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OPTIMASI MULTIVARIABEL DENGAN KENDALA KESAMAAN

Fungsi kontinu  Min f(x) Kendala g j = 0, di mana j = 1, 2, ... m. Vektor R n , syarat m  n. OPTIMASI MULTIVARIABEL DENGAN KENDALA KESAMAAN. Penyelesaian dengan 3 cara :. Metode Substitusi Langsung a. nyatakan n variabel dengan (n-m) variabel lain

cynthia-nolan
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OPTIMASI MULTIVARIABEL DENGAN KENDALA KESAMAAN

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  1. Fungsi kontinu  Min f(x) Kendala gj = 0, di mana j = 1, 2, ... m Vektor Rn, syarat m  n ... OPTIMASI MULTIVARIABEL DENGAN KENDALA KESAMAAN

  2. Penyelesaian dengan 3 cara : • Metode Substitusi Langsung a. nyatakan n variabel dengan (n-m) variabel lain b. substitusikan m kendala ke fungsi tujuan, fungsi tujuan mengandung n-m variabel c. selesaikan optimasi n-m variabel tanpa kendala Contohnya

  3. = ½(4x1+2(1-2x1)(-2)) = 2x1-2(1-2x1) = 0  x1-(1-2x1)= 0  x1-1+2x1= 0  3x1= 1 x1= 1/3, x2= 1/3 x3= 1-2/3 = 1/3 = 2+4=6>0, f(1/3,1/3,1/3)= 1/6  Optimum Minimum Contoh: Minimumkan f(x1, x2, x3)= ½(x12+ x22+ x32) Kendala g1= x1- x2 = 0 g2= x1+x2+x3-1 Jawab: g1= 0 x1= x2 g2= 0 x1+x1+x3-1=0  x3=1-2x1 f(x1, x2, x3)= ½(x12+ x12+ (1-2x1)2)=½(2x12+(1-2x1)2) f(x1)

  4. 2. Metode Constrained Variation a. Untuk n = 2, m = 1 agar f(x1*, x2*) merupakan optimum x1*, x2* harus memenuhi ... ... b. Syarat perlu agar f(x*) merupakan optimum Contohnya

  5. Minimumkan f = 9-8x1- 6x2- 4x3+ 2x12+2x22+ x32+ 2x1 x2+2x1x3 Kendala x1+x2+2x3= 3 Jawab: n = 3, m = 1, Ambil y3 = x3, y2 = x2 sehingga y1= x1 k = m + 1 = 2 -6+4x2+ 2x1 -8+ 4x1+2x2+2x3 1 1 = -6+4x2+2x1+8-4x1-2x2-2x3 = 2+2x2+2x1-2x3 = 0 x1 – x2 + x3 – 1 = 0 ...(1) Lanjutkan

  6. k = m + 2 = 3 = n -4+2x3+2x1 -8+ 4x1+2x2+2x3 2 1 = -4+2x3+2x1+16-8x1-4x2-4x3 = 12-2x3-6x1-4x2 = 2(6-x3-3x1-2x2)= 0 3x1+2x2+x3-6= 0 ... (2) 3x1+2x2+x3-6 = 0 ... (2) x1 – x2+x3–1 = 0 ...(1) (-) 2x1+ 3x2 - 6 = 0 x2= 5 - 2x1 3

  7. x1 – x2+x3–1 = 0 x1 – +x3–1 = 0 3x1–5+2x1+3x3–3= 0 5x1+3x3– 8= 0 x1 – x2+2x2 = 3 x1 – +2 = 3 3x1 + 5 - 2x1 + 16 - 10x1 = 9 5 - 2x1 5 - 2x1 8 - 5x1 x3 = 8 - 5x1 9x1= 12 3 9 9 3 3 3 3 x1 = 4 ; x3 = 4 x2 = 7 ;

  8. < 0  max > 0  min Catatan: Metode tersebut juga berlaku untuk n variabel bebas dan m restriksi (n + m persamaan) 3. Metode Multiplikator Jika titik-titik ekstrem dari fungsi Z = f(x;y) harus ditentukan dengan restriksi (x;y)=0, maka berlaku persyaratan sebagai berikut: (x;y)=0; Penentuan Titik Ekstrem Contohnya

  9. Jawab : Xy – 9 = 0 2x + y + y= 0 x + 2y + x= 0 x1;2 =  3 y1;2 =  3 = - 3 Nilai ekstrem adalah P1 (3;3;27) & P2 (-3;-3;-27) Contoh: Fungsi Z = f(x;y) = x2 + xy + y2 Restriksi : (x;y) = xy – 9 = 0 Tentukan titik ekstrimnya!

