CSC 445/545 Notes 5.2: Eta Factorizations

The Revised Simplex Method:

• solve B^Ty = CB, and
• B * d = a.

Computer Implementations: use a device to facilitate solution.

Ch. 24: L, U-factorization of basis.

Ch. 7: Eta-factorization of basis.

Let B_k-1 be the basis matrix at step k-1, and
let B_k be the basis matrix at step k.

where a is the column of A corresponding to the entering variable.

Note: B_k-1 * d = a
So,

where E_k-1 is the Eta matrix with the column corresponding to the leaving variable replaced by d.

Furniture Example

Product	Hours	Metal	Wood	Profit
Bookcase	3	1	4	19
Desk	2	1	3	13
Chair	1	1	3	12
Bed frame	2	1	4	17
Total	225	117	420	Max

After two steps of the revised Simplex Method:

        X1 X3 X7
       [ 3  1  0]     [0.5]       [2]
B   =  [ 1  1  0]  d= [0.5]    a= [1]
 2     [ 4  3  1]     [0.5]       [4]


X7 leaves and X4 enters.

        X1 X3 X7              X1 X3 X4
       [ 3  1  0] [1 0 0.5]   [3  1  2]
B   =  [ 1  1  0] [0 1 0.5] = [1  1  1]
 3     [ 4  3  1] [0 0 0.5]   [4  3  4]

B0= I= initial basis from slack variables.
B1= E1
B2= E1 * E2
:
:
Bk= E1 * E2 * ... * Ek

This is called an eta-factorization of B_k.

Solving y^T B_k = CB^T is equivalent to:

(... ((y^T E₁) E₂) ... ) E_k = CB^T

Example:
(((y^T E₁) E₂) E₃) E₄ = CB^T

1. y₄^T E₄= CB^T or equivalently E₄^T y₄= CB.
2. y₃^T E₃= y₄^T or equivalently E₃^T y₃= y₄.
3. y₂^T E₂= y₃^T or equivalently E₂^T y₂= y₃.
4. y₁^T E₁= y₂^T or equivalently E₁^T y₁= y₂.
5. The answer is y= y₁

1. Set y_k+1= CB
2. For i from k down to 1 do
   • Solve y_i^T * E_i= y _i+1^T or E_i^T y_i = y_i+1.
3. The answer is y= y₁.

These systems of equations are easy to solve because E_i is an eta matrix.

For example:

[ 1 0 d1 0] [X1] [b1]
[ 0 1 d2 0] [X2]=[b2]
[ 0 0 d3 0] [X3] [b3]
[ 0 0 d4 1] [X4] [b4]

To solve:

1. Set X3= b3/d3.
2. for i not equal to 3, Xi = bi - di * X3.

In general:

Eta matrix with column e= [d1 d2 d3 ... dn]^T..

To Solve E * x = b:

1. Set Xe= be/de.
2. for i not equal to e, Xi = bi - di * Xe.

Let s be the number of non-zero entries in column e.

This process involves one division, s-1 multiplications, and s-1 subtractions.

Back to the furniture example:

A= [3 2 1 2 1 0 0]   b= [225]
   [1 1 1 1 0 1 0]      [117]
   [4 3 3 4 0 0 1]      [420]

c= [19 13 12 17 0 0 0]T

HB=[X5 X6 X7]T

B0= I, 

x* = [225 117 420]T,

CB=[0 0 0]T

Step 1.1: yT * B0= CBT => y = [0 0 0]T

Step 1.2: Look for yT AN < CNT:
[0 0 0 0] < [19 13 12 17]
True for all non-basic variables. 
Choose X3 to enter.

Step 1.3: Choose the pivot row:

B0 * d = a  = [1 1 3]T

so d= [1 1 3]T.


225  - t  >= 0  t <= 225
117  - t  >= 0  t <= 117        (*) tightest 
420  -3t  >= 0  t <= 420/3= 140

X6 leaves.

X5= 225 - 117= 108
X3= 117
X7= 420 - 3(117)= 69

B1=  I * E1 where E1 =
 X5 X3  X7
[ 1  1  0]
[ 0  1  0]
[ 0  3  1]

Iteration #2:

Step 2.1: yT B1= CBT
      =>  B1T y = CB

HB=[X5 X3 X7]T

[ 1 0 0] [y1]     [ 0]
[ 1 1 3] [y2]  =  [12]
[ 0 0 1] [y3]     [ 0]

y3=0, y1=0, y2= 12- y1 -3y3= 12.

y= [0 12 0]T

Step 2.2: Look for yT AN < CNT:
     X1 X2 X4 X6 
AN= [ 3  2  2  0]  
    [ 1  1  1  1]           
    [ 4  3  4  0] 

yT AN = [12 12 12 12] < [19 13 17 0]

Any variable except X6 can enter.

Let X1 enter.

Step 2.3: Find the pivot row. 

