# College of Management NCTU Operation Research I Fall 2008 Chap 4 The Simplex Method The Essence of the Simplex Met

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```					College of Management, NCTU               Operation Research I                          Fall, 2008

Chap 4 The Simplex Method

The Essence of the Simplex Method
Recall the Wyndor problem
Max Z = 3x1 + 5x2
S.T. x1              ≤ 4
2x2 ≤ 12
3x1 + 2x2 ≤ 18
x1, x2 ≥ 0

8 corner point solutions. 5 out of
them are CPF solutions.
Each corner-point solution lies at the intersection of two constraint boundaries.
General speaking, for a LP problem with n variables, each of its
corner-point solutions lies at the intersections of n constraint boundaries.
For any LP with n decision variables, two CPF solutions are adjacent to each
other if they share n – 1 constraint boundaries.
For example, (0, 0) and (0, 6) are adjacent. (0, 0) and (4, 3) are not.
Two adjacent CPF solutions are connected by a line segment that lies on these
same shared constraint boundaries, call an edge of the feasible region.
Note that two edges emanate from each CPF solution. Thus, each CPF solution
has two adjacent CPF solutions (in the Wyndor example).
How many adjacent CPF are there for an n decision variable case?

Optimality Test—Consider a LP problem that possesses at least one optimal
solution. If a CPF solution has no adjacent CPF solutions that are better, then it
must be an optimal solution.
(2, 6) is the optimal solution in the Wyndor Example.

Solving the Wyndor Example (from a geometric viewpoint)
Initialization—Choose (0, 0) as the initial CPF solution for convenient reason.
Optimality Test—(0, 0) is not an optimal solution (adjacent solutions are
better).

Jin Y. Wang                                                                              Chap4-1
College of Management, NCTU                Operation Research I                     Fall, 2008

Iteration 1—Move to a better adjacent CPF solution, (0, 6).
Between the two edges of the feasible region that emanate from (0, 0),
choose to move along the edge that leads up the x2 axis.
Why choose this one?

Stop at the first new constraint boundary: 2x2 = 12.
Solve for the intersection of the new set of constraint boundaries: (0, 6).
Optimality Test: still not an optimal

Iteration 2—Following the same procedures, move to a better adjacent CPF
solution (2, 6).
Optimality Test—(2, 6) is an optimal solution.

Key Solution Concepts Behind the Simplex Method
Concept 1: For any problem with at least one optimal solution, finding one
requires only finding a best CPF solution. Thus, we focus solely on CPF
solutions.
The number of CPF solutions is finite. This simplifies the searching space
tremendously.
Concept 2: The simplex method is an iterative algorithm.

Find an initial
CPF solution

Yes       Is Current
Stop            CPF solution
optimal?

No
Perform an
iteration to find a
better CPF

Jin Y. Wang                                                                          Chap4-2
College of Management, NCTU                Operation Research I                      Fall, 2008

Concept 3: Whenever possible, chooses the origin to be the initial CPF solution.

Concept 4: Given a CPF solution, it is much quicker computationally to gather
While moving from the current CPF solution to a better one, it always
chooses a CPF solution that is adjacent to the current one.
The entire path followed to eventually reach an optimal solution is along the
edges of the feasible region.
Concept 5: After the current CPF solution is identified, the simplex method
examines each of the edges of the feasible region that emanate from the current
CPF solution.
Among the edges with a positive rate of improvement in Z, it chooses to
move along the one with the largest improvement rate.
Concept 6: How the optimality test is performed efficiently?
Check the improvement rate of each edge emanate from the current CPF
solution.
If all of them are negative, the current CPF solution is optimal.
Solving the Wyndor Example (a algebraic procedure)
Max Z = 3x1 + 5x2
S.T. x1              ≤ 4
2x2 ≤ 12
3x1 + 2x2 ≤ 18
x1, x2 ≥ 0

Setting up the Simplex method
Introducing slack variables to transfer the functional inequalities into
equalities and obtain the augmented form of the model.
Max Z = 3x1 + 5x2
St x1           + x3           =4
2x2       + x4      = 12
3x1 + 2x2             + x5 = 18
x1, x2, x3, x4, x5 ≥ 0

If a slack variable equals 0 in the current solution, then this solution lies on
the constraint boundary for the corresponding functional constraint.
A value greater than 0 means that the solution lies on the feasible side of this
constraint boundary.

