1.Riset Operasi adalah suatu cara yang digunakan untuk
mengambil suatu solusi atau keputusan yang
terbaik dari permasalahan yang kita hadapi sehari-hari.
2.Misalkan :
X1 = Roti Pia
X2 = Roti Kismis
X3 = Roti Cokelat Keju.
Fungsi tujuan :
Z
maksimum = 150X1 + 400X2 + 600X3
Fungsi kendala :
4X1 + 2X2 + 6X3 ≤ 130 (
Penyimpanan Bahan )
3X1 + 4X2 + 9X3 ≤ 170 ( Peracikan
)
X1 + 2X2 + 4X3 ≤ 52 ( Pengovenan
).
Formula standar :
MakZ = 150X1 + 400X2 + 600X3
Fungsi kendala :
4X1
+ 2X2 + 6X3 + X4 + = 130
3X1
+ 4X2 + 9X3 +
X5 =170
X1
+ 2X2 + 4X3 + +
X6 =52
Awal
|
Basis
|
X1
|
X2
|
X3
|
X4
|
X5
|
X6
|
RK
|
Rasio
|
X4
|
4
|
2
|
6
|
1
|
0
|
0
|
130
|
||
X5
|
3
|
4
|
9
|
0
|
1
|
0
|
170
|
||
X6
|
1
|
2
|
4
|
0
|
0
|
1
|
52
|
||
Zj-cj
|
-150
|
-400
|
-600
|
0
|
0
|
0
|
0
|
2.Memilih kolom kunci :
Rasio
= nilai kanan
Nilai
Kk
Awal
|
Basis
|
X1
|
X2
|
X3
|
X4
|
X5
|
X6
|
RK
|
Rasio
|
X4
|
4
|
2
|
6
|
1
|
0
|
0
|
130
|
21.66
|
|
X5
|
3
|
4
|
9
|
0
|
1
|
0
|
170
|
18.18
|
|
X3
|
1
|
2
|
4
|
0
|
0
|
1
|
52
|
13
|
|
Zj-cj
|
-150
|
-400
|
-600
|
0
|
0
|
0
|
0
|
-
|
3.Baris kunci = Baris lama
Angka
kunci
Awal
|
Basis
|
X1
|
X2
|
X3
|
X4
|
X5
|
X6
|
RK
|
Rasio
|
X4
|
4
|
2
|
6
|
1
|
0
|
0
|
130
|
||
X5
|
3
|
4
|
9
|
0
|
1
|
0
|
170
|
||
X3
|
¼
|
½
|
1
|
0
|
0
|
¼
|
13
|
||
Zj-cj
|
-150
|
-400
|
-600
|
0
|
0
|
0
|
0
|
4.Mengubah nilai-nilai selain baris kunci sehingga
nilai-nilai kolom kunci = 0
Baris baru = baris lama – koefisien angka kolom kunci
x nilai baris baru kunci
-150 -400 -600 0 0 0 0
-600| ¼ ½ 1 0 0 ¼ 13
_
0 -100 0 0 0 150 7800
3 4 9 0 1 0 170
9| ¼ ½ 1 0 0 ¼ 13 _ _
0.75 -0.5 0 0 1 -2.25 53
4 2 6 1 0 0 130
6| ¼ ½ 1 0 0 ¼ 13
__ _
2.5 -1 0 1 0 -1.5 52
Awal
|
Basis
|
X1
|
X2
|
X3
|
X4
|
X5
|
X6
|
RK
|
Rasio
|
X4
|
2.5
|
-1
|
0
|
1
|
0
|
-1.5
|
52
|
-
|
|
X5
|
0.75
|
-0.5
|
0
|
0
|
1
|
-2.25
|
53
|
-
|
|
X3
|
¼
|
½
|
1
|
0
|
0
|
¼
|
13
|
6.5
|
|
Zj-cj
|
0
|
-100
|
0
|
0
|
150
|
7800
|
0
|
-
|
5.Baris kunci = Baris lama
Angka kunci
Awal
|
Basis
|
X1
|
X2
|
X3
|
X4
|
X5
|
X6
|
RK
|
Rasio
|
X4
|
2.5
|
-1
|
0
|
1
|
0
|
-1.5
|
52
|
-
|
|
X5
|
0.75
|
-0.5
|
0
|
0
|
1
|
-2.25
|
53
|
-
|
|
X3
|
0.5
|
1
|
2
|
0
|
0
|
0.5
|
6.5
|
6.5
|
|
Zj-cj
|
0
|
-100
|
0
|
0
|
150
|
7800
|
0
|
-
|
6. Mengubah nilai-nilai selain baris kunci sehingga nilai-nilai
kolom kunci = 0
Baris baru =
baris lama – koefisien angka kolom kunci x nilai baris baru kunci.
