Tugas 2 Riset Operasi
Tugas II
Selesaikan
tabel simpleks berikut hingga mencapai nilai optimal
|
Cj
|
80
|
100
|
0
|
0
|
0
|
|
Basis
|
|
X1
|
X2
|
S1
|
S2
|
S3
|
bj
|
S1
|
|
3
|
2
|
1
|
0
|
0
|
18
|
S2
|
|
2
|
4
|
0
|
1
|
0
|
20
|
S3
|
|
0
|
1
|
0
|
0
|
0
|
4
|
Zj
|
|
|
|
|
|
|
|
Cj-Zj
|
|
|
|
|
|
|
Penyelesaian:
|
Cj
|
80
|
100
|
0
|
0
|
0
|
|
Basis
|
|
X1
|
X2
|
S1
|
S2
|
S3
|
bj
|
S1
|
0
|
3
|
2
|
1
|
0
|
0
|
18
|
S2
|
0
|
2
|
4
|
0
|
1
|
0
|
20
|
S3
|
0
|
0
|
1
|
0
|
0
|
0
|
4
|
Zj
|
0
|
0
|
0
|
0
|
0
|
|
|
Cj-Zj
|
80
|
100
|
0
|
0
|
0
|
|
Ratio
18/2 = 9 → ratio terbesar
20/4 = 5
4/1 = 4 → BK (baris khusus diambil dari nilai bj
dengan ratio terkecil)
·
KK
(kolom khusus) = 100 (nilai terbesar dari Cj)
·
Pivot
= 1 (didapatkan setelah memplotingkan nilai KK dan BK)
·
Faktor
pengali = 2 (diambil dari ratio paling besar yaitu 9)
|
Cj
|
80
|
100
|
0
|
0
|
0
|
|
Basis
|
|
X1
|
X2
|
S1
|
S2
|
S3
|
bj
|
S1
|
0
|
3
|
0
|
1
|
0
|
0
|
10
|
S2
|
0
|
2
|
2
|
0
|
1
|
0
|
12
|
S3
|
100
|
0
|
1
|
0
|
0
|
0
|
4
|
Zj
|
0
|
100
|
0
|
0
|
0
|
400
|
|
Cj-Zj
|
80
|
0
|
0
|
0
|
0
|
|
Ratio
10/3 = 3,33 → BK
12/2 = 6
4/0 = ∞ → ratio terbesar
·
Pivot
= 3
·
Faktor
pengali = 0
|
Cj
|
80
|
100
|
0
|
0
|
0
|
|
Basis
|
|
X1
|
X2
|
S1
|
S2
|
S3
|
bj
|
X1
|
80
|
1
|
0
|
1/3
|
0
|
0
|
10/3
|
S2
|
0
|
2
|
2
|
0
|
0
|
0
|
12
|
S3
|
100
|
0
|
1
|
0
|
0
|
0
|
4
|
Zj
|
80
|
100
|
80/3
|
0
|
0
|
666,67
|
|
Cj-Zj
|
0
|
0
|
-80/30
|
0
|
0
|
|
Karena (Cj-Zj)
≤ 0 , maka sudah di dapatkan nilai yang optimal dengan hasil 666,67.
Perhitungan
tabel 2
S1
baris (1,1)
= 3 ─ 2.0 = 3
baris (1,2)
= 2 ─ 2.1 = 0
baris (1,3)
= 1 ─ 2.0 = 1
baris (1,4)
= 0 ─ 2.0 = 0
baris (1,5)
= 0 ─ 2.0 = 0
baris (1,6)
= 18 ─ 2.4 = 10
S2
baris (2,1)
= 2 ─ 2.0 = 2
baris (2,2)
= 4 ─ 2.1 = 2
baris (2,3)
= 0 ─ 2.0 = 0
baris (2,4)
= 1 ─ 2.0 = 1
baris (2,5)
= 0 ─ 2.0 = 0
baris (2,6)
= 20 ─ 2.4 = 12
Perhitungan
tabel 3
X2
baris (3,1)
= 0 ─ 0.1 = 0
baris (3,2)
= 1 ─ 0.0 = 1
baris (3,3)
= 0 ─ 0.0,33 = 0
baris (3,4)
= 0 ─ 0.0 = 0
baris (3,5)
= 0 ─ 0.0 = 0
baris (3,6)
= 4 ─ 0.3,33 = 4
S2
baris (2,1)
= 2 ─ 0.1 = 2
baris (2,2)
= 2 ─ 0.0 = 2
baris (2,3)
= 0 ─ 0.0,33 = 0
baris (2,4)
= 1 ─ 0.0 = 1
baris (2,5)
= 0 ─ 0.0 = 0
baris (2,6)
= 12 ─ 0.3,33 = 12
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