Jin Y. Wang                                                                           Chap4-3
College of Management, NCTU               Operation Research I                       Fall, 2008

What if the value of a slack variable is less than 0?

More terminologies
An augmented solution is a solution for the original variables and the slack
variables (feasible or infeasible).

A basic solution is an augmented corner-point solution (feasible or
infeasible). For example, (4, 6) is a corner-point solution and (4, 6, 0, 0, -6)
is the corresponding basic solution.
A basic feasible (BF) solution is an augmented CPF solution. For example
(0, 6) is a CPF solution and (0, 6, 4, 0, 6) is a BF solution.
There are 3 equations with 5 variables. So, we can have 2 free variables. That is,
if two of the variables are set, the value of the other three variables can be
calculated.
The simplex method uses zero for this arbitrary value.
That is, two of the variables (called the nonbasic variables) are set to zero,
and then the simultaneous solution of the three equations for the other three
variables (called basic variables) is a basic solution.
Some properties of a basic solution
Each variable is designated as either a basic or a non-basic variable.
The number of basic variables equals to the number of functional
constraints.
Therefore, number of non-basic variables equals the total number of
variables (including slack variables) minus number of functional
constraints.
The non-basic variables are set to zero.
The values of the basic variables are obtained as the simultaneous solution
of the system of equations.
The set of basic variables is often referred to as the basis.
If the basic variables satisfy the non-negativity constraints, the basic
solution is a BF solution.
Let x1 = x4 = 0 (nonbasic variables), obtain x3 =4, x2 = 6, and x5 = 6 (basic
variables). It is a BF solution.

Jin Y. Wang                                                                              Chap4-4
College of Management, NCTU                 Operation Research I                      Fall, 2008

Two BF solutions are adjacent if all but one of their basic (nonbasic) variables
are the same. (Note: not the values.)
That is, moving from the current BF solution to an adjacent one involves
switching one variable from nonbasic to basic and vice versa.

(0, 6, 4, 0, 6)            (2, 6, 2, 0, 0)

(4, 3, 0, 6, 0)

(4, 0, 0, 12, 6)
(0, 0, 4, 12, 18)
We could deal with the objective function equation at the same time as the new
constraint equation. Note that Z is not really a decision variable.
Max Z
S.T. Z - 3x1 - 5x2 = 0
x1      =4
2x2 = 12
3x1 + 2x2 = 18
x1, x2 ≥ 0
The Algebra of the Simplex Method
Recall the Wyndor Problem again.
Initialization
Max Z
S.T. Z - 3x1 - 5x2                =0     (0)
x1         + x3       =4     (1)
2x2      + x4      = 12 (2)
3x1 + 2x2            + x5 = 18 (3)
x1, x2, x3, x4, x5 ≥ 0
Pick x1 and x2 be the nonbasic variables (starting from the origin). x1 = x2 = 0.
Easily obtain x3 = 4, x4 = 12, and x5 = 18 (x3, x4, and x5 are basic variables).
Why it is easy?
Each equation has just one basic variable, which has a coefficient 1,
and this basic variable does not appear in any other equation (and the
objective function).

Jin Y. Wang                                                                            Chap4-5
College of Management, NCTU               Operation Research I                     Fall, 2008

Optimality test
Z - 3x1 - 5x2 = 0       Z (to be maximized) = 3x1 + 5x2
Can we improve the objective value any more (Z = 0 currently)?
Determining the direction of movement
If we change the value of x1, or x2, can we increase the value of objective
function? Which one looks more promising?

Pick        to increase, called the entering basic variable.
Determining where to stop. (How far could we increase the entering variable?)
x1 remain zero.
x1       + x3          =4        x3 = 4             no bound on x2
2x2     + x4      = 12      x4 = 12 –2x2        x2 ≤ 6     minimum
3x1 + 2x2          + x5 = 18       x5 = 18 – 2x2      x2 ≤ 9
x2 can be increased to 6, where x4 will drop to zero.
x4 is called the leaving basic variable (since it becomes zero).
These procedures are referred to as the minimum ratio test.
Only consider the equations where the coefficient of the entering basic
variable is strictly positive (>0).
Solving for the New BF Solution
Initial BF solution: Nonbasic: x1 = x2 = 0, Basic: x3 = 4, x4 = 12, x5 = 18.
New BF solution: Nonbasic: x1 = x4 = 0, Basic: x3 = ?, x2 = ?, x5 = ?
We would like to convert the system of equations to a more convenient form
for obtaining the variable values and conducting the optimality test.
Objective function equation only contains the nonbasic variables.
Each functional constraint equation contains only one basic variable
and its coefficient equals to 1.
The original algebra form
Z – 3x1 – 5x2              =0               (0)
x1       + x3         =4               (1)
2x2    + x4      = 12             (2)
3x1 + 2x2         + x5 = 18             (3)

