... equations
x
1
+ x
2
−x
3
= 5
2x
1
−3x
2
+x
3
= 3
−x
1
+2x
2
−x
3
= 1
To obtain an original basis, we form the augmented 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 ... 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...
... sake of
safety.
54 Chapter 3 The Simplex 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 ... problem
x
2
x
3
x
4
x
5
b
01 11 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
Fi...
... row.
a
1
a
2
a
3
··b
1
12 10 3
213 0 15
−3 −2 50 0−8
1/ 203/2 ···
Second tableau
Minimizing the new associated 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 ... tableau:
a
1
a
2
a
3
··b
011 2 11
10 1 11 2
00 011 0
0 01 ··
Final tableau
∗
4.7 Reduction of Linear Inequalities 10 1
By subtracting the second equati...
... +1
=
1+
1
m
2
1
m 1 /2
1
1
m +1
< exp
1
2m +1
−
1
m +1
=exp
−
1
2m +1
Convergence
The ellipsoid method is initiated by selecting y
0
and R such that condition (A1) ... S
1
we have
d
x
+S
1
Xd
s
=S
1
1−x
Then, premultiplying by A and using Ad
x
=0, we have
AS
1
Xd
s
=AS
1
1−Ax =AS
1
1−b
Using d
s
=−A
T
d
y
we have
AS
1
XA
T...
...
1
1
+
n
n
, with
1
+
n
= 1. Using the relation
1
/
1
+
n
/
n
=
1
+
n
−
1
1
−
n
n
/
1
n
, an appropriate bound is
lim
1
n
1/
1
+
n
−/
1
n
The ... large, we have
steepest descent rate =
r 1
r +1
2
1 1/ r
4
proposed method rate =
r
2
1
r
2
+1
2
1 1/ r
2
4
Since 1 1/ r
2
r
1 1/ r...
... lies below the
line 1 − 2 /a +b. Thus we conclude that
q 1−
2
a +b
on 0a+b /2 and that
q
a +b
2
−
2
a +b
We can see that on a +b /2 b
q 1−
2
a +b
since for q ... Qx
0
−x
∗
=
1
1
e
1
+
2
2
e
2
++
n
n
e
n
and since the eigenvectors are mutually orthogonal, we have
Ex
0
=
1
2
x
0
−x
∗
T
Qx
0
−x
∗
=
1
2
n
i=1
i...
... method and greatly reduces the computation
required at each step.
Example. Consider the 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
... −310
0000
⎤
⎥
⎥
⎦
(22 )
The gradient at the point (2, 2, 1, 0) is g = 2 4 2 −3 and hence we find
d =−Pg =
1
11
−8 24 −8 0
12. 4 The Gradient Proje...