

This example demonstrates how you can use the network simplex solver to find the minimum-cost flow in a directed graph. Consider the directed graph in Output 6.5.1, which appears in Ahuja, Magnanti, and Orlin (1993).
Output 6.5.1: Minimum Cost Network Flow Problem: Data

You can use the following SAS statements to create the input data sets nodedata and arcdata:
data nodedata; input _node_ $ _sd_; datalines; 1 10 2 20 3 0 4 -5 5 0 6 0 7 -15 8 -10 ; data arcdata; input _tail_ $ _head_ $ _lo_ _capac_ _cost_; datalines; 1 4 0 15 2 2 1 0 10 1 2 3 0 10 0 2 6 0 10 6 3 4 0 5 1 3 5 0 10 4 4 7 0 10 5 5 6 0 20 2 5 7 0 15 7 6 8 0 10 8 7 8 0 15 9 ;
You can use the following call to PROC OPTMODEL to find the minimum-cost flow:
proc optmodel;
set <str> NODES;
num supply_demand {NODES};
set <str,str> ARCS;
num arcLower {ARCS};
num arcUpper {ARCS};
num arcCost {ARCS};
read data arcdata into ARCS=[_tail_ _head_]
arcLower=_lo_ arcUpper=_capac_ arcCost=_cost_;
read data nodedata into NODES=[_node_] supply_demand=_sd_;
var flow {<i,j> in ARCS} >= arcLower[i,j] <= arcUpper[i,j];
min obj = sum {<i,j> in ARCS} arcCost[i,j] * flow[i,j];
con balance {i in NODES}:
sum {<(i),j> in ARCS} flow[i,j] - sum {<j,(i)> in ARCS} flow[j,i]
= supply_demand[i];
solve with lp / algorithm=ns scale=none logfreq=1;
print flow;
quit;
%put &_OROPTMODEL_;
The optimal solution is displayed in Output 6.5.2.
Output 6.5.2: Network Simplex Solver: Primal Solution Output
| Problem Summary | |
|---|---|
| Objective Sense | Minimization |
| Objective Function | obj |
| Objective Type | Linear |
| Number of Variables | 11 |
| Bounded Above | 0 |
| Bounded Below | 0 |
| Bounded Below and Above | 11 |
| Free | 0 |
| Fixed | 0 |
| Number of Constraints | 8 |
| Linear LE (<=) | 0 |
| Linear EQ (=) | 8 |
| Linear GE (>=) | 0 |
| Linear Range | 0 |
| Constraint Coefficients | 22 |
| Performance Information | |
|---|---|
| Execution Mode | On Client |
| Number of Threads | 1 |
| Solution Summary | |
|---|---|
| Solver | LP |
| Algorithm | Network Simplex |
| Objective Function | obj |
| Solution Status | Optimal |
| Objective Value | 270 |
| Iterations | 8 |
| Iterations2 | 