{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Solutions

# Solutions - CSA 273 Solutions to Homework Set#9 Problem#1...

This preview shows pages 1–2. Sign up to view the full content.

CSA 273 Solutions to Homework Set #9 Problem #1 Adding artificial variables a 1 and a 2 to each of the constraints and a surplus variable e 2 to the second constraint, the "artificial" problem in standard form becomes: Minimize Z=3x 1 + 2x 2 + 4x 3 + Ma 1 + Ma 2 such that 2x 1 + x 2 + 3x 3 + a 1 = 10 3x 1 +3x 2 + 5x 3 -e 2 + a 2 = 20 x 1 ,x 2 ,x 3 ,a 1 ,a 2 ,e 2 0 Start out with the tableau: Z x 1 x 2 x 3 e 2 a 1 a 2 RHS Row 0 1 -3 -2 -4 0 -M -M 0 Row 1 0 2 1 3 0 1 0 10 Row 2 0 3 3 5 -1 0 1 20 To get an initial feasible solution with basic variables z,a 1 and a 2 , the coefficients of a 1 and a 2 must be cleared in row 0. This can be done by applying the eros: New Row 0 = Old Row 0 + M*Old Row1 + M*Old Row 2 The other rows would remain unchanged. Our Initial Simplex Tableau: Basic Z x 1 x 2 x 3 e 2 a 1 a 2 RHS Solution Ratio Row 0 1 5M-3 4M-2 8M-4 -M 0 0 30M Z=30M Row 1 0 2 1 3 0 1 0 10 a 1 =10 10/3=3.33 Row 2 0 3 3 5 -1 0 1 20 a 2 =20 20/5=4 has an initial basic feasible solution of a 1 =10, a 2 =20, x i =0 for i=1,2,3, e 2 =0 , and Z=30M to the artificial problem. The entering basic variable should be x 3 since it has the largest coefficient in Row 0. The leaving basic variable should be a 1 by the ratio test. Problem #2

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 3

Solutions - CSA 273 Solutions to Homework Set#9 Problem#1...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online