0 -100 0 0 150 7800 0
-100|0.5 1 2 0 0 0.5 6.5
_
200 0 200 0 150 8000 650
0.75 -0.5 0 0 1 -2.25 53
-0.5|0.5 1 2 0 0 0.5 6.5
_
1 0 2 0 1 -2 56.25
2.5 -1 0 1 0 -1.5 52
-1|0.5 1 2 0 0 0.5 6.5
_
3 0 2 1 0 -1 58.5
Awal
|
Basis
|
X1
|
X2
|
X3
|
X4
|
X5
|
X6
|
RK
|
Rasio
|
X4
|
3
|
0
|
2
|
1
|
0
|
-1
|
58.5
|
-
|
|
X5
|
1
|
0
|
2
|
0
|
1
|
-2
|
56.25
|
-
|
|
X(3,2)
|
0.5
|
1
|
2
|
0
|
0
|
0.5
|
6.5
|
6.5
|
|
Zj-cj
|
200
|
0
|
200
|
0
|
150
|
8000
|
650
|
-
|
Sehingga di
peroleh : X1 = 0 (Roti pia)
X2 =
6.5 (Roti Kismis)
X3 = 6.5 (Roti cokelat keju)
Total pendapatan
Z = 150(0) + 400(6.5) + 600(6.5)
=
6500
3.Penyelesaian
Z = 4000X1 + 3000X2
Z – 4000X1 – 3000X2 =
0
Fungsi Kendala :
4X1 +
6X2 + X3 =
1200
4X1 +
2X2 X4 = 800
X1 X5 =
250
X1 X6 = 300
Menyusun
persamaan kendala table :
Var.Dasar
|
Z
|
X1
|
X2
|
X3
|
X4
|
X5
|
X6
|
NK
|
Index
|
Z
|
1
|
-400
|
-300
|
0
|
0
|
0
|
0
|
0
|
|
X3
|
0
|
4
|
6
|
1
|
0
|
0
|
0
|
1200
|
|
X4
|
0
|
4
|
2
|
0
|
1
|
0
|
0
|
800
|
|
X5
|
0
|
1
|
0
|
0
|
0
|
1
|
0
|
250
|
|
X6
|
0
|
0
|
1
|
0
|
0
|
0
|
1
|
300
|
Memilih kolom
kunci :
Index = Nilai
Kanan
Nilai Kk
Var.Dasar
|
Z
|
X1
|
X2
|
X3
|
X4
|
X5
|
X6
|
NK
|
Index
|
Z
|
1
|
-400
|
-300
|
0
|
0
|
0
|
0
|
0
|
-
|
X3
|
0
|
4
|
6
|
1
|
0
|
0
|
0
|
1200
|
300
|
X1
|
0
|
4
|
2
|
0
|
1
|
0
|
0
|
800
|
200
|
X5
|
0
|
1
|
0
|
0
|
0
|
1
|
0
|
250
|
250
|
X6
|
0
|
0
|
1
|
0
|
0
|
0
|
1
|
300
|
-
|
5.Baris Kunci = Baris
Lama
Angka Knci
Var.Dasar
|
Z
|
X1
|
X2
|
X3
|
X4
|
X5
|
X6
|
NK
|
Index
|
Z
|
1
|
-400
|
-300
|
0
|
0
|
0
|
0
|
0
|
|
X3
|
0
|
4
|
6
|
1
|
0
|
0
|
0
|
1200
|
300
|
X1
|
0
|
1
|
½
|
0
|
¼
|
0
|
0
|
200
|
200
|
X5
|
0
|
1
|
0
|
0
|
0
|
1
|
0
|
250
|
250
|
X6
|
0
|
0
|
1
|
0
|
0
|
0
|
1
|
300
|
6.Mengubah
nilai-nilai selain baris kunci sehingga nilai-nilai kolom kunci = 0
Baris baru =
baris lama – koefisien angka kolom kunci x nilai baris baru kunci.
