Operations Research
Instructor: Mohamed Omar
Handout : Simplex Method
Math 187
Assumptions: The LP problem is of the form
max
z = cT x
s.t.
Ax = b
x≥0
where A is an m × n matrix, n ≥ m, rank(A)=m. A feasible basis B is known. We denote
the set {1, 2, . . . , n}\B by N .
The Algorithm
1. Construct the feasible tableau for the basis B. This is of the form:
z
−
X
cj xj
=v
aij xj
= bi
j∈N
xi
+
X
i∈B
j∈N
2. (Check optimality) If cj ≤ 0 for all j ∈ N , STOP. The basic solution determined by
B is optimal.
3. (Choose an entering variable) Choose k ∈ N with ck > 0.
4. (Check unboundedness) If aik ≤ 0 for all i ∈ B, STOP. The problem is unbounded.
5. (Choose a leaving variable) Choose r ∈ B such that ark > 0 and
br
bi
= min
i∈B, aik >0 aik
ark
6. (Pivot) Pivot on (r, k). That is, replace B by (B ∪ {k})\{r}. Return to STEP 1 with
the new B.
Example. Consider the linear programming problem
max
2x1
+ 3x2
s.t.
2x1
+ x2
x1
+ x2
− x1
+ x2
x1 ,
x2 ,
+ x3
= 10
+ x4
x3 ,
x4 ,
1
=6
+ x5
=4
x5
≥0
Observe that B = {3, 4, 5} is a basis with corresponding basic feasible solution x∗ =
(0 0 10 6 4)T . The corresponding tableau is
z
− 2x1
2x1
− 3x2
+ x2
x1
+ x2
− x1
+ x2
=0
= 10
+ x3
+ x4
=6
+ x5
=4
Observe that c1 > 0, so x1 will enter the basis. To determine what leaves the basis, we
compute
b3 b4
10 6
,
= min
,
= 5,
min
a31 a41
2 1
(recall this comes from the fact that if we
x1 by t and leave the other non-basic
 increase
t
 0 


∗

variables 0, we have the solution x (t) = 
10 − 2t which is feasible only for nonnegative
 6−t 
4+t
values of t for which t ≤ 4). This means x3 leaves the basis. Our new basis is B = {1, 4, 5}.
The corresponding tableau is
Row z + Row 1 :
1
Row1 :
2
1
Row 2 − Row 1 :
2
1
Row 3 + Row 1 :
2
z
x1
− 2x2
1
+ x2
2
1
+ x2
2
3
+ x2
2
+ x3
1
+ x3
2
1
− x3
2
1
+ x3
2
= 10
=5
+ x4
=1
+ x5
=9
Observe that c2 > 0, so x2 will enter the basis. To determine what leaves the basis, we
compute
1
9
5
min
,
,
= 2,
1/2 1/2 3/2
so x4 becomes nonbasic. Our new basis is then B = {1, 2, 5}. We need to compute the
tableau for this basis. This is
z
+ x1
+ x2
− x3
+ x3
+ 4x4
− x4
= 14
=4
− x3
+ 2x4
=2
+ 2x3
− 3x4
2
x5
=6
Observe c3 > 0, so x3 will enter the basis. To determine what leaves the basis, we compute
4 6
min
= 3,
,
1 2
Thus x5 leaves the basis. The new basis is thus {1, 2, 3}. The corresponding tableau is
5
+ x4
2
1
+ x4
2
1
+ x4
2
3
− x4
2
z
+ x1
+ x2
+ x3
1
+ x5
2
1
− x5
2
1
+ x5
2
1
+ x5
2
= 17
=1
=5
=3
Since cj < 0 for every j ∈ N , we have that the basic feasible solution corresponding to the
basis B = {1, 2, 3} gives an optimal solution to the linear program with optimal value 17.
3