output_sgi.txt

Opening the transcript file "output".
 
MATMAN, version 1.61
Last modified on 25 January 2000.
 
An interactive program which carries out
elementary row operations on a matrix, or
the simplex method of linear programming.
 
Developed by Charles Cullen and John Burkardt.
All rights reserved by the authors.  This program may
not be reproduced in any form without written permission.
 
Send comments to burkardt@psc.edu.
 
Enter command? ("H" for short menu, "HELP" for full menu, ? for full help)
#
# Turn off paging.
#
$
Paging turned OFF.
 
Enter command? ("H" for short menu, "HELP" for full menu, ? for full help)
#
# Get a brief list of legal commands.
#
h
A(I,J)=S  Set matrix entry to S.
CHECK checks if the matrix is row reduced.
E     enters a matrix to work on.
HELP  for full help.
L     switches to linear programming.
Q     quits.
Z     automatic row reduction (requires password).
?     for interactive help.
 
R1 <=> R2  interchanges two rows
R1 <= S R1 multiplies a row by S.
R1 <= R1 + S R2 adds a multiple of another row.
 
 
Enter command? ("H" for short menu, "HELP" for full menu, ? for full help)
#
# Get the full list of legal commands.
#
help
 
Here is a list of all MATMAN commands:
 
A(I,J)=S  Set matrix entry to S.
ADD    Add S times row I to row J.
B      Set up sample problem.
BASIC I, J changes basic variable I to J.
CHECK  Check matrix for reduced row echelon form.
CHECK  Check linear program table for optimality.
D/M    Divide/Multiply row I by S.
DEC    Use decimal arithmetic.
DEC_DIGIT Set the number of decimal digits.
DET    Print the determinant of the matrix.
E      Enter matrix with I rows and J columns.
E      Enter a linear programming problem, I constraints, J variables.
EDET   Print ERO determinant.
F      Choose arithmetic (Real, Fraction, or Decimal).
G      Add/delete a row or column of the matrix.
H      for quick help.
HELP   for full help (this list).
I      Interchange rows I and J.
I_BIG  Set size of largest integer for fractions.
INIT   Initialize data.
J      Jacobi rotation in (I,J) plane.
K      Open/close the transcript file.
L      To switch between linear algebra and linear programming.
P      Pivot linear program, entering I, departing J.
Q      Quit.
RES    Restore a saved matrix or table
RAT    Use rational arithmetic.
REAL   Use real arithmetic.
S      Store the current matrix or table.
T      Type out the matrix
TR     Transpose the matrix.
TS     Type linear programming solution.
U      Undo last operation.
V      Remove LP artificial variables.
W/X    Write/read example to/from file.
Y      Turn automatic printing ON or OFF.
Z      Automatic operation (requires password).
#      Begins a comment line.
<      Get input from a file.
%/$    Turn paging on/off.
 
R1 <=> R2  interchanges two rows
R1 <= S R1 multiplies a row by S.
R1 <= R1 + S R2 adds a multiple of another row.
 
Enter command? ("H" for short menu, "HELP" for full menu, ? for full help)
#
# Work in fractional arithmetic.
#
rational
You are already using the arithmetic type that
you have requested.
 
Enter command? ("H" for short menu, "HELP" for full menu, ? for full help)
#
# Enter a 3 by 3 matrix.
#
e
Enter number of rows, number of columns.
3,3
Enter entries 1 to 3 of row 1
1,2,3
Enter entries 1 to 3 of row 2
4,5,6
Enter entries 1 to 3 of row 3
7,8,11
A copy of this matrix is being saved.
 
  The current matrix:
 
    1    2    3
 
    4    5    6
 
    7    8   11
 
Enter command? ("H" for short menu, "HELP" for full menu, ? for full help)
#
# Interchange rows 1 and 3
#
r1 <=> r3
  ERO: Row 1 <=> Row 3
 
  The current matrix:
 
    7    8   11
 
    4    5    6
 
    1    2    3
 
Enter command? ("H" for short menu, "HELP" for full menu, ? for full help)
#
# Divide row 1 by 7
#
r1 <= r1/7
  ERO: Row 1 <=  Row 1 / 7
 
  The current matrix:
 
    1    8   11
         7    7
 
    4    5    6
 
    1    2    3
 
Enter command? ("H" for short menu, "HELP" for full menu, ? for full help)
#
# Add -4 times row 1 to row 2,
# Add -1 times row 1 to row 3.
#
r2 <= r2 - 4 r1
  ERO: Row 2 <=  Row 2 - 4 Row 1
 
  The current matrix:
 
    1    8   11
         7    7
 
    0    3   -2
         7    7
 
    1    2    3
 
Enter command? ("H" for short menu, "HELP" for full menu, ? for full help)
r3 <= r3 - 1 r1
  ERO: Row 3 <=  Row 3 - Row 1
 
  The current matrix:
 
    1    8   11
         7    7
 
    0    3   -2
         7    7
 
    0    6   10
         7    7
 
Enter command? ("H" for short menu, "HELP" for full menu, ? for full help)
#
# Interchange rows 2 and 3
# and divide row 2 by 6/7.
#
r2 <=> r3
  ERO: Row 2 <=> Row 3
 
  The current matrix:
 
    1    8   11
         7    7
 
    0    6   10
         7    7
 
    0    3   -2
         7    7
 
Enter command? ("H" for short menu, "HELP" for full menu, ? for full help)
r2 <= r2 / 6/7
  ERO: Row 2 <=  Row 2 * 7/6
 
  The current matrix:
 
    1    8   11
         7    7
 
    0    1    5
              3
 
    0    3   -2
         7    7
 
Enter command? ("H" for short menu, "HELP" for full menu, ? for full help)
#
# Add -8/7 times row 2 to row 1,
# Add -3/7 times row 2 to row 3.
#
r1 <= r1 - 8/7 r2
  ERO: Row 1 <=  Row 1 - 8/7 Row 2
 
  The current matrix:
 
   1   0  -1
           3
 
   0   1   5
           3
 
   0   3  -2
       7   7
 
Enter command? ("H" for short menu, "HELP" for full menu, ? for full help)
r3 <= r3 - 3/7 r2
  ERO: Row 3 <=  Row 3 - 3/7 Row 2
 
  The current matrix:
 
   1   0  -1
           3
 
   0   1   5
           3
 
   0   0  -1
 
Enter command? ("H" for short menu, "HELP" for full menu, ? for full help)
#
# Check for row reduced echelon form.
#
check
 
Checking the matrix for row echelon form...
 
  The matrix is NOT in row echelon form.
  The first nonzero entry in row 3
  which occurs in column 3
  is -1 rather than 1.
 
Enter command? ("H" for short menu, "HELP" for full menu, ? for full help)
#
# Divide row 3 by -1
#
r3 <= - r3
  ERO: Row 3 <= - Row 3
 
  The current matrix:
 
   1   0  -1
           3
 
   0   1   5
           3
 
   0   0   1
 
Enter command? ("H" for short menu, "HELP" for full menu, ? for full help)
#
# Add 1/3 row 3 to row 1
# Add -5/3 row 3 to row 2
#
r1 <= r1 + 1/3 r3
  ERO: Row 1 <=  Row 1 + 1/3 Row 3
 
  The current matrix:
 
   1   0   0
 
   0   1   5
           3
 
   0   0   1
 
Enter command? ("H" for short menu, "HELP" for full menu, ? for full help)
r2 <= r2 -5/3 r3
  ERO: Row 2 <=  Row 2 - 5/3 Row 3
 
  The current matrix:
 
   1   0   0
 
   0   1   0
 
   0   0   1
 
Enter command? ("H" for short menu, "HELP" for full menu, ? for full help)
#
# Consider another matrix
#
e
Enter number of rows, number of columns.
3,3
Enter entries 1 to 3 of row 1
1,2,3
Enter entries 1 to 3 of row 2
4,5,6
Enter entries 1 to 3 of row 3
7,8,9
A copy of this matrix is being saved.
 
  The current matrix:
 
   1   2   3
 
   4   5   6
 
   7   8   9
 
Enter command? ("H" for short menu, "HELP" for full menu, ? for full help)
#
# Request automatic conversion to row reduced echelon form.
#
z 31984
 
  ERO: Row 3 <=> Row 1
  ERO: Row 1 <=  Row 1 / 7
  ERO: Row 2 <=  Row 2 - 4 Row 1
  ERO: Row 3 <=  Row 3 - Row 1
 
  ERO: Row 3 <=> Row 2
  ERO: Row 2 <=  Row 2 * 7/6
  ERO: Row 1 <=  Row 1 - 8/7 Row 2
  ERO: Row 3 <=  Row 3 - 3/7 Row 2
 
  The current matrix:
 
   1   0  -1
 
   0   1   2
 
   0   0   0
 
Enter command? ("H" for short menu, "HELP" for full menu, ? for full help)
#
# Restore the original matrix.
#
restore
The saved matrix has been restored.
 
