metode dualitas
TRANSCRIPT
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DUALITAS
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Teori Dualitas• Konsep Program Linier yang penting dan
menarik• Setiap persoala Program Linier (primal)
mempunyai suatu Proram Lnier Lain yang saling berkaitan yang disebut dengan dual
• Pemecahan Program Linier bisa didekati secara Primal maupun Dual. Maka hasilnya Primal harus = Dual
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Cara merubah primal dual
• Pada primal jadikan bentuk normal– Jk f.tujuan = max, maka seluruh pembatas jadikan
<=– Jk f. tujuan = min, maka seluruh pembatas jadikan
>=
• F. tujuan berubah bentuk– primal max, maka dual min– Primal min, maka dual max
• Kons. Kanan primal koef.f.tujuan dual• Koef.f.tujuan primal kons. Kanan dual
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Cara merubah primal dual
• Untuk tiap pembatas primal ada 1 var dual
• Untuk tiap var.primal ada 1 pembatas dual
• Tanda pembatas pada dual akan tergantung pada f.tujuannya– f.tujuan max, maka pembatas <=– F.tujuan min, maka pembatas >=
• Dual dari dual primal
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Primal Perusahaan PT Sayang Anak
• max Z = 3x1 + 2x2• batasan :
2x1 + x2 <= 100X1 + x2 <= 80X1 <= 40X1, x2 >= 0
• X1 = boneka• X2 = kereta api• Waktu poles max 100 jam• Waktu kayu max 80 jam
Primal mencari keuntungan max
Berapa boneka dan kereta api yangHarus di produksi agar keuntungan max
Dual mencari berapa kebutuhan optimalDari sumber daya yang ada
Berapa jam waktu poles dan waktu kayuYang dipakai sesungguhnya, agar hasil tetapoptimal
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Dengan grafis
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Dengan simpleks
1. Jadikan standard– max Z = 3x1 + 2x2 + 0S1 + 0S2 + 0S3
z – 3x1 – 2x2 = 0
– batasan 2x1 + x2 + S1 = 100
X1 + x2 + S2 = 80
X1 + S3 = 40
X1, x2, S1, S2, S3 >= 0
2. BV S1, S2, S3
NBV x1, x2
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3. table
pivotEV
pivot
pivot
LV
1 -1 0 20
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• Optimal x1 = 20X2 = 60 Z = 3.20 + 2.60 = 180
• Keuntungan maksimum adalah 180• X1 = boneka yang diproduksi sebanyak 20
buah• X2 = kereta api yang di produksi sebanyak
60 buah
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Dual dari Primal
• Min w =100 y1 + 80 y2 + 40 y3
• Batasan
2 y1 + y2 + y3 >= 3
y1 + y2 >= 2
y1, y2, y3 >= 0
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Dual TORA
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Bahasan
• Min w =100 y1 + 80 y2 + 40 y3• Batasan
2 y1 + y2 + y3 >= 3y1 + y2 >= 2
Jawaban :y1 = 1y2 =1
Maka min w = 100.1 + 80.1 = 180Hasil sama dengan max Z = 180Artinya sumber daya poles perlu 100 jam dan sumber
daya kayu perlu 80 jam agar mendapat hasil optimum
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Contoh SoalPrimalMaks Z = 60X1 + 30X2 + 20X3
Kendala 8X1 + 6X2 + X3 ≤ 48 4X1 + 2X2 + 1,5X3 ≤ 20
2X1 + 1,5X2 + 0,5X3 ≤ 8
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Primal dengan TORA
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Primal dengan TORA
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Dual dengan TORA
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Dual dengan TORA
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Dual dengan TORA
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Model Primal :
memaksimumkan Z = $160x1 + 200x2
terbatas pada
2x1 + 4x2 ≤ 40 jam tenaga kerja
18x1 + 18x2 ≤ 216 pon kayu
24x1 + 12x2 ≤ 240 M2 tempat penyimpanan
x1, x2 ≥ 0
dimana
x1 = jumlah meja yang diproduksi
x2 = jumlah kursi yang diproduksi
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• Hasil optimum
• X1 = meja 4
• X2 = kursi 8
• Jadi keuntungan max adalah :– Max Z =2.240
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Model primal : maks Model dual : min • Model DUAL :
meminimumkan Z = 40y1 + 216y2 + 240y3
terbatas pada
2y1+ 18y2 + 24y3 ≥ 160
4y1 + 18y2 + 12y3 ≥ 200
y1, y2, y3 > 0
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• Hasil optimum
• y1 20
• y2 6,67
• Jadi min w =2.240
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Contoh 2 :
meminimumkan Z = 6x1 + 3x2
terbatas pada
2x1 + 4x2 ≥ 16 pon nitrogen
4x1 + 3x2 ≥ 24 pon phospate
x1, x2 ≥0
dimana
x1 = jumlah sak pupuk Super‑gro
