Chapter 7. The Revised Simplex Method
 Recall Theorem 3.1, same basis  same dictionary
Entire dictionary can be constructed using original data as long as we know
which variables are basic.
 Suppose we have the following form after adding slack variables to the
standard LP. (or any LP with equality constraints and nonnegativity, will be
discussed later in Chapter 8)
maximize 𝑐 ′ 𝑥
subject to 𝐴𝑥 = 𝑏
𝑥≥0
𝑎11
Where, 𝐴 = ⋮
𝑎𝑚1
OR-1 2015
… 𝑎1𝑛
⋱
⋮
… 𝑎𝑚𝑛
1 … 0
⋮ ⋱ ⋮
0 … 1
1
 ex) max 19𝑥1 + 13𝑥2 + 12𝑥3 + 17𝑥4
s.t. 3𝑥1 + 2𝑥2 + 𝑥3 + 2𝑥4 ≤ 225
𝑥1 + 𝑥2 + 𝑥3 + 𝑥4 ≤ 117
4𝑥1 + 3𝑥2 + 3𝑥3 + 4𝑥4 ≤ 320
𝑥1 , 𝑥2 , 𝑥3 , 𝑥4 ≥ 0
Add slack variables and, after two iterations of the simplex method, we obtain
𝑧 = 1782 − 2.5𝑥2 + 1.5𝑥4 − 3.5𝑥5 − 8.5𝑥6
𝑥1 = 54 − 0.5𝑥2 − 0.5𝑥4 − 0.5𝑥5 + 0.5𝑥6
𝑥3 = 63 − 0.5𝑥2 − 0.5𝑥4 + 0.5𝑥5 − 1.5𝑥6
𝑥7 = 15 + 0.5𝑥2 − 0.5𝑥4 + 0.5𝑥5 + 2.5𝑥6
This dictionary can be represented compactly using matrix notation.
OR-1 2015
2
 We write the constraints as 𝐴𝑥 = 𝑏 with
3 2 1
𝐴= 1 1 1
4 3 3
2 1 0
1 0 1
4 0 0
0
225
0 , 𝑏 = 117 ,
1
420
𝑥1
𝑥2
𝑥3
𝑥 = 𝑥4 .
𝑥5
𝑥6
𝑥7
We express 𝐴𝑥 as 𝐵𝑥𝐵 + 𝑁𝑥𝑁 with
3 1 0
2 2
𝐵= 1 1 0 , 𝑁= 1 1
4 3 1
3 4
𝑥2
𝑥1
1 0
𝑥4
𝑥
0 1 , 𝑥𝐵 = 3 , 𝑥𝑁 = 𝑥 .
5
𝑥7
0 0
𝑥6
(In the text, 𝐴𝐵 , 𝐴𝑁 are used instead of 𝐵, 𝑁.)
Also 𝑐′ = 19 13 12 17 0 0 0
Express 𝑐 ′ 𝑥 as 𝑐𝐵 ′𝑥𝐵 + 𝑐𝑁 ′𝑥𝑁 , with
𝑐𝐵 ′ = 19 12 0 , 𝑐𝑁 ′ = 13 17 0 0
OR-1 2015
3
 Then 𝐴𝑥 = 𝑏  𝐵𝑥𝐵 + 𝑁𝑥𝑁 = 𝑏  𝐵𝑥𝐵 = 𝑏 − 𝑁𝑥𝑁
If 𝐵 is nonsingular (which is true for the example), we can premultiply 𝐵−1 on
both sides.
𝑥𝐵 = 𝐵−1 𝑏 − 𝐵−1 𝑁𝑥𝑁
(matrix form of constraints in current dictionary with basis 𝐵)
For objective row,
𝑧 = 𝑐 ′ 𝑥 = 𝑐𝐵 ′𝑥𝐵 + 𝑐𝑁 ′𝑥𝑁 = 𝑐𝐵 ′ 𝐵−1 𝑏 − 𝐵−1 𝑁𝑥𝑁 + 𝑐𝑁 ′𝑥𝑁
= 𝑐𝐵 ′𝐵−1 𝑏 + 𝑐𝑁 ′ − 𝑐𝐵 ′𝐵−1 𝑁 𝑥𝑁
= 𝑐𝐵 ′𝐵−1 𝑏 +
𝑗∈𝑁
= 𝑦′𝑏 +
𝑐𝑗 − 𝑦′𝐴𝑗 𝑥𝑗
𝑗∈𝑁
𝑐𝑗 − 𝑐𝐵 ′𝐵−1 𝐴𝑗 𝑥𝑗
( if we let 𝑦′ ≡ 𝑐𝐵 ′𝐵−1 )
 Matrix representation of a dictionary
z = 𝑐𝐵 ′𝐵−1 𝑏 + 𝑐𝑁 ′ − 𝑐𝐵 ′𝐵−1 𝑁 𝑥𝑁
𝑥𝐵 = 𝐵−1 𝑏 − 𝐵−1 𝑁𝑥𝑁
OR-1 2015
4
 Note that matrix 𝐵 corresponding to basic variables in a dictionary is
nonsingular:
Pf) Can show that 𝐵 is nonsingular by showing that 𝐵𝑥𝐵 = 𝑏 has a unique
solution. The existence of a solution is evident: since the basic feasible solution
𝑥 ∗ satisfies 𝐴𝑥 ∗ = 𝑏 and 𝑥𝑁∗ = 0, it satisfies 𝐵𝑥𝐵∗ = 𝐴𝑥 ∗ − 𝑁𝑥𝑁∗ = 𝑏.
To verify that there are no other solutions, consider an arbitrary vector 𝑥𝐵 such
that 𝐵𝑥𝐵 = 𝑏 and set 𝑥𝑁 = 0. Since the resulting vector 𝑥 satisfies 𝐴𝑥 =
𝐵𝑥𝐵 + 𝑁𝑥𝑁 = 𝑏, it must satisfy the bottom 𝑚 equations in the dictionary
representing 𝑥 ∗ . But then 𝑥𝑁 = 0 implies 𝑥𝐵 = 𝑥𝐵∗ . Hence the solution to
𝐵𝑥𝐵 = 𝑏 is unique and 𝐵 is nonsingular.

