Opening the transcript file "output". MATMAN, version 1.59 Last modified on 06 July 1998. 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 C I,J,S changes matrix entry I, J to S. E enters a matrix to work on. HELP for full help. L switches to linear programming. O checks if the matrix is row reduced. 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 Add S times row I to row J. B Set up sample problem. BASIC I, J changes basic variable I to J. C Change entry I, J to S. D/M Divide/Multiply row I by S. DEC Use decimal arithmetic. 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. J Jacobi rotation in (I,J) plane. K Open/close the transcript file. L To switch between linear algebra and linear programming. MAXINT Set the maximum integer. N Set the number of decimal digits. O Check matrix for reduced row echelon form. O Check linear program tableau for optimality. P Pivot linear program, entering I, departing J. Q Quit. R Restore a saved matrix or tableau RAT Use rational arithmetic. REAL Use real arithmetic. S Store the current matrix or tableau. 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 "R" command can bring it back. 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 <= -4 Row 1 + Row 2 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 <= -1 Row 1 + Row 3 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 / (6/7) 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 <= -8/7 Row 2 + Row 1 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 <= -3/7 Row 2 + 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) # # Check for row reduced echelon form. # o This 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 <= -1 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 <= 1/3 Row 3 + Row 1 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 <= -5/3 Row 3 + Row 2 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 "R" command can bring it back. 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 Enter authorization key for "Z" command. 31984 Authorization confirmed. ERO: Row 3 <=> Row 1 ERO: Row 1 <= Row 1 / 7 ERO: Row 2 <= -4 Row 1 + Row 2 ERO: Row 3 <= -1 Row 1 + Row 3 ERO: Row 3 <=> Row 2 ERO: Row 2 <= Row 2 / (6/7) ERO: Row 1 <= -8/7 Row 2 + Row 1 ERO: Row 3 <= -3/7 Row 2 + Row 3 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. # r Your ERO command could not be understood. The first row number "R1" did not make sense. Because of an error, your command was not completed. We return to the main menu. 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 0 -1 0 1 2 61 31 63 10 5 10 0 0 0 Enter command? ("H" for short menu, "HELP" for full menu, ? for full help) # # Change entry 3,2 to 77 # c 3,2 77 Change entry 3, 2 to 77 Warning! Changing the matrix has probably made the ERO determinant incorrect. The current matrix: 1 0 -1 0 1 2 61 77 63 10 10 0 0 0 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 -1 0 2 61 63 10 10 0 0 Enter command? ("H" for short menu, "HELP" for full menu, ? for full help) # # Get the determinant test matrix # b d The following examples are available: "D" for determinant; "E" for eigenvalues; "I" for inverse; "S" for linear solve. "C" to cancel. Enter number of rows desired. 4 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 Converting to real arithmetic. Note: Real arithmetic can be inaccurate. 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 "R" command can bring it back. 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 <= -4 Row 1 + Row 2 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 <= -1 Row 1 + Row 3 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 <= -0.428571 Row 2 + Row 3 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 / -1.000000 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) # o This 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. # c 3,2 0.0 Change entry 3, 2 to 0 Warning! Changing the matrix has probably made the ERO determinant incorrect. 