Tugas pengganti mid
MANAJEMEN PRODUKSI
DI SUSUN OLEH :
I PUTU SUMARIANTO
2009 12 032
UNIVERSITAS ANDI DJEMMA
PALOPO
2012
Soal
1. Fungsi tujuan
Minimumkan C = 2000 X + 3000 Y
Fungsi batasan
Zat 01 30x + 15Y ≥ 225
Zat 02 15x + 30y ≥ 225
Penyelesaian :
Langkah 1
Fungsi tujuan
C = 2000X + 3000Y + 0S1 + 0S2 atau
Z – 200X - 300Y + 0S1 + 0S2 = 0
Fungsi batasan
Zat 01 30X + 15Y + S1 = 225
Zat 02 15X + 30Y +S2 = 225
Langkah 2 menyusun persamaan-persamaan dalam tabel
| VB | X | Y | S1 | S2 | NK | R |
| C | -2000 | -3000 | 0 | 0 | 0 | |
| S1 | 30 | 15 | 1 | 0 | 225 | |
| S2 | 15 | 30 | 0 | 1 | 225 | |
Langkah 3 memilih kolom kunci
| VB | X | Y | S1 | S2 | NK | R |
| C | -2000 | -3000 | 0 | 0 | 0 | |
| S1 | 30 | 15 | 1 | 0 | 225 | 7,5 |
| S2 | 15 | 30 | 0 | 1 | 225 | 15 |
Langkah 4 memilih baris kunci
| VB | X | Y | S1 | S2 | NK | R |
| C | -2000 | -3000 | 0 | 0 | 0 | |
| S1 | 30 | 15 | 1 | 0 | 225 | 7,5 |
| S2 | 15 | 30 | 0 | 1 | 225 | 15 |
Langkah 5 mengubah nilai-nilai baris kunci
| VB | X | Y | S1 | S2 | NK | R |
| C | -2000 | -300 | 0 | 0 | 0 | |
| X | 1 | ½ | 1/30 | 0 | 7,5 | |
| S2 | 15 | 30 | 0 | 1 | 225 | |
Langkah 6 perhitungan nilai baris kunci
Baris S2 : 15 30 0 1 225
15 ( 1 ½ 1/30 0 7,5 ) 0 22,5 ½ 1 112,5
Baris C -2000 -3000 0 0 0
-2000( 1 ½ 1/30 0 7,5 ) 0 -2000 66,67 0 15000
| VB | X | Y | S1 | S2 | NK | R |
| Z | 0 | -2000 | 66,67 | 0 | 15000 | |
| X | 1 | 1/2 | 1/30 | 0 | 7,5 | |
| S2 | 0 | 22,5 | ½ | 1 | 112,5 | |
Langkah 7 langkah perbaikan
| VB | X | Y | S1 | S2 | NK | R |
| C | 0 | -2000 | 66,67 | 0 | 15000 | 7,5 |
| X | 1 | 1/2 | 1/30 | 0 | 7,5 | 15 |
| S2 | 0 | 22,5 | ½ | 1 | 112,5 | 5 |
| VB | X | Y | S1 | S2 | NK | R |
| C | 0 | -2000 | 66,67 | 0 | 15000 | |
| X | 1 | 1/2 | 1/30 | 0 | 7,5 | |
| Y | 0 | 1 | 0,22 | 1/22,5 | 5 | |
Perhitungan nilai baris kunci
Baria C 0 -2000 66,67 0 15000
-2000( 0 1 0,22 1/22,5 5 ) 0 0 506,67 88,89 5000
Baris X 1 ½ 1/30 0 7,5
½( 0 1 0,22 1/22,5 5 ) 1 0 0,077 0,022 5
| VB | X | Y | S1 | S2 | NK | R |
| C | 0 | 0 | 506,67 | 88,89 | 5000 | |
| X | 1 | 0 | 0,077 | 0,022 | 5 | |
| Y | 0 | 1 | 0,22 | 1/22,5 | 5 | |
Di peroleh hasil x = 5, y=5
Jadi total biaya
C = 2000 X + 3000 Y
C=2000(5)+3000(5)
C=Rp.25.000,-