 Recall that a basic feasible solution for a dictionary is equivalent to an extreme
point of the polyhedron 𝑃 = {𝑥 ∈ 𝑅𝑛+𝑚 : 𝐴𝑥 = 𝑏, 𝑥 ≥ 0}.
OR-1 2015
5
 Matrix representation of dictionary
z = 𝑐𝐵 ′𝐵−1 𝑏 + 𝑐𝑁 ′ − 𝑐𝐵 ′𝐵−1 𝑁 𝑥𝑁
𝑥𝐵 = 𝐵−1 𝑏 − 𝐵−1 𝑁𝑥𝑁
In tableau form,
−𝑧 + 0′ 𝑥𝐵 + 𝑐𝑁 ′ − 𝑐𝐵 ′𝐵−1 𝑁 𝑥𝑁 = −𝑐𝐵 ′𝐵−1 𝑏
𝐼𝑥𝐵
+ 𝐵−1 𝑁𝑥𝑁 = 𝐵−1 𝑏
 Note that 𝑐𝐵 ′𝐵−1 𝑁 = 𝑐𝐵 ′𝐵−1 𝐴𝑁1 , 𝐴𝑁2 , … , 𝐴𝑁𝑛 = 𝑦′ 𝐴𝑁1 , 𝐴𝑁2 , … , 𝐴𝑁𝑛
We frequently use 𝑦 ′ = 𝑐𝐵 ′𝐵−1 , i.e. 𝑦 is the solution of 𝑦 ′ 𝐵 = 𝑐𝐵 ′.
Hence 𝑐𝑁 ′ − 𝑐𝐵 ′𝐵−1 𝑁 𝑥𝑁 =
𝑗∈𝑁
𝑐𝑗 − 𝑦′𝐴𝑗 𝑥𝑗
and −𝑐𝐵 ′𝐵−1 𝑏 = −𝑦 ′ 𝑏
OR-1 2015
6
 Current tableau in matrix notation is:
−𝑧 + 0′ 𝑥𝐵 + 𝑐𝑁 ′ − 𝑐𝐵 ′𝐵−1 𝑁 𝑥𝑁 = −𝑐𝐵 ′𝐵−1 𝑏
𝐼𝑥𝐵
+ 𝐵−1 𝑁𝑥𝑁 = 𝐵−1 𝑏
𝑐𝐵 ′𝐵−1 𝑁 = 𝑐𝐵 ′𝐵−1 𝐴𝑁1 , 𝐴𝑁2 , … , 𝐴𝑁𝑛 = 𝑦′ 𝐴𝑁1 , 𝐴𝑁2 , … , 𝐴𝑁𝑛
 Here, 𝑐𝐵 ′𝐵−1 𝑁 can be viewed differently:
𝑐𝐵 ′𝐵−1 𝑁 = 𝑐𝐵 ′𝐵−1
𝑎1 ′
𝑎1 ′
⋮ = 𝑦′ ⋮
𝑎𝑚 ′
𝑎𝑚 ′
i.e., we take linear combination of rows of 𝑁 using weights −𝑦𝑖 to 𝑖 − 𝑡ℎ
constraint and add it to 𝑐𝑁 ′ in the 𝑧 −row. (or subtract 𝑦 ′ 𝑁 from 𝑐𝑁 ′)
Similarly, for basic variables, we subtract 𝑦 ′ 𝐵 from 𝑐𝐵 ′ in the 𝑧 −row.
Then, 𝑐𝐵 ′ − 𝑐𝐵 ′𝐵−1 𝐵 = 0. Also, the right-hand side constant in the
𝑧 −row can be obtained by subtracting 𝑦 ′ 𝐵 from 0 (initial value).
OR-1 2015
7
 Hence given the initial tableau, we obtain zeroth equation of the updated
tableau by taking a linear combination of rows of the initial tableau using
weights from 𝑦 −vector and subtract it from zeroth equation.
It is the net effect of many elementary row operations performed on the
tableau.
 The value 𝑐𝑗 − 𝑦′𝐴𝑗 is called reduced cost. It is 0 for every basic variable.
OR-1 2015
8
 Initially, we have
z
z
0
0
A
I
b
cB '
cN '

