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Operation Research
 Metode Simpleks
    SOEDARMANTO
     JANUARI 2013
Pertanyaan

    DIKETAHUI
 FUNGSI TUJUAN
    Z = 2X₁ + 3X₂
 FUNGSI KENDALA
   5X ₁ + 6X₂ ≤ 60
    X ₁ +2X₂ ≤ 16
    X₁       ≤ 10
         X₂       ≤6
Menyisipkan Slack


      2X ₁ + 3X ₂ + OS ₁ + OS ₂ + OS ₃ + OS₄    =0
   5X ₁ + 6X ₂ + S ₁                              = 60
   X ₁ + 2X ₂           +S₂                       = 16
X₁                                 +S₃                 = 10
X₂                                           + S₄      =6
Iterasi 0


     Ci Vb X ₁   X₂   S₁   S₂   S₃   S₄
bi
  0 Cj ₁ 2
     S    5      36   01   00   00   00
60
  0 S₂ 1         2    0    1    0    0
16
  0 S₃ 1         0    0    0    1    0
10
  0 S₄ 0         1    0    0    0    1
6
     Zj 0        0    0    0    0    0
Langkah - Langkah

             PENENTUAN KOLOM KUNCI
 Konsep Opportunity cost (Cj-Zj) terbesar
              PENENTUAN BARIS KUNCI
 Rasio= bi/aij (pilih nilai positif terkecil)
PENGISIAN ELEMEN-ELEMEN YG KOSONG DENGAN ASB
       (ATURAN SUDUT YANG BERLAWANAN)
Aturan Sudut Berlawanan
Elemen Sudut                                       Elemen Lama
   yang
Berlawanan
                     E₂        EX                  Elemen Baru

                                                   Elemen Sudut
Elemen Kunci
                     EK        E₁                     yang
                                                   Berlawanan


                                    E₁ xE₂
         EX (baru) = EX (lama) -   -------------
                                      EK
Iterasi 0

     Ci Vb X ₁   X₂   S₁   S₂   S₃   S₄
bi
  0Cj ₁ 25
    S            36   01   00   00   00
60
  0 S₂ 1         2    0    1    0    0
16
  0 S₃ 1         0    0    0    1    0
10
  0 S₄ 0         1    0    0    0    1
6
    Zj 0         0    0    0    0    0
0
Iterasi 1


Ci Vb 2 X ₁ 3 X ₂ 0 S ₁ 0 S ₂ 0 S ₃ 0 S₄
 Cj
   bi
   S₁         0
   S₂         0
   S₃         0
   X₂ 0       1     0     0     0     1
   6
0 Zj
 Cj-Zj
Iterasi 1
     Ci Vb X ₁   X₂   S₁   S₂   S₃    S₄
bi
     0 S ₁ 2 5 30
     Cj               01   00   00   0-6
24
     0S ₂   1    0    0    1    0     -2
4
     0S ₃   1    0    0    0    1     0
10
     3X ₂   0    1    0    0    0     1
6
     0 Zj   0    3    0    0    0     3
Iterasi 2

Ci Vb 2 X ₁ 3 X ₂ 0 S ₁ 0 S ₂ 0 S ₃ 0 S₄
 Cj
   bi
   S₁ 0
   X₁ 1       0     0     1     0     -2
   4
   S₃ 0
   X₂ 0
   Zj
 Cj-Zj
Iterasi 2
     Ci Vb X ₁   X₂   S₁    S₂   S₃   S₄
bi
     0 S ₁ 20
     Cj          30   01   0-5 00     04
4
     2X ₁   1    0    0     1    0    -2
4
     0S ₃   0    0    0     -1   1    2
6
     3X ₂   0    1    0     0    0    1
6
       Zj   2    3    0     2    0    -1
Iterasi 3
Ci Vb X ₁ X ₂ S ₁ S ₂ S ₃ S₄
 Cjbi 2  3   0   0   0   0
0S ₁ 0    0   1   -5 0    1
   4
2X ₁                      0

0S ₃                      0

3X ₂                      0

  Zj
Iterasi 3
     Ci Vb X ₁    X₂   S₁    S₂   S₃   S₄
bi
     0 S ₁ 20
     Cj          30    01   0-5 00     01
4
     2X ₁   1     0    ½ -1½ 0         0
6
     0S ₃   0     0    -½    1½ 1      0
4
     3X ₂   0     1    -¼    1¼ 0      0
5
       Zj   2     3    ¼     ¾ 0       0
Terimakasih
  BAHAN AJAR