Jin Y. Wang                                                                         Chap4-6
College of Management, NCTU              Operation Research I                       Fall, 2008

Perform the elementary algebraic operations (as learned in linear algebra) to
produce the convenient form (called Gaussian elimination).
Multiply (or divide) an equation by a nonzero constant.
Add (or subtract) a multiple of one equation to (or from) another
equation.
Note that the coefficients of x2 are -5, 0, 2, and 2. We want these
coefficients to become 0, 0, 1, and 0.

Optimality test (for the new BF solution)
Z – 3x1 + 5/2x4 = 30        Z = 30 + 3x1 – 5/2x4
Do we need to continue?

Determine the direction of movement

Determine where to stop

Jin Y. Wang                                                                          Chap4-7
College of Management, NCTU                 Operation Research I                     Fall, 2008

Solving for the new BF solution

Optimality Test
Do we need to continue?

Optimal solution is

Simplex Method in Tabular Form
Logic is identical to that of the algebraic form.
Only records the essential information. It can significantly reduce the effort of
calculations.
Z – 3x1 – 5x2               =0
x1       + x3          =4
2x2      + x4      = 12
3x1 + 2x2           + x5 = 18

Basic                    Coefficient of:                      Right Side
Variable        Z      x1    x2      x3              x4   x5
Z               1      -3   -5       0               0    0        0
x3              0      1     0       1               0    0        4
x4              0      0     2       0               1    0       12
x5              0      3     2       0               0    1       18
Optimality test
If every coefficient in Z row (row 0) is nonnegative ( ≥ 0).

Jin Y. Wang                                                                           Chap4-8
College of Management, NCTU                Operation Research I                       Fall, 2008

Iteration
Determining the entering basic variable by selecting the variable with the
negative coefficient having the largest absolute value in row 0. Put a box
around the column below this coefficient, and call this the pivot column.
Basic                    Coefficient of:                   Right Side
Variable      Z        x1      x2      x3     x4      x5
Z          1        -3     -5       0       0      0           0
x3         0        1       0       1       0      0           4
x4         0        0       2       0       1      0          12
x5         0        3       2       0       0      1          18

Determining the leaving basic variable by applying the minimum ratio test.
Pick out each coefficient in the pivot column that is strictly positive
(>0).
Divide each of these coefficients into the right side entry for the same
row.
Identify the row that has the smallest of these ratios.
Put a box around this row and call it the pivot row. Also, call the
number that is in both boxes the pivot number.
Basic                Coefficient of:             Right              Ratio
Variable       Z     x1 x2 x3 x4            x5    Side
Z         1     -3 -5 0          0     0       0     0
x3        0     1    0    1      0     0       4     4
x4        0     0    2    0      1     0      12     12/2=6      min
x5        0     3    2    0      0     1      18     18/2=9

The basic variable for the pivot row is the leaving basic variable, so replace
that variable by the entering basic variable in the basic variable column of
the next simplex tableau.
Use elementary row operations to construct a new simplex tableau in proper
form by using Gaussian elimination.
Divide the pivot row by the pivot number.

