Topic III The Simplex Method

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Topic IIIThe Simplex Method

• Setting up the Method
• Tabular Form
• Chapter(s) 4

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Key Concepts

• The simplex method focuses solely on CPF
  solutions
• For any problem with at least one optimal
  solution, finding one only requires finding a
  best CPF solution
• The simplex method is an iterative algorithm
• Initialization
• Optimality Test
• If no, perform an iteration to find a better
  solution
• If yes, stop
• Whenever possible, the initialization of the
  simplex method chooses the origin to be initial
  CPF solution

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Key Concepts

• Given a CPF solution, it is much quicker
  computationally to gather information about its
  adjacent CPF solutions that about other solutions
• After the current CPF solution is identified, the
  method identifies the rate of improvement in Z
  that would be obtained by moving along an edge to
  an adjacent solution
• Chooses to move along the one with the largest
  rate of improvement in Z
• If none of the edges give a positive rate of
  improvement, then the current CPF solution is
  optimal

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Setting up the Simplex Method

• Convert the functional inequality constraints to
  equivalent equality constraints
• Accomplished by introducing slack variables
• An augmented solution is a solution for the
  original (decision) variables that has been
  augmented by the corresponding values of the
  slack variables
• Example
• x1 4
• Adding slack variable gives x1 x3 4
• Note that these are equivalent iff x3 0

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Setting up the Simplex Method

• Original Model (from Topic II)
• Maximizing Total profit, Z
• Maximize Z 3x1 5x2
• Constraints
• x1 4
• 2x2 12
• 3x1 2x2 18
• Other constraints
• x1 0
• x2 0

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Setting up the Simplex Method

• Augmented form of the model
• Maximizing Total profit, Z
• Maximize Z 3x1 5x2
• Constraints
• x1 x3 4
• 2x2 x4 12
• 3x1 2x2 x5 18
• Other constraints
• xj 0, for j 1, 2, 3, 4, 5

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Setting up the Simplex Method

• The system of functional constraints has 5
  variables and 3 equations
• Number of variables number of equations 5 3
  2
• 2 Degrees of freedom in solving the system (as
  long as there arent any redundant equations)
• Set any two variables to an arbitrary value to
  solve the three equation system
• The simplex method uses zero for this arbitrary
  value
• The two variables set to zero are the nonbasic
  variables
• The other three variables are the basic variables

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Basic Solution

• A basic solution is an augmented corner-point
  solution
• Properties of a basic solution
• Each variable is designated as either a nonbasic
  variable or a basic variable
• The number of basic variables equals the number
  of functional constraints
• The number of nonbasic variables equals the total
  number of variables minus the number of
  functional constraints

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Basic Solution

• A basic solution is an augmented corner-point
  solution
• Properties of a basic solution
• The nonbasic variables are set to zero
• The values of the basic variables are obtained as
  the simultaneous solution of the system of
  equations (functional constrains in augmented
  form)
• The set is often referred to as the basis
• If the basic variables satisfy the nonnegativity
  constraints, the basic solution is a BF solution
• A basic feasible (BF) solution is an augmented
  CPF solution

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Basic Feasible (BF) Solutions

• Two BF solutions are adjacent if all but one of
  their nonbasic variables are the same
• Note that all but one of their basic variables
  are also the same
• Moving from the current BF solution to an
  adjacent one involves switching one variable from
  nonbasic to basic (and vice versa for one other
  variable)
• Adjust the values of the basic variables to
  satisfy the system of equations

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The Simplex Method

• Step 1 Initialization
• Choose x1 and x2 to be the nonbasic variables
  (the variables set to zero)
• Using system of equations, x3, x4, x5 equal 4,
  12, 18
• Thus, the initial BF solution is (0, 0, 4, 12, 18)

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The Simplex Method

• Step 2 Optimality Test
• The objective function is Z 3x1 5x2
• Z 0 for the initial BF solution
• Rate of improvement for x2 is more than x1 (5 gt
  3)
• Increase x2