0
B
N

b
After reordering
of columns
OR-1 2015
c'
9
Updated tableau
z
(cB’B-1 = y’ )
z
cB '
cN '

0
B
N

b
= cB’ – cB’B-1B
0
= cN’ – cB’B-1N
c N 'c B ' B 1 N
B 1 N
I
( = B-1 [ B : N ] )
OR-1 2015
= 0 – cB’B-1b
  c B 1b
B
 B 1b
(= B-1 b)
10
 Also 𝐵−1 gives information on what elementary row operations have been
performed on the constraints. Updated 𝑖 − 𝑡ℎ constraint = (𝑖 − 𝑡ℎ row of
𝐵−1 ) 𝐵: 𝑁|𝑏 , so 𝑗 − 𝑡ℎ component of (𝑖 − 𝑡ℎ row of 𝐵−1 ) is the weight we
multiply to the 𝑗 − 𝑡ℎ original constraint.
We take linear combination of rows of 𝐵: 𝑁|𝑏 .
 Hence updated tableau carries information on what elementary operations
have been performed (the net effect) to obtain the current tableau.
zeroth equation obtained from 𝑐 − 𝑦 ′ 𝐴 (objective value: 0 − 𝑦 ′ 𝑏)
𝑖 − 𝑡ℎ constraint : (𝑖 − 𝑡ℎ row of 𝐵 −1 ) [B : N] (r.h.s: ( 𝑖 − 𝑡ℎ row of 𝐵 −1 ) 𝑏 )
OR-1 2015
11
 Recall the proof of strong duality theorem, in which we claimed that, at the optimal
tableau, the negative of the coefficients of slack variables in zeroth equation is an
optimal dual solution. It is the 𝑦 −vector obtained from 𝑦′ = 𝑐𝐵 ′𝐵−1 .
 The updated tableau may be given using original sequence of variables. Then,
z
z
OR-1 2015
c'
0
0
A
I
b
c  y' A
0  y' I
0  y' b
B 1 A
B 1
B 1b
12
 In an optimal tableau, we have all coefficients in 𝑧 −row are ≤ 0.
Therefore, 𝑐 ′ − 𝑦 ′ 𝐴 ≤ 0, −𝑦′ ≤ 0. Here, 𝑦 ′ = 𝑐𝐵 ′𝐵−1 .
With this 𝑦, we have 𝑦 ′ 𝐴 ≥ 𝑐 ′ , 𝑦′ ≥ 0, hence 𝑦 is a dual feasible solution.
Also, −𝑦 ′ 𝑏 in the r.h.s of 𝑧 −row is the negative of the current primal
objective value, which is also the negative of the objective value of the dual
feasible solution 𝑦. So we have primal feasible solution 𝑥 and dual feasible
solution 𝑦 which give the same objective value. Hence 𝑥 is a primal optimal
solution and 𝑦 is a dual optimal solution.
 Since the dual constraint for the 𝑗 − 𝑡ℎ primal structural variable is 𝑦′𝐴𝑗 ≥ 𝑐𝑗 ,
we obtain 𝑦′𝐴𝑗 − 𝑦𝑚+𝑗 = 𝑐𝑗 if we subtract nonnegative surplus variable to
convert it into an equation. Then 𝑐𝑗 − 𝑦′𝐴𝑗 = −𝑦𝑚+𝑗 . So the coefficient of the
𝑗 − 𝑡ℎ primal structural variable in the zeroth equation can be viewed as the
negative of 𝑗 − 𝑡ℎ surplus variable of the dual problem. At an optimal tableau,
𝑦𝑚+𝑗 ≥ 0 for 𝑗 = 1, … , 𝑛.
OR-1 2015
13
 In summary, each tableau that appears in the simplex iterations gives primal
basic feasible solution and, as a by-product, gives a dual solution (it is a dual
