The simplex method in matrix form
EXAMPLE
maximize 4x1 + 3x2
subject to x1 − x2 ≤ 1
2x1 − x2 ≤ 3
x2 ≤ 5
x1 , x2 ≥ 0.
Form the initial dictionary:
ζ = 4x1 + 3x2
x1 − x2 + w1
=1
2x1 − x2
+ w2
=3
x2
+ w3 = 5
The initial basic indices are B = (3, 4, 5) , the initial nonbasic indices are N = (1, 2).
The coefficient matrices are:






1 −1 1 0 0
1 0 0
1 −1
A = 2 −1 0 1 0 , B = 0 1 0 , N = 2 −1 .
0 1 0 0 1
0 0 1
0 1
 ∗  
∗ x3
1
z
−4
∗
∗
∗




.
The primal basic entries are: xB = x4 = 3 . The dual basic entries are: zN = 1∗ =
z
−3
2
x∗5
5
Step 1
Primal side
∗
zN
New basic index: j = 1
 
1
1
Step direction: ∆ = B −1 N
= 2
0
0
n
Step size: t = max−1
∆3
x∗3
= 11 ,
∆4
x∗4
= 23 ,
∆5
x∗5
=
0
5
o
=1
New nonbasic index: i = 3
     
1
1
0
x∗B − t∆ = 3 − 2 = 1
5
0
5    
x∗1
1
∗
∗



Updated basic vector: xB = x4 = 1
x∗5
5
Updated index lists: B = (1, 4,

1 0
Updated matrices: B = 2 1
0 0
Dual side
−4
=
has negative entries
−3
5) , N = (3, 2)



0
1 −1
0 , N = 0 −1
1
0 1
 
1
−1
>
−>


0 =
Step direction: ∆ = −N B
1
0
Step size: s =
z1∗
∆1
=
−4
−1
=4
−4
1
0
− s∆ =
−4
=
−3
−1
−7
∗ z
4
∗
Updated basic vector: zN = 3∗ =
z2
−7
∗
zN
Step 2
Primal side
Dual side
∗ =
zN
New basic index: j = 2
 
−1
0
Step direction: ∆ = B −1 N
= 1 
1
1
n
Step size: t = max−1
∆1
x∗1
=
−1 ∆4
1 , x∗4
= 11 ,
∆5
x∗5
=
1
5
o
Step size: s =
∗
zN
New basic index: j = 3
−1 ∆2
2 , x∗2
=
−2 ∆5
1 , x∗5
=
New nonbasic index: i = 5
 
   
2
−1
4
x∗B − t∆ = 1 − 2 −2 = 5
4
2
0 ∗   
4
x1
Updated basic vector: x∗B = x∗2  = 5
x∗3
2
Updated index lists: B = (1, 2, 3) , N = (5, 4)
ζmax = 4x∗1 + 3x∗2 = 4 × 4 + 3 × 5 = 31
2
4
o
=
−7
−1
=7
4
2
−10
− s∆ =
−7
=
−7
−1
0
∗ z3
−10
∗
Updated basic vector: zN = ∗ =
z4
7
Primal side
=
z2∗
∆2
∗
zN
Updated index lists: B = (1, 2, 5) , N = (3, 4)




1 −1 0
1 0
Updated matrices: B = 2 −1 0 , N = 0 1
0 1 1
0 0
Step 3
∆1
x∗1
has negative entries
 
0
2
>
−>
Step direction: ∆ = −N B 1 =
−1
0
     
1
−1
2
x∗B − t∆ = 1 −  1  = 0
5
1
4   
x∗1
2
∗

Updated basic vector: xB = x∗2  = 1
x∗5
4
Step size: t = max−1
4
−7
=1
New nonbasic index: i = 4
 
−1
1
Step direction: ∆ = B −1 N
= −2
0
2
n
Dual side
−10
=
has negative entries
7
=2
 
0
−2
>
−>
Step direction: ∆ = −N B 0 =
1
1
Step size: s =
z3∗
∆3
=
−10
−2
=5
−10
−2
0
− s∆ =
−5
=
7
1
2
∗ ∗ = z5 = 5
Updated basic vector: zN
z4∗
2
∗
zN
∗ ≥0
zN