Lectures on Simplex Algorithm ENGR 300 Lecture

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Lectures on Simplex AlgorithmENGR 300Lecture
2LP Graphical Method SolutionElif Kongar,
Ph.D.April 6, 2006Thursday 100 215, CARH
253 - CarlsonSchool of EngineeringUniversity of
BridgeportSpring 2006
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A Typical LP Problem

• Two types of products P1 and P2
• By two workstations W1 and W2
• Each station is visited by both product types
• If W1 is dedicated completely to P1, it can
  process 40 units per day
• If W1 is dedicated completely to P2, it can
  process 60 units per day.
• Similarly, W2 can produce daily 50 units of
  product P1 and 50 units of product P2
• Profit from one unit of product P1 is 200
• Profit from one unit of P2 is 400
• Assuming that the company can dispose its entire
  production
• How many units of each product should the company
  produce on a daily basis to maximize its profit?

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LP Model

• max f(X1, X2) 200 X1 400 X2
• s.t.
• 1/40 X1 1/60 X2 1.0
• 1/50 X1 1/50 X2 1.0
• Xi 0 i 1,2

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Feasible Regions of Two-Var LP's

• max f(X1, X2) 200 X1 400 X2
• s.t.
• 1/40 X1 1/60 X2 1.0
• 1/50 X1 1/50 X2 1.0
• Xi 0 i 1,2

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Objective Function
a
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Optimal Solution
1/50 X1 1/50 X2 0 with the X2 axis X1opt 0
X2opt 50. The maximal daily profit is f(X1opt ,
X2 opt) 200 . 0 400 . 50 20,000 Notice
that the optimal point is one of the corner
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Extreme point cont 4

• In a similar fashion, in the n-dim space, a point
  is uniquely defined by n linear equations which
  are linearly independent, i.e.,
• a11X1 a12X2 a1nXn b1
• a21X1 a22X2 a2nXn b2
• an1X1 an2X2 annXn bn

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LP's in standard form

• To further exploit the previous characterization
  of extreme points as the solution to n binding
  linearly independent constraints, we must define
  the concept of LP's in standard form. An LP is
  said to be in standard form, if
• (i) all technological constraints are equality
  constraints, and
• (ii) all the variables have a nonnegativity sign
  restriction.

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LP in Standard Form cont

• AX b (1)
• X 0
• with m technological constraints and n variables.
  In standard form, it becomes
• AX IS b (2)
• X, S 0

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Standard form example

• max f(X1, X2) 200 X1 400 X2
• s.t.
• 1/40 X1 1/60 X2 S1 1.0
• 1/50 X1 1/50 X2 S2 1.0
• Xi , Si 0 i 1,2

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Basic Solutions

• max f(X1, X2) 200 X1 400 X2
• s.t.
• 1/40 X1 1/60 X2 1.0
• 1/50 X1 1/50 X2 1.0
• Xi 0 i 1,2
• A and B are adjacent points

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The Simplex Algorithm
If an LP has a bounded optimal solution, then
there exists an extreme point of the feasible
region which is optimal. Extreme points of the
feasible region of an LP correspond to basic
feasible solutions of its standard form
representation.
N variables, m technological constraints
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Why do we need the Simplex Algorithm

• 10 variables (in standard form) and 3
  technological constraints can have up to 120
  extreme points, while an LP with 100 variables
  and 20 constraints can have up to 5.36x1020
  extreme points. And yet, this is a rather small
  LP!

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Simplex Algorithm
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Obtaining the canonical form with respect to the
new bfs Pivoting the entering variable

• To perform this transformation, first, we rewrite
  the original LP equations in the form
• z 200 X1 - 400X2 0.0
• 1/40 X1 1/60 X2 S1 1.0
• 1/50 X1 1/50 X2 S2 1.0

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Simplex Algorithm

• Testing for the optimality Since the objective
  function row includes negative values the
  solution is not optimal
• Selecting the entering variable Since x2 has the
  most negative coefficient in the objective
  function row, x2 is selected as the entering
  variable (enters the basis).
• Selecting the leaving variable (minimum ratio
  test) Since the ratio of s2 has the minimum
  positive value, s2 is selected as the leaving
  variable (leaves the basis).

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Simplex Algorithm cont

• Algebraic calculations
• We divide s2 row by 1/50 (which is the pivot
  value marked with a red circle) and place it in
  a New Simplex Tableau (Tableau 2).
• Multiply old s2 row (new x2 row) with (-1/60) and
  to s1 row.
• Multiply old s2 row (new x2 row) with (400) and
  to z row.

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Initial Simplex Tableau (Tableau 1)
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Optimal Simplex Tableau (Tableau 2)
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Testing for the optimality

• Testing for the optimality Since the objective
  function row does not include negative values the
  solution is optimal.
• The optimal values for the basic variables are
  s1 1/6 and x2 50, and the optimal objective
  value is z 20,000.
• Hence, the optimal solution is x1 0 and x2
  50.0, with a corresponding optimal objective
  value of z 20,000.

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LP TOOLS

• LINGO
• Solver

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