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9 Matrix form of the simplex algorithm.
9.1 Initial tableau.
The primal problem
max when
T
Ax b
c x
x
with
n
x ∈ \ ,
m
b ∈ \ , so that A is m × n , yields an augmented system of the form
x
A I b
y
To keep the notation simple, continue to write 0 for each matrix with all entries 0, and also I
for each identity matrix. The order of any matrix can be deduced from its relative position in the
system. For instance, the I in the augmented system must be m × m. With this in mind, the
initial tableau has the form
T
A I b
c
The simplex algorithm assumes that
b
is a feasible solution of the augmented system. Each
i
x in x is a non-basic variable, and each
j
y
in y is a basic variable. It is of course in general not true that b ≥ 0 , but in this case the first
phase of the two-phase method generates a tableau in the form consistent with the simplex
algorithm.
9.2 Subsequent tableaux.
The application of elementary row operations to convert a non-basic variable to a basic variable is
referred to as pivoting. Each elementary row operation corresponds to a matrix, and if several
operations are applied in succession, then some matrix again represents the combined effect. To
avoid loosing track of the meaning of the various matrices use the following example.
Example
Consider the primal problem
1
2
1
2 1
2
max 2 1 when
x
x x
x
x
x
Here 2 1
T
c
b
, and
A
The vertices of the feasible set are
( ) 0, 0 ,
( ) 3, 0 ,
( ) 2, 3 and
( ) 0, 5. The initial tableau
T
A I b
c
is given by
Here
1 2
x , x are non-basic variables, and the slack variables
1 2
y , y are basic variables. The values
are given by
1 2
x = 0, x = 0 , and
1 2
y = 9, y = 5. Four different pivots are possible. Two of the
pivots turn
1
x into a basic variable. The pivot on the first row yields
Now
1 2
x , y are basic variables and
2 1
x , y are non-basic variables. The values are given by
1 2
x = 3, x = 0 , and
1 2
y = 0, y = 2. Geometrically the pivot turns the attention away from the
vertex
( ) 0, 0 and directs it to the vertex
( ) 3, 0. Since the identity matrix has been replaced by
it follows that this matrix represents the combined effect of the elementary row operations. A
quick check confirms that
Instead of observing the change of the identity matrix in the initial tableau, examine how the
identity matrix in the new tableau is created. It must be that
and hence the combined effect of the elementary row operations is represented by the inverse of
the matrix
This matrix is given by the columns in the initial tableau corresponding to the new basic
variables in the appropriate order. Let B denote this matrix and let N denote the matrix
consisting of all the remaining columns in A I
in order from left to right. In the present
example N is given by
N
variables is negative shows that one of the constraints is violated. Since 9 / 3 < 5 /1, the pivot
rule selects the first row as opposed to the second row. The row operations are represented by
1
1
B
−
−
Observe how the order of the columns in the inverse matches the order of the basic variables in
the new tableau. The corresponding columns form the identity matrix in that order.
This time
N
and 0 2
T
B
c
T
N
c
. It follows that
[ ]
1
T
B
c B b
−
and
1
T T
N B
c c B N
−
Exercise
Compute the next tableau by honoring the pivot rule as it applies to the second column of
Extract all the information as in the preceding example. Next, compare this with the tableau and
the conclusions reached when the pivot rule is violated.
9.3 Optimality condition.
Given a tableau with { }
1
m
k … k , such that 1
j
≤ k ≤ n + m , the indices of the basic variables in
the order corresponding to the identity matrix. Let B be the m × m matrix created from A I
by selecting the columns { }
1
m
k … k in this order. Let N be the matrix formed by the remaining
columns in order from left to right. Let 0
T
BN
c c
be of order
( ) 1 × n + m. Define
T
B
c by
selecting from
T
BN
c the entries corresponding to { }
1
m
k … k preserving this order. Let
T
N
c be the
matrix formed by the remaining entries in order from left to right. Let
x
z
y
and define
N
z in
the same manner. Again, since
B N
Bz + Nz = b , it follows that
1 1
B N
z B b B Nz
− −
= − , and
T T 1 T T 1 T 1
B N B N B
c x c B b c c B N z c B b
− − −
= + − =. The current solution has 0
N
z = , and each
competing solution has 0
N
z ≥. It follows that if
1
T T
N B
c c B N
−
− ≤ , then the current solution is
optimal. In the tableau this corresponds to no positive values in the row below the horizontal line.
9.4 Duality.
The optimality condition is equivalent to
T 1 T
B N
c B N c
−
≥. A feasible point λ in the dual problem
satisfies
T
A λ ≥ c and λ ≥ 0. This is equivalent to
T T
λ A ≥ c and 0
T
λ ≥. Using the augmented
matrix this is equivalent to 0
T T
λ A I c
. A reordering leads to
T T T
B N
λ B N c c
This last inequality is equivalent to
T T T T
B N
λ B λ N c c
. When this is compared with
T 1 T
B N
c B N c
−
≥ , the choice
T T 1
B
λ c B
−
= stands out. It is only a matter of backtracking through
the equivalent inequalities to see that
T T 1
B
λ c B
−
= is feasible in the dual problem. The value of
the primal function at the basic solution
1
B
z B b
−
N
z = is given by
T 1
B
c B b
−
. The value of
the dual function at
T T 1
B
λ c B
−
= is given by
T 1
B
c B b
−
. The two values are equal and by the
Verification Theorem,
T T 1
B
λ c B
−
= is a global minimum of the dual problem.
The initial tableau is given by
T
A I b
c
Each subsequent tableau has the form
1 1 1
T T 1 T 1 T 1
B B B
B A B B b
c c B A c B c B b
− − −
− − −
This is the same as
1 1
B
T
T T
z
B A B
b
μ
λ λ
− −
where
T
μ = A λ − c. Also note that
T T T
B B
λ b = c z = c x. It is only if 0
B
z ≥ , λ ≥ 0 , and μ ≥ 0
that λ an optimal solution of the dual problem.
