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Bagian 2

Metode Komputasi. Bagian 2. Metode Gradient Descent/Ascent. Dosen :. Deni Saepudin : Ruang C114 Telp . +628122086193. Gradien Descent. Diketahui permukaan z = f( x,y ) dengan kurva ketinggian dinyatakan pada gambar

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Bagian 2

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  1. MetodeKomputasi Bagian2 Metode Gradient Descent/Ascent Dosen: DeniSaepudin : Ruang C114 Telp. +628122086193

  2. Gradien Descent • Diketahuipermukaan z = f(x,y) dengankurva ketinggiandinyatakan padagambar • Berangkatdarititik (x0, y0), nilai f menurun paling cepatdalamarah-f(x0,y0) • Titik minimum (x1,y1) tercapaisaatf(x1,y1)=(0, 0)

  3. Langkah-langkah • Berdasarkanprinsipkalkulus, pencariantitik minimum dapatdilakukandenganlangkah-langkahberikut: • Masukan : f(x,y), titikawal (x0,y0), ukuranlangkah, stopping kriteria  • Langkah-langkah (untukpeminimuman): Selama |f(x0, y0)|> masihberlaku (i) Hitungf(x0, y0) (ii) (x1,y1)= (x0,y0) - f(x0, y0) (iii) (x0,y0)=(x1,y1)

  4. Contoh 1: • Mencarititik minimum dari f(x,y) = x2+2y2 • Titikawal P0(x0,y0)=(2,1), ukuranlangkah = 0.1 • ∇f(x,y) = (2x, 4y) • Nilaifungsi di titikawalf(2,1) = 5. • Arahgerak agar nilai f menurun paling cepat di titik P0adalah -∇f(2,1) = (-4,-4) • Titikbaru (x1,y1) = (2,1) + (-4,-4) (iterasi 1) = (2,1) + 0.1(-4,-4) = (1.6, 0.6) • Nilaif(1.6, 0.6) = 2.56 + 2 0.36 = 2.56 + 0.72 = 3.28 • (x0,y0) = (1.6, 0.6)

  5. Contoh1 (lanjutan) • (x1,y1) = (1.6, 0.6) - ∇f(1.6,0.6) (iterasi 2) • = (1.6, 0.6) – 0.1(3.2, 2.4) = (1.28, 0.36) • Nilai f(1.28, 0.36) = 1.8976 • (x0,y0) = (1.28, 0.36) • Ulangiterus proses sampaikriteriapenghentiandicapai

  6. Latihan: • TerapkanmetodeGradient Descent untukmencarititik minimum darifungsi • f(x,y) = x2 - 4x + y2 + 6y + 8 • TitikAwal : (x0, y0) = (10, 5)

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