  The current matrix:
 
   1   2   3
 
   4   5   6
 
   7   8   9
 
Enter command? ("H" for short menu, "HELP" for full menu, ? for full help)
#
# Insert a new row at row position 3.
#
g +r3
Enter entries 1 to 3 of row 3
6.1, 6.2, 6.3
 
  The current matrix:
 
    1    2    3
 
    4    5    6
 
   61   31   63
   10    5   10
 
    7    8    9
 
Enter command? ("H" for short menu, "HELP" for full menu, ? for full help)
#
# Change entry 3,2 to 77
#
a(3,2) = 77
 
  A(3,2) = 77
 
  The current matrix:
 
    1    2    3
 
    4    5    6
 
   61   77   63
   10        10
 
    7    8    9
 
Enter command? ("H" for short menu, "HELP" for full menu, ? for full help)
#
# Remove column 2.
#
g -c2
The column has been deleted!
 
  The current matrix:
 
    1    3
 
    4    6
 
   61   63
   10   10
 
    7    9
 
Enter command? ("H" for short menu, "HELP" for full menu, ? for full help)
#
# Get a determinant test matrix of order 4.
#
b d 4
 
The following examples are available:
  "D" for determinant;
  "E" for eigenvalues;
  "I" for inverse;
  "S" for linear solve.
 
  "C" to cancel.
 
 
  The current matrix:
 
    0    6   -4   -8
 
   -3  -10   -4   -5
 
    9   -5   -1   -8
 
   -1    5    1    9
 
Enter command? ("H" for short menu, "HELP" for full menu, ? for full help)
det
 
The determinant is -4350
 
Enter command? ("H" for short menu, "HELP" for full menu, ? for full help)
#
# Repeat work in real arithmetic.
#
real
 
  Switching to REAL arithmetic.
 
  A note about REAL arithmetic:
 
  The representation is approximate;
  Arithmetic operations are approximate.
 
  In particular, a singular matrix may be
  incorrectly found to be nonsingular.
 
  The current matrix:
 
    0.    6.   -4.   -8.
   -3.  -10.   -4.   -5.
    9.   -5.   -1.   -8.
   -1.    5.    1.    9.
 
Enter command? ("H" for short menu, "HELP" for full menu, ? for full help)
e
Enter number of rows, number of columns.
3,3
Enter entries 1 to 3 of row 1
1,2,3
Enter entries 1 to 3 of row 2
4,5,6
Enter entries 1 to 3 of row 3
7,8,11
A copy of this matrix is being saved.
 
  The current matrix:
 
    1.    2.    3.
    4.    5.    6.
    7.    8.   11.
 
Enter command? ("H" for short menu, "HELP" for full menu, ? for full help)
#
r1 <=> r3
  ERO: Row 1 <=> Row 3
 
  The current matrix:
 
    7.    8.   11.
    4.    5.    6.
    1.    2.    3.
 
Enter command? ("H" for short menu, "HELP" for full menu, ? for full help)
r1 <= r1/7
  ERO: Row 1 <=  Row 1 / 7
 
  The current matrix:
 
   1.0000000   1.1428572   1.5714285
   4.0000000   5.0000000   6.0000000
   1.0000000   2.0000000   3.0000000
 
Enter command? ("H" for short menu, "HELP" for full menu, ? for full help)
r2 <= r2 -4 r1
  ERO: Row 2 <=  Row 2 - 4 Row 1
 
  The current matrix:
 
   1.0000000   1.1428572   1.5714285
   0.0000000   0.4285712  -0.2857141
   1.0000000   2.0000000   3.0000000
 
Enter command? ("H" for short menu, "HELP" for full menu, ? for full help)
r3 <= r3 - r1
  ERO: Row 3 <=  Row 3 - Row 1
 
  The current matrix:
 
   1.0000000   1.1428572   1.5714285
   0.0000000   0.4285712  -0.2857141
   0.0000000   0.8571428   1.4285715
 
Enter command? ("H" for short menu, "HELP" for full menu, ? for full help)
 
 
Enter command? ("H" for short menu, "HELP" for full menu, ? for full help)
r2 <=> r3
  ERO: Row 2 <=> Row 3
 
  The current matrix:
 
   1.0000000   1.1428572   1.5714285
   0.0000000   0.8571428   1.4285715
   0.0000000   0.4285712  -0.2857141
 
Enter command? ("H" for short menu, "HELP" for full menu, ? for full help)
r2 <= r2 / 0.8571428
  ERO: Row 2 <=  Row 2 / 0.857143
 
  The current matrix:
 
   1.0000000   1.1428572   1.5714285
   0.0000000   1.0000000   1.6666669
   0.0000000   0.4285712  -0.2857141
 
Enter command? ("H" for short menu, "HELP" for full menu, ? for full help)
r3 <= r3 - 0.4285712 r2
  ERO: Row 3 <=  Row 3 - 0.428571 Row 2
 
  The current matrix:
 
   1.0000000   1.1428572   1.5714285
   0.0000000   1.0000000   1.6666669
   0.0000000   0.0000000  -0.9999996
 
Enter command? ("H" for short menu, "HELP" for full menu, ? for full help)
 
 
Enter command? ("H" for short menu, "HELP" for full menu, ? for full help)
r3 <= r3 / -0.9999996
  ERO: Row 3 <=  Row 3 / -1.00000
 
  The current matrix:
 
   1.0000000   1.1428572   1.5714285
   0.0000000   1.0000000   1.6666669
   0.0000000   0.0000000   1.0000000
 
Enter command? ("H" for short menu, "HELP" for full menu, ? for full help)
#
check
 
Checking the matrix for row echelon form...
 
  The matrix is NOT in row echelon form.
  The first nonzero entry in row 3
  which occurs in column 2
  is -0.298023E-07 rather than 1.
 
Enter command? ("H" for short menu, "HELP" for full menu, ? for full help)
#
# OK, cheat and force the entry to be zero.
#
a(3,2) = 0.0
 
  A(3,2) = 0
 
  The current matrix:
 
   1.0000000   1.1428572   1.5714285
   0.0000000   1.0000000   1.6666669
   0.0000000   0.0000000   1.0000000
 
Enter command? ("H" for short menu, "HELP" for full menu, ? for full help)
check
 
Checking the matrix for row echelon form...
 
  The matrix is in row echelon form.
 
Checking the matrix for row reduced echelon form...
 
  The matrix is NOT in reduced row echelon form.
  Row 2 has its leading 1 in column 2.
  This means that all other entries of that column should be zero.
  But the entry in row 1 is 1.14286
 
Enter command? ("H" for short menu, "HELP" for full menu, ? for full help)
#
e
Enter number of rows, number of columns.
3,3
Enter entries 1 to 3 of row 1
1,2,3
Enter entries 1 to 3 of row 2
4,5,6
Enter entries 1 to 3 of row 3
7,8,9
A copy of this matrix is being saved.
 
  The current matrix:
 
   1.   2.   3.
   4.   5.   6.
   7.   8.   9.
 
Enter command? ("H" for short menu, "HELP" for full menu, ? for full help)
z
 
  ERO: Row 3 <=> Row 1
  ERO: Row 1 <=  Row 1 / 7
  ERO: Row 2 <=  Row 2 - 4 Row 1
  ERO: Row 3 <=  Row 3 - Row 1
 
  ERO: Row 3 <=> Row 2
  ERO: Row 2 <=  Row 2 / 0.857143
  ERO: Row 1 <=  Row 1 - 1.14286 Row 2
  ERO: Row 3 <=  Row 3 - 0.428571 Row 2
 
  ERO: Row 3 <=  Row 3 / 0.357628E-06
  ERO: Row 1 <=  Row 1 + 1.00000 Row 3
  ERO: Row 2 <=  Row 2 - 2.00000 Row 3
 
  The current matrix:
 
   1.   0.   0.
   0.   1.   0.
   0.   0.   1.
 
Enter command? ("H" for short menu, "HELP" for full menu, ? for full help)
#
r
#
# Insert a row.
#
g +r3
Enter entries 1 to 3 of row 3
6.1, 6.2, 6.3
 
  The current matrix:
 
   1.0000000   0.0000000   0.0000000
   0.0000000   1.0000000   0.0000000
   6.0999999   6.1999998   6.3000002
   0.0000000   0.0000000   1.0000000
 
Enter command? ("H" for short menu, "HELP" for full menu, ? for full help)
#
# Change A(3,2) to 77.
#
a(3,2) = 77
 
  A(3,2) = 77
 
  The current matrix:
 
    1.0000000    0.0000000    0.0000000
    0.0000000    1.0000000    0.0000000
    6.0999999   77.0000000    6.3000002
    0.0000000    0.0000000    1.0000000
 
Enter command? ("H" for short menu, "HELP" for full menu, ? for full help)
#
# Delete column 2.
#
g -c2
The column has been deleted!
 
