{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Linear Prog COP4355_Part_8

# Linear Prog COP4355_Part_8 - 36 4.3 Chapter 4 Linear...

This preview shows pages 1–3. Sign up to view the full content.

36 Chapter 4 Linear Programming: The Simplex Algorithm 4.3 (continued) Initial Tableau X Y S1 A1 S2 Cj 510- M 0 Bi Θ A1 -1 M 1 1 -1 1 0 6 6 S2 0 2 1 0 0 1 8 4 Zj - M - M M - M 0- 6 M Zj M M - M 00 Entering: X Leaving: Slack 2 Tableau 2 X Y S1 A1 S2 Cj 51 0 - M 0 Bi Θ A1 -1 M 0 0.50 -1 1 -0.50 2 4 X 5 1 0.50 0 0 0.50 4 8 Zj 5 2.5-.5 M M - M 2.5+.5 M -2 M Zj 0 -1.5+.5 M - M 0 -2.5-.5 M Entering: Y Leaving: Artificial 1 Optimal Tableau X Y S1 A1 S2 Cj 51 0 - M 0 Bi Y 1 0 1 -2 2 -1 4 X 5 1 0 1 -1 1 2 Zj 5 1 3 -3.00 4 14 Zj 0 0 -3 3- M -4 Optimal Solution: X = 2, Y = 4, Z MAX = 14

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Solutions Manual and Workbook 37 4.4 MAX Z: 3X 1 + 5X 2 + 7X 3 + 0S 1 + 0S 2 – MA 2 – MA 3 Subject to: 2X 1 – 4X 2 + 5X 3 + 1S 1 + 0S 2 + 0A 2 + 0A 3 = 12 X 1 + 6X 2 – 3X 3 + 0S 1 – 1S 2 + 1A 2 + 0A 3 = 16 5X 1 + 2X 2 + 1X 3 + 0S 1 + 0S 2 + 0A 2 + 1A 3 = 7 (X 1 , X 2 , X 3 , S 1 , S 2 A 2 , A 3 0) Initial Tableau X1 X2 X3 S1 S2 A2 A3 Cj 3 5 7 0 0 - M - M Bi Θ S1 0 2 -4 5 1 0 0 0 12 -3 A2 - M 1 6 -3 0 -1 1 0 16 2.67 A3 - M 5 2 1 0 0 0 1 7 3.50 Zj -6 M -8 M 2 M 0 M - M - M -23 M Zj 3+6 M 5+8 M 7-2 M 0 - M 0 0 Entering: X2 Leaving: A2 4.5 MIN Z: 4X 1 – 12X 2 + 5X 3 + MA 1 + 0S 2 + 0S 3 + MA
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 4

Linear Prog COP4355_Part_8 - 36 4.3 Chapter 4 Linear...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online