basic solution although we do not prove it here).
The dual solution is not necessarily dual feasible, but always gives an objective
value of the dual problem which is the same as the primal objective value of
the current primal solution. Also the primal and the dual solution pairs satisfy
the complementary slackness condition (see the next slide)
If we obtain a dual feasible solution, i.e. the coefficients in the zeroth equation
are all nonpositive, we obtain primal, dual feasible solution with the same
objective value. Hence optimality of the primal solution is established.
 Other types of algorithms can be designed using the complementary slackness
optimality conditions. (e.g. dual simplex method, some algorithms for network
flow problems, ...)
OR-1 2015
14
 Complementary slackness between primal and dual solutions during simplex
iterations.
(recall CS conditions)
𝑚
∗
𝑖=1 𝑎𝑖𝑗 𝑦𝑖
𝑛
∗
𝑗=1 𝑎𝑖𝑗 𝑥𝑗
= 𝑐𝑗 or 𝑥𝑗∗ = 0 (or both), for every 𝑗 = 1,2, … , 𝑛
= 𝑏𝑖 or 𝑦𝑖∗ = 0 (or both), for every 𝑖 = 1,2, … , 𝑚
Structural variable 𝑥𝑗∗ > 0  𝑥𝑗∗ basic  𝑐𝑗 − 𝑦 ∗ ′𝐴𝑗 = 0
𝑦 ∗ ′𝐴𝑗 ≠ 𝑐𝑗  𝑐𝑗 − 𝑦 ∗ ′𝐴𝑗 ≠ 0  𝑥𝑗∗ nonbasic  𝑥𝑗∗ = 0
∗
𝑎𝑖′ 𝑥 ∗ ≠ (<)𝑏𝑖  𝑥𝑛+𝑖
> 0, hence basic  𝑐𝑛+𝑖 − 𝑦′𝐴𝑛+𝑖 = 0 − 𝑦𝑖∗ = 0
∗
𝑦𝑖∗ ≠ 0  𝑐𝑛+𝑖 − 𝑦 ∗ ′𝐴𝑛+𝑖 ≠ 0  𝑥𝑛+𝑖 nonbasic, hence 𝑥𝑛+𝑖
=0 
𝑎𝑖′ 𝑥 ∗ = 𝑏𝑖
OR-1 2015
15
 A basis 𝐵 is called an optimal basis if 𝑐𝑗 − 𝑐𝐵 ′𝐵−1 𝐴𝑗 ≤ 0 for all 𝑗 ∈ 𝑁 (we
have 𝑐𝑗 − 𝑐𝐵 ′𝐵−1 𝐴𝑗 = 0 for all 𝑗 ∈ 𝐵) and 𝑥𝐵∗ = 𝐵−1 𝑏 ≥ 0 (we have 𝑥𝑁∗ = 0).
 If 𝑏𝑖 → 𝑏𝑖 ± 1, then the objective value 𝑦 ′ 𝑏 → 𝑦 ′ 𝑏 ± 𝑦𝑖 (as long as we have
the same basis). Hence 𝑦𝑖 represent the amount of change in the objective
value (with current basis 𝐵) when 𝑏𝑖 changes by 1 unit.
Note that if we change the value of 𝑏𝑖 while 𝑥𝐵∗ = 𝐵−1 𝑏 ≥ 0, current basis is
still the optimal basis, hence optimal objective value changes by 𝑦𝑖 .
But, if the nonnegativity of the basic variables is violated by changing 𝑏𝑖 , the
current basis is no longer an optimal basis for the changed problem.
(more in Chapter 10. Sensitivity Analysis later.)
OR-1 2015
16
 Ex)
max 19 x1  13x2
s.t. 3x1
2 x2
x1
 x2
4 x1  3x2
 12 x3
 x3
 x3
 3x3
 17 x4
 2 x4
 1x4
 4 x4
 225
 117
 420
x1, x2 , x3 , x4  0
OR-1 2015
17
 Current dictionary
z
x1
x3
x7
 1782  2.5 x2
 54  0.5 x2
 63  0.5 x2
 15  0.5 x2
 1.5 x4
 0.5 x4
 0.5 x4
 0.5 x4
 3.5 x5
 0.5 x5
 0.5 x5
 0.5 x5
 8.5 x6
 0.5 x6
 1.5 x6
 2.5 x6
Basic variables are {x1, x3 , x7 }
 3 1 0
 1 / 2  1 / 2 0
B  1 1 0, B 1   1 / 2 3 / 2 0