  The current matrix:
 
   1.0000000   0.0000000
   0.0000000   0.0000000
   6.0999999   6.3000002
   0.0000000   1.0000000
 
Enter command? ("H" for short menu, "HELP" for full menu, ? for full help)
#
# Get a determinant test matrix of order 4.
#
b d 4
 
The following examples are available:
  "D" for determinant;
  "E" for eigenvalues;
  "I" for inverse;
  "S" for linear solve.
 
  "C" to cancel.
 
 
  The current matrix:
 
  -2.   7.  -2.  -8.
  -5.   8.  -2.  -5.
   3.  -6.  -3.   3.
   2.  -2.   8.  -7.
 
Enter command? ("H" for short menu, "HELP" for full menu, ? for full help)
det
 
The determinant is 891
 
Enter command? ("H" for short menu, "HELP" for full menu, ? for full help)
#
# Repeat work in decimal arithmetic.
#
#
# Set number of decimal digits to 5
#
dec_digit 5
How many decimal places should be used in
converting real results to a decimal?
 
 1 means 123.45 becomes 1 * 10**2
 2 means 123.45 becomes 12 * 10**1
 3 means 123.45 becomes 123
and so on.
 
The number of decimal digits will now be 5
 
Enter command? ("H" for short menu, "HELP" for full menu, ? for full help)
#
# Change to decimal arithmetic.
#
decimal
 
  Switching to DECIMAL arithmetic.
  The current number of decimal places is 5
 
  A note about DECIMAL arithmetic:
 
  The representation of decimals is exact.
  The arithmetic of decimals is approximate,
  because only a certain number of decimal
  places are stored.
  Also, the representation can break down
  if any exponent becomes too large or small.
 
  The current matrix:
 
       -2    7   -2   -8
       -5    8   -2   -5
        3   -6   -3    3
        2   -2    8   -7
 
Enter command? ("H" for short menu, "HELP" for full menu, ? for full help)
#
# Enter the test matrix.
#
e
Enter number of rows, number of columns.
3,3
Enter entries 1 to 3 of row 1
1,2,3
Enter entries 1 to 3 of row 2
4,5,6
Enter entries 1 to 3 of row 3
7,8,11
A copy of this matrix is being saved.
 
  The current matrix:
 
        1    2    3
        4    5    6
        7    8   11
 
Enter command? ("H" for short menu, "HELP" for full menu, ? for full help)
r1 <= r3
  ERO: Row 1 <=> Row 3
 
  The current matrix:
 
        7    8   11
        4    5    6
        1    2    3
 
Enter command? ("H" for short menu, "HELP" for full menu, ? for full help)
r1 <= r1/7
  ERO: Row 1 <=  Row 1 / 7
 
  The current matrix:
 
            1   1.1429   1.5714
            4        5        6
            1        2        3
 
Enter command? ("H" for short menu, "HELP" for full menu, ? for full help)
r2 <= r2 - 4 r1
  ERO: Row 2 <=  Row 2 - 4 Row 1
 
  The current matrix:
 
             1    1.1429    1.5714
             0    0.4284   -0.2856
             1         2         3
 
Enter command? ("H" for short menu, "HELP" for full menu, ? for full help)
r3 <= r3 - r1
  ERO: Row 3 <=  Row 3 - Row 1
 
  The current matrix:
 
             1    1.1429    1.5714
             0    0.4284   -0.2856
             0    0.8571    1.4286
 
Enter command? ("H" for short menu, "HELP" for full menu, ? for full help)
 
 
Enter command? ("H" for short menu, "HELP" for full menu, ? for full help)
r2 <=> r3
  ERO: Row 2 <=> Row 3
 
  The current matrix:
 
             1    1.1429    1.5714
             0    0.8571    1.4286
             0    0.4284   -0.2856
 
Enter command? ("H" for short menu, "HELP" for full menu, ? for full help)
r2 <= r2 / 0.8571
  ERO: Row 2 <=  Row 2 / 0.8571
 
  The current matrix:
 
             1    1.1429    1.5714
             0         1    1.6668
             0    0.4284   -0.2856
 
Enter command? ("H" for short menu, "HELP" for full menu, ? for full help)
r3 <= r3 - 0.4284 r2
  ERO: Row 3 <=  Row 3 - 0.4284 Row 2
 
  The current matrix:
 
              1     1.1429     1.5714
              0          1     1.6668
              0          0   -0.99965
 
Enter command? ("H" for short menu, "HELP" for full menu, ? for full help)
 
 
Enter command? ("H" for short menu, "HELP" for full menu, ? for full help)
r3 <= r3 / -0.99965
  ERO: Row 3 <=  Row 3 / -0.99965
 
  The current matrix:
 
            1   1.1429   1.5714
            0        1   1.6668
            0        0        1
 
Enter command? ("H" for short menu, "HELP" for full menu, ? for full help)
 
 
Enter command? ("H" for short menu, "HELP" for full menu, ? for full help)
#
check
 
Checking the matrix for row echelon form...
 
  The matrix is in row echelon form.
 
Checking the matrix for row reduced echelon form...
 
  The matrix is NOT in reduced row echelon form.
  Row 2 has its leading 1 in column 2.
  This means that all other entries of that column should be zero.
  But the entry in row 1 is 1.1429
 
Enter command? ("H" for short menu, "HELP" for full menu, ? for full help)
#
# Enter a new matrix.
#
e
Enter number of rows, number of columns.
3,3
Enter entries 1 to 3 of row 1
1,2,3
Enter entries 1 to 3 of row 2
4,5,6
Enter entries 1 to 3 of row 3
7,8,9
A copy of this matrix is being saved.
 
  The current matrix:
 
       1   2   3
       4   5   6
       7   8   9
 
Enter command? ("H" for short menu, "HELP" for full menu, ? for full help)
#
# Automatically process it.
#
z
 
  ERO: Row 3 <=> Row 1
  ERO: Row 1 <=  Row 1 / 7
  ERO: Row 2 <=  Row 2 - 4 Row 1
  ERO: Row 3 <=  Row 3 - Row 1
 
  ERO: Row 3 <=> Row 2
  ERO: Row 2 <=  Row 2 / 0.8571
  ERO: Row 1 <=  Row 1 - 1.1429 Row 2
  ERO: Row 3 <=  Row 3 - 0.4284 Row 2
 
  ERO: Row 3 <=  Row 3 / 0.00036
  ERO: Row 1 <=  Row 1 + 1.0002 Row 3
  ERO: Row 2 <=  Row 2 - 2.0001 Row 3
 
  The current matrix:
 
       1   0   0
       0   1   0
       0   0   1
 
Enter command? ("H" for short menu, "HELP" for full menu, ? for full help)
#
# Restore the original matrix.
#
r
#
# Insert a row.
#
g +r3
Enter entries 1 to 3 of row 3
6.1, 6.2, 6.3
 
  The current matrix:
 
         1     0     0
         0     1     0
       6.1   6.2   6.3
         0     0     1
 
Enter command? ("H" for short menu, "HELP" for full menu, ? for full help)
#
# Change entry (3,2) to 77.
#
a(3,2) = 77
 
  A(3,2) = 77
 
  The current matrix:
 
         1     0     0
         0     1     0
       6.1    77   6.3
         0     0     1
 
Enter command? ("H" for short menu, "HELP" for full menu, ? for full help)
#
# Remove column 2.
#
g -c2
The column has been deleted!
 
  The current matrix:
 
         1     0
         0     0
       6.1   6.3
         0     1
 
Enter command? ("H" for short menu, "HELP" for full menu, ? for full help)
#
# Get a determinant test matrix of order 4.
#
b d 4
 
The following examples are available:
  "D" for determinant;
  "E" for eigenvalues;
  "I" for inverse;
  "S" for linear solve.
 
  "C" to cancel.
 
 
  The current matrix:
 
         7     5    -8   -10
         5     4     0   -10
        -5    -4    -1     2
        -9    -4    -7    -1
 
Enter command? ("H" for short menu, "HELP" for full menu, ? for full help)
det
 
The determinant is -1199
 
Enter command? ("H" for short menu, "HELP" for full menu, ? for full help)
#
# Work in real arithmetic
#
real
 
  Switching to REAL arithmetic.
 
  The current matrix:
 
    7.    5.   -8.  -10.
    5.    4.    0.  -10.
   -5.   -4.   -1.    2.
   -9.   -4.   -7.   -1.
 
Enter command? ("H" for short menu, "HELP" for full menu, ? for full help)
#
# Get a sample matrix for the Jacobi method.
#
b e 4
 
The following examples are available:
  "D" for determinant;
  "E" for eigenvalues;
  "I" for inverse;
  "S" for linear solve.
 
  "C" to cancel.
 
We have a symmetric matrix, whose eigenvalues
can be found with the "J" command.
 
