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METODE ENUMERASI IMPLISIT
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Pendahuluan Merupakan metode integer programming (IP) yang pada dasarnya hampir mirip dengan metode knapsack. Semua variabel keputusan harus berharga 0 atau 1
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Perbandingan Metode IP
Branch and Bound Knapsack Enumerasi Implisit Variabel Keputusan = 2 Variabel Keputusan ≥ 2 Fungsi Pembatas ≥ 1 Fungsi Pembatas = 1 Nilai VK bernilai semua bilangan real dan memiliki arti sebenarnya Nilai VK bernilai 0 atau 1
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Prosedur Metode Enumerasi Implisit (1)
Melakukan penyempurnaan terbaik bagi suatu node : Input harga setiap variabel keputusan kepada fungsi pembatas untuk menentukan apakah fisible atau tidak
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Prosedur Metode Enumerasi Implisit (2)
2. Menguji fisibilitas dari semua fungsi pembatas Jenis Pembatas Tanda pada koefesien variabel pada pembatas Nilai pada variabel pembatas ≤ + - 1 ≥
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Aturan pencabangan node
Jika langkah 1 didapatkan hasil fisible dan langkah 2 tidak fisible atau sebaliknya,maka lakukan pencabangan pada node tersebut Jika langkah 1 dan 2 fisible, maka node berhenti (calon solusi) Jika langkah 1 dan 2 tidak fisible, maka node berhenti (fathomed)
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Contoh : Maks Z = -7X1 – 3X2 – 2X3 – X4 – 2X5 s/t -4X1 – 2X2 + X3 – 2X4 – X5 ≤ -3 -4X1 – 2X2 - 4X3 + X4 + 2X5 ≤ -7 Xi = 0 atau 1
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Node 1 Langkah 1 : Penyempurnaan terbaik : X1 = 0 X2 = 0 X3 = 0 X4 = 0 X5 = 0 P1 : -4(0) – 2(0) + (0) – 2(0) – (0) ≤ -3 0 ≤ -3 (TF) P2 : -4(0) – 2(0) – 4(0) + (0) + 2(0) ≤ -7 0 ≤ -7 (TF)
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Node 1 Langkah 2 : Uji Fisibilitas Pembatas P1 : X1 = 1 X2 = 1 X3 = 0 X4 = 1 X5 = 1 -4(1) – 2(1) + (0) – 2(1) – (1) ≤ ≤ -3 (F) P2 : X1 = 1 X2 = 1 X3 = 1 X4 = 0 X5 = 0 -4(1) – 2(1) – 4(1) + (0) + 2(0) ≤ ≤ -7 (F)
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Node 1 1 X1 = 0 X1 = 1 2 3
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Node 2 Langkah 1 : Penyempurnaan terbaik :
X1 = 0 X2 = 0 X3 = 0 X4 = 0 X5 = 0 P1 : -4(0) – 2(0) + (0) – 2(0) – (0) ≤ -3 0 ≤ -3 (TF) P2 : -4(0) – 2(0) – 4(0) + (0) + 2(0) ≤ -7 0 ≤ -7 (TF)
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Node 2 Langkah 2 : Uji Fisibilitas Pembatas
P1 : X1 = 0 X2 = 1 X3 = 0 X4 = 1 X5 = 1 -4(0) – 2(1) + (0) – 2(1) – (1) ≤ -3 -5 ≤ -3 (F) P2 : X1 = 0 X2 = 1 X3 = 1 X4 = 0 X5 = 0 -4(0) – 2(1) – 4(1) + (0) + 2(0) ≤ -7 -6 ≤ -7 (TF)
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Node 3 Langkah 1 : Penyempurnaan terbaik :
X1 = 1 X2 = 0 X3 = 0 X4 = 0 X5 = 0 P1 : -4(1) – 2(0) + (0) – 2(0) – (0) ≤ -3 -4 ≤ -3 (F) P2 : -4(1) – 2(0) – 4(0) + (0) + 2(0) ≤ -7 -4 ≤ -7 (TF)
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Node 3 Langkah 2 : Uji Fisibilitas Pembatas
P1 : X1 = 1 X2 = 1 X3 = 0 X4 = 1 X5 = 1 -4(1) – 2(1) + (0) – 2(1) – (1) ≤ -3 -9 ≤ -3 (F) P2 : X1 = 1 X2 = 1 X3 = 1 X4 = 0 X5 = 0 -4(1) – 2(1) – 4(1) + (0) + 2(0) ≤ -7 -10 ≤ -7 (F)
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Node 3 3 X2 = 0 X2 = 1 4 5
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Node 4 Langkah 1 : Penyempurnaan terbaik :
X1 = 1 X2 = 0 X3 = 0 X4 = 0 X5 = 0 P1 : -4(1) – 2(0) + (0) – 2(0) – (0) ≤ -3 -4 ≤ -3 (F) P2 : -4(1) – 2(0) – 4(0) + (0) + 2(0) ≤ -7 -4 ≤ -7 (TF)
