METODE SIMPLEKS

Penyelesaian Linear Program dengan Metode Simpleks

Contoh soal :

Selesaikan tabel simpleks berikut hingga mencapai nilai optimal!

	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
(Cj-Zj)

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	0	4	4 : 1 = 4
Zj		0	0	0	0	0	0
(Cj-Zj)		80	100	0	0	0

	Cj	80	100	0	0	0
Basis		X1	X2	S1	S2	S3	bj	Ratio
S1	0	3	0	1	0	0	10	10 : 3 = 3,33
S2	0	2	2	0	1	0	12	12 : 2 = 6
X2	0	0	1	0	0	0	4	4 : 0 =
Zj		0	100	0	0	0	400
(Cj-Zj)		80	0	0	0	0

Untuk S1 : (1,1) = 3 - 2.0 = 3 (1,2) = 2 - 2.1 = 0 (1,3) = 1 - 2.0 = 1 (1,4) = 0 - 2.0 = 0 (1,5) = 0 - 2.0 = 0 (1,6) = 18 - 2.4 = 10

Untuk S2 : (2,1) = 2 - 2.0 = 2 (2,2) = 4 - 2.1 = 2 (2,3) = 0 - 2.0 = 0 (2,4) = 1 - 2.0 = 1 (2,5) = 0 - 2.0 = 0 (2,6) = 20 - 2.4 = 12

Untuk X2 : (3,1) = 0 : 1 = 0 (3,2) = 1 : 1 = 1 (3,3) = 0 : 1 = 0 (3,4) = 0 : 1 = 0 (3,5) = 0 : 1 = 0 (3,6) = 4 : 1 = 4

	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	1	0	12
X2	100	0	1	0	0	0	4
Zj		80	100	26,4	0	0	666,7
(Cj-Zj)		0	0	-26,4	0	0

Untuk X1 : (1,1) = 3 : 3 = 1 (1,2) = 0 : 3 = 0 (1,3) = 1 : 3 = 0,33 (1,4) = 0 : 3 = 0 (1,5) = 0 : 3 = 0 (1,6) = 10 : 3 = 3,33

Untuk S2 : (2,1) = 2 - 0.1 = 2 (2,2) = 2 - 0.2 = 2 (2,3) = 0 - 0.0,33 = 0 (2,4) = 1 - 0.0 = 1 (2,5) = 0 - 0.0 = 0 (2,6) = 12 - 0.3,33 = 12

Untuk X2 : (3,1) = 0 - 0.1 = 0 (3,2) = 1 - 0.0 = 1 (3,3) = 0 - 0.0,33 = 0 (3,4) = 0 - 0.0 = 0 (3,5) = 0 - 0.0 = 0 (3,6) = 4 - 0.3,33 = 0

Sudah dapat disebut optimal karena (C_j-Z_j) pada X₁ dan X₂ bernilai 0. Nilai optimal yang didapat adalah 666,7.