  The current matrix:
 
  -7.5036054   0.5010364  -0.3891510   1.7822232
   0.5010364  -4.6027484   0.4247353  -2.7617321
  -0.3891510   0.4247353  -3.5915809  -2.0459721
   1.7822232  -2.7617321  -2.0459721   5.6979346
 
Enter command? ("H" for short menu, "HELP" for full menu, ? for full help)
#
# Use Jacobi rotations.
#
j 1,2
 
The current matrix
 
  -7.5036054   0.5010364  -0.3891510   1.7822232
   0.5010364  -4.6027484   0.4247353  -2.7617321
  -0.3891510   0.4247353  -3.5915809  -2.0459721
   1.7822232  -2.7617321  -2.0459721   5.6979346
 
 
The single step Q transformation matrix
 
   0.9862034   0.1655380   0.0000000   0.0000000
  -0.1655380   0.9862034   0.0000000   0.0000000
   0.0000000   0.0000000   1.0000000   0.0000000
   0.0000000   0.0000000   0.0000000   1.0000000
 
The approximate eigenvector matrix (Q-Transpose)
 
   0.9862034   0.1655380   0.0000000   0.0000000
  -0.1655380   0.9862034   0.0000000   0.0000000
   0.0000000   0.0000000   1.0000000   0.0000000
   0.0000000   0.0000000   0.0000000   1.0000000
 
The current matrix
 
  -7.5877061   0.0000001  -0.4540919   2.2148061
   0.0000000  -4.5186472   0.3544562  -2.4286041
  -0.4540919   0.3544562  -3.5915809  -2.0459721
   2.2148061  -2.4286041  -2.0459721   5.6979346
 
Enter row I, column J, or "Q" to quit.
3,2
 
The single step Q transformation matrix
 
   1.0000000   0.0000000   0.0000000   0.0000000
   0.0000000   0.9471976   0.3206506   0.0000000
   0.0000000  -0.3206506   0.9471976   0.0000000
   0.0000000   0.0000000   0.0000000   1.0000000
 
The approximate eigenvector matrix (Q-Transpose)
 
   0.9862034   0.1567971   0.0530798   0.0000000
  -0.1655380   0.9341295   0.3162267   0.0000000
   0.0000000  -0.3206506   0.9471976   0.0000000
   0.0000000   0.0000000   0.0000000   1.0000000
 
The current matrix
 
  -7.5877061   0.1456050  -0.4301147   2.2148061
   0.1456048  -4.6386399   0.0000000  -1.6443256
  -0.4301147  -0.0000001  -3.4715884  -2.7166731
   2.2148061  -1.6443256  -2.7166731   5.6979346
 
Enter row I, column J, or "Q" to quit.
4,2
 
The single step Q transformation matrix
 
   1.0000000   0.0000000   0.0000000   0.0000000
   0.0000000   0.9881632   0.0000000  -0.1534069
   0.0000000   0.0000000   1.0000000   0.0000000
   0.0000000   0.1534069   0.0000000   0.9881632
 
The approximate eigenvector matrix (Q-Transpose)
 
  0.9862034  0.1549412  0.0530798 -0.0240538
 -0.1655380  0.9230723  0.3162267 -0.1433019
  0.0000000 -0.3168551  0.9471976  0.0491900
  0.0000000  0.1534069  0.0000000  0.9881632
 
The current matrix
 
  -7.5877061   0.4836479  -0.4301147   2.1662531
   0.4836478  -4.8939128  -0.4167563  -0.0000001
  -0.4301147  -0.4167564  -3.4715884  -2.6845164
   2.1662531  -0.0000001  -2.6845164   5.9532080
 
Enter row I, column J, or "Q" to quit.
4,3
 
The single step Q transformation matrix
 
   1.0000000   0.0000000   0.0000000   0.0000000
   0.0000000   1.0000000   0.0000000   0.0000000
   0.0000000   0.0000000   0.9666696  -0.2560271
   0.0000000   0.0000000   0.2560271   0.9666696
 
The approximate eigenvector matrix (Q-Transpose)
 
  0.9862034  0.1549412  0.0451523 -0.0368419
 -0.1655380  0.9230723  0.2689976 -0.2194882
  0.0000000 -0.3168551  0.9282211 -0.1949578
  0.0000000  0.1534069  0.2529966  0.9552273
 
The current matrix
 
  -7.5877061   0.4836479   0.1388407   2.2041721
   0.4836478  -4.8939128  -0.4028657   0.1067008
   0.1388406  -0.4028658  -4.1825957  -0.0000001
   2.2041721   0.1067008  -0.0000004   6.6642151
 
Enter row I, column J, or "Q" to quit.
3,2
 
The single step Q transformation matrix
 
   1.0000000   0.0000000   0.0000000   0.0000000
   0.0000000   0.9115425  -0.4112061   0.0000000
   0.0000000   0.4112061   0.9115425   0.0000000
   0.0000000   0.0000000   0.0000000   1.0000000
 
The approximate eigenvector matrix (Q-Transpose)
 
  0.9862034  0.1598023 -0.0225546 -0.0368419
 -0.1655380  0.9520331 -0.1343702 -0.2194882
  0.0000000  0.0928633  0.9764057 -0.1949578
  0.0000000  0.2438706  0.1675353  0.9552273
 
The current matrix
 
  -7.5877061   0.4979577  -0.0723198   2.2041721
   0.4979576  -5.0756507   0.0000001   0.0972623
  -0.0723198  -0.0000001  -4.0008597  -0.0438761
   2.2041721   0.0972622  -0.0438764   6.6642151
 
Enter row I, column J, or "Q" to quit.
4,2
 
The single step Q transformation matrix
 
   1.0000000   0.0000000   0.0000000   0.0000000
   0.0000000   0.9999657   0.0000000   0.0082839
   0.0000000   0.0000000   1.0000000   0.0000000
   0.0000000  -0.0082839   0.0000000   0.9999657
 
The approximate eigenvector matrix (Q-Transpose)
 
  0.9862034  0.1601020 -0.0225546 -0.0355169
 -0.1655380  0.9538187 -0.1343702 -0.2115941
  0.0000000  0.0944751  0.9764057 -0.1941818
  0.0000000  0.2359492  0.1675353  0.9572147
 
The current matrix
 
  -7.5877061   0.4796814  -0.0723198   2.2082217
   0.4796813  -5.0764570   0.0003636   0.0000000
  -0.0723198   0.0003633  -4.0008597  -0.0438746
   2.2082217   0.0000000  -0.0438749   6.6650205
 
Enter row I, column J, or "Q" to quit.
4,3
 
The single step Q transformation matrix
 
   1.0000000   0.0000000   0.0000000   0.0000000
   0.0000000   1.0000000   0.0000000   0.0000000
   0.0000000   0.0000000   0.9999915  -0.0041135
   0.0000000   0.0000000   0.0041135   0.9999915
 
The approximate eigenvector matrix (Q-Transpose)
 
  0.9862034  0.1601020 -0.0227005 -0.0354238
 -0.1655380  0.9538187 -0.1352395 -0.2110396
  0.0000000  0.0944751  0.9755987 -0.1981966
  0.0000000  0.2359492  0.1714713  0.9565175
 
The current matrix
 
  -7.5877061   0.4796814  -0.0632358   2.2085006
   0.4796813  -5.0764570   0.0003636  -0.0000015
  -0.0632358   0.0003633  -4.0010400   0.0000001
   2.2085006  -0.0000015  -0.0000001   6.6652012
 
Enter row I, column J, or "Q" to quit.
3,2
 
The single step Q transformation matrix
 
   1.0000000   0.0000000   0.0000000   0.0000000
   0.0000000   1.0000000   0.0003380   0.0000000
   0.0000000  -0.0003380   1.0000000   0.0000000
   0.0000000   0.0000000   0.0000000   1.0000000
 
The approximate eigenvector matrix (Q-Transpose)
 
  0.9862034  0.1601097 -0.0226463 -0.0354238
 -0.1655380  0.9538644 -0.1349171 -0.2110396
  0.0000000  0.0941454  0.9756306 -0.1981966
  0.0000000  0.2358912  0.1715511  0.9565175
 
The current matrix
 
  -7.5877061   0.4797028  -0.0630737   2.2085006
   0.4797027  -5.0764575   0.0000001  -0.0000015
  -0.0630736  -0.0000001  -4.0010405   0.0000001
   2.2085006  -0.0000015  -0.0000001   6.6652012
 
Enter row I, column J, or "Q" to quit.
q
 
I_READ - Fatal error!
  S_TO_I returned error flag!
 
Enter command? ("H" for short menu, "HELP" for full menu, ? for full help)
#
# Switch to Linear Programming mode,
# and don't save the current matrix.
#
l
Switching to linear programming mode.
Enter "Y" to use current matrix in linear programming.
no
 
Enter command? ("H" for short menu, "HELP" for full menu, ? for full help)
#
# Get a brief list of linear programming commands.
#
h
 
A(I,J)=S  Set table entry to S.
CHECK checks if the solution is optimal.
E     Enters a table to work on.
HELP  for full help.
L     switches to linear algebra.
P     I, J performs a pivot operation.
Q     quits.
TS    types the linear programming solution.
V     removes artificial variables.
Z     automatic solution (requires password).
?     interactive help.
 
