Network Model 1 DR Rahma Fitriani, S.Si., M.Sc., Riset Operasi 2011 Semester Genap 2011/2012

Shortest Path Problem Pengiriman dari titik ke titik Supply, transhipment (substation), dan demand nodes Shortest path problem – Biaya proportional dengan jarak – Masalah pemilihan jarak terpendek (biaya minimum)

Contoh: Sumber Tujuan

Algoritma Djikstra Check the file in MS words..

∞∞ ∞ ∞∞ Distance label Temporary ={1, 2, 3, 4, 5, 6} Permanent={ }

∞∞ ∞ ∞∞ Distance label Permanen Temporary ={2, 3, 4, 5, 6} Permanent={1 }

∞∞ ∞ ∞∞ Distance label Permanen Temporary ={2, 3, 4, 5, 6} Permanent={1 }

∞ ∞ 3∞ Temporary Distance label Permanen Temporary ={2, 3, 4, 5, 6} Permanent={1 }

∞ ∞ 3∞ Temporary Distance label Permanen Temporary ={2, 4, 5, 6} Permanent={1, 3 }

∞ ∞ 3∞ Temporary Distance label Permanen Temporary ={2, 4, 5, 6} Permanent={1, 3 }

∞ ∞ 36 Temporary Distance label Permanen Temporary ={2, 4, 5, 6} Permanent={1, 3 }

∞ ∞ 36 Temporary Distance label Permanen Temporary ={4, 5, 6} Permanent={1, 2,3 }

∞ ∞ 36 Temporary Distance label Permanen Temporary ={4, 5, 6} Permanent={1, 2,3 }

∞ 36 Temporary Distance label Permanen Temporary ={4, 5, 6} Permanent={1, 2,3 }

∞ 36 Temporary Distance label Permanen Temporary ={4, 5, 6} Permanent={1, 2,3 }

∞ 36 Temporary Distance label Permanen Temporary ={4, 6} Permanent={1, 2,3, 5 }

Temporary Distance label Permanen Temporary ={4, 6} Permanent={1, 2,3, 5 }

Temporary Distance label Permanen Temporary ={4, 6} Permanent={1, 2,3, 5 }

Temporary Distance label Permanen Temporary ={6} Permanent={1, 2,3, 4, 5 }

Min (9,8)=8 36 Temporary Distance label Permanen Temporary ={6} Permanent={1, 2,3, 4, 5 }

Temporary Distance label Permanen Temporary ={ } Permanent={1, 2,3, 4, 5, 6 }

Temporary Distance label Permanen Shortest path: 1 – 2 – 5 – 6

Shortest Path sebagai Transhipment Problem Transhipment problem dengan setiap demand dan supply sama dengan 1 Jalur yang tidak terdefinisi dikenai biaya besar Biaya nol untuk jalur dari node i ke node i

Cost23456Supply Demand

Model LP shortest path sbg transhipment problem

Solusi optimal Contoh: Sumber Tujuan Total distance (cost) = 8

Max Flow Problem Model network di mana kapasitas jalur diperhitungkan Tujuan: Memaksimumkan jumlah pengiriman dari source ke destination dengan kendala kapasitas setiap jalur

Contoh: dengan kapasitas setiap jalur S D a0 a0 jalur buatan untuk conservation flow, outflow = inflow

LP untuk max flow problem S D a0 See Excell Transhipme nt.xlsx Transhipme nt.xlsx

Solusi optimal max flow S D 2(1) 3(2) 3(0) 4(1) 1(1) 2(2) a0 x0xs1xs2x12x13x2dx3d Dari Excel