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METODE SIMPLEKS

  • 1. Operation Research Metode Simpleks SOEDARMANTO JANUARI 2013
  • 2. Pertanyaan DIKETAHUI FUNGSI TUJUAN Z = 2X₁ + 3X₂ FUNGSI KENDALA 5X ₁ + 6X₂ ≤ 60 X ₁ +2X₂ ≤ 16 X₁ ≤ 10 X₂ ≤6
  • 3. Menyisipkan Slack 2X ₁ + 3X ₂ + OS ₁ + OS ₂ + OS ₃ + OS₄ =0 5X ₁ + 6X ₂ + S ₁ = 60 X ₁ + 2X ₂ +S₂ = 16 X₁ +S₃ = 10 X₂ + S₄ =6
  • 4. Iterasi 0 Ci Vb X ₁ X₂ S₁ S₂ S₃ S₄ bi 0 Cj ₁ 2 S 5 36 01 00 00 00 60 0 S₂ 1 2 0 1 0 0 16 0 S₃ 1 0 0 0 1 0 10 0 S₄ 0 1 0 0 0 1 6 Zj 0 0 0 0 0 0
  • 5. Langkah - Langkah PENENTUAN KOLOM KUNCI Konsep Opportunity cost (Cj-Zj) terbesar PENENTUAN BARIS KUNCI Rasio= bi/aij (pilih nilai positif terkecil) PENGISIAN ELEMEN-ELEMEN YG KOSONG DENGAN ASB (ATURAN SUDUT YANG BERLAWANAN)
  • 6. Aturan Sudut Berlawanan Elemen Sudut Elemen Lama yang Berlawanan E₂ EX Elemen Baru Elemen Sudut Elemen Kunci EK E₁ yang Berlawanan E₁ xE₂ EX (baru) = EX (lama) - ------------- EK
  • 7. Iterasi 0 Ci Vb X ₁ X₂ S₁ S₂ S₃ S₄ bi 0Cj ₁ 25 S 36 01 00 00 00 60 0 S₂ 1 2 0 1 0 0 16 0 S₃ 1 0 0 0 1 0 10 0 S₄ 0 1 0 0 0 1 6 Zj 0 0 0 0 0 0 0
  • 8. Iterasi 1 Ci Vb 2 X ₁ 3 X ₂ 0 S ₁ 0 S ₂ 0 S ₃ 0 S₄ Cj bi S₁ 0 S₂ 0 S₃ 0 X₂ 0 1 0 0 0 1 6 0 Zj Cj-Zj
  • 9. Iterasi 1 Ci Vb X ₁ X₂ S₁ S₂ S₃ S₄ bi 0 S ₁ 2 5 30 Cj 01 00 00 0-6 24 0S ₂ 1 0 0 1 0 -2 4 0S ₃ 1 0 0 0 1 0 10 3X ₂ 0 1 0 0 0 1 6 0 Zj 0 3 0 0 0 3
  • 10. Iterasi 2 Ci Vb 2 X ₁ 3 X ₂ 0 S ₁ 0 S ₂ 0 S ₃ 0 S₄ Cj bi S₁ 0 X₁ 1 0 0 1 0 -2 4 S₃ 0 X₂ 0 Zj Cj-Zj
  • 11. Iterasi 2 Ci Vb X ₁ X₂ S₁ S₂ S₃ S₄ bi 0 S ₁ 20 Cj 30 01 0-5 00 04 4 2X ₁ 1 0 0 1 0 -2 4 0S ₃ 0 0 0 -1 1 2 6 3X ₂ 0 1 0 0 0 1 6 Zj 2 3 0 2 0 -1
  • 12. Iterasi 3 Ci Vb X ₁ X ₂ S ₁ S ₂ S ₃ S₄ Cjbi 2 3 0 0 0 0 0S ₁ 0 0 1 -5 0 1 4 2X ₁ 0 0S ₃ 0 3X ₂ 0 Zj
  • 13. Iterasi 3 Ci Vb X ₁ X₂ S₁ S₂ S₃ S₄ bi 0 S ₁ 20 Cj 30 01 0-5 00 01 4 2X ₁ 1 0 ½ -1½ 0 0 6 0S ₃ 0 0 -½ 1½ 1 0 4 3X ₂ 0 1 -¼ 1¼ 0 0 5 Zj 2 3 ¼ ¾ 0 0