4 3 1
 1 / 2  5 / 2 1
 3 1 0 1 0 0 1

 
 1 1 0 0 1 0   0
  4 3 1 0 0 1  0
 

OR-1 2015
1
2
5
3
3
3
0 13 0 0 1 0 0 1 2  1 2 0 




3
1
1
0  3 1 0   0 1 0  2
0 
2
1  4 3 0 1 0 0 1  1 2  5 2 1 
18
 1 2  1 2 0  2 2 1 0  1 2
B 1N   1 2 3 2 0 1 1 0 1   1 2


 1 2  5 2 1 3 4 0 0  1 2
 1 2  1 2 0  225 54
B 1b   1 2 3 2 0 117   63

  
 1 2  5 2 1 429 15
1
1
1
2
2
2
1


2
1
2
1
2
 12 
3 
2 
 5 2 
y'
 1 2  1 2 0 225
225
cB ' B 1b  19 12 0 1 2 3 2 0 117  3.5 8.5 0117  1782




5
1
 2  2 1 429
429
2 2 1
𝑐𝑁 − 𝑐𝐵 𝐵−1 𝑁 = 13, 17, 0, 0 − 3.5, 8.5, 0 1 1 0
3 4 0
−2.5, 1.5, −3.5, −8.5
OR-1 2015
0
1 =
1
19
x1 x2
z
x3
x4
x5
x6
cB
x7
19 13 12 17 0 0 0   0 
 3 2 1 2 1 0 0   225

 

 1 1 1 1 0 1 0  117 

 

 4 3 3 4 0 0 1   420
x1 x3
z
x7
x2
x4
19 12 0 13 17
3 1 0 2 2

1 1 0 1 1

4 3 1 3 4
B
x5
x6
0
1
0
0
0
0
1
0
cN
 0 
   225
 

  117 
 


420
 

N
Updated tableau
y'  (3.5, 8.5, 0)
z
c j  y' A j
B 1 B
OR-1 2015
x1 x3
0
1

0

0
0
0
1
0
B 1 N
x7
x2
0  2.5
0 0.5
0 0.5
1  0.5
x4
x5
x6
1.5  3.5  8.5    1782
0.5 0.5  0.5   54 
 

0.5  0.5 1.5    63 
 

0.5  0.5  2.5   15 
 y' b
B 1b
20
x1 x3
z
0
1

0

0
0
0
1
0
x7
x2
0  2.5
0 0.5
0 0.5
1  0.5
x4
x5
x6
1.5  3.5  8.5    1782
0.5 0.5  0.5   54 
 