R1 <=> R2  interchanges two rows
R1 <= S R1 multiplies a row by S.
R1 <= R1 + S R2 adds a multiple of another row.
 
 
Enter command? ("H" for short menu, "HELP" for full menu, ? for full help)
#
# Work in fractional arithmetic
#
rational
 
  Switching to RATIONAL arithmetic.
  with maximum integer = 2147483647
 
Enter command? ("H" for short menu, "HELP" for full menu, ? for full help)
#
# Get the "simple" linear programming example.
#
b s
 
The following examples are available:
  "S" a simple linear programming problem;
  "A" an advanced linear programming problem.
 
  "C" to cancel.
 
 
Simple linear programming problem:
 
Maximize:
  Z = 120 X + 100 Y + 70
subject to
  2 X + 2 Y < 8
  5 X + 3 Y < 15
 
 
  The linear programming table:
 
         1     2     3     4     P     C
 
X3       2     2     1     0     0     8
 
X4       5     3     0     1     0    15
 
Obj   -120  -100     0     0     1    70
 
Enter command? ("H" for short menu, "HELP" for full menu, ? for full help)
#
# Pivot on the first variable and second row.
#
p 1,2
 
Objective row
 
         1     2     3     4     P     C
 
Obj   -120  -100     0     0     1    70
 
Variable with most negative objective coefficient?
 
The entering variable is 1
 
Variable with smallest nonnegative feasibility ratio?
 
Nonnegative feasibility ratios:
 
Row 1, variable 3, ratio = 4.00000
Row 2, variable 4, ratio = 3.00000
 
 
The departing variable is 4 with feasibility ratio 3
 
  ERO: Row 2 <=  Row 2 / 5
  ERO: Row 1 <=  Row 1 - 2 Row 2
  ERO: Row 3 <=  Row 3 + 120 Row 2
 
The objective changed from 70
to 430
 
  The linear programming table:
 
         1     2     3     4     P     C
 
X3       0     4     1    -2     0     2
               5           5
 
X1       1     3     0     1     0     3
               5           5
 
Obj      0   -28     0    24     1   430
 
Enter command? ("H" for short menu, "HELP" for full menu, ? for full help)
#
# Pivot on the second variable, first row.
#
p 2,1
 
Objective row
 
         1     2     3     4     P     C
 
Obj      0   -28     0    24     1   430
 
Variable with most negative objective coefficient?
 
The entering variable is 2
 
Variable with smallest nonnegative feasibility ratio?
 
Nonnegative feasibility ratios:
 
Row 1, variable 3, ratio = 2.50000 = 5/2
Row 2, variable 1, ratio = 5.00000
 
 
The departing variable is 3 with feasibility ratio 2.50000
 
  ERO: Row 1 <=  Row 1 * 5/4
  ERO: Row 2 <=  Row 2 - 3/5 Row 1
  ERO: Row 3 <=  Row 3 + 28 Row 1
 
The objective changed from 430
to 500
 
  The linear programming table:
 
         1     2     3     4     P     C
 
X2       0     1     5    -1     0     5
                     4     2           2
 
X1       1     0    -3     1     0     3
                     4     2           2
 
Obj      0     0    35    10     1   500
 
Enter command? ("H" for short menu, "HELP" for full menu, ? for full help)
#
# Check for optimality.
#
check
 
Optimality test
 
 
Are all objective entries nonnegative?
 
Yes.  The current solution is optimal.
 
The linear programming solution:
 
       1   2   3   4
 
       3   5   0   0
       2   2
 
Objective = 500
 
Enter command? ("H" for short menu, "HELP" for full menu, ? for full help)
#
# Repeat work in real arithmetic
#
real
 
  Switching to REAL arithmetic.
 
  The linear programming table:
 
                 1             2             3             4             P
 
X2       0.0000000     1.0000000     1.2500000    -0.5000000     0.0000000
X1       1.0000000     0.0000000    -0.7500000     0.5000000     0.0000000
Obj      0.0000000     0.0000000    35.0000000    10.0000000     1.0000000
 
                 C
 
X2       2.5000000
X1       1.5000000
Obj    500.0000000
 
Enter command? ("H" for short menu, "HELP" for full menu, ? for full help)
#
# Get the "simple" linear programming example.
#
b s
 
The following examples are available:
  "S" a simple linear programming problem;
  "A" an advanced linear programming problem.
 
  "C" to cancel.
 
 
Simple linear programming problem:
 
Maximize:
  Z = 120 X + 100 Y + 70
subject to
  2 X + 2 Y < 8
  5 X + 3 Y < 15
 
 
  The linear programming table:
 
          1      2      3      4      P      C
 
X3       2.     2.     1.     0.     0.     8.
X4       5.     3.     0.     1.     0.    15.
Obj   -120.  -100.     0.     0.     1.    70.
 
Enter command? ("H" for short menu, "HELP" for full menu, ? for full help)
p 1,2
 
Objective row
 
          1      2      3      4      P      C
 
Obj   -120.  -100.     0.     0.     1.    70.
 
Variable with most negative objective coefficient?
 
The entering variable is 1
 
Variable with smallest nonnegative feasibility ratio?
 
Nonnegative feasibility ratios:
 
Row 1, variable 3, ratio = 4.00000
Row 2, variable 4, ratio = 3.00000
 
 
The departing variable is 4 with feasibility ratio 3
 
  ERO: Row 2 <=  Row 2 / 5
  ERO: Row 1 <=  Row 1 - 2 Row 2
  ERO: Row 3 <=  Row 3 + 120 Row 2
 
The objective changed from 70 to 430
 
  The linear programming table:
 
                 1             2             3             4             P
 
X3       0.0000000     0.8000000     1.0000000    -0.4000000     0.0000000
X1       1.0000000     0.6000000     0.0000000     0.2000000     0.0000000
Obj      0.0000000   -28.0000000     0.0000000    24.0000000     1.0000000
 
                 C
 
X3       2.0000000
X1       3.0000000
Obj    430.0000000
 
Enter command? ("H" for short menu, "HELP" for full menu, ? for full help)
p 2,1
 
Objective row
 
          1      2      3      4      P      C
 
Obj      0.   -28.     0.    24.     1.   430.
 
Variable with most negative objective coefficient?
 
The entering variable is 2
 
Variable with smallest nonnegative feasibility ratio?
 
Nonnegative feasibility ratios:
 
Row 1, variable 3, ratio = 2.50000
Row 2, variable 1, ratio = 5.00000
 
 
The departing variable is 3 with feasibility ratio 2.50000
 
  ERO: Row 1 <=  Row 1 / 0.800000
  ERO: Row 2 <=  Row 2 - 0.600000 Row 1
  ERO: Row 3 <=  Row 3 + 28 Row 1
 
The objective changed from 430 to 500
 
  The linear programming table:
 
                 1             2             3             4             P
 
X2       0.0000000     1.0000000     1.2500001    -0.5000001     0.0000000
X1       1.0000000     0.0000000    -0.7500001     0.5000001     0.0000000
Obj      0.0000000     0.0000000    35.0000038     9.9999981     1.0000000
 
                 C
 
X2       2.5000002
X1       1.4999998
Obj    500.0000000
 
Enter command? ("H" for short menu, "HELP" for full menu, ? for full help)
check
 
Optimality test
 
 
Are all objective entries nonnegative?
 
Yes.  The current solution is optimal.
 
The linear programming solution:
 
               1           2           3           4
 
       1.4999998   2.5000002   0.0000000   0.0000000
 
Objective = 500
 
Enter command? ("H" for short menu, "HELP" for full menu, ? for full help)
#
# Repeat work in decimal arithmetic
#
decimal
 
  Switching to DECIMAL arithmetic.
  The current number of decimal places is 5
 
  The linear programming table:
 
           1       2       3       4       P       C
 
X2         0       1    1.25    -0.5       0     2.5
X1         1       0   -0.75     0.5       0     1.5
Obj        0       0      35      10       1     500
 
Enter command? ("H" for short menu, "HELP" for full menu, ? for full help)
#
# Get the "simple" linear programming example.
#
b s
 
The following examples are available:
  "S" a simple linear programming problem;
  "A" an advanced linear programming problem.
 
  "C" to cancel.
 
 
Simple linear programming problem:
 
Maximize:
  Z = 120 X + 100 Y + 70
subject to
  2 X + 2 Y < 8
  5 X + 3 Y < 15
 
 
  The linear programming table:
 
          1      2      3      4      P      C
 
X3        2      2      1      0      0      8
X4        5      3      0      1      0     15
Obj    -120   -100      0      0      1     70
 
Enter command? ("H" for short menu, "HELP" for full menu, ? for full help)
p 1,2
 
Objective row
 
          1      2      3      4      P      C
 
Obj    -120   -100      0      0      1     70
 
Variable with most negative objective coefficient?
 
