The OPTLP Procedure

Example 11.8 Using the Network Simplex Solver

This example demonstrates how to use the network simplex solver to find the minimum-cost flow in a directed graph. Consider the directed graph in Output 11.8.1, which appears in Ahuja, Magnanti, and Orlin (1993).

Output 11.8.1: Minimum Cost Network Flow Problem: Data


You can use the following SAS statements to create the input data set ex8:

data ex8;
   input field1 $8. field2 $13. @25 field3 $13. field4  @53 field5 $13. field6;
   datalines;
NAME    .               .                    .      .                   .
ROWS    .               .                    .      .                   .
N       obj             .                    .      .                   .
E       balance['1']    .                    .      .                   .
E       balance['2']    .                    .      .                   .
E       balance['3']    .                    .      .                   .
E       balance['4']    .                    .      .                   .
E       balance['5']    .                    .      .                   .
E       balance['6']    .                    .      .                   .
E       balance['7']    .                    .      .                   .
E       balance['8']    .                    .      .                   .
COLUMNS .               .                    .      .                   .
.       x['1','4']      obj                  2      balance['1']        1
.       x['1','4']      balance['4']        -1      .                   .
.       x['2','1']      obj                  1      balance['1']       -1
.       x['2','1']      balance['2']         1      .                   .
.       x['2','3']      balance['2']         1      balance['3']       -1
.       x['2','6']      obj                  6      balance['2']        1
.       x['2','6']      balance['6']        -1      .                   .
.       x['3','4']      obj                  1      balance['3']        1
.       x['3','4']      balance['4']        -1      .                   .
.       x['3','5']      obj                  4      balance['3']        1
.       x['3','5']      balance['5']        -1      .                   .
.       x['4','7']      obj                  5      balance['4']        1
.       x['4','7']      balance['7']        -1      .                   .
.       x['5','6']      obj                  2      balance['5']        1
.       x['5','6']      balance['6']        -1      .                   .
.       x['5','7']      obj                  7      balance['5']        1
.       x['5','7']      balance['7']        -1      .                   .
.       x['6','8']      obj                  8      balance['6']        1
.       x['6','8']      balance['8']        -1      .                   .
.       x['7','8']      obj                  9      balance['7']        1
.       x['7','8']      balance['8']        -1      .                   .
RHS     .               .                    .      .                   .
.       .RHS.           balance['1']        10      .                   .
.       .RHS.           balance['2']        20      .                   .
.       .RHS.           balance['4']        -5      .                   .
.       .RHS.           balance['7']       -15      .                   .
.       .RHS.           balance['8']       -10      .                   .
BOUNDS  .               .                    .      .                   .
UP      .BOUNDS.        x['1','4']          15      .                   .
UP      .BOUNDS.        x['2','1']          10      .                   .
UP      .BOUNDS.        x['2','3']          10      .                   .
UP      .BOUNDS.        x['2','6']          10      .                   .
UP      .BOUNDS.        x['3','4']           5      .                   .
UP      .BOUNDS.        x['3','5']          10      .                   .
UP      .BOUNDS.        x['4','7']          10      .                   .
UP      .BOUNDS.        x['5','6']          20      .                   .
UP      .BOUNDS.        x['5','7']          15      .                   .
UP      .BOUNDS.        x['6','8']          10      .                   .
UP      .BOUNDS.        x['7','8']          15      .                   .
ENDATA  .               .                    .      .                   .
;

You can use the following call to PROC OPTLP to find the minimum-cost flow:

proc optlp
  presolver  = none
  printlevel = 2
  logfreq    = 1
  data       = ex8
  primalout  = ex8out
  algorithm  = ns;
run;

The optimal solution is displayed in Output 11.8.2.

Output 11.8.2: Network Simplex Solver: Primal Solution Output

Primal Solution

Obs Objective
Function ID
RHS ID Variable Name Variable
Type
Objective
Coefficient
Lower
Bound
Upper Bound Variable
Value
Variable
Status
Reduced
Cost
1 obj .RHS. x['1','4'] D 2 0 15 10 B 0
2 obj .RHS. x['2','1'] D 1 0 10 0 L 2
3 obj .RHS. x['2','3'] D 0 0 10 10 B 0
4 obj .RHS. x['2','6'] D 6 0 10 10 B 0
5 obj .RHS. x['3','4'] D 1 0 5 5 B 0
6 obj .RHS. x['3','5'] D 4 0 10 5 B 0
7 obj .RHS. x['4','7'] D 5 0 10 10 U -5
8 obj .RHS. x['5','6'] D 2 0 20 0 L 0
9 obj .RHS. x['5','7'] D 7 0 15 5 B 0
10 obj .RHS. x['6','8'] D 8 0 10 10 B 0
11 obj .RHS. x['7','8'] D 9 0 15 0 L 6


The optimal solution is represented graphically in Output 11.8.3.

Output 11.8.3: Minimum Cost Network Flow Problem: Optimal Solution


The iteration log is displayed in Output 11.8.4.

Output 11.8.4: Log: Solution Progress

NOTE: The problem has 11 variables (0 free, 0 fixed).                           
NOTE: The problem has 8 constraints (0 LE, 8 EQ, 0 GE, 0 range).                
NOTE: The problem has 22 constraint coefficients.                               
NOTE: The LP presolver value NONE is applied.                                   
NOTE: The LP solver is called.                                                  
NOTE: The Network Simplex algorithm is used.                                    
NOTE: The network has 8 rows (100.00%), 11 columns (100.00%), and 1 component.  
NOTE: The network extraction and setup time is 0.00 seconds.                    
                        Primal         Primal           Dual                    
      Iteration      Objective  Infeasibility  Infeasibility     Time           
              1   0.000000E+00   2.000000E+01   8.900000E+01     0.00           
              2   0.000000E+00   2.000000E+01   8.900000E+01     0.00           
              3   5.000000E+00   1.500000E+01   8.400000E+01     0.00           
              4   5.000000E+00   1.500000E+01   8.300000E+01     0.00           
              5   7.500000E+01   1.500000E+01   8.300000E+01     0.00           
              6   7.500000E+01   1.500000E+01   7.900000E+01     0.00           
              7   1.300000E+02   1.000000E+01   7.600000E+01     0.00           
              8   2.700000E+02   0.000000E+00   0.000000E+00     0.00           
NOTE: The Network Simplex solve time is 0.00 seconds.                           
NOTE: The total Network Simplex solve time is 0.00 seconds.                     
NOTE: Optimal.                                                                  
NOTE: Objective = 270.                                                          
NOTE: The data set WORK.EX8OUT has 11 observations and 10 variables.