0.5  0.5 1.5    63 
 

0.5  0.5  2.5   15 
Apply same elementary row operations
x1 x2
x3
x4
x5
x6
x7
19 13 12 17 0 0 0   0 
 3 2 1 2 1 0 0   225

 

 1 1 1 1 0 1 0  117 

 

 4 3 3 4 0 0 1   420
A
OR-1 2015
I
b
x1
x2
x3
0
1

0

0
 2.5
0.5
0.5
 0.5
0
0
1
0
B 1 A
c j  y' A j
x4
x5
x6
1.5  3.5  8.5
0.5 0.5  0.5
0.5  0.5 1.5
0.5  0.5  2.5
B 1
0  y' b
x7
0    1782
0   54 
 

0   63 
 

1   15 
B 1b
21
c j  y' A j
x1 x2
x3
x4
19 13 12 17
3 2 1 2

1 1 1 1

4 3 3 4
x1
x5
x6
x7
0
1
0
0
0
0
1
0
0  0 
0    225
 

0   117 
 

1   420
0
1

0

0
x2
- 2.5
0.5
0.5
 0.5
x3
0
0
1
0
B 1 A
x4
x6
1.5 - 3.5 - 8.5
0.5 0.5 - 0.5
0.5 - 0.5 1.5
0.5 - 0.5 - 2.5
x7
0   1782
0   54 
 

0   63 
 

1   15 
B 1b
B 1
cj – y’Aj = - ym+j
cj – y’ej = - yj
x1
0
1

0

0
OR-1 2015
x5
x2
- 2.5
0.5
0.5
 0.5
x3
0
0
1
0
x4
x5
x6
1.5 - 3.5 - 8.5
0.5 0.5 - 0.5
0.5 - 0.5 1.5
0.5 - 0.5 - 2.5
x7
0   1782
0   54 
 

0   63 
 

1   15 
22
The Revised Simplex Method
 Efficient implementation of the simplex method.
 Recall the needed operations for the simplex method:
Find the entering and the leaving variable, update dictionary. (If we know the
basis, entire tableau can be constructed. Hence we do not really need to
update the dictionary.)
 Find the entering variable xj : (𝑐𝑗 − 𝑐𝐵 ′𝐵−1 𝐴𝑗 > 0, 𝑗 ∈ 𝑁)
1) Find 𝑦 ′ = 𝑐𝐵 ′𝐵−1 by solving 𝑦 ′ 𝐵 = 𝑐𝐵 ′
2) Evaluate 𝑐𝑗 − 𝑦′𝐴𝑗 for all 𝑗 ∈ 𝑁
Choose a nonbasic variable with 𝑐𝑗 − 𝑦′𝐴𝑗 > 0 as the entering nonbasic
variable.
 Find the leaving variable:
𝑥𝐵 = 𝑥𝐵∗ − 𝐵−1 𝑁𝑥𝑁 = 𝑥𝐵∗ − 𝐵−1 𝐴𝑁1 𝐵−1 𝐴𝑁2 … 𝐵−1 𝐴𝑁𝑛 𝑥𝑁
= 𝑥𝐵∗ −
OR-1 2015
𝑗∈𝑁
𝐵−1 𝐴𝑗 𝑥𝑗
23
Suppose 𝑥𝑗 , 𝑗 ∈ 𝑁 is chosen as the entering nonbasic variable. As we
increase 𝑥𝑗 from 0 to 𝑡 ≥ 0 while keeping other nonbasic variables at 0, we
get
𝑥𝐵 ← 𝑥𝐵∗ − 𝑡𝑑
(𝑑 = 𝐵−1 𝐴𝑗 ) (Find 𝑑 by solving 𝐵𝑑 = 𝐴𝑗 )
Then determine the largest value 𝑡 which makes 𝑥𝐵∗ − 𝑡𝑑 ≥ 0, and find the
basic variable which becomes 0, say 𝑥𝑙 . (minimum ratio test)
i.e., 𝑙 = 𝑎𝑟𝑔𝑚𝑖𝑛𝑖∈𝐵
𝑥𝑖∗
𝑑𝑖
| 𝑑𝑖 > 0 and set
𝑡∗
= 𝑚𝑖𝑛𝑖∈𝐵
𝑥𝑖∗
𝑑𝑖
| 𝑑𝑖 > 0
 Update:
Set 𝑥𝑗 = 𝑡 ∗
𝑥𝐵∗ ← 𝑥𝐵∗ − 𝑡 ∗ 𝑑
Replace the leaving column 𝐴𝑙 of 𝐵 by entering column 𝐴𝑗 , and 𝑥𝑗 enters the
basis in place of 𝑥𝑙 . (position in the basis is important)
OR-1 2015
24
ex)
max 19 x1  13x2
s.t. 3x1
2 x2
x1
 x2
4 x1  3x2
 12 x3
 x3
 x3
 3x3
 17 x4
 2 x4
 1x4
 4 x4
 225
 117
 420
x1, x2 , x3 , x4  0
Current dictionary
z
x1
x3
x7
 1782  2.5 x2
 54  0.5 x2
 63  0.5 x2
 15  0.5 x2
 1.5 x4
 0.5 x4
 0.5 x4
 0.5 x4
Basic variables are {x1, x3 , x7 }
OR-1 2015
 3.5 x5
 0.5 x5
 0.5 x5
 0.5 x5
 8.5 x6
 0.5 x6
 1.5 x6
 2.5 x6
 x1*  54
 3 1 0
 *  
*
xB   x3   63 , B  1 1 0
 