Solve 
B1 d = a where a=  [3 1 4 ]T 

[ 1 1 0] [d1]     [ 3]
[ 0 1 0] [d2]  =  [ 1]
[ 0 3 1] [d3]     [ 4]

d2= 1, d1= 3-d2=2, d3=4-3d2=1 

d=[2 1 1]T

Basis var. X5:  108 - 2t >= 0,   t <= 54 (*)
Basis var. X3:  117 -  t >= 0,   t <=117
Basis var. X7:   69 -  t >= 0,   t <= 69

X5 leaves.

X1= 54
X3= 117-54= 63
X7= 69-54= 15

HB=[X1 X3 X7]T

B2= E1 * E2 where 

E1 =         E2 =
[ 1 1 0]        [ 2 0 0]
[ 0 1 0]        [ 1 1 0]
[ 0 3 1]        [ 1 0 1]

Iteration #3:
Step 3.1: E2T * E1T * y= CB

HB=[X1 X3 X7]T

First, E2T z = CB

[ 2 1 1]  [z1]    [19]
[ 0 1 0]  [z2]=   [12]
[ 0 0 1]  [z3]    [ 0]

z= [7/2  12  0]T

Next, E1T y = z
[ 1 0 0]  [y1]   [7/2]
[ 1 1 3]  [y2] = [ 12]
[ 0 0 1]  [y3]   [  0]

y=[7/2  17/2  0]T

Step 3.2: Look for yT AN < CNT:

               X2 X4 X5 X6
[7/2 17/2 0]  [ 2  2  1  0]  
              [ 1  1  0  1]           
              [ 3  4  0  0] 

= [15.5 15.5 7/2 17/2] < [13 17 0 0]

X4 enters.

Step 3.3: Find the pivot row. 

Solve B2 d = E1 * E2 d = [2 1 4 ]T 

Solve E1 f = a :
[ 1 1 0]  [f1]    [ 2]
[ 0 1 0]  [f2]=   [ 1]
[ 0 3 1]  [f3]    [ 4]

f2=1, f1=2-f2= 1, f3= 4 -3f2= 1.  f= [1 1 1]T

Solve E2 d = f :
[ 2 0 0]  [d1] = [ 1]
[ 1 1 0]  [d2] = [ 1]
[ 1 0 1]  [d3] = [ 1]

d1= 1/2,  d2= 1/2, d3= 1/2.

Basis var. X1:   54 - 0.5 t  >= 0,   t <=108     
Basis var. X3:   63 - 0.5 t  >= 0,   t <=126
Basis var. X7:   15 - 0.5 t  >= 0,   t <= 30 (*)

X4 enters, X7 leaves.

X1= 54 -15 = 39
X3= 63 -15 = 48
X4= 30

E3=
[1 0 0.5]
[0 1 0.5]
[0 0 0.5] 

Iteration #4:

Step 4.1: yT B3= CBT => E3T * E2T * E1T y= CB

HB=[X1 X3 X4]T

E3T w = CB
[1   0   0  ] [w1]   [19]
[0   1   0  ] [w2] = [12]
[0.5 0.5 0.5] [w3]   [ 7]

w1= 19, w2= 12, w3= 3.  w= [19 12  3]T

E2T z =  w 
[ 2 1 1]  [z1]    [19]
[ 0 1 0]  [z2]=   [12]
[ 0 0 1]  [z3]    [ 3]

z2= 12, z3= 3, z1= 2. z= [2  12  3]T

Next, E1T y = z
[ 1 0 0]  [y1]   [  2]
[ 1 1 3]  [y2] = [ 12]
[ 0 0 1]  [y3]   [  3]

y1= 2, y3=3, y2= 12 -y1 -3y3= 12 -2 -9= 1

y=[2  1  3]T

Step 4.2: Look for yT AN < CNT:

          X2 X5 X6 X7
[2 1 3]  [ 2  1  0 0]  
         [ 1  0  1 0]
         [ 3  0  0 1] 

= [14 2 1 3] < [13 0 0 0 ]

STOP: Solution is optimal. 

The dictionaries to check correctness:

After   2 pivots:
X1 =  54 - 0.50 X2 - 0.50 X4 - 0.50 X5 + 0.50 X6 
X3 =  63 - 0.50 X2 - 0.50 X4 + 0.50 X5 - 1.50 X6 
X7 =  15 + 0.50 X2 - 0.50 X4 + 0.50 X5 + 2.50 X6 
-------------------------------------------------
z=  1782 - 2.50 X2 + 1.50 X4 - 3.50 X5 - 8.50 X6 

X4  enters.   X7  leaves.  z = 1782 

After   3 pivots:
X1 =  39 - 1 X2 - 1 X5 - 2 X6 + 1 X7 
X3 =  48 - 1 X2 + 0 X5 - 4 X6 + 1 X7 
X4 =  30 + 1 X2 + 1 X5 + 5 X6 - 2 X7 
-------------------------------------------------
z=  1827 - 1 X2 - 2 X5 - 1 X6 - 3 X7 

The optimal solution: 1827
X1  = 39  X2  =  0  X3  = 48  X4  = 30  
X5  =  0  X6  =  0  X7  =  0