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) o This matrix is in row echelon form. This 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 "R" command can bring it back. 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 <= -4 Row 1 + Row 2 ERO: Row 3 <= -1 Row 1 + Row 3 ERO: Row 3 <=> Row 2 ERO: Row 2 <= Row 2 / 0.857143 ERO: Row 1 <= -1.14286 Row 2 + Row 1 ERO: Row 3 <= -0.428571 Row 2 + Row 3 ERO: Row 3 <= Row 3 / 0.357628E-06 ERO: Row 1 <= 1.00000 Row 3 + Row 1 ERO: Row 2 <= -2.00000 Row 3 + Row 2 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 Your ERO command could not be understood. The first row number "R1" did not make sense. Because of an error, your command was not completed. We return to the main menu. Enter command? ("H" for short menu, "HELP" for full menu, ? for full help) 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) c 3,2 77 Change entry 3, 2 to 77 Warning! Changing the matrix has probably made the ERO determinant incorrect. 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) 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 the determinant test matrix # b d The following examples are available: "D" for determinant; "E" for eigenvalues; "I" for inverse; "S" for linear solve. "C" to cancel. Enter number of rows desired. 4 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 digits to 5 # n 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 Converting to decimal arithmetic. Note: The representation of decimals is exact. However, this representation will break down if any exponent becomes too large or small. Calculations with decimals are NOT exact. 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 "R" command can bring it back. 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 <= -4 Row 1 + Row 2 The current matrix: 1 1.1429 1.5714 0 0.4284 5.9993 1 2 3 Enter command? ("H" for short menu, "HELP" for full menu, ? for full help) r3 <= r3 - r1 ERO: Row 3 <= -1 Row 1 + Row 3 The current matrix: 1 1.1429 1.5714 0 0.4284 5.9993 0 0.8571 2.9998 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 2.9998 0 0.4284 5.9993 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 3.4999 0 0.4284 5.9993 Enter command? ("H" for short menu, "HELP" for full menu, ? for full help) r3 <= r3 - 0.4284 r2 ERO: Row 3 <= -0.4284 Row 2 + Row 3 The current matrix: 1 1.1429 1.5714 0 1 3.4999 0 0 4.4999 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 / -0.99966 ERO: Row 3 <= Row 3 / -0.99966 The current matrix: 1 1.1429 1.5714 0 1 3.4999 0 0 -4.5014 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) # o This matrix is NOT in row echelon form. The first nonzero entry in row 3 which occurs in column 3 is -4.5014 rather than 1. 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 "R" command can bring it back. 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 <= -4 Row 1 + Row 2 ERO: Row 3 <= -1 Row 1 + Row 3 ERO: Row 3 <=> Row 2 ERO: Row 2 <= Row 2 / 0.8571 ERO: Row 1 <= -1.1429 Row 2 + Row 1 ERO: Row 3 <= -0.4284 Row 2 + Row 3 ERO: Row 3 <= Row 3 / 4.5 ERO: Row 1 <= 2.7143 Row 3 + Row 1 ERO: Row 2 <= -3.4999 Row 3 + Row 2 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 Your ERO command could not be understood. The first row number "R1" did not make sense. Because of an error, your command was not completed. We return to the main menu. Enter command? ("H" for short menu, "HELP" for full menu, ? for full help) 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) c 3,2 77 Change entry 3, 2 to 77 Warning! Changing the matrix has probably made the ERO determinant incorrect. 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) 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 the determinant test matrix # b d The following examples are available: "D" for determinant; "E" for eigenvalues; "I" for inverse; "S" for linear solve. "C" to cancel. Enter number of rows desired. 