The entering variable is 1
 
Variable with smallest nonnegative feasibility ratio?
 
Nonnegative feasibility ratios:
 
Row 1, variable 3, ratio = 4.00000
Row 2, variable 4, ratio = 3.00000
 
 
The departing variable is 4 with feasibility ratio 3
 
  ERO: Row 2 <=  Row 2 / 5
  ERO: Row 1 <=  Row 1 - 2 Row 2
  ERO: Row 3 <=  Row 3 + 120 Row 2
 
The objective changed from 70 to 430
 
  The linear programming table:
 
          1      2      3      4      P      C
 
X3        0    0.8      1   -0.4      0      2
X1        1    0.6      0    0.2      0      3
Obj       0    -28      0     24      1    430
 
Enter command? ("H" for short menu, "HELP" for full menu, ? for full help)
p 2,1
 
Objective row
 
         1     2     3     4     P     C
 
Obj      0   -28     0    24     1   430
 
Variable with most negative objective coefficient?
 
The entering variable is 2
 
Variable with smallest nonnegative feasibility ratio?
 
Nonnegative feasibility ratios:
 
Row 1, variable 3, ratio = 2.50000
Row 2, variable 1, ratio = 5.00000
 
 
The departing variable is 3 with feasibility ratio 2.50000
 
  ERO: Row 1 <=  Row 1 / 0.8
  ERO: Row 2 <=  Row 2 - 0.6 Row 1
  ERO: Row 3 <=  Row 3 + 28 Row 1
 
The objective changed from 430 to 500
 
  The linear programming table:
 
           1       2       3       4       P       C
 
X2         0       1    1.25    -0.5       0     2.5
X1         1       0   -0.75     0.5       0     1.5
Obj        0       0      35      10       1     500
 
Enter command? ("H" for short menu, "HELP" for full menu, ? for full help)
check
 
Optimality test
 
 
Are all objective entries nonnegative?
 
Yes.  The current solution is optimal.
 
The linear programming solution:
 
         1     2     3     4
 
       1.5   2.5     0     0
 
Objective = 500
 
Enter command? ("H" for short menu, "HELP" for full menu, ? for full help)
#
# Work in fractional arithmetic
#
rational
 
  Switching to RATIONAL arithmetic.
  with maximum integer = 2147483647
 
  The linear programming table:
 
         1     2     3     4     P     C
 
X2       0     1     5    -1     0     5
                     4     2           2
 
X1       1     0    -3     1     0     3
                     4     2           2
 
Obj      0     0    35    10     1   500
 
Enter command? ("H" for short menu, "HELP" for full menu, ? for full help)
#
# Get the "advanced" linear programming example.
#
b a
 
The following examples are available:
  "S" a simple linear programming problem;
  "A" an advanced linear programming problem.
 
  "C" to cancel.
 
 
Advanced linear programming problem:
 
Maximize
  Z=40 X + 30 Y
subject to
  X + 2 Y > 6
2 X +   Y > 4
  X +   Y < 5
2 X +   Y < 8
 
 
  The linear programming table:
 
        1    2    3    4    5    6    7    8    P    C
 
X7      1    2   -1    0    0    0    1    0    0    6
 
X8      2    1    0   -1    0    0    0    1    0    4
 
X5      1    1    0    0    1    0    0    0    0    5
 
X6      2    1    0    0    0    1    0    0    0    8
 
Obj2    0    0    0    0    0    0    1    1    1    0
 
Obj   -40  -30    0    0    0    0    0    0    1    0
 
Enter command? ("H" for short menu, "HELP" for full menu, ? for full help)
#
# Automatically handle it.
#
z
 
The objective entry in column 7 is not zero,
but this corresponds to a basic variable.
 
  ERO: Row 5 <=  Row 5 - Row 1
 
The objective entry in column 8 is not zero,
but this corresponds to a basic variable.
 
  ERO: Row 5 <=  Row 5 - Row 2
 
The entering variable is 2
 
The departing variable is 7 with feasibility ratio 3
 
  ERO: Row 1 <=  Row 1 / 2
  ERO: Row 2 <=  Row 2 - Row 1
  ERO: Row 3 <=  Row 3 - Row 1
  ERO: Row 4 <=  Row 4 - Row 1
  ERO: Row 5 <=  Row 5 + 3 Row 1
 
The objective changed from -10
to -1
 
The entering variable is 1
 
The departing variable is 8 with feasibility ratio 0.666667
 
  ERO: Row 2 <=  Row 2 * 2/3
  ERO: Row 1 <=  Row 1 - 1/2 Row 2
  ERO: Row 3 <=  Row 3 - 1/2 Row 2
  ERO: Row 4 <=  Row 4 - 3/2 Row 2
  ERO: Row 5 <=  Row 5 + 3/2 Row 2
 
The objective changed from -1
to 0
 
Optimality test
 
 
Are all objective entries nonnegative?
 
Yes.  The current solution is optimal.
 
The linear programming solution:
 
       1   2   3   4   5   6   7   8
 
       2   8   0   0   5   4   0   0
       3   3           3
 
Objective = 0
 
This problem has artificial variables.
Use the "V" command to remove them.
 
  The linear programming table:
 
        1    2    3    4    5    6    7    8    P    C
 
X2      0    1   -2    1    0    0    2   -1    0    8
                  3    3              3    3         3
 
X1      1    0    1   -2    0    0   -1    2    0    2
                  3    3              3    3         3
 
X5      0    0    1    1    1    0   -1   -1    0    5
                  3    3              3    3         3
 
X6      0    0    0    1    0    1    0   -1    0    4
 
Obj2    0    0    0    0    0    0    1    1    1    0
 
Obj   -40  -30    0    0    0    0    0    0    1    0
 
Enter command? ("H" for short menu, "HELP" for full menu, ? for full help)
#
# Remove artificial variables.
#
v
 
All the artificial variables were deleted.
The original objective function is restored.
 
You must now use the "A" command to zero out
objective row entries for all basic variables.
 
  The linear programming table:
 
        1    2    3    4    5    6    P    C
 
X2      0    1   -2    1    0    0    0    8
                  3    3                   3
 
X1      1    0    1   -2    0    0    0    2
                  3    3                   3
 
X5      0    0    1    1    1    0    0    5
                  3    3                   3
 
X6      0    0    0    1    0    1    0    4
 
Obj   -40  -30    0    0    0    0    1    0
 
Enter command? ("H" for short menu, "HELP" for full menu, ? for full help)
#
# Eliminate the nonzero objective row entries associated with
# the basic variables, X1 and X2.
#
r5 <= r5 + 40 r2
  ERO: Row 5 <=  Row 5 + 40 Row 2
 
  The linear programming table:
 
        1    2    3    4    5    6    P    C
 
X2      0    1   -2    1    0    0    0    8
                  3    3                   3
 
X1      1    0    1   -2    0    0    0    2
                  3    3                   3
 
X5      0    0    1    1    1    0    0    5
                  3    3                   3
 
X6      0    0    0    1    0    1    0    4
 
Obj     0  -30   40  -80    0    0    1   80
                  3    3                   3
 
Enter command? ("H" for short menu, "HELP" for full menu, ? for full help)
r5 <= r5 + 30 r1
  ERO: Row 5 <=  Row 5 + 30 Row 1
 
  The linear programming table:
 
         1     2     3     4     5     6     P     C
 
X2       0     1    -2     1     0     0     0     8
                     3     3                       3
 
X1       1     0     1    -2     0     0     0     2
                     3     3                       3
 
X5       0     0     1     1     1     0     0     5
                     3     3                       3
 
X6       0     0     0     1     0     1     0     4
 
Obj      0     0   -20   -50     0     0     1   320
                     3     3                       3
 
Enter command? ("H" for short menu, "HELP" for full menu, ? for full help)
#
# Reapply the simplex method to the new tableau.
#
z
 
The entering variable is 4
 
The departing variable is 6 with feasibility ratio 4
 
  ERO: Row 1 <=  Row 1 - 1/3 Row 4
  ERO: Row 2 <=  Row 2 + 2/3 Row 4
  ERO: Row 3 <=  Row 3 - 1/3 Row 4
  ERO: Row 5 <=  Row 5 + 50/3 Row 4
 
The objective changed from 320/3 = 106.667
to 520/3 = 173.333
 
The entering variable is 3
 
The departing variable is 5 with feasibility ratio 1
 
  ERO: Row 3 <=  Row 3 * 3
  ERO: Row 1 <=  Row 1 + 2/3 Row 3
  ERO: Row 2 <=  Row 2 - 1/3 Row 3
  ERO: Row 5 <=  Row 5 + 20/3 Row 3
 
The objective changed from 520/3 = 173.333
to 180
 
Optimality test
 
 
Are all objective entries nonnegative?
 
Yes.  The current solution is optimal.
 