  10. Oleh karena itu: Untuk P1 berlaku:  = 2*9 – 2(1-3)*3*3 + 2*9 = 72 > 0  Minimum Untuk P2 berlaku:  = 2*9 – 2(1+3)*(-3)*(-3) + 2*9 = -36 > 0  Maksimum Penentuan Jenis Titik Ekstrem :

  11. Metode Lagrange Prinsipnya adalah menambahkan satu variabel  • Syarat perlu agar f(x1*,x2*) merupakan jawaban masalah optimasi Minimasi f(x1*,x2*) Kendala g(x1,x2) = 0 adalah 

  12. Definit Positif Dimana i = 1, 2, ... n, j  0, j  ji J1 = kendala aktif J2 = kendala tidak aktif • Misalkan x suatu vektor masalah optimasi f(x) terhadap kendala g(x) = 0 didapat dengan f(x) = g(x), dan g(x) = 0 L(xi,) = f(xi) + g(xi) Vektor x, y memenuhi persamaan tersebut = titik kritis • Syarat cukup agar f(x*) merupakan minimum relatif • Optimasi multivariabel dengan kendala pertidaksamaan • Prinsipnya adalah menambah variable slack tak negatif yj2 sehinggaminimum f(x) dan kendala gj(x)  0, j = 1, 2, ... m menjadi Gj(x,y) = gj(x) + yj2 = 0, j = 1, 2, ... m • Titik x* dimana f(x*) minimum dengan syarat Kuhn-Tucker

  13. , i = 1, 2, ... n , j = 1, 2, ... m , j = 1, 2, ... m , j = 1, 2, ... m Jika kumpulan kendala aktif tidak diketahui, maka: Contoh: • Minimasi f(x1, x2, x3) = x12+x22+ x3+40x1+20 x2-3000 • Kendala g1= x1-50  0 • g2= x1+x2-100  0 • g3= x1+x2+x3-150  0

  14. Syarat Kuhn-Tucker • 2x1+40+ 1+ 2+ 3 = 0 2x2+20+ 2+ 3 = 0 2x3+ 3 = 0 jgj = 0, j = 1, 2, 3 • 1(x1-50) = 0 2(x1+x2-100) = 0 3(x1+x2+x3-150) = 0 j  0, j = 1, 2, 3, 1  0, 2  0, 3  0 Dari 1(x1-50) = 0  1 = 0 atau x1 = 50 Contoh: • Minimasi f(x1, x2, x3) = x12+x22+ x3+40x1+20 x2-3000 • Kendala g1= x1-50  0 • g2= x1+x2-100  0 • g3= x1+x2+x3-150  0

  15. (i) Jika x1 = 50  2x1+40+ 1+ 2+ 3 = 0 2x2+20+ 2+ 3 = 0 2x3+ 3 = 0 • 3 = -2x3 2 = -20-2x2-3 = -20-2x2+2x3 1 = -40-2x1-2-3 = -120+2x2 Substitusi: 2(x1+x2-100) = 0 3(x1+x2+x3-150) = 0 Sehingga: (-20-2x2+2x3)(x1+x2-100) -2x3 (x1+x2+x3-150) = 0 • -20-2x2+2x3 = 0, x1+x2+x3-150 • -10-x2+x3 = 0 • 50+x2+x3-150 = 0 • 90-2x2= 0 • x2= 0 • x1= 50, x2= 45 melanggar x1+x2 100 x3 = 150-x1-x2 x3 = 150-50-45 = 55 x1= 50, x2= 45, x3 = 55 Sistem ini mempunyai 4 jawaban, yaitu: -

  16. 3= -2x3 = 100, 2= -20-100+100= -20, 1= -120+100 = -20 • 1= -20, 2= -20, 3= -100 x1= 50, x2= 50, x3= 50 Sehingga titik optimum : x1*= x2* = x3* = 50 • 2. -20-2x2+2x3 = 0, -2x3 = 0  x3 = 0, x2 = -10 • x1= 50, x2= -10, x3 = 0, melanggar x1+x2  100 • x1+x2-100 = 0, -2x3 = 0  x3 = 0, 50+x2 = 100  x2 = 50 • x1= 50, x2= 50, x3 = 0, melanggar x1+x2+x3  150 • x1+x2-100 = 0, x1+x2+x3 – 150 = 0 • 50+x2 = 100, 50+50+x3 =150 • x2 = 50 x3 = 50 • x1= 50, x2= 50, x3 = 50 Jawaban ini memenuhi kendala

  17. 2. Maksimumkan f(x1, x2) = x1x2, x1>0, x2>0 Kendala g(x1,x2)= x12+x22-4 = 0 Jawab: L(x1,x2,) = x1x2 + (x12+x22-4) (1) dan (2) -x2/x1 = -x1/x2 x22 = x12 (3) x12+x22-4 = 0 2x12 = 4 x12 = 2 x1 = 2, x2 = 2  = 1/2

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