 x7*  15
4 3 1
 
25
Find entering nonbasic variable 𝑥𝑗 such that 𝑐𝑗 − 𝑦′𝐴𝑗 > 0 (𝑦 = 𝑐𝐵 ′𝐵−1 )
First solve 𝑦′𝐵 = 𝑐𝐵 ′.
 y1
y2
 3 1 0
y3 1 1 0  19 12 0  y '  3.5 8.5 0


4 3 1
 2
c2  y ' A2  13  3.5 8.5 01   2.5
 
3
 2
c4  y ' A4  17  3.5 8.5 01   1.5
 
4
1 
c5  y ' A5  0  3.5 8.5 00  3.5
 
0
0 
c6  y ' A6  0  3.5 8.5 01  8.5
 
0
x4 enters the basis
OR-1 2015
26
From 𝑥𝐵 = 𝑥𝐵∗ − 𝐵−1 𝐴𝑁 𝑥𝑁 , consider 𝑥𝐵∗ − 𝑡𝐵−1 𝐴4
Solve 𝐵𝑑 = 𝐴4 to find 𝐵−1 𝐴4 .
3 1 0  d1  2
1 1 0   d   1 

 2   
4 3 1  d3  4
x1  54  0.5t
x3  63  0.5t
x7  15  0.5t
 d1  0.5
 d 2   0.5
   
 d3  0.5
( x4 : 0  t )
54
0.5 39
x*B  63  300.5  48
 
   
15
0.5  0 
OR-1 2015
 54 63 15 
ratio test : min 
,
,
  30
 0.5 0.5 0.5 
t *  30,
( x4* becomes 30)
x7 leaves the basis
 x1*  39
 3 1 2
 *  
*
new xB   x3   48 , new B  1 1 1
 


*
 x4  30
4 3 4
 
27
Remarks
 Recall the economic significance of dual variables discussed in Chapter 5. In
the revised simplex method, the vector 𝑦 obtained from 𝑦 ′ 𝐵 = 𝑐𝐵′ (𝑦′𝐴𝑗 =
𝑐𝑗 , 𝑗 ∈ 𝐵) may be interpreted as a dual solution (may not be dual feasible) and
it gives tentative values of the resources. Let 𝐴𝑗 be the column of 𝐴
corresponding to the variable 𝑥𝑗 . It represents the amounts of resources
consumed when we perform one unit of activity 𝑗 (𝐴𝑥 = 𝐴𝑗 𝑥𝑗 = 𝑏).
Then, for a nonbasic variable 𝑥𝑗 , it compares the profit obtained from
performing one unit of activity 𝑗 (𝑐𝑗 ) with the value of the resources consumed
when we do one unit of activity 𝑗 (𝑦′𝐴𝑗 ). If 𝑐𝑗 − 𝑦 ′ 𝐴𝑗 > 0, it means
performing activity 𝑗 gives more profit than the value of the resources
consumed. Hence it is desirable to increase the level of activity 𝑗 (make 𝑥𝑗
basic). If all reduced costs are ≤ 0, there are no profitable activities remaining
under current values of resources, hence current solution is optimal.
OR-1 2015
28