Lecture 5

# T x1x2s1 40 2x1x2s260 x1x2s1s20

This preview shows page 1. Sign up to view the full content.

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

Unformatted text preview: LP not in standard form is: max z = 4x1 + 3x2 s.t. x1 + x2 ≤ 40 2x1 + x2 ≤ 60 x1, x2 ≥ 0 (leather constraint) (labor constraint) SoluWon: x1=20, x2=20, z = 140 The same LP in standard form is: max z = 4x1 + 3x2 s.t. x1 + x2 + s1 = 40 2x1 + x2 + s2 = 60 x1, x2, s1, s2 ≥ 0 SoluWon: x1=20, x2=20, s1=0, s2=0, z = 140 In summary, if a constraint i of an LP is a ≤ constraint, convert it to an equality constraint by adding a slack variable si to the ith constraint and adding the sign restricWon si ≥ 0. 4.1 – How to Convert an LP to Standard Form A ≥ constraint can be converted to an equality constraint. Consider the formulaWon below: min z = 50 x1 + 20x2 + 30X2 + 80 x4 s.t. 400x1 + 200x2 + 150 x3 + 500x4 ≥ 500 3x1 + 2x2 ≥ 6 2x1 + 2x2 + 4x3 + 4x4 ≥ 10 2x1 + 4x2 + x3 + 5x4 ≥ 8 x1, x2, x3, x4 ≥ 0 To convert the ith ≥ constraint to an equality constraint, deﬁne an excess variable (someWmes called a surplus variable) ei (ei will always be the excess variable for the ith ≥ constraint. We deﬁne ei to be the amount by which ith constraint is over saWsﬁed. 4.1 – How to Convert an LP to Standard Form Transforming the LP on the previous slide to standard form yields: min z = 50 x1 + 20x2 + 30X2 + 80 x4 s.t. 400x1 + 200x2 + 150 x3 + 500x4 – e1 = 500 3x1 + 2x2 ‐ e2 = 6 2x1 + 2x2 + 4x3 + 4x4 ‐ e3 = 10 2x1 + 4x2 + x3 + 5x4 ‐ e4 = 8 xi, ei > 0 (i = 1,2,3,4) In summary, if the i th constraint of an LP is a ≥ constraint, it can be converted to an equality constraint by subtracWng the excess variable ei from...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online