4 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 30214 Enter command? ("H" for short menu, "HELP" for full menu, ? for full help) # # Work in real arithmetic # real Converting 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 the sample matrix for the Jacobi method. # b e The following examples are available: "D" for determinant; "E" for eigenvalues; "I" for inverse; "S" for linear solve. "C" to cancel. Enter number of rows desired. # Because of an error, your command was not completed. We return to the main menu. Enter command? ("H" for short menu, "HELP" for full menu, ? for full help) # Use Jacobi rotations. # j 1,2 The value of I, the row index, is illegal. The value of I, the row index, is illegal. Enter row I, column J, or "Q" to quit. 3,2 The value of I, the row index, is illegal. The value of I, the row index, is illegal. Enter row I, column J, or "Q" to quit. 4,2 The value of I, the row index, is illegal. The value of I, the row index, is illegal. Enter row I, column J, or "Q" to quit. 4,3 The value of I, the row index, is illegal. The value of I, the row index, is illegal. Enter row I, column J, or "Q" to quit. 3,2 The value of I, the row index, is illegal. The value of I, the row index, is illegal. Enter row I, column J, or "Q" to quit. 4,2 The value of I, the row index, is illegal. The value of I, the row index, is illegal. Enter row I, column J, or "Q" to quit. 4,3 The value of I, the row index, is illegal. The value of I, the row index, is illegal. Enter row I, column J, or "Q" to quit. 3,2 The value of I, the row index, is illegal. The value of I, the row index, is illegal. Enter row I, column J, or "Q" to quit. q 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. n Enter command? ("H" for short menu, "HELP" for full menu, ? for full help) # # Work in fractional arithmetic # rational Converting to fractional arithmetic. 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 tableau: 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) # # Get a brief list of linear programming commands. # h C I, J, S changes tableau entry I, J to S. E Enters a tableau to work on. HELP for full help. L switches to linear algebra. O checks if the solution is optimal. 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) # # 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 <= -2 Row 2 + Row 1 ERO: Row 3 <= 120 Row 2 + Row 3 The objective changed from 70 to 430 The linear programming tableau: 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 / (4/5) ERO: Row 2 <= -3/5 Row 1 + Row 2 ERO: Row 3 <= 28 Row 1 + Row 3 The objective changed from 430 to 500 The linear programming tableau: 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. # o 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 Converting to real arithmetic. The linear programming tableau: 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) 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 tableau: 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 <= -2 Row 2 + Row 1 ERO: Row 3 <= 120 Row 2 + Row 3 The objective changed from 70 to 430 The linear programming tableau: 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 <= -0.600000 Row 1 + Row 2 ERO: Row 3 <= 28 Row 1 + Row 3 The objective changed from 430 to 500 The linear programming tableau: 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) o 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 Converting to decimal arithmetic. The linear programming tableau: 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) 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 tableau: 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 <= -2 Row 2 + Row 1 ERO: Row 3 <= 120 Row 2 + Row 3 The objective changed from 70 to 430 The linear programming tableau: 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 24.0 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 24.0 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 <= -0.6 Row 1 + Row 2 ERO: Row 3 <= 28 Row 1 + Row 3 The objective changed from 430 to 500 The linear programming tableau: 1 2 3 4 P C X2 0 1 1.25 -0.5 0 2.5 X1 1 0 -0.750 0.203 0 -12 Obj 0 0 35.00 23.86 1 500 Enter command? ("H" for short