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Minimum Ratio Test

• Step 2 Optimality Test
• Minimum Ratio Test
• Objective is to determine which basic variable
  drops to zero first as the entering basic
  variable is increased
• The system of equations
• x1 x3 4
• No upper bound on increasing x2
• 2x2 x4 12
• x4 12 2x2
• Thus, x2 6
• 3x1 2x2 x5 18
• x5 18 2x2
• Thus, x2 9
• Since the 2nd equation restricts x2 to 6, x4 is
  the leaving basic variable for this iteration

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Solve for New Solution

• Step 3 Solving for the new BF Solution
• Original System
• Z 3x1 5x2 0
• x1 x3 4
• 2x2 x4 12
• 3x1 2x2 x5 18
• x2 has replaced x4 as the basic variable
• The pattern of coefficients of x4 (0, 0, 1, 0)
  need to become the coefficients of x2

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Solve for New Solution

• Step 3 Solving for the new BF Solution
• How
• Divide constraint equation 2 by 2
• x2 ½x4 6
• Add 5 times this new equation to the objective
  function
• Z 3x1 5/2 x4 30
• Subtract 2 times new equation to constraint
  equation 3
• 3x1 x4 x5 6

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Solve for New Solution

• Step 3 Solving for the new BF Solution
• New System
• Z 3x1 5/2 x4 30
• x1 x3 4
• x2 ½x4 6
• 3x1 x4 x5 6
• New BF Solution
• (0, 6, 4, 0, 6)

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Next Iteration

• Next Iteration Return to Step 2
• Z 30 3x1 5/2 x4
• Z can be increased by increasing x1, but not x4
• Thus, x1 needs to be the next entering basic
  variable
• Minimum Ratio Test
• x1 x3 4
• x1 4
• x2 ½x4 6
• No upper bound on x1
• 3x1 x4 x5 6
• x1 2
• x5 is the leaving basic variable

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Next Iteration

• The pattern of coefficients of x5 (0, 0, 0, 1)
  needs to become the pattern for x1
• Divide constraint equation 3 by 3
• x1 1/3 x4 1/3 x5 2
• Add 3 times this equation to objective function
• Z 3/2 x4 x5 36
• Subtract new equation from constraint equation 1
• x3 1/3 x4 1/3 x5 2

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Next Iteration

• The pattern of coefficients of x5 (0, 0, 0, 1)
  needs to become the pattern for x1
• New System
• Z 3/2 x4 x5 36
• x3 1/3 x4 1/3 x5 2
• x2 ½x4 6
• x1 1/3 x4 1/3 x5 2
• New BF Solution
• (2, 6, 2, 0, 0)

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Final Iteration

• Next iteration
• Z 36 3/2 x4 x5
• If either nonbasic variable x4 or x5 is
  increased, Z would decrease
• Thus, the current BF solution is optimal
• Original variables x1 and x2
• x1 2
• x2 6
• Maximum value of Z 36

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Tabular Form
Start with initial equations
Basic Var Eq Z x1 x2 x3 x4 x5 Right Side
Z 0 1 -3 -5 0 0 0 0
x3 1 0 1 0 1 0 0 4
x4 2 0 0 2 0 1 0 12
x5 3 0 3 2 0 0 1 18
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Tabular Form

• Iterations
• Determine entering basic variable
• Select variable with negative coefficient with
  largest absolute value
• If none, the algorithm is finished
• Draw box around column below this variable as the
  pivot column

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Tabular Form

• Iterations
• Minimum ratio test
• Select each coefficient in pivot column that is
  positive
• Divide each coefficient into corresponding right
  side entry
• Identify smallest ratio
• Basic variable for that row is leaving basic
  variable
• Replace it by entering basic variable column of
  table
• Box the row and call it the pivot row
• The number in both pivot row and pivot column is
  pivot number

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Tabular Form

• Iterations
• New BF solution
• Divide pivot row by pivot number (use this total
  in next two steps)
• For each other row (including row 0) that has a
  negative coefficient in the pivot column
• Add to this row the product of absolute value of
  this coefficient and new pivot row
• For each other row that has a positive
  coefficient in the pivot column
• Subtract from this the product of its coefficient
  and the new pivot row