Практикум по методам оптимизации. Семушин И.В. - 123 стр.

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1 2 3 4
(1) b
0T
= 1 2 1/2 7/2
k = 2.
a
0T
k
(2) 0 4
=
β
T
k
0 1 0 1
G
2 3
2 6
1 1
2 2
; a
0
k
[NF (2)] = 4 < 0.
min
j=NF (2)
(c
0T
, π
T
)
| a
0
kj
|
= min
1
| 4 |
=
1
4
l = 2, s = 2, c
0
s
= 1.
(3) a
0
s
= 1 0 0 1
0 1 0 1
0 0 1 1/2
0 0 0 1/2
3
6
1
2
=
1
4
0
1
; a
0
ks
= 4.
1 2 3 4
(a) NB = 3 2 5 1
1 2
NF = 6 4
1 2 3 4
(b) b
0T
= 3/2 1/2 1/2 3
z
+
= 7/2 1 1/2 = 4
(d) B
1
= 1 1/4 0 3/4
0 1/4 0 1/4
0 0 1 1/2
0 1/4 0 3/4