menu, "HELP" for full menu, ? for full help) o Optimality test Are all objective entries nonnegative? Yes. The current solution is optimal. The linear programming solution: 1 2 3 4 -12 2.5 0 0 Objective = 500 Enter command? ("H" for short menu, "HELP" for full menu, ? for full help) # # Work in fractional arithmetic # rational Converting to fractional arithmetic. The linear programming tableau: 1 2 3 4 P C X2 0 1 5 -1 0 5 4 2 2 X1 1 0 -3 203 0 -12 4 1000 Obj 0 0 35 1193 1 500 50 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 tableau: 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 <= -1 Row 1 + Row 5 The objective entry in column 8 is not zero, but this corresponds to a basic variable. ERO: Row 5 <= -1 Row 2 + Row 5 The entering variable is 2 The departing variable is 7 with feasibility ratio 3 ERO: Row 1 <= Row 1 / 2 ERO: Row 2 <= -1 Row 1 + Row 2 ERO: Row 3 <= -1 Row 1 + Row 3 ERO: Row 4 <= -1 Row 1 + Row 4 ERO: Row 5 <= 3 Row 1 + Row 5 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 / (3/2) ERO: Row 1 <= -1/2 Row 2 + Row 1 ERO: Row 3 <= -1/2 Row 2 + Row 3 ERO: Row 4 <= -3/2 Row 2 + Row 4 ERO: Row 5 <= 3/2 Row 2 + Row 5 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 tableau: 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 tableau: 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 <= 40 Row 2 + Row 5 The linear programming tableau: 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 <= 30 Row 1 + Row 5 The linear programming tableau: 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 <= -1/3 Row 4 + Row 1 ERO: Row 2 <= 2/3 Row 4 + Row 2 ERO: Row 3 <= -1/3 Row 4 + Row 3 ERO: Row 5 <= 50/3 Row 4 + Row 5 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 / (1/3) ERO: Row 1 <= 2/3 Row 3 + Row 1 ERO: Row 2 <= -1/3 Row 3 + Row 2 ERO: Row 5 <= 20/3 Row 3 + Row 5 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 tableau: 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 Converting to real arithmetic. The linear programming tableau: 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 tableau: 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 <= -1 Row 1 + Row 5 The objective entry in column 8 is not zero, but this corresponds to a basic variable. ERO: Row 5 <= -1 Row 2 + Row 5 The entering variable is 2 The departing variable is 7 with feasibility ratio 3 ERO: Row 1 <= Row 1 / 2 ERO: Row 2 <= -1 Row 1 + Row 2 ERO: Row 3 <= -1 Row 1 + Row 3 ERO: Row 4 <= -1 Row 1 + Row 4 ERO: Row 5 <= 3 Row 1 + Row 5 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 <= -0.500000 Row 2 + Row 1 ERO: Row 3 <= -0.500000 Row 2 + Row 3 ERO: Row 4 <= -1.50000 Row 2 + Row 4 ERO: Row 5 <= 1.50000 Row 2 + Row 5 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 tableau: 1 2 3 4 5 6 X2 0.0000000 1.0000000 -0.6666667 0.3333333 0.0000000 0.0000000 X1 1.0000000 0.0000000 0.3333333 -0.6666667 0.0000000 0.0000000 X5 0.0000000 0.0000000 0.3333333 0.3333333 1.0000000 0.0000000 X6 0.0000000 0.0000000 0.0000000 1.0000000 0.0000000 1.0000000 Obj2 0.0000000 0.0000000 0.0000000 0.0000000 0.0000000 0.0000000 Obj -40.0000000 -30.0000000 0.0000000 0.0000000 0.0000000 0.0000000 7 8 P C X2 0.6666667 -0.3333333 0.0000000 2.6666667 X1 -0.3333333 0.6666667 0.0000000 0.6666667 X5 -0.3333333 -0.3333333 0.0000000 1.6666666 X6 0.0000000 -1.0000000 0.0000000 4.0000000 Obj2 1.0000000 1.0000000 1.0000000 0.0000000 Obj 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 tableau: 1 2 3 4 5 6 X2 0.0000000 1.0000000 -0.6666667 0.3333333 0.0000000 0.0000000 X1 1.0000000 0.0000000 0.3333333 -0.6666667 0.0000000 0.0000000 X5 0.0000000 0.0000000 0.3333333 0.3333333 1.0000000 0.0000000 X6 0.0000000 0.0000000 0.0000000 1.0000000 0.0000000 1.0000000 Obj -40.0000000 -30.0000000 0.0000000 0.0000000 0.0000000 0.0000000 P C X2 0.0000000 2.6666667 X1 0.0000000 0.6666667 X5 0.0000000 1.6666666 X6 0.0000000 4.0000000 Obj 1.0000000 0.0000000 Enter command? ("H" for short menu, "HELP" for full menu, ? for full help) r5 <= r5 + 40 r2 ERO: Row 5 <= 40 