Ларин Р.М., Пяткин А.В., Плясунов А.В. Методы оптимизации примеры и задачи
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A
σ → min,
< c, p > −σ ≡ p
3
− σ ≤ 0,
< ϕ
0
3
(x
2
), p > −σ ≡ −0.3p
1
− 4p
2
− p
3
− σ ≤ 0,
−p
1
− 1 ≤ 0, p
1
− 1 ≤ 0, −p
2
− 1 ≤ 0, p
2
− 1 ≤ 0.
p
2
= (1, 1, −2.15)
>
, σ
2
= −2.15 < −δ
2
, δ
3
= δ
2
= 0.5
α
2
α
2
= 0.36 ϕ
2
(x
2
+ α
2
p
2
) ≈ 0
x
3
= x
2
+ α
2
p
2
= (−0.29, 1.01, −0.57)
>
, J(x
3
, δ
3
) = {1, 2},
ϕ
1
(x
3
) = −0.32 > −δ
3
, ϕ
2
(x
3
) = −0.03 > −δ
3
, ϕ
3
(x
3
) = −0.68 < −δ
3
A
σ → min,
< c, p > −σ ≡ p
3
− σ ≤ 0,
< ϕ
0
1
(x
3
), p > −σ ≡ −0.56p
1
+ 2.46p
2
− p
3
− σ ≤ 0,
< ϕ
0
2
(x
3
), p > −σ ≡ −1.58p
1
+ 3.02p
2
− p
3
− σ ≤ 0,
−p
1
− 1 ≤ 0, p
1
− 1 ≤ 0, −p
2
− 1 ≤ 0, p
2
− 1 ≤ 0.
p
3
= (1, −1, −1.51)
>
, σ
3
= −1.51 < −δ
3
δ
4
= δ
3
= 0.5
α
3
α
3
= 0.12 ϕ
3
(x
3
+ α
3
p
3
) = 0
x
4
= x
3
+ α
3
p
3
= (−0.17, 0.89, −0.75)
>
, J(x
4
, δ
4
) = {1, 2, 3},
ϕ
1
(x
4
) = −0.49 > −δ
4
, ϕ
2
(x
4
) = −0.37 > −δ
4
, ϕ
3
(x
4
) = 0
A
σ → min,
< c, p > −σ ≡ p
3
− σ ≤ 0,
< ϕ
0
1
(x
4
), p > −σ ≡ −0.56p
1
+ 2.22p
2
− p
3
− σ ≤ 0,
< ϕ
0
2
(x
4
), p > −σ ≡ −1.34p
1
+ 2.78p
2
− p
3
− σ ≤ 0,
< ϕ
0
3
(x
4
), p > −σ ≡ 0.66p
1
− 4p
2
− p
3
− σ ≤ 0,
−p
1
− 1 ≤ 0, p
1
− 1 ≤ 0, −p
2
− 1 ≤ 0, p
2
− 1 ≤ 0.
σ
4
= 0 B
I(x
4
) = {3}
σ → min,
< c, p > −σ ≡ p
3
− σ ≤ 0,
< ϕ
0
3
(x
4
), p > −σ ≡ 0.66p
1
− 4p
2
− p
3
− σ ≤ 0,
−p
1
− 1 ≤ 0, p
1
− 1 ≤ 0, −p
2
− 1 ≤ 0, p
2
− 1 ≤ 0.
p
4
= (−1, 1, −2.33)
>
σ
4
= −2.33 < 0 δ
5
= δ
4
/2 =
0.25, p
4
= p
4
α
4
x
1
= −0.17 − α, x
2
= 0.89 + α, x
3
= −0.75 − 2.33α (α > 0).
ϕ
j
(x
4
+ αp
4
) = 0 (j = 1, 2, 3)
α
1
= 0.06 ϕ
2
(x
4
+ α
4
p
4
) = 0
x
5
= x
4
+ α
4
p
4
= (−0.23, 0.95, −0.88)
>
, J(x
5
, δ
5
) = {1, 2, 3},
ϕ
1
(x
5
) = −0.19 > −δ
5
, ϕ
2
(x
5
) = 0 > −δ
5
, ϕ
3
(x
5
) = −0.1 > −δ
5
A
σ → min,
< c, p > −σ ≡ p
3
− σ ≤ 0,
< ϕ
0
1
(x
5
), p > −σ ≡ −0.56p
1
+ 2.34p
2
− p
3
− σ ≤ 0,
< ϕ
0
2
(x
5
), p > −σ ≡ −1.46p
1
+ 2.90p
2
− p
3
− σ ≤ 0,
< ϕ
0
3
(x
5
), p > −σ ≡ −0.54p
1
− 4p
2
− p
3
− σ ≤ 0,
−p
1
− 1 ≤ 0, p
1
− 1 ≤ 0, −p
2
− 1 ≤ 0, p
2
− 1 ≤ 0.
p
5
= (1, 0.18, −0.08)
>
, σ
5
= −0.08 > −δ
5
δ δ
6
= δ
5
/2 = 0.125
α
5
x
1
= −0.23 + α, x
2
= 0.95 + 0.18α, x
3
= −0.88 − 0.08α (α > 0).
ϕ
j
(x
5
+ αp
5
) = 0 (j = 1, 2, 3)
α
5
= 0.34 ϕ
3
(x
5
+ α
5
p
5
) = 0
x
6
= x
5
+ α
5
p
5
= (0.11, 1.01, −0.91)
>
, J(x
6
, δ
6
) = {1, 3},
ϕ
1
(x
6
) = −0.05 > −δ
5
, ϕ
2
(x
6
) = −0.16 < −δ
5
, ϕ
3
(x
6
) = 0 > −δ
5
p
6
= (−1, −0.15, −0.26)
>
δ
7
= δ
6
= 0.125 α
6
= 0.2 x
7
= (−0.09, 0.98, −0.96)
>
p
7
= (0.3, −0.03, −0.18)
>
δ
8
= δ
7
= 0.125 α
7
= 0.09
x
8
= (−0.06, 0.98, −0.98)
>
x
∗
= (0, 1, −1)
>
{x
k
}
ϕ
1
(x
∗
) = 0, ϕ
2
(x
∗
) = 0, ϕ
3
(x
∗
) = 0,
x
∗
B
σ → min,
< c, p > −σ ≡ p
3
− σ ≤ 0,
< ϕ
0
1
(x
∗
), p > −σ ≡ 3p
2
− p
3
− σ ≤ 0,
< ϕ
0
2
(x
∗
), p > −σ ≡ −p
1
+ 3p
2
− p
3
− σ ≤ 0,
< ϕ
0
3
(x
∗
), p > −σ ≡ p
1
− 4p
2
− p
3
− σ ≤ 0,
−p
1
− 1 ≤ 0, p
1
− 1 ≤ 0, −p
2
− 1 ≤ 0, p
2
− 1 ≤ 0.
min
σ,p
σ = σ
∗
= 0 x
∗