x2 = jumlah sak pupuk Crop‑quik
Z = total biaya pembelian pupuk
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Dual :
memaksimumkan Zd = 16y1 + 24y2
terbatas pada
2y1 + 4y2 ≤ 6, biaya dari Super‑gro
4y1 + 3y2 ≤ 3, biaya dari Crop‑quik
y1 , y2 ≥0
dimana
y1 = nilai marjinal nitrogen
y2 = nilai marjinal phospate
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Contoh 3:
memaksimumkan Z = 10x1 + 6x2
terbatas pada
x1 + 4x2≤ 40
3x1 + 2x2 = 60
2x1 + x2 ≥ 25
x1, x2 ≥0
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Shg Perubahan Batasan :memaksimumkan Zp = 10x1 + 6x2 terbatas pada
x1 + 4x2 ≤ 40
3x1 + 2x2 ≤ 60
‑3x1 – 2x2 ≤ ‑60
‑2x1 – x2 ≤ -25
x1, x2 ≥ 0
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• Bentuk dual :
meminimumkan
Zd = 40y1 + 60y2 – 60y3 – 25y4
terbatas pada
y1 + 3y2 –3y3 – 2y4 ≥ 10
4y1 + 2y2 –2y3 – y4 ≥ 6
y1, y2, y3, y4 ≥ 0
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primal
• Max z = 5 x1 + 12 x2 + 4 x3
• Batasan x1 + 2 x2 + x3 <= 10
2 x1 – x2 + 3 x3 = 8
x1, x2, x3 >= 0
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Primal normal
• Max z = 5 x1 + 12 x2 + 4 x3
• Batasan x1 + 2 x2 + x3 <= 10
2 x1 – x2 + 3 x3 <= 8
- 2 x1 + x2 - 3 x3 <= - 8
x1, x2, x3 >= 0
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dual
• Min W = 10 y1 + 8 y2 - 8 y3’
• Batasan y1 + 2 y2 - 2 y3 >= 5
2 y1 – y2 + y3 >= 12
y1 + 3 y2 - 3 y3 >= 4
y1 , y2 , y3 >= 0
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Primal dengan TORA
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Dual dengan TORA
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Primal dengan manual
Itr BV x1 x2 x3 s1 R2 Solusi
0 z -(2M+5) M-12 -(3M+4) 0 0 -8M
s1 1 2 1 1 0 10
R2 2 -1 3 0 1 8
1 z -7/3 -40/3 0 0 4/3+M 32/3
x2 1/3 7/3 0 1 -1/3 22/3
s2 2/3 -1/3 1 0 1/3 8/3
2 z -3/7 0 0 40/7 -4/7+M 368/7
x2 1/7 1 0 3/7 -1/7 22/7
x1 5/7 0 1 1/7 2/7 26/7
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Itr BV x1 x2 x3 s1 R2 Solusi
3 z 0 0 3/5 29/5 -2/5+M 54 4/5
x2 0 1 -1/5 2/5 -1/5 12/5
x1 1 0 7/5 1/5 2/5 26/5
Primal (cont’d)
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Dual dengan manual
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Contoh Lain
Reddy Mikks model :Primal Dual
max z = 5x1+4x2 min w = 24y1+6y2+y3+2y4
st: st:
6x1 + 4x2 ≤ 24 (M1) 6y1+y2-y3 ≥5
x1 + 2x2 ≤ 6 (M2) 4y1+2y2+y3+y4 ≥4
-x1 + x2 ≤ 1 y1, y2, y3, y4 ≥ 0
x2 ≤ 2
x1, x2 ≥ 0Optimal solution : Optimal solution :
x1 = 3, x2 = 1.5, z = 21 y1=0.75, y2=0.5, y3=y4=0, w=21
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PR
• Cari bentuk dual dari primal berikut ini dan cari jawaban primal dan dual
• Max Z = -5x1 + 2x2
• Batasan
-x1 + x2 <= -2
2x1 + 3x2 <= 5
X1, x2 >=0
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Prinsip Complementary Slack
• Jika suatu kendala adalah tidak aktif, sehingga terdapat variabel slack atau surplus, maka variabel dualnya adalah 0 dalam solusi optimal dual.
• Jika suatu kendala adalah aktif sehingga tidak terdapat variabel slack atau surplus, maka variabel dualnya adalah positif dalam solusi otpimal model dual
Variabel Model Primal Variabel Model Dual
Basis ≥ Non Basic = 0
Nons Basic = 0 Basis ≥
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Contoh:PrimalMaks Z = 3X1 + 4X2
Kendala -4X1 + 5X2 ≤ 10 5X1 + 2X2 ≤ 50
X1 + X2 ≤ 10DualMin Z = 10X1 + 50X2 + 10X3
Kendala -4Y1 + 5Y2 + Y3 ≥ 3 5y1 + 2Y2 + Y3 ≥ 4
Solusi optimal dari soal primal adalah:X1=4,44; X2=5,56; X3=0.00; X4 = 16,67; X5 = 0,00; Z = 35,56Solusi optimal dari dual adalah:Y1=0,11; Y2=0,00; Y3=3.44; Y4 = 0,00; Y5 = 0,00; Z = 35,56
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Untuk kedua solusi optimal, hubungan dalam prinsip comlementary slackness Terpenuhi, yaitu
Variabel Model Primal Variabel Model DualBasis Non BasisX1 = 4,44 Y1 = 0 (variabel surplus)X2 = 5,56 Y1 = 0 (variabel surplus)Non Basis BasisX3 = 0 (variabel slack) Y1 = 0,11X4 = 0 (variabel slack) Y2 = 0,00X5 = 0 (variabel slack Y3 = 3,44
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KESIMPULAN
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Model Primal Model DualMaksimasi MinimasiKoefisien fungsi tujuan Koefisien sisi sebelah kananKendala ke i ≥ Variabel ke yi ≤ 0Kendala ke i ≤ Variabel ke yi ≥ 0Kendala ke i = Variabel yi (unrestricted)Variabel ke i ≥ Kendala ke I ≥Variabel ke i ≤ Kendala ke I ≤