Linear Programming Simplex Method, Lecture Notes - Mathematics, Study notes of Linear Programming

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Linear Programming: Chapter 2

The Simplex Method

Robert J. Vanderbei

October 17, 2007

Operations Research and Financial Engineering Princeton University Princeton, NJ 08544 http://www.princeton.edu/∼rvdb

Simplex Method

• maximize −x 1 + 3 x 2 − 3 x An Example.
• subject to 3 x 1 − x 2 − 2 x 3 ≤
  • − 2 x 1 − 4 x 2 + 4 x 3 ≤ - x 1 − 2 x 3 ≤
  • − 2 x 1 + 2 x 2 + x 3 ≤
    • 3 x 1 ≤ - x 1 , x 2 , x 3 ≥

Dictionary Solution is Feasible

maximize ζ = −x 1 + 3 x 2 − 3 x 3 subject to w 1 = 7 − 3 x 1 + x 2 + 2 x 3 w 2 = 3 + 2 x 1 + 4 x 2 − 4 x 3 w 3 = 4 − x 1 + 2 x 3 w 4 = 8 + 2 x 1 − 2 x 2 − x 3 w 5 = 5 − 3 x 1 x 1 , x 2 , x 3 , w 1 , w 2 , w 3 w 4 w 5 ≥ 0.

Notes:

• All the variables in the current dictionary solution are nonnegative.
• Such a solution is called feasible.
• The initial dictionary solution need not be feasible—we were just lucky above.

Simplex Method—First Iteration

• If x 2 increases, obj goes up.
• How much can x 2 increase? Until w 4 decreases to zero.
• Do it. End result: x 2 > 0 whereas w 4 = 0.
• That is, x 2 must become basic and w 4 must become nonbasic.
• Algebraically rearrange equations to, in the words of Jean-Luc Picard, ”Make it so.”
• This is a pivot.

Simplex Method—Second Pivot

Here’s the dictionary after the first pivot:

• Now, let x 1 increase.
• Of the basic variables, w 5 hits zero first.
• So, x 1 enters and w 5 leaves the basis.
• New dictionary is...

Simplex Method—Final Dictionary

• It’s optimal (no pink)!
• Click here to practice the simplex method.
• For instructions, click here.

Unboundedness

Consider the following dictionary:

• Could increase either x 1 or x 3 to increase obj.
• Consider increasing x 1.
• Which basic variable decreases to zero first?
• Answer: none of them, x 1 can grow without bound, and obj along with it.
• This is how we detect unboundedness with the simplex method.

Initialization

Consider the following problem:

maximize − 3 x 1 + 4 x 2 subject to − 4 x 1 − 2 x 2 ≤ − 8 − 2 x 1 ≤ − 2 3 x 1 + 2 x 2 ≤ 10 −x 1 + 3 x 2 ≤ 1 − 3 x 2 ≤ − 2 x 1 , x 2 ≥ 0.

Phase-I Problem

• Modify problem by subtracting a new variable, x 0 , from each constraint and
• replacing objective function with −x 0

Initialization—First Pivot

Applet depiction shows both the Phase-I and the Phase-II objectives:

• Dictionary is infeasible even for Phase-I.
• One pivot needed to get feasible.
• Entering variable is x 0.
• Leaving variable is one whose current value is most negative, i.e. w 1.
• After first pivot...

Initialization—Second Pivot

Going into second pivot:

• Feasible!
• Focus on the yellow highlights.
• Let x 1 enter.
• Then w 5 must leave.
• After second pivot...

End of Phase-I

Current dictionary:

• Optimal for Phase-I (no yellow highlights).
• obj = 0, therefore original problem is feasible.

Phase-II

Current dictionary:

For Phase-II:

• Ignore column with x 0 in Phase-II.
• Ignore Phase-I objective row.

w 5 must enter. w 4 must leave...