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PERSOALAN TRANSPORTASI
Riset Operasi (7)
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Tujuan
Setelah menyelesaikan pertemuan ini mahasiswa diharapkan mampu :
Membuat model transportasi Menyelesaikan solusi awal model
transportasi dengan menggunakan metode northwest corner (pojok kiri atas), biaya/ongkos terkecil, vogel
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Persoalan Transportasi
Membahas: Pendistribusian produk dari sumber (supply) ke tujuan
(demand) Dengan tujuan meminimalkan ongkos
Ciri-ciri persoalan transportasi: Ada sejumlah sumber dan tujuan Ada ongkos pengangkutan dari sumber ke tujuan
Contoh:10 9 4
6 11 9
Pabrik roti A
Pabrik roti B
Kota P Kota Q Kota R
Sumber
Tujuan
100 200 300
400
200
Permintaan (demand)
Persediaan (supply)
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Model Transportasi
Keseimbangan Model Transportasi: Seimbang ∑ sumber = ∑ tujuan Jika tidak seimbang , maka
Ditambah dummy dengan ongkos = 0
10 9 4 0
6 11 9 0
Pabrik roti A
Pabrik roti B
Kota P Kota Q Kota R
Sumber
Tujuan
100 200 100
400
200
Permintaan (demand)
Persediaan (supply)
Dummy
200
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Contoh Kasus
MG Auto mempunyai 3 plants produksi:
Los Angeles memproduksi 1000 mobilDetroit memproduksi 1500 mobilNew Orleans memproduksi 1200 mobil
Pusat Distribusi :Denver membutuhkan 2300 mobilMiami membutuhkan 1400 mobil
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Contoh Kasus (2)
Jika kasus diatas dibentuk dalam tabel:
Denver (1) Miami (2)
Los Angeles (1) $80 $215
Detroit (2) $100 $108
New Orleans (3) $102 $68
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Contoh Kasus (3)
Tabel TransportasiDenver Miami Supply
Los Angeles
$80x11
$215x12
1000
Detroit $100x21
$108x22
1500
New Orleans
$102x31
$68x32
1200
Demand 2300 1400
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Penyelesaian dengan LP
Minimize z = 80x11+ 215x12+ 100x21+ 108x22+ 102x31 + 68x32
x11 + x12 = 1000 (Los Angeles)
x21 + x22 = 1500 (Detroit)
x31 + x32 = 1200 (New Orleans)
x11 + x21 + x31 = 2300 (Denver)
x12+ x22 + x32 = 1400 (Miami)
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Penyelesaian dengan Trans. Model Memasukkan Data dari Transportation
Tableau ke TORA. Hasil yang diperoleh sama.
x11 = 1000, x21 = 1300,
x22 = 200, x32 = 1200 Biaya Transportasi yang minimum =
(1000 x $80) + (1300 x $100) + (200 x $108) + (1200 x $68) = $313.200
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Optimal Solution
1000
1500
1200 1200
1300Los Angeles
Detroit
New Orleans
Denver
Miami
2300
1400200
1000
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TORA
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TORA
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Merubah Linier jadi transportasi Karena rumitnya persoalan diatas jika di
jawab dengan linier, maka muncul metode baru yaitu Transportasi
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Unbalanced Trans. Mode
Jika modelnya unbalanced (tidak seimbang), maka kita menambahkan dummy source atau dummy destination untuk menyeimbangkan
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Contoh Unbalanced Trans. Mode Bila supply < demand.
Misal Detroit hanya mensupply 1300 mobil. Maka Dummy source = 3700 – 3500 = 200
Denver Miami Supply
Los Angeles $80 $215 1000
Detroit $100 $108 1300
New Orleans $102 $68 1200
Dummy Source
$0 $0 200
Demand 2300 1400
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Penyelesaian Unbalanced Trans. Mode
Denver Miami Supply
Los Angeles $801000
$2151000
Detroit $1001300
$1081300
New Orleans $102 $681200 1200
Dummy Plant $0 $0200 200
Demand 2300 1400
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Contoh 2 Unbalanced Trans. Mode Bila supply > demand.
Misal Denver hanya membutuhkan 1900 mobil. Maka Dummy = 3700 – 3300 = 400
Denver Miami Dummy destination
Supply
Los Angeles $80 $215 $0 1000
Detroit $100 $108 $0 1500
New Orleans $102 $68 $0 1200
Demand 1900 1400 400
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Penyelesaian Unbalanced Trans. Mode (2)
Denver Miami Dummy Supply
Los Angeles
$801000
$215 $01000
Detroit $100900
$108200
$0400 1300
New Orleans
$102 $681200
$01200
Demand 2300 1400 400
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Metode Penyelesaian TransportasiMacam-macam metode penyelesaian Metode Pojok Kiri Atas (Northwest
Corner) Metode Ongkos Terkecil Metode Vogel
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Metode Pojok Kiri Atas
10 0 20 11
12 7 9 20
0 14 16 18
1
2
1 2 3
Sumber
Tujuan
5 15 10
25
15
demand
supply
4
15
3 5
5 10
5 15 5
5
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Metode Pojok Kiri Atas
Total Ongkos =(5x10) + (20x0) + (5x7) + (15x9) +
(5x20) + (5x18) = 410
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Metode Ongkos Terkecil
10 0 20 11
12 7 9 20
0 14 16 18
1
2
1 2 3
Sumber
Tujuan
5 15 10
25
15
demand
supply
4
15
3 5
15
15 10
5
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Metode Ongkos Terkecil
Total Ongkos= (15x0) + (5x0) + (15x9) + (10x20)= 335
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Vogel Approximation Method1. Hitung Penalty selisih ongkos dari 2 ongkos
terkecil
2. Cari penalty terbesar Jika ada yang sama :
Bandingkan antar ongkos terkecil Ambil ongkos terkecil diantara yang terkecil
3. Alokasikan pada ongkos terkecil4. Sesuaikan sumber dan tujuan5. Tandai kolom/baris yang sudah terpenuhi6. If tinggal 1 kolom/baris yang belum
ditandai, then stop Else back to 1
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Metode Vogel
10 0 20 11
12 7 9 20
0 14 16 18
1
2
1 2 3
Sumber
Tujuan
5 15 10
25
15
4
15
3 5
15
15 10
5
Penalty
10
2
14
Penalty10 7 7 7
11
2
–
– 7 11 9
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Metode Vogel
Total ongkos= (15x0) + (15x9) + (10x20) + (5x0)= 335