Imroatul Hudati (07111750020003) Tugas 1
Imroatul Hudati (07111750020003) Tugas 1
Imroatul Hudati (07111750020003) Tugas 1
TUGAS 1
1.1-2 Find the minimum value of
−2𝑥1 + 4𝑥2 = 0
−2𝑥1 + 4(−1) = 0
−2𝑥1 − 4 = 0
−2𝑥1 = 4
𝑥1 = −2
Diperoleh
𝑥2 −1
𝑥 ∗ = [𝑥 ] = [ ]
1 −2
Curvatur matrix didapatkan dari Jacobian matrix
2 1
𝐻(𝑥) = [ ]
2 4
|𝐻1 | = 2 > 0
|𝐻2 | = (2)(4) − (2)(1) = 6 > 0
Nilai tiap minor dari matriks Hessian adalah definit posistif sehingga menunjukkan titik
optimal tersebut merupakan titik minimum.
Imroatul Hudati (07111750020003)
0
X2
-1
-2
-3
-4
-5
-6
-6 -5 -4 -3 -2 -1 0 1 2 3 4
X1
Contour
100
80
60
40
20
-20
5
4
0
2
0
-5 -2
-4
X2 -10 -6
X1
Gradient
Imroatul Hudati (07111750020003)
1 0
𝑥 ∗ = − ([ ]
0 1
−2 −2 2 1
−[ ] ([ ]
−1 0 1 1
−1
−2 −2 𝑇 1 0 −2 −2 −2 −2 𝑇 1 0 −1
+[ ] [ ][ ]) [ ] [ ]) [ ]
−1 0 0 2 −1 0 −1 0 0 2 −3
1 0 −2 −2 2 1 6 4 −1 −2 −2 −1
𝑥 ∗ = − ([ ]−[ ] ([ ]+[ ]) [ ]) [ ]
0 1 −1 0 1 1 4 4 −2 0 −3
1 1
−
1 0 −2 −2 3 3] [−2 −2]) [−1]
𝑥 ∗ = − ([ ]−[ ][
0 1 −1 0 1 8 −2 0 −3
−
3 15
2
0 −
1 0 5] [−2 −2]) [−1]
𝑥 ∗ = − ([ ]−[
0 1 1 1 −2 0 −3
−
3 3
4
0
1 0 −1
𝑥 ∗ = − ([ ] − [5 ]) [ ]
0 1 2 −3
0
3
∗ −0,2
𝑥 =[ ]
−1
3. 𝜆∗ = (𝑄 − 𝑄𝐵(𝑅 + 𝐵𝑇 𝑄𝐵)−1 𝐵𝑇 𝑄)𝑐
1 0 1 0 −2 −2 2 1
𝜆∗ = ([ ]−[ ][ ] ([ ]
0 2 0 2 −1 0 1 1
−1
−2 −2 𝑇 1 0 −2 −2 −2 −2 𝑇 1 0 −1
+[ ] [ ][ ]) [ ] [ ]) [ ]
−1 0 0 2 −1 0 −1 0 0 2 −3
1 0 −2 −2 2 1 6 4 −1 −2 −2 −1
𝜆∗ = ([ ]−[ ] ([ ]+[ ]) [ ]) [ ]
0 2 −2 0 1 1 4 4 −2 0 −3
1 1
−
1 0 −6 −4 3 3] [−2 −2]) [−1]
𝜆∗ = ([ ]−[ ][
0 2 −4 −2 1 8 −2 0 −3
−
3 15
2 2
− −
1 0 15] [−2 −2]) [−1]
𝜆∗ = ([ ]−[ 3
0 2 2 4 −2 0 −3
−
3 15
8 4
1 0 −1
𝜆∗ = ([ ] − [5 3]) [ ]
0 2 4 4 −3
5 3
0,2
𝜆∗ = [ ]
2
1
4. 𝐿∗ = 2 𝑐 𝑇 λ
1 −1 𝑇 0,2
𝐿∗ = [ ] [ ]
2 −3 2
𝐿∗ = 3,1