David G Luenberger, Yinyu Ye - Linear and Nonlinear Programming International Series Episode 2 Part 3 pot
David G. Luenberger, Yinyu Ye - Linear and Nonlinear Programming International Series Episode 2 Part 3 pot
... Self-scaling
1 20 0 .33 3 20 0 .33 3 20 0 .33 3 20 0 .33 3
2 2.7 32 7 89 93. 65457 93. 65457 2. 811061
33 836 899×10
2
56. 929 99 56. 929 99 3 5 627 69×10
2
46 37 6461×10
−4
1. 620 688 1. 620 688 4 20 0600 ×10
−4
51 21 9515×10
−5
5 25 1115×10
−1
5 25 1115×10
−1
4 726 918×10
−6
62 457944 ... ×10
−4
51 21 9515×10
−5
5 25 1115×10
−1
5 25 1115×10
−1
4 726 918×10
−6
62 457944 ×1...
David G. Luenberger, Yinyu Ye - Linear and Nonlinear Programming International Series Episode 2 Part 1 pot
... 34 198 02
6 31 3 36 19 3 42 9865
8 − 32 4 9978 3 42 9998
9 − 32 9 0408 34 30000
15 33 96 124
20 34 19 022
25 3 42 6004
30 3 42 83 72
35 3 42 927 5
40 3 42 9650
45 3 42 9 825
50 3 42 9909
55 −
3 42 9951
60 3 42 9971
Solution
x ... −1446460
4 2 130 796 2 0 529 49 2 0 529 49
5 2 1 635 86 2 149690 2 06 0 23 4
6 2 17 027 2 2 1496 93 2 06 0 23 7
7 2 1 727 86 2 1679 83 2...
David G. Luenberger, Yinyu Ye - Linear and Nonlinear Programming International Series Episode 2 Part 5 potx
... 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
0i=1 2 3 4
Suppose that given the feasible point x = 2 2 1 0 we ... 14 73
⎤
⎦
and finally
P =
1
11
⎡
⎢
⎢
⎣
1 31 0
39 30
1 31 0
0000
⎤
⎥
⎥
⎦
(22 )
The gradient at the point (2, 2, 1, 0) is g = 2 4 2 3 and hence we find
d =−Pg =
1...
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 +cx (66)
We rewrite (66) as
2x
2
+2xy +y
2
−2y +cx
=2x
2
+2xy +cx+y −1
2
−1
=2x
2
+2x +cx+y −1
2
+2xy −1 ... charac-
terization given here is a generalization of that in Luenberger [L10]. For the geometric inter-
pretation, see Luenberger [L8]. The central path for nonlinear programm...
David G. Luenberger, Yinyu Ye - Linear and Nonlinear Programming International Series Episode 2 Part 10 potx
... programming 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 ... higher than the minimal
objective cost.
Example 2 (Linear Programming) . To see that the problem (SDP) (that is,
(56)) generalizes linear programing define C = diagc
1
...
David G. Luenberger, Yinyu Ye - Linear and Nonlinear Programming International Series Episode 1 Part 3 ppsx
... decreasing objective, the algorithm must reach a basis
satisfying one of the two terminating conditions.
Example 1. Maximize 3x
1
+x
2
+3x
3
subject to
2x
1
+ x
2
+ x
3
2
x
1
+2x
2
+3x
3
5
2x
1
+2x
2
+ ... yields:
21 2104
33 10 13
r
T
−5 −4 30 0−7
First tableau
Pivoting in the column having the most negative bottom row component as
indicated, we obtain:
0 −1
4 /31 2/...
David G. Luenberger, Yinyu Ye - Linear and Nonlinear Programming International Series Episode 1 Part 10 pot
... 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. ... −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...
David G. Luenberger, Yinyu Ye - Linear and Nonlinear Programming International Series Episode 2 Part 2 ppt
... 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
... g
0
g
1
g
k
= g
0
Qg
0
Q
k
g
0
b) d
0
d
1
d
k
= g
0
Qg
0
Q
k
g
0
c) d
T
k
Qd
i
=0 for i k −1
d)
k
=g
T
k
g
k
/d
T
k
Qd
k...
David G. Luenberger, Yinyu Ye - Linear and Nonlinear Programming International Series Episode 2 Part 4 pps
... 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 necessary conditions are
1+ x
1
=0
2x
2
+x
3
+x
2
=0
x
2
+4x
3
+x
3
=0
One solution to ... 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...
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