The linear programming solution:
 
       1   2   3   4   5   6
 
       3   2   1   4   0   0
 
Objective = 180
 
  The linear programming table:
 
         1     2     3     4     5     6     P     C
 
X2       0     1     0     0     2    -1     0     2
 
X1       1     0     0     0    -1     1     0     3
 
X3       0     0     1     0     3    -1     0     1
 
X4       0     0     0     1     0     1     0     4
 
Obj      0     0     0     0    20    10     1   180
 
Enter command? ("H" for short menu, "HELP" for full menu, ? for full help)
#
# Repeat work in real arithmetic
#
real
 
  Switching to REAL arithmetic.
 
  The linear programming table:
 
          1      2      3      4      5      6      P      C
 
X2       0.     1.     0.     0.     2.    -1.     0.     2.
X1       1.     0.     0.     0.    -1.     1.     0.     3.
X3       0.     0.     1.     0.     3.    -1.     0.     1.
X4       0.     0.     0.     1.     0.     1.     0.     4.
Obj      0.     0.     0.     0.    20.    10.     1.   180.
 
Enter command? ("H" for short menu, "HELP" for full menu, ? for full help)
#
b a
 
The following examples are available:
  "S" a simple linear programming problem;
  "A" an advanced linear programming problem.
 
  "C" to cancel.
 
 
Advanced linear programming problem:
 
Maximize
  Z=40 X + 30 Y
subject to
  X + 2 Y > 6
2 X +   Y > 4
  X +   Y < 5
2 X +   Y < 8
 
 
  The linear programming table:
 
         1     2     3     4     5     6     7     8     P     C
 
X7      1.    2.   -1.    0.    0.    0.    1.    0.    0.    6.
X8      2.    1.    0.   -1.    0.    0.    0.    1.    0.    4.
X5      1.    1.    0.    0.    1.    0.    0.    0.    0.    5.
X6      2.    1.    0.    0.    0.    1.    0.    0.    0.    8.
Obj2    0.    0.    0.    0.    0.    0.    1.    1.    1.    0.
Obj   -40.  -30.    0.    0.    0.    0.    0.    0.    1.    0.
 
Enter command? ("H" for short menu, "HELP" for full menu, ? for full help)
z
 
The objective entry in column 7 is not zero,
but this corresponds to a basic variable.
 
  ERO: Row 5 <=  Row 5 - Row 1
 
The objective entry in column 8 is not zero,
but this corresponds to a basic variable.
 
  ERO: Row 5 <=  Row 5 - Row 2
 
The entering variable is 2
 
The departing variable is 7 with feasibility ratio 3
 
  ERO: Row 1 <=  Row 1 / 2
  ERO: Row 2 <=  Row 2 - Row 1
  ERO: Row 3 <=  Row 3 - Row 1
  ERO: Row 4 <=  Row 4 - Row 1
  ERO: Row 5 <=  Row 5 + 3 Row 1
 
The objective changed from -10 to -1
 
The entering variable is 1
 
The departing variable is 8 with feasibility ratio 0.666667
 
  ERO: Row 2 <=  Row 2 / 1.50000
  ERO: Row 1 <=  Row 1 - 0.500000 Row 2
  ERO: Row 3 <=  Row 3 - 0.500000 Row 2
  ERO: Row 4 <=  Row 4 - 1.50000 Row 2
  ERO: Row 5 <=  Row 5 + 1.50000 Row 2
 
The objective changed from -1 to 0
 
Optimality test
 
 
Are all objective entries nonnegative?
 
Yes.  The current solution is optimal.
 
The linear programming solution:
 
               1           2           3           4           5           6
 
       0.6666667   2.6666667   0.0000000   0.0000000   1.6666666   4.0000000
 
               7           8
 
       0.0000000   0.0000000
 
Objective = 0
 
This problem has artificial variables.
Use the "V" command to remove them.
 
  The linear programming table:
 
                1            2            3            4            5
 
X2      0.0000000    1.0000000   -0.6666667    0.3333333    0.0000000
X1      1.0000000    0.0000000    0.3333333   -0.6666667    0.0000000
X5      0.0000000    0.0000000    0.3333333    0.3333333    1.0000000
X6      0.0000000    0.0000000    0.0000000    1.0000000    0.0000000
Obj2    0.0000000    0.0000000    0.0000000    0.0000000    0.0000000
Obj   -40.0000000  -30.0000000    0.0000000    0.0000000    0.0000000
 
                6            7            8            P            C
 
X2      0.0000000    0.6666667   -0.3333333    0.0000000    2.6666667
X1      0.0000000   -0.3333333    0.6666667    0.0000000    0.6666667
X5      0.0000000   -0.3333333   -0.3333333    0.0000000    1.6666666
X6      1.0000000    0.0000000   -1.0000000    0.0000000    4.0000000
Obj2    0.0000000    1.0000000    1.0000000    1.0000000    0.0000000
Obj     0.0000000    0.0000000    0.0000000    1.0000000    0.0000000
 
Enter command? ("H" for short menu, "HELP" for full menu, ? for full help)
v
 
All the artificial variables were deleted.
The original objective function is restored.
 
You must now use the "A" command to zero out
objective row entries for all basic variables.
 
  The linear programming table:
 
                1            2            3            4            5
 
X2      0.0000000    1.0000000   -0.6666667    0.3333333    0.0000000
X1      1.0000000    0.0000000    0.3333333   -0.6666667    0.0000000
X5      0.0000000    0.0000000    0.3333333    0.3333333    1.0000000
X6      0.0000000    0.0000000    0.0000000    1.0000000    0.0000000
Obj   -40.0000000  -30.0000000    0.0000000    0.0000000    0.0000000
 
                6            P            C
 
X2      0.0000000    0.0000000    2.6666667
X1      0.0000000    0.0000000    0.6666667
X5      0.0000000    0.0000000    1.6666666
X6      1.0000000    0.0000000    4.0000000
Obj     0.0000000    1.0000000    0.0000000
 
Enter command? ("H" for short menu, "HELP" for full menu, ? for full help)
r5 <= r5 + 40 r2
  ERO: Row 5 <=  Row 5 + 40 Row 2
 
  The linear programming table:
 
                1            2            3            4            5
 
X2      0.0000000    1.0000000   -0.6666667    0.3333333    0.0000000
X1      1.0000000    0.0000000    0.3333333   -0.6666667    0.0000000
X5      0.0000000    0.0000000    0.3333333    0.3333333    1.0000000
X6      0.0000000    0.0000000    0.0000000    1.0000000    0.0000000
Obj     0.0000000  -30.0000000   13.3333340  -26.6666679    0.0000000
 
                6            P            C
 
X2      0.0000000    0.0000000    2.6666667
X1      0.0000000    0.0000000    0.6666667
X5      0.0000000    0.0000000    1.6666666
X6      1.0000000    0.0000000    4.0000000
Obj     0.0000000    1.0000000   26.6666679
 
Enter command? ("H" for short menu, "HELP" for full menu, ? for full help)
r5 <= r5 + 30 r1
  ERO: Row 5 <=  Row 5 + 30 Row 1
 
  The linear programming table:
 
                 1             2             3             4             5
 
X2       0.0000000     1.0000000    -0.6666667     0.3333333     0.0000000
X1       1.0000000     0.0000000     0.3333333    -0.6666667     0.0000000
X5       0.0000000     0.0000000     0.3333333     0.3333333     1.0000000
X6       0.0000000     0.0000000     0.0000000     1.0000000     0.0000000
Obj      0.0000000     0.0000000    -6.6666660   -16.6666679     0.0000000
 
                 6             P             C
 
X2       0.0000000     0.0000000     2.6666667
X1       0.0000000     0.0000000     0.6666667
X5       0.0000000     0.0000000     1.6666666
X6       1.0000000     0.0000000     4.0000000
Obj      0.0000000     1.0000000   106.6666718
 
Enter command? ("H" for short menu, "HELP" for full menu, ? for full help)
z
 
The entering variable is 4
 
The departing variable is 6 with feasibility ratio 4
 
  ERO: Row 1 <=  Row 1 - 0.333333 Row 4
  ERO: Row 2 <=  Row 2 + 0.666667 Row 4
  ERO: Row 3 <=  Row 3 - 0.333333 Row 4
  ERO: Row 5 <=  Row 5 + 16.6667 Row 4
 
The objective changed from 106.667 to 173.333
 
The entering variable is 3
 
The departing variable is 5 with feasibility ratio 1.00000
 
  ERO: Row 3 <=  Row 3 / 0.333333
  ERO: Row 1 <=  Row 1 + 0.666667 Row 3
  ERO: Row 2 <=  Row 2 - 0.333333 Row 3
  ERO: Row 5 <=  Row 5 + 6.66667 Row 3
 
The objective changed from 173.333 to 180.000
 
Optimality test
 
 
Are all objective entries nonnegative?
 
Yes.  The current solution is optimal.
 