Row 2 + Row 5 The linear programming tableau: 1 2 3 4 5 6 X2 0.0000000 1.0000000 -0.6666667 0.3333333 0.0000000 0.0000000 X1 1.0000000 0.0000000 0.3333333 -0.6666667 0.0000000 0.0000000 X5 0.0000000 0.0000000 0.3333333 0.3333333 1.0000000 0.0000000 X6 0.0000000 0.0000000 0.0000000 1.0000000 0.0000000 1.0000000 Obj 0.0000000 -30.0000000 13.3333340 -26.6666679 0.0000000 0.0000000 P C X2 0.0000000 2.6666667 X1 0.0000000 0.6666667 X5 0.0000000 1.6666666 X6 0.0000000 4.0000000 Obj 1.0000000 26.6666679 Enter command? ("H" for short menu, "HELP" for full menu, ? for full help) r5 <= r5 + 30 r1 ERO: Row 5 <= 30 Row 1 + Row 5 The linear programming tableau: 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 <= -0.333333 Row 4 + Row 1 ERO: Row 2 <= 0.666667 Row 4 + Row 2 ERO: Row 3 <= -0.333333 Row 4 + Row 3 ERO: Row 5 <= 16.6667 Row 4 + Row 5 The objective changed from 106.667 to 173.333 The entering variable is 3 The departing variable is 5 with feasibility ratio 1.000000 ERO: Row 3 <= Row 3 / 0.333333 ERO: Row 1 <= 0.666667 Row 3 + Row 1 ERO: Row 2 <= -0.333333 Row 3 + Row 2 ERO: Row 5 <= 6.66667 Row 3 + Row 5 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 tableau: 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 Converting to decimal arithmetic. The linear programming tableau: 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 tableau: 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 <= -1 Row 1 + Row 5 The objective entry in column 8 is not zero, but this corresponds to a basic variable. ERO: Row 5 <= -1 Row 2 + Row 5 The entering variable is 2 The departing variable is 7 with feasibility ratio 3 ERO: Row 1 <= Row 1 / 2 ERO: Row 2 <= -1 Row 1 + Row 2 ERO: Row 3 <= -1 Row 1 + Row 3 ERO: Row 4 <= -1 Row 1 + Row 4 ERO: Row 5 <= 3 Row 1 + Row 5 The objective changed from -10 to -1 The entering variable is 3 The departing variable is 8 with feasibility ratio 2 ERO: Row 2 <= Row 2 / 0.5 ERO: Row 1 <= 0.5 Row 2 + Row 1 ERO: Row 3 <= -0.5 Row 2 + Row 3 ERO: Row 4 <= -0.5 Row 2 + Row 4 ERO: Row 5 <= 0.5 Row 2 + Row 5 The objective changed from -1 to 9 The entering variable is 7 Cannot find a departing variable. Presumably, the feasible set is unbounded. Because of an error, your command was not completed. We return to the main menu. 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 tableau: 1 2 3 4 5 6 P C X2 15.5 1 0 -1.0 0 0 0 13 X3 3 0 1 -2 0 0 0 2 X5 -14.5 0 0 1.0 1 0 0 -8 X6 -13.5 0 0 1.0 0 1 0 -5 Obj -40 -30 0 0 0 0 1 0 Enter command? ("H" for short menu, "HELP" for full menu, ? for full help) r5 <= r5 + 40 r2 ERO: Row 5 <= 40 Row 2 + Row 5 The linear programming tableau: 1 2 3 4 5 6 P C X2 15.5 1 0 -1.0 0 0 0 13 X3 3 0 1 -2 0 0 0 2 X5 -14.5 0 0 1.0 1 0 0 -8 X6 -13.5 0 0 1.0 0 1 0 -5 Obj -40 -30 40 -80 0 0 1 8 Enter command? ("H" for short menu, "HELP" for full menu, ? for full help) r5 <= r5 + 30 r1 ERO: Row 5 <= 30 Row 1 + Row 5 The linear programming tableau: 1 2 3 4 5 6 P C X2 15.5 1 0 -1.0 0 0 0 13 X3 3 0 1 -2 0 0 0 2 X5 -14.5 0 0 1.0 1 0 0 -8 X6 -13.5 0 0 1.0 0 1 0 -5 Obj -40 0 40 -83 0 0 1 47 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 <= -40 Row 2 + Row 5 The entering variable is 1 The departing variable is 3 with feasibility ratio 0.666667 ERO: Row 2 <= Row 2 / 3 ERO: Row 1 <= -15.5 Row 2 + Row 1 ERO: Row 3 <= 14.5 Row 2 + Row 3 ERO: Row 4 <= 13.5 Row 2 + Row 4 ERO: Row 5 <= 40 Row 2 + Row 5 The objective changed from 39 to 41.666 Optimality test Are all objective entries nonnegative? Yes. The current solution is optimal. The linear programming solution: 1 2 3 4 5 6 0.66667 -90.333 0 0 88.667 85 Objective = 41.666 The linear programming tableau: 1 2 3 4 5 6 X2 0 1 -51.66615 -0.13997 0 0 X1 1 0 0.33333 -0.66667 0 0 X5 0 0 48.33285 0.14003 1 0 X6 0 0 44.99955 0.1401 0 1 Obj 0 0 1.33332 0.0064504 0 0 P C X2 0 -90.333 X1 0 0.66667 X5 0 88.667 X6 0 85 Obj 1 41.666 Enter command? ("H" for short menu, "HELP" for full menu, ? for full help) # # Close the disk file. # k Closing the transcript file "output".