Jin Y. Wang                                                                            Chap4-9
College of Management, NCTU             Operation Research I                      Fall, 2008

Use elementary row operations to make other coefficients in pivot column
to be zero.
Basic           Coefficient of:       Right          Ratio
Variable Z      x1 x2 x3 x4 x5 Side
Z        1  -3 0       0 5/2 0        30
x3       0   1    0    1      0  0    4            4/1=4
x2       0   0    1    0 1/2 0        6
x5       0   3    0    0 -1 1         6            6/3=2

Basic                 Coefficient of:     Right            Ratio
Variable      Z      x1 x2 x3        x4 x5 Side
Z        1      0    0    0 3/2    1   36
x3       0      0    0    1 1/3 -1/3    2
x2       0      0    1    0 1/2    0    6
x1       0      1    0    0 -1/3 1/3    2

Tie for the entering basic variable
Z = 3x1 + 3x2
Tie for the leaving variable—degeneracy
Z – 3x1 – 5x2 = 0
x1       + x3 = 4
2x2      + x4 = 12
3x1 + 2x2            + x5 = 12

x2 is the entering variable, x4 and x5 reach zero simultaneously.
All the tied basic variables reach zero simultaneously. The one not chosen to be
the leaving basic variable also will have a value of zero in the new BF solution.
If x4 is chosen as the leaving basic variable

Z – 3x1     + 5/2x4      = 30
x1     + x3              =4
x2     + 1/2x4       =6
3x1            – x4 + x5 = 0
So, the value of basic variable x5 is zero (called a degenerate basic variable).

Jin Y. Wang                                                                        Chap4-10
College of Management, NCTU                  Operation Research I                   Fall, 2008

If the degenerate basic variable retains its value of zero until it is chosen to be a
leaving basic variable, the corresponding entering basic variable also must
remain zero, so the value of Z must remain unchanged.
If x1 is the entering basic variable and x5 is the leaving basic variable, x1 will
remain zero because any increment of x1 will result in a negative value of x5.

If Z may remain the same rather than increase at each iteration, the simplex
method may then go around a loop.
Rarely happen and ease to avoid.

No Leaving basic variable—unbounded Z
x1 is the entering basic variable
Z –3x1 + 5x5 = 30
-x1 + x3 = 4
-2x1 + x4 = 9
Multiple optimal solution
If there are multiple optimal solutions, two of them must be CPF.
How to tell and how to find the other optimal solutions?
At least one of the nonbasic variables has a coefficient of zero in the final
objective row. Increasing any such variable will not change the value of Z.
Just perform one of iteration of simplex method will find it.
Z – 0x3 + x5                         = 18
x1 +         x3                 =4
3x3 + x4     – x5 = 6
x2 – 3/2x3       + 1/2x5 = 3

Jin Y. Wang                                                                          Chap4-11
College of Management, NCTU                 Operation Research I                Fall, 2008

Rewrite the algebraic form

All optimal solution can be written as linear combination of these two CPF.
(x1, x2, x3, x4, x5 ) = w1(2, 6, 2, 0, 0) + w2(4, 3, 0, 6, 0) or
(x1, x2) = w1(2, 6) + w2(4, 3)
w1 + w2 = 1, w1, w2 ≥ 0

Not all models can be stated in standard form
It causes the difficulty of finding the initial BF solution.
Equality constraints
Adopt 3x1 + 2x2 = 18 instead of 3x1 + 2x2 ≤ 18 in the Wyndor Problem.
Replace this equality by 3x1 + 2x2 ≤ 18 and 3x1 + 2x2 ≥ 18 is workable. But it is
not a good idea since the increment of number of constraints.
After introducing slack variables, we have
Max Z = 3x1 + 5x2
S.T. x1           + x3      =4
2x2       + x4 = 12
3x1 + 2x2           = 18
x1, x2, x3, x4 ≥ 0
It is not easy to get the initial feasible solution anymore.
The origin point is not feasible.
Introduce an artificial variable x5 and construct the artificial problem
Max Z = 3x1 + 5x2
S.T. x1           + x3           =4
2x2        + x4      = 12
3x1 + 2x2             + x5 = 18
x1, x2, x3, x4, x5 ≥ 0
Assign an overwhelming penalty to having x5 > 0 by changing the objective
function Z = 3x1 + 5x2 to Z = 3x1 + 5x2 – Mx5 (M is a huge positive number, Big M
method).

Jin Y. Wang                                                                        Chap4-12
College of Management, NCTU               Operation Research I                     Fall, 2008

x5 needs to be zero to assure the feasibility. So, impose a huge penalty when
it becomes non-zero.
Max Z = 3x1 + 5x2 – Mx5
S.T. x1        + x3                    =4
2x2      + x4               = 12
3x1 + 2x2           + x5          = 18
x1, x2, x3, x4, x5 ≥ 0
Simplex method will try very hard not to have x5 > 0 in the optimal solution.
The initial basic variables are x3, x4, and x5. So, we have to rewrite the objective
function in term of nonbasic variables.