The linear programming solution:
 
               1           2           3           4           5           6
 
       3.0000002   2.0000000   0.9999998   4.0000000   0.0000000   0.0000000
 
Objective = 180.000
 
  The linear programming table:
 
                 1             2             3             4             5
 
X2       0.0000000     1.0000000     0.0000000     0.0000000     2.0000002
X1       1.0000000     0.0000000     0.0000000     0.0000000    -1.0000001
X3       0.0000000     0.0000000     1.0000000     0.0000000     3.0000002
X4       0.0000000     0.0000000     0.0000000     1.0000000     0.0000000
Obj      0.0000000     0.0000000     0.0000000     0.0000000    20.0000000
 
                 6             P             C
 
X2      -1.0000001     0.0000000     2.0000000
X1       1.0000000     0.0000000     3.0000002
X3      -1.0000001     0.0000000     0.9999998
X4       1.0000000     0.0000000     4.0000000
Obj     10.0000010     1.0000000   180.0000153
 
Enter command? ("H" for short menu, "HELP" for full menu, ? for full help)
#
# Repeat work in decimal arithmetic
#
decimal
 
  Switching to DECIMAL arithmetic.
  The current number of decimal places is 5
 
  The linear programming table:
 
         1     2     3     4     5     6     P     C
 
X2       0     1     0     0     2    -1     0     2
X1       1     0     0     0    -1     1     0     3
X3       0     0     1     0     3    -1     0     1
X4       0     0     0     1     0     1     0     4
Obj      0     0     0     0    20    10     1   180
 
Enter command? ("H" for short menu, "HELP" for full menu, ? for full help)
#
# Get the advanced linear programming example.
#
b a
 
The following examples are available:
  "S" a simple linear programming problem;
  "A" an advanced linear programming problem.
 
  "C" to cancel.
 
 
Advanced linear programming problem:
 
Maximize
  Z=40 X + 30 Y
subject to
  X + 2 Y > 6
2 X +   Y > 4
  X +   Y < 5
2 X +   Y < 8
 
 
  The linear programming table:
 
         1     2     3     4     5     6     7     8     P     C
 
X7       1     2    -1     0     0     0     1     0     0     6
X8       2     1     0    -1     0     0     0     1     0     4
X5       1     1     0     0     1     0     0     0     0     5
X6       2     1     0     0     0     1     0     0     0     8
Obj2     0     0     0     0     0     0     1     1     1     0
Obj    -40   -30     0     0     0     0     0     0     1     0
 
Enter command? ("H" for short menu, "HELP" for full menu, ? for full help)
z
 
The objective entry in column 7 is not zero,
but this corresponds to a basic variable.
 
  ERO: Row 5 <=  Row 5 - Row 1
 
The objective entry in column 8 is not zero,
but this corresponds to a basic variable.
 
  ERO: Row 5 <=  Row 5 - Row 2
 
The entering variable is 2
 
The departing variable is 7 with feasibility ratio 3
 
  ERO: Row 1 <=  Row 1 / 2
  ERO: Row 2 <=  Row 2 - Row 1
  ERO: Row 3 <=  Row 3 - Row 1
  ERO: Row 4 <=  Row 4 - Row 1
  ERO: Row 5 <=  Row 5 + 3 Row 1
 
The objective changed from -10 to -1
 
The entering variable is 1
 
The departing variable is 8 with feasibility ratio 0.666667
 
  ERO: Row 2 <=  Row 2 / 1.5
  ERO: Row 1 <=  Row 1 - 0.5 Row 2
  ERO: Row 3 <=  Row 3 - 0.5 Row 2
  ERO: Row 4 <=  Row 4 - 1.5 Row 2
  ERO: Row 5 <=  Row 5 + 1.5 Row 2
 
The objective changed from -1 to 0
 
The entering variable is 3
 
The departing variable is 1 with feasibility ratio 2.00003
 
  ERO: Row 2 <=  Row 2 / 0.33333
  ERO: Row 1 <=  Row 1 + 0.66666 Row 2
  ERO: Row 3 <=  Row 3 - 0.33334 Row 2
  ERO: Row 4 <=  Row 4 - 0.00001 Row 2
  ERO: Row 5 <=  Row 5 + 0.00001 Row 2
 
The objective changed from 0 to 0.00002
 
The entering variable is 4
 
The departing variable is 5 with feasibility ratio 0.999920
 
  ERO: Row 1 <=  Row 1 + 0.99997 Row 3
  ERO: Row 2 <=  Row 2 + 2 Row 3
  ERO: Row 4 <=  Row 4 - Row 3
  ERO: Row 5 <=  Row 5 + 0.00002 Row 3
 
The objective changed from 0.00002 to 0.000039998
 
Optimality test
 
 
Are all objective entries nonnegative?
 
Yes.  The current solution is optimal.
 
The linear programming solution:
 
             1         2         3         4         5         6         7
 
             0    4.9997    3.9998   0.99992         0    2.9999         0
 
             8
 
             0
 
Objective = 0.000039998
 
This problem has artificial variables.
Use the "V" command to remove them.
 
  The linear programming table:
 
                 1             2             3             4             5
 
X2         0.99993             1             0             0       0.99997
X3               1             0             1             0             2
X4              -1             0             0             1             1
X6         0.99997             0             0             0            -1
Obj2       0.00001             0             0             0       0.00002
Obj            -40           -30             0             0             0
 
                 6             7             8             P             C
 
X2               0             0             0             0        4.9997
X3               0            -1             0             0        3.9998
X4               0             0            -1             0       0.99992
X6               1             0             0             0        2.9999
Obj2             0       0.99999       0.99998             1   0.000039998
Obj              0             0             0             1             0
 
Enter command? ("H" for short menu, "HELP" for full menu, ? for full help)
v
The phase 1 objective function is nonzero.
Hence, this problem may have no solution.
 
All the artificial variables were deleted.
The original objective function is restored.
 
You must now use the "A" command to zero out
objective row entries for all basic variables.
 
  The linear programming table:
 
             1         2         3         4         5         6         P
 
X2     0.99993         1         0         0   0.99997         0         0
X3           1         0         1         0         2         0         0
X4          -1         0         0         1         1         0         0
X6     0.99997         0         0         0        -1         1         0
Obj        -40       -30         0         0         0         0         1
 
             C
 
X2      4.9997
X3      3.9998
X4     0.99992
X6      2.9999
Obj          0
 
Enter command? ("H" for short menu, "HELP" for full menu, ? for full help)
r5 <= r5 + 40 r2
  ERO: Row 5 <=  Row 5 + 40 Row 2
 
  The linear programming table:
 
             1         2         3         4         5         6         P
 
X2     0.99993         1         0         0   0.99997         0         0
X3           1         0         1         0         2         0         0
X4          -1         0         0         1         1         0         0
X6     0.99997         0         0         0        -1         1         0
Obj          0       -30        40         0        80         0         1
 
             C
 
X2      4.9997
X3      3.9998
X4     0.99992
X6      2.9999
Obj     159.99
 
Enter command? ("H" for short menu, "HELP" for full menu, ? for full help)
r5 <= r5 + 30 r1
  ERO: Row 5 <=  Row 5 + 30 Row 1
 
  The linear programming table:
 
             1         2         3         4         5         6         P
 
X2     0.99993         1         0         0   0.99997         0         0
X3           1         0         1         0         2         0         0
X4          -1         0         0         1         1         0         0
X6     0.99997         0         0         0        -1         1         0
Obj     29.997         0        40         0    109.99         0         1
 
             C
 
X2      4.9997
X3      3.9998
X4     0.99992
X6      2.9999
Obj     309.98
 
Enter command? ("H" for short menu, "HELP" for full menu, ? for full help)
z
 
The objective entry in column 3 is not zero,
but this corresponds to a basic variable.
 
  ERO: Row 5 <=  Row 5 - 40 Row 2
 
The entering variable is 1
 
The departing variable is 6 with feasibility ratio 2.99999
 
  ERO: Row 4 <=  Row 4 / 0.99997
  ERO: Row 1 <=  Row 1 - 0.99993 Row 4
  ERO: Row 2 <=  Row 2 - Row 4
  ERO: Row 3 <=  Row 3 + Row 4
  ERO: Row 5 <=  Row 5 + 10.003 Row 4
 
The objective changed from 149.99 to 179.99
 
Optimality test
 
 
Are all objective entries nonnegative?
 
Yes.  The current solution is optimal.
 
The linear programming solution:
 
            1        2        3        4        5        6
 
            3        2   0.9998   3.9999        0        0
 
Objective = 179.99
 
  The linear programming table:
 
              1          2          3          4          5          6
 
X2            0          1          0          0     1.9999   -0.99993
X3            0          0          1          0          3         -1
X4            0          0          0          1          0          1
X1            1          0          0          0         -1          1
Obj           0          0          0          0     19.987     10.003
 
              P          C
 
X2            0          2
X3            0     0.9998
X4            0     3.9999
X1            0          3
Obj           1     179.99
 
Enter command? ("H" for short menu, "HELP" for full menu, ? for full help)
#
# Close the disk file.
#
k
Closing the transcript file "output".