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Node 4 Langkah 2 : Uji Fisibilitas Pembatas
P1 : X1 = 1 X2 = 0 X3 = 0 X4 = 1 X5 = 1 -4(1) – 2(0) + (0) – 2(1) – (1) ≤ -3 -7 ≤ -3 (F) P2 : X1 = 1 X2 = 0 X3 = 1 X4 = 0 X5 = 0 -4(1) – 2(0) – 4(1) + (0) + 2(0) ≤ -7 -8 ≤ -7 (F)
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Node 4 4 X3 = 0 X3 = 1 6 7
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Node 6 Langkah 1 : Penyempurnaan terbaik :
X1 = 1 X2 = 0 X3 = 0 X4 = 0 X5 = 0 P1 : -4(1) – 2(0) + (0) – 2(0) – (0) ≤ -3 -4 ≤ -3 (F) P2 : -4(1) – 2(0) – 4(0) + (0) + 2(0) ≤ -7 -4 ≤ -7 (TF)
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Node 6 Langkah 2 : Uji Fisibilitas Pembatas
P1 : X1 = 1 X2 = 0 X3 = 0 X4 = 1 X5 = 1 -4(1) – 2(0) + (0) – 2(1) – (1) ≤ -3 -7 ≤ -3 (F) P2 : X1 = 1 X2 = 0 X3 = 0 X4 = 0 X5 = 0 -4(1) – 2(0) – 4(0) + (0) + 2(0) ≤ -7 -4 ≤ -7 (TF)
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Node 7 Langkah 1 : Penyempurnaan terbaik : X1 = 1 X2 = 0 X3 = 1 X4 = 0 X5 = 0 P1 : -4(1) – 2(0) + (1) – 2(0) – (0) ≤ ≤ -3 (F) P2 : -4(1) – 2(0) – 4(1) + (0) + 2(0) ≤ ≤ -7 (F)
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Node 7 Langkah 2 : Uji Fisibilitas Pembatas
P1 : X1 = 1 X2 = 0 X3 = 1 X4 = 1 X5 = 1 -4(1) – 2(0) + (1) – 2(1) – (1) ≤ -3 -6 ≤ -3 (F) P2 : X1 = 1 X2 = 0 X3 = 1 X4 = 0 X5 = 0 -4(1) – 2(0) – 4(1) + (0) + 2(0) ≤ -7 -8 ≤ -7 (F)
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Node 7 Merupakan calon solusi yang didapatkan
Masukkan nilai variabel keputusan pada penyempurnaan terbaik X1 = 1 X2 = 0 X3 = 1 X4 = 0 X5 = 0 Z = -7(1) – 3(0) – 2(1) – (0) – 2(0) = -9
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Node 5 Langkah 1 : Penyempurnaan terbaik :
X1 = 1 X2 = 1 X3 = 0 X4 = 0 X5 = 0 P1 : -4(1) – 2(1) + (0) – 2(0) – (0) ≤ -3 -6 ≤ -3 (F) P2 : -4(1) – 2(1) – 4(0) + (0) + 2(0) ≤ -7 -6 ≤ -7 (TF)
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Node 5 Langkah 2 : Uji Fisibilitas Pembatas P1 : X1 = 1 X2 = 1 X3 = 0 X4 = 1 X5 = 1 -4(1) – 2(1) + (0) – 2(1) – (1) ≤ ≤ -3 (F) P2 : X1 = 1 X2 = 1 X3 = 1 X4 = 0 X5 = 0 -4(1) – 2(1) – 4(1) + (0) + 2(0) ≤ ≤ -7 (F)
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Node 5 5 X3 = 0 X3 = 1 8 9
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Node 8 Langkah 1 : Penyempurnaan terbaik :
X1 = 1 X2 = 1 X3 = 0 X4 = 0 X5 = 0 P1 : -4(1) – 2(1) + (0) – 2(0) – (0) ≤ -3 -6 ≤ -3 (F) P2 : -4(1) – 2(1) – 4(0) + (0) + 2(0) ≤ -7 -6 ≤ -7 (TF)
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Node 8 Langkah 2 : Uji Fisibilitas Pembatas
P1 : X1 = 1 X2 = 1 X3 = 0 X4 = 1 X5 = 1 -4(1) – 2(1) + (0) – 2(1) – (1) ≤ -3 -9 ≤ -3 (F) P2 : X1 = 1 X2 = 1 X3 = 0 X4 = 0 X5 = 0 -4(1) – 2(1) – 4(0) + (0) + 2(0) ≤ -7 -6 ≤ -7 (TF)
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Node 9 Langkah 1 : Penyempurnaan terbaik : X1 = 1 X2 = 1 X3 = 1 X4 = 0 X5 = 0 P1 : -4(1) – 2(1) + (1) – 2(0) – (0) ≤ ≤ -3 (F) P2 : -4(1) – 2(1) – 4(1) + (0) + 2(0) ≤ ≤ -7 (F)
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Node 9 Langkah 2 : Uji Fisibilitas Pembatas P1 : X1 = 1 X2 = 1 X3 = 1 X4 = 1 X5 = 1 -4(1) – 2(1) + (1) – 2(1) – (1) ≤ ≤ -3 (F) P2 : X1 = 1 X2 = 1 X3 = 1 X4 = 0 X5 = 0 -4(1) – 2(1) – 4(1) + (0) + 2(0) ≤ ≤ -7 (F)
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Node 9 Merupakan calon solusi yang didapatkan
Masukkan nilai variabel keputusan pada penyempurnaan terbaik X1 = 1 X2 = 1 X3 = 1 X4 = 0 X5 = 0 Z = -7(1) – 3(1) – 2(1) – (0) – 2(0) = -12
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1 X1= 0 X1= 1 3 2 X2= 0 X2= 1 4 5 X3= 1 X3= 0 X3= 1 X3= 0 6 7 8 9
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