METODE ENUMERASI IMPLISIT

Pendahuluan Merupakan metode integer programming (IP) yang pada dasarnya hampir mirip dengan metode knapsack. Semua variabel keputusan harus berharga 0 atau 1

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

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

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 ≥

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)

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

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)

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 -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)

Node 1 1 X1 = 0 X1 = 1 2 3

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)

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)

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)

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)

Node 3 3 X2 = 0 X2 = 1 4 5

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)

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)

Node 4 4 X3 = 0 X3 = 1 6 7

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)

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)

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 -3 ≤ -3 (F) P2 : -4(1) – 2(0) – 4(1) + (0) + 2(0) ≤ -7 -10 ≤ -7 (F)

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)

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

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)

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 -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)

Node 5 5 X3 = 0 X3 = 1 8 9

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)

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)

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 -5 ≤ -3 (F) P2 : -4(1) – 2(1) – 4(1) + (0) + 2(0) ≤ -7 -10 ≤ -7 (F)

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 -8 ≤ -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)

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

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