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Topic III The Simplex Method

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Title: Topic III The Simplex Method


1
Topic IIIThe Simplex Method
  • Setting up the Method
  • Tabular Form
  • Chapter(s) 4

2
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

3
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

4
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

5
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

6
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

7
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

8
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

9
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

10
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

11
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)

12
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

13
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

14
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

15
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

16
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)

17
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

18
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

19
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)

20
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

21
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
22
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

23
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

24
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
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