Applied Optimization with MATLAB Programming
P.Venkataraman
John Wiley (2002)

page 100

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page 100

The third row will be obtained by

row 3 - {3 * (pivot row 2 )}

The fourth row  .....
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page 133

Table 3.14 is obtained incorrectly.  Table 3.15 must be newly developed.  The current Table 3.15 will become the new  Table 3.15a .
The complete problem is redone below

Example 3.3    Using Simplex Method

A two phase Simplex method is necessary to solve the problem.  Table 1 below represents the representation of the standard model.  It is not in canonical form

Initial Table     (Table 3.11)

 x1 x21 x22 s1 a1 s2 a2 b 1 1 -1 1 0 0 0 5 2 1 -1 0 1 0 0 4 1 1 -1 0 0 -1 1 1 -2 1 -1 0 0 0 0 f 0 0 0 0 1 0 1 Af

Canonical form can be obtained by
row 5 = row5 - (row2 +row3)

Phase I
Phase I:    Table 1    (Table 3.12)
 x1 x21 x22 s1 a1 s2 a2 b 1 1 -1 1 0 0 0 5 2 1 -1 0 1 0 0 4 1 1 -1 0 0 -1 1 1 -2 1 -1 0 0 0 0 f -3 -2 2 0 0 1 0 Af-5

Canonical Form:    yes
Solution:  x1 = 0;    x21 = 0;     x22 = 0;    s1 = 5;    a1 = 4;    s2 = 0;    a2 = 1;
f = 0;    Af = 5

Stop:  No - Af = 5  (it must be zero to stop)
EBV = x1
LBV = a2
Pivot row = row 3

Table 2 (Table 3.13) is obtained by
row1 = row1 : pivot row
row2 = row2 - 2*pivot row
row3 = row3
row4 = row4 + 2*pivot row
row5 = row5 + 3*pivot row

Phase I:    Table 2    (Table 3.13)

 x1 x21 x22 s1 a1 s2 a2 b 0 0 0 1 0 1 -1 4 0 -1 1 0 1 2 -2 2 1 1 -1 0 0 -1 1 1 0 3 -3 0 0 -2 2 f+2 0 1 -1 0 0 -2 3 Af-2

Canonical Form:    yes
Solution:  x1 = 1;    x21 = 0;     x22 = 0;    s1 = 4;    a1 = 2;    s2 = 0;    a2 = 0;
f = -2;    Af = 2

Stop:  No - Af = 2 (it must be zero to stop)
EBV = s2
LBV = a1
Pivot row = row 2

Table 3  is obtained by
row2 = row2/2  :(pivot row)
row1 = row1 - pivot row
row3 = row3 + pivot row
row4 = row4 + 2*pivot row
row5 = row5 + 2*pivot row

Phase I:    Table 3  (Table 3.14)
 x1 x21 x22 s1 a1 s2 a2 b 0 0.5 -0.5 1 -0.5 0 0 3 0 -0.5 0.5 0 0.5 1 -1 1 1 0.5 -0.5 0 0.5 0 0 2 0 2 -2 0 1 0 0 f+4 0 0 0 0 1 0 1 Af-0

Canonical Form:    yes
Solution:  x1 = 2;    x21 = 0;     x22 = 0;    s1 = 3;    a1 = 0;    s2 = 1;    a2 = 0;
f = -4;     Af = 0 : Phase I over

Prepare Table for Phase II :  delete last row, columns a1 and a2

Phase II
Phase II:    Table 1 (new Table 3.15)
 x1 x21 x22 s1 s2 b 0 0.5 -0.5 1 0 3 0 -0.5 0.5 0 1 1 1 0.5 -0.5 0 0 2 0 2 -2 0 0 f+4

Canonical Form:    yes
Solution:  x1 = 2;    x21 = 0;     x22 = 0;    s1 = 3;      s2 = 1;
f = -4;

Stop:  No  (negative coefficient in last row)
EBV = x22
LBV = s2
Pivot row = row 2

Table 2  is obtained by:
row2 = 2*row2 :(pivot row)
row1 = row1  +0.5*pivot row
row3 = row3  + 0.5*pivot row
row4 = row4 + 2*pivot row

Phase II::    Table 2
(new Table 3.15a)
 x1 x21 x22 s1 s2 b 0 0 0 1 1 4 0 -1 1 0 2 2 1 0 0 0 1 3 0 0 0 0 4 f+8

Canonical Form:    yes
Solution:  x1 = 3;    x21 = 0;     x22 = 2;    x2 = (0 -2 ) = -2;    s1 = 4;      s2 = 0;
f = -8;

Stop:  Yes:  No negative coefficient in last row

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page 152

3.3    Solve the following linear programming problem

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page 178

Last line (equation 4.47a)

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page 192

Above  equation (4.68a)

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page 204

printing is squished together
unconstrained problem.  The side constraints accompany all problems in this book to
convey the idea that there are no truly unconstrained problems.  Side constraints also

define .....

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page 212

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page 222

Equation (5.26)

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page 239

First line on page 239

page 240

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page 277

in Step 3 - identifying the SUMT iteration  count - and some format change ( for clarity)

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page  293

For consistency  The subscripts  are redefined with i+1  instead of i

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page 388
Example 10.3

values are different with different MATLAB version 6.1
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page  391
Example 10.4

values are different with version 6.0
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