PENYELESAIAN LP DENGAN METODE SIMPLEKS
PENYELESAIAN LP DENGAN METODE SIMPLEKS
Nama : Irvan Taufik Arifianto (13315464)
Kelas : 2ta01
Riset Operasi
Nama : Irvan Taufik Arifianto (13315464)
Kelas : 2ta01
Riset Operasi
a.
Persoalan :
Cj
|
80
|
100
|
0
|
0
|
0
|
||
Basis
|
X1
|
X2
|
S1
|
S2
|
S3
|
bj
|
|
3
|
2
|
1
|
0
|
0
|
18
|
||
2
|
4
|
0
|
1
|
0
|
20
|
||
0
|
1
|
0
|
0
|
0
|
4
|
||
Zj
|
|||||||
(Cj - Zj)
|
|||||||
b.
Penyelesaian :
Cj
|
80
|
100
|
0
|
0
|
0
|
||||
Basis
|
X1
|
X2
|
S1
|
S2
|
S3
|
bj
|
Ratio
|
||
S1
|
0
|
3
|
2
|
1
|
0
|
0
|
18
|
18/2 = 9
|
|
S2
|
0
|
2
|
4
|
0
|
1
|
0
|
20
|
20/4 = 5
|
|
S3
|
0
|
0
|
1
|
0
|
0
|
4
|
BK
|
4/1 = 4
|
|
Zj
|
0
|
0
|
0
|
0
|
0
|
Pivot
= 1
|
|||
(Cj - Zj)
|
80
|
100
|
0
|
0
|
0
|
||||
KK
|
Cj
|
80
|
100
|
0
|
0
|
0
|
||||
Basis
|
X1
|
X2
|
S1
|
S2
|
S3
|
bj
|
Ratio
|
||
S1
|
0
|
3
|
0
|
1
|
0
|
0
|
10
|
BK
|
10/3 = 3,33
|
S2
|
0
|
2
|
2
|
0
|
0
|
0
|
12
|
12/2 = 6
|
|
X2
|
100
|
0
|
1
|
0
|
0
|
0
|
4
|
4/0 = ꝏ
|
|
Zj
|
0
|
100
|
0
|
0
|
0
|
400
|
Pivot
= 3
|
||
(Cj - Zj)
|
80
|
0
|
0
|
0
|
0
|
||||
KK
|
Cj
|
80
|
100
|
0
|
0
|
0
|
||
Basis
|
X1
|
X2
|
S1
|
S2
|
S3
|
bj
|
|
X1
|
80
|
1
|
0
|
0,33
|
0
|
0
|
3,33
|
S2
|
0
|
2
|
2
|
0
|
0
|
0
|
12
|
X2
|
100
|
0
|
1
|
0
|
0
|
0
|
4
|
Zj
|
80
|
100
|
26,7
|
0
|
0
|
666,67
|
|
(Cj - Zj)
|
0
|
0
|
-26,7
|
0
|
0
|
Cj – Zj ≤ 0 , maka nilai optimum yang didapatkan sebesar 666,67
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