
Quadratic Programming 279
and
α
max
z
= min{1, min
i:(d
z
)
i
<0
{−
z
(0)
i
(d
z
)
i
}}
= min{1,
1
0.6133
,
1
0.9202
} = 1,
and so
α = min{1, 0.95α
max
x
, 0.95α
max
z
} = 0.8854,
then
x
(1)
= x
(0)
+ αd
x
=
1
1
1
+ (0.8854)
−1.0730
−0.6265
−0.3005
=
0.0499
0.4453
0.7339
π
(1)
= π
(0)
+ αd
π
=
1
1
+ (0.8854)
−26.2540
0.2771
=
−22.2453
1.2453
z
(1)
= z
(0)
+ αd
z
=
1
1
1
+ (0.8854)
0.1796
−0.6133
−0.9202
=
1.1590
0.4570
0.1853
.
Check the stopping condition for point (x
(1)
, π
(1)
, z
(0)
)
T
max
Ax
(1)
− b
−Qx
(1)
+ A
T
π
(1)
+ z
(1)
(x
(1)
)
T
z
(1)
= max
0.2300
0.4083
0.3974
= 0.4083 > ε,
so another iteration should be performed. After 8 iterations the tolerance is
satisfied and so the interior point metho