Z = –18M + (3M+3)x1 + (2M+5)x2
Please finish the rest procedures (in tabular or algebraic form). See page 128 for
details.
What does it mean by having at least one non-zero artificial variable when the
simplex method stops?

Negative right hand side
Multiply both sides by –1.

Functional constraints in ≥ form
0.6x1 + 0.4x2 ≥ 6
Introduce a surplus variable x5 to make it be equality. Then, introduce an
artificial variable x6 and treat it as before.
0.6x1 + 0.4x2            ≥ 6
0.6x1 + 0.4x2 – x5       = 6
0.6x1 + 0.4x2 – x5 + x6 = 6
Minimization
Multiply the objective function by –1
Min Z = 0.4x1 + 0.5x2
Max –Z = – 0.4x1 – 0.5x2

Jin Y. Wang                                                                         Chap4-13
College of Management, NCTU                   Operation Research I             Fall, 2008

Min Z = 0.4x1 + 0.5x2
S.T. 0.3x1 + 0.1x2 ≤ 2.7
0.5x1 + 0.5x2 = 6
0.6x1 + 0.4x2 ≥ 6
x1, x2 ≥ 0
Apply above procedures
Max –Z = –0.4x1 – 0.5x2 – Mx4 – Mx6
S.T. 0.3x1 + 0.1x2 + x3                          = 2.7
0.5x1 + 0.5x2         + x4              =6
0.6x1 + 0.4x2                – x5 + x6 = 6
x1, x2, x3, x4, x5, x6 ≥ 0
Rewrite the objective function in term of nonbasic variables and solve by the
simplex method

Jin Y. Wang                                                                     Chap4-14
College of Management, NCTU               Operation Research I                        Fall, 2008

The big M method requires extra calculation after the initial BF solution (for the
original problem) is found.
When the artificial variables are equal to 0, they remain zero for the rest simplex
iterations.
Summary of the Two-Phase Method
Initialization: Revise the constraints by introducing artificial variables.
Phase 1: The objective is to find a BF solution for the real problem.
Min Z = sum of artificial variables, subject to revised constraints
Phase 2: The objective is to find an optimal solution for the real problem.
Phase 1 problem for radiation therapy example

Jin Y. Wang                                                                            Chap4-15
College of Management, NCTU                Operation Research I                     Fall, 2008

Phase 2 problem for radiation therapy example
Drop the artificial variables and restore the original objective function.

Variables with a bound or with no bound
xj ≥ L
Let xj’ = xj – L so xj’ ≥ 0 and can be treated as usual

xj ≤ U
Let xj’ = U – xj so xj’ ≥ 0 and can be treated as usual

xj does not have a bound
xj = xj+ – xj-   , xj+ , xj- ≥ 0

Jin Y. Wang                                                                          Chap4-16
College of Management, NCTU                  Operation Research I                 Fall, 2008

Shadow Price—Find managerial decisions on resource allocations.
The shadow price for resource i (denoted by yi* ) measures the marginal value of
this resource.
The rate which Z could be increased by (slightly) increasing the amount of this
Shadow price yi* equals to the coefficient (positive value) of the ith slack
variable in the final objective row.
The final iteration of the Wyndor example
Z            +3/2x4    +x5    = 36
x3 +1/3x4 –1/3x5 = 2
x2      +1/2x4         =6
x1            –1/3x4 +1/3x5 = 2

If the unit cost for resource 2 is 1, do you want to go for it?

Notice that the increment of resources is not unlimited.

Jin Y. Wang                                                                        Chap4-17
College of Management, NCTU           Operation Research I                    Fall, 2008

Binding constraints are:

Sensitivity Analysis—Evaluate estimates of model parameters
Define the allowable range of parameters without changing the optimal solution
(or optimal basis).
Sensitive parameters: The parameters that cannot be changed without changing
the optimal solution.
As for bi:

As for ci:

Jin Y. Wang                                                                    Chap4-18
College of Management, NCTU            Operation Research I   Fall, 2008

As for aij:

Sensitivity report provided by the Excel Solver

Jin Y. Wang                                                    Chap4-19

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