-400 -300 0 0 0 0 0
-400|1 ½ 0 ¼ 0 0 200 _
0 -100 0 100 0 0 80000
4 6 1 0 0 0 1200
4|1 ½ 0 ¼ 0 0 200
_
0 2 1 -1 0 0 400
1 0 0 0 1 0 250
1|1 ½ 0 ¼ 0 0 200
_
0 -1/2 0 -¼ 0 0 50
0 1 0 0 0 1 300
0|1 ½ 0 ¼ 0 0 200
_
0 1 0 0 0 1 300
Var.Dasar
|
Z
|
X1
|
X2
|
X3
|
X4
|
X5
|
X6
|
NK
|
Index
|
Z
|
1
|
0
|
-100
|
0
|
100
|
0
|
0
|
80000
|
|
X3
|
0
|
0
|
2
|
1
|
-1
|
0
|
0
|
400
|
|
X1
|
0
|
1
|
½
|
0
|
¼
|
0
|
0
|
200
|
|
X5
|
0
|
0
|
-1/2
|
0
|
-1/4
|
0
|
0
|
50
|
|
X6
|
0
|
1
|
0
|
0
|
0
|
0
|
1
|
300
|
7.Melanjutkan
perbaikan-perbaikan (langkah 3 sampai langkah 6) sampai baris Z tidak ada nilai
negative.
Var.Dasar
|
Z
|
X1
|
X2
|
X3
|
X4
|
X5
|
X6
|
NK
|
Index
|
Z
|
1
|
0
|
-100
|
0
|
100
|
0
|
0
|
80000
|
|
X2
|
0
|
0
|
2
|
1
|
-1
|
0
|
0
|
400
|
200
|
X1
|
0
|
1
|
½
|
0
|
¼
|
0
|
0
|
200
|
400
|
X5
|
0
|
0
|
-1/2
|
0
|
-1/4
|
0
|
0
|
50
|
|
X6
|
0
|
1
|
0
|
0
|
0
|
0
|
1
|
300
|
300
|
8.Mengubah nilai
baris kunci dengan membaginya dengan angka kunci.
Var.Dasar
|
Z
|
X1
|
X2
|
X3
|
X4
|
X5
|
X6
|
NK
|
Index
|
Z
|
1
|
0
|
-100
|
0
|
100
|
0
|
0
|
80000
|
|
X2
|
0
|
0
|
1
|
1/2
|
-1/2
|
0
|
0
|
200
|
200
|
X1
|
0
|
1
|
½
|
0
|
¼
|
0
|
0
|
200
|
400
|
X5
|
0
|
0
|
-1/2
|
0
|
-1/4
|
0
|
0
|
50
|
|
X6
|
0
|
1
|
0
|
0
|
0
|
0
|
1
|
300
|
300
|
9. Mengubah
nilai-nilai selain baris kunci sehingga nilai-nilai kolom kunci = 0
Baris baru =
baris lama – koefisien angka kolom kunci x nilai baris baru kunci
0 -100 0 100 0 0 80000
-100|0 1 1/2 -1/2 0 0 200 _
0 0 50 50 0 0 100000
1 ½ 0 1/4 0 0 200
1/2|0 1 ½ -1/2 0 0 200_ _
1 0 -1/4 1/2 0 0 100
0 -1/2 0 -1/4 0 0 50
-1/2|0 1 1/2 -1/2 0 0 200 _
0 0 1/4 -1/2 0 0 150
1 0 0 0 0 1 300
0 |0 1 1/2 -1/2 0 0 200 _
1 0 0 0 0 0 300
Var.Dasar
|
Z
|
X1
|
X2
|
X3
|
X4
|
X5
|
X6
|
NK
|
Index
|
Z
|
1
|
0
|
0
|
50
|
50
|
0
|
0
|
100000
|
|
X2
|
0
|
0
|
1
|
½
|
-1/2
|
0
|
0
|
200
|
200
|
X1
|
0
|
1
|
0
|
-1/4
|
1/2
|
0
|
0
|
100
|
400
|
X5
|
0
|
0
|
0
|
1/4
|
-1/2
|
0
|
0
|
150
|
|
X6
|
0
|
1
|
0
|
0
|
0
|
0
|
0
|
100
|
300
|
Sehingga
ddiperoleh hasil :
X1 = 100
X2 =200
Jadi Z Maksimum =
4000X1 + 3000X2
=4000(100) + 3000(200)
=400000 + 300000
=700000
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