1 + x2 − x3 + x4 + x5 = −x1 + 2x2 + x3 + 2x4 + x5 = 1 −x1 − x2 x1 x2 − 3x4 + 2x5 = 1 x3 x4 x5 x5 26 Reduce to minimal size x1 + x2 +...
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David G. Luenberger, Yinyu Ye - Linear and Nonlinear Programming International Series Episode 1 Part 6 pptx

David G. Luenberger, Yinyu Ye - Linear and Nonlinear Programming International Series Episode 1 Part 6 pptx
... Ek +1 is defined to be the minimal-volume ellipsoid containing 1/ 2 Ek It is constructed as follows Define = m +1 = yk m2 m2 − 1/ 2 E Fig 5 .1 A half-ellipsoid =2 ∗ 5.3 The Ellipsoid Method 11 7 Then ... m +1 The reduction in volume is the product of the square roots m2 1 of these, giving the equality in the theorem Then using + x p exp , we have m2 m2 − m 1 /2 m = 1+ m +1 m 1 < exp m 1 /2 1 1 ... 5.5 The Central Path 5.5 12 1 THE CENTRAL PATH The concept underlying interior-point methods for linear programming is to use nonlinear programming techniques of analysis and methodology The analysis...
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David G. Luenberger, Yinyu Ye - Linear and Nonlinear Programming International Series Episode 1 Part 7 pps

David G. Luenberger, Yinyu Ye - Linear and Nonlinear Programming International Series Episode 1 Part 7 pps
... constraint equations in standard form: x 11 + x12 + · · · + x1n = a1 = a2 x 21 + x22 + · · · + x2n xm1 + xm2 + · · · + xmn = am + x 21 x 11 xm1 + x22 x12 x1n + xm2 + x2n = b1 = b2 (3) + xmn = bn The ... Goldfarb and Xiao [G 11] , Gonzaga and Todd [G14], Todd [T4], Tun el [T10], Tutuncu [T 11] , and others The homogeneous and self-dual embedding method can be found in Ye et al [Y2], Luo et al [L18], Andersen ... Basic Feasible Solution x 11 x 21 x12 x22 x13 x23 ··· ··· x1n x2n 14 9 a1 a2 (7) xm1 b1 xm2 b2 xm3 b3 ··· ··· xmn bn am The individual elements of the array appear in cells and represent a solution...
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David G. Luenberger, Yinyu Ye - Linear and Nonlinear Programming International Series Episode 1 Part 8 pdf

David G. Luenberger, Yinyu Ye - Linear and Nonlinear Programming International Series Episode 1 Part 8 pdf
... (1, 2) (3, 1) (b) (4, 1) (3, 1) 2 (–, ∞) 1 1 1 3 (1, 1) (2, 1) (c) (d) (–, ∞) (5, 1) 1 2 (e) Fig 6.6 Example of maximal flow problem (5, 1) 17 1 17 2 Chapter Transportation and Network Flow Problems ... source, and node is the sink The original network with capacities indicated on the 6 .8 Maximal Flow (1, 1) (2, 1) 2 (–, ∞) 1 (4, 1) (1, 2) (2, 1) (a) (4, 1) (3, 2) 2 (–, ∞) 1 1 (1, 2) (3, 1) (b) ... T + sources and T destinations.) b) Using the values T = s = 200 r1 = 10 0 r2 = 13 0 r3 = 15 0 r4 = 14 0 c1 = c2 = c0 = 12 , solve the problem 11 The marriage problem A group of n men and n women...
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David G. Luenberger, Yinyu Ye - Linear and Nonlinear Programming International Series Episode 1 Part 9 ppsx

David G. Luenberger, Yinyu Ye - Linear and Nonlinear Programming International Series Episode 1 Part 9 ppsx
... at x0 = 10 0, each of the sequences below might be generated from iterative application of this algorithm 10 0 50 25 12 −6 −2 1/ 2 10 0 −40 20 −5 −2 1/ 4 1/ 8 10 0 10 1 1 /16 1/ 100 1/ 1000 1/ 10 000 The ... , alternatively y = x1 or y = x2 , we have Setting x = x1 + − x2 and f x1 f x1 + − (10 ) f x2 Multiplying (10 ) by f x + f x x1 − x f x + f x x2 − x (11 ) and (11 ) by (1 − ) and adding, we obtain ... Proposition Let f1 and f2 be convex functions on the convex set the function f1 + f2 is convex on Proof f1 Let x1 , x2 ∈ x1 + − , and < x2 +f2 < Then x1 + − x2 f1 x1 + f2 x1 + − f x + f2 x Proposition...
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David G. Luenberger, Yinyu Ye - Linear and Nonlinear Programming International Series Episode 1 Part 10 pot

David G. Luenberger, Yinyu Ye - Linear and Nonlinear Programming International Series Episode 1 Part 10 pot
... write (13 ) as xk +1 − x∗ = xk − x∗ xk 1 − x∗ g xk 1 xk x∗ g xk 1 xk Now, by the mean value theorem with remainder, we have (see Exercise 2) g xk 1 xk = g (14 ) k and g xk 1 xk x∗ = g where k and tively ... Q 1 x n i =1 n i =1 xi2 i xi n i =1 xi2 / i which can be written as xT x 1/ n = n i =1 i i ≡ xT Qx xT Q 1 x i =1 i / i where i = xi2 / n xi2 We have converted the expression to the ratio of two i =1 ... xk 1 xk x∗ = g where k and tively Thus k k (15 ) are convex combinations of xk , xk 1 and xk , xk 1 , x∗ , respec- xk +1 − x∗ = g 2g k xk − x∗ xk 1 − x∗ (16 ) k It follows immediately that the process...
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