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ANALISIS SISTEM REKAYASA KONSTRUKSI
1
Soal : Maksimumkan Z
=
3X1
+
2X2
X1
+
X2
≤
2X1
+
X2
≤
28
X1
+
2X2
≤
20
≥
0
Kendala ;
X1, X2
Z 03 = S2 S3
=
3X1 0
-
2X2
-
0
X1
+
X2
+
S1
+
X2
=
2X1 28
+
2X2
=
X1 20
15
S1
-
0
S2
-
15
1
2
Penyelesaian : Tabel 1. Tabel Simpleks Awal Basis
X1
X2
S1
S2
S3
Solusi
Z
-3
-2
0
0
0
0
S1
1
1
1
0
0
15
S2
2
1
0
1
0
28
S3
1
2
0
0
1
20
2
1
0
1
0
28
1
1/2
0
1/2
0
14
-3
-2
0
0
0
0
1
1/2
0
1/2
0
14
0
-1/2
0
3/2
0
42
Baris X1 : :2
Baris Z :
+3
2
3
Baris S3 :
1
2
0
0
1
20
1
1/2
0
1/2
0
14
0
-3/2
0
1/2
1
6
-
Tabel 2. Tabel Simpleks Iterasi Pertama
Basis
X1
X2
S1
S2
S3
Solusi
Ratio
Z
0
-1/2
0
3/2
0
42
0
S1
0
1/2
1
-1/2
0
1
2
X1
1
1/2
0
1/2
0
14
28
S3
0
3/2
0
-1/2
1
6
4
3
4
Baris X2 :
0
1/2
1
-1/2
0
1
0
1
2
-1
0
2
0
-1/2
0
3/2
0
42
0
1
2
-1
0
2
0
0
1
1
0
43
0
3/2
0
-1/2
1
6
0
1
2
-1
0
2
0
0
-3
1
1
3
x2
Baris Z :
+
:2
-
x 3/2
Baris S3 :
4
5
Baris S1 :
1
1
1
0
0
15
1
1/2
0
1/2
0
14
0
1/2
1
-1/2
0
1
1
1/2
0
1/2
0
14
0
1
2
-1
0
2
1
0
-1
1
0
13
-
Baris X1 :
-
x 1/2
Tabel 3. Tabel Simpleks Optimum Basis
X1
X2
S1
S2
S3
Solusi
Z
0
0
1
1
0
43
X1
1
0
-1
1
0
13
X2
0
1
2
-1
0
2
S3
0
0
-3
1
1
3
5
6
X1
=
13
X2
=
2
Z
=
43
6
1
Soal : Maksimumkan Z
=
3X1
+
2X2
X1
+
X2
≤
2X1
+
X2
≤
28
X1
+
2X2
≤
20
≥
0
Kendala ;
X1, X2
Z 03 = S2 S3
=
3X1 0
-
2X2
-
0
X1
+
X2
+
S1
+
X2
=
2X1 28
+
2X2
=
X1 20
15
S1
-
0
S2
-
15
1
2
Penyelesaian : Tabel 1. Tabel Simpleks Awal Basis
X1
X2
S1
S2
S3
Solusi
Z
-3
-2
0
0
0
0
S1
1
1
1
0
0
15
S2
2
1
0
1
0
28
S3
1
2
0
0
1
20
2
1
0
1
0
28
1
1/2
0
1/2
0
14
-3
-2
0
0
0
0
1
1/2
0
1/2
0
14
0
-1/2
0
3/2
0
42
Baris X1 : :2
Baris Z :
+3
2
3
Baris S3 :
1
2
0
0
1
20
1
1/2
0
1/2
0
14
0
-3/2
0
1/2
1
6
-
Tabel 2. Tabel Simpleks Iterasi Pertama
Basis
X1
X2
S1
S2
S3
Solusi
Ratio
Z
0
-1/2
0
3/2
0
42
0
S1
0
1/2
1
-1/2
0
1
2
X1
1
1/2
0
1/2
0
14
28
S3
0
3/2
0
-1/2
1
6
4
3
4
Baris X2 :
0
1/2
1
-1/2
0
1
0
1
2
-1
0
2
0
-1/2
0
3/2
0
42
0
1
2
-1
0
2
0
0
1
1
0
43
0
3/2
0
-1/2
1
6
0
1
2
-1
0
2
0
0
-3
1
1
3
x2
Baris Z :
+
:2
-
x 3/2
Baris S3 :
4
5
Baris S1 :
1
1
1
0
0
15
1
1/2
0
1/2
0
14
0
1/2
1
-1/2
0
1
1
1/2
0
1/2
0
14
0
1
2
-1
0
2
1
0
-1
1
0
13
-
Baris X1 :
-
x 1/2
Tabel 3. Tabel Simpleks Optimum Basis
X1
X2
S1
S2
S3
Solusi
Z
0
0
1
1
0
43
X1
1
0
-1
1
0
13
X2
0
1
2
-1
0
2
S3
0
0
-3
1
1
3
5
6
X1
=
13
X2
=
2
Z
=
43
6
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