Soal Metode Simplex (Maksimisasi)
Misalkan roti yang diproduksi adalah X1 dan X2
Fungsi Tujuan : Max Z = 400X1 + 300 X2
Fungsi Pembatas : 3 X1 + 2 X2 ≤ 400
2 X1 + 2 X2 ≤ 300
Dimana X1, X2 ≥ 0
Ditanya :
Berapa roti 1 dan roti 2 yang harus diproduksi dan berapa keuntungan maksimum yang dapat diperoleh ?
Jawab:
Z - 400 X1 - 300 X2+ 0X3+ 0X4 = 0
3 X1+ 2 X2 + X3 = 400
2 X1 + 2 X2 + X4 = 300
Tabel simplex yang pertama
Variabel Dasar
|
Z
|
X1
|
X2
|
X3
|
X4
|
NK
|
Z
|
1
|
- 400
|
- 300
|
0
|
0
|
0
|
X3
|
0
|
3
|
2
|
1
|
0
|
400
|
X4
|
0
|
2
|
2
|
0
|
1
|
300
|
Tabel simpleks: pemilihan kolom kunci pada tabel pertama
Variabel Dasar
|
Z
|
X1
|
X2
|
X3
|
X4
|
NK
|
Keterangan
(Index)
|
Z
|
1
|
- 400
|
- 300
|
0
|
0
|
0
| |
X3
|
0
|
3
|
2
|
1
|
0
|
400
|
= 400/3
|
X4
|
0
|
2
|
2
|
0
|
1
|
300
|
= 300/2
|
Tabel simpleks: Mengubah nilai baris kunci
Variabel Dasar
|
Z
|
X1
|
X2
|
X3
|
X4
|
NK
|
Keterangan
(Index)
|
Z
|
1
|
- 400
|
- 300
|
0
|
0
|
0
| |
X3
|
0
|
3
|
2
|
1
|
0
|
400
|
= 400/3
|
X4
|
0
|
2
|
2
|
0
|
1
|
300
|
= 150
|
Z
| |||||||
X3
|
0
|
1
|
2/3
|
1/3
|
0
|
400/3
|
Pengali
|
X4
|
Mengubah nilai-nilai selain pada baris kunci
Rumus:
Baris baru = baris lama – (koefisien pada kolom kunci) x nilai baru baris kunci
Baris pertama (Z)
- 400
|
- 300
|
0
|
0
|
0
| |||
(-400)
|
1
|
2/3
|
1/3
|
0
|
400/3
|
( - )
| |
Nilai baru
|
=
|
0
|
- 100/3
|
400/3
|
0
|
160.000/3
|
Baris ketiga (batasan 1)
2
|
2
|
0
|
1
|
300
| |||
(2)
|
1
|
2/3
|
1/3
|
0
|
400/3
|
( - )
| |
Nilai baru
|
=
|
0
|
2/3
|
-1/3
|
1
|
100/3
|
Tabel pertama nilai lama dan tabel kedua nilai baru
Variabel Dasar
|
Z
|
X1
|
X2
|
X3
|
X4
|
NK
|
Z
|
1
|
- 400
|
- 300
|
0
|
0
|
0
|
X3
|
0
|
3
|
2
|
1
|
0
|
400
|
X4
|
0
|
2
|
2
|
0
|
1
|
300
|
Tabel Simplex Baru
| ||||||
Z
|
1
|
0
|
- 100/3
|
400/3
|
0
|
160.000/3
|
X1
|
0
|
1
|
2/3
|
1/3
|
0
|
400/3
|
X4
|
0
|
0
|
2/3
|
-1/3
|
1
|
100/3
|
Melanjutkan perbaikan
Variabel Dasar
|
Z
|
X1
|
X2
|
X3
|
X4
|
NK
|
Keterangan
(Indeks)
|
Z
|
1
|
0
|
-100/3
|
400/3
|
0
|
160.000/3
| |
X3
|
0
|
1
|
2/3
|
1/3
|
0
|
400/3
|
400/3 : 2/3 = 200
|
X4
|
0
|
0
|
2/3
|
-1/3
|
1
|
100/3
|
100/3 : 2/3 = 50
|
Z
| |||||||
X1
| |||||||
X2
|
0
|
0
|
1
|
1/2
|
2/3
|
50
|
Baris pertama (Z)
0
|
- 100/3
|
400/3
|
0
|
160.000/3
| |||
-100/3
|
0
|
1
|
1/2
|
2/3
|
50
|
( - )
| |
Nilai baru
|
=
|
0
|
0
|
150
|  |
55.000
|
Baris kedua (batasan 3)
1
|
2/3
|
1/3
|
0
|
400/3
| |||
2/3
|
0
|
1
|
1/2
|
2/3
|
50
|
( - )
| |
Nilai baru
|
=
|
1
|
0
|
0
|
-1
|
100
|
Tabel simpleks final hasil perubahan
Variabel Dasar
|
Z
|
X1
|
X2
|
X3
|
X4
|
NK
|
Z
|
1
|
0
|
-100/3
|
400/3
|
0
|
160.000/3
|
X3
|
0
|
1
|
2/3
|
1/3
|
0
|
400/3
|
X4
|
0
|
0
|
2/3
|
-1/3
|
1
|
100/3
|
Tabel Simplex Baru
| ||||||
Z
|
1
|
0
|
0
|
150
| |
55.000
|
X1
|
0
|
1
|
0
|
0
|
-1
|
100
|
X2
|
0
|
0
|
1
|
1/2
|
2/3
|
50
|
Tabel simpleks final hasil perubahan
Variabel Dasar
|
Z
|
X1
|
X2
|
X3
|
X4
|
NK
|
Z
|
1
|
0
|
0
|
150
| |
55.000
|
X1
|
0
|
1
|
0
|
0
|
-1
|
100
|
X2
|
0
|
0
|
1
|
1/2
|
2/3
|
50
|
Dari hasil perhitungan diatas, maka dapat diperoleh:
X1 = 100
X2 = 50
Zmaksimum = 55.000
Roti 1 yaitu 100, roti 2 yaitu 50 yang harus diproduksi, dan keuntungan maksimum yang dapat diperoleh yaitu sebesar 55.000
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