Model Assignment

a. Persoalan

Tenaga (SPG)	I	II	III	IV	V
A	20	13	18	19	23
B	15	12	19	12	17
C	14	19	17	18	15
D	17	13	13	21	20
E	16	15	12	18	14

1)      Bagaimana menempatkan SPG agar ongkos maksimal?

2)      Berapa ongkos total?

b. Penyelesaian

I	II	III	IV	V
20	13	18	19	23
15	12	19	12	17
14	19	17	18	15
17	13	13	21	20
16	15	12	18	14

I	II	III	IV	V
7	0	5	6	10
3	0	7	0	5
0	5	3	4	1
4	0	0	8	7
4	3	0	6	2

I	II	III	IV	V
7	0	5	6	10
3	0	7	0	5
0	5	3	4	1
4	0	0	8	7
4	3	0	6	2

I	II	III	IV	V
6	0	5	5	8
3	0	7	0	3
0	6	4	4	0
3	0	0	7	5
3	3	0	5	0

1)      A = II

B = IV

C = I

D = III

E = V

2)      14 + 13 + 13 +12 + 14 = 66

Penyelesaian LP dengan Metode Simpleks

a.      Persoalan

	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

b.      Penyelesaian

	Cj	80	100	0	0	0
Basis		X1	X2	S1	S2	S3	bj	Ratio
S1	0	3	2	1	0	0	18	9
S2	0	2	4	0	1	0	20	5
S3	0	0	1	0	0	0	4	4	BK
Zj		0	0	0	0	0
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	3,333
S2	0	2	0	0	1	0	4	2	BK
X2	100	0	1	0	0	0	4	tidak terhingga
Zj		0	100	0	0	0	400
Cj - Zj		80	0	0	0	0
		KK
	Cj	80	100	0	0	0
Basis		X1	X2	S1	S2	S3	bj
S1	0	0	0	1	-1,5	0	4
X1	80	1	0	0	0,5	0	2
X2	100	0	1	0	0	0	4
Zj		80	100	0	40	0	560
Cj - Zj		0	0	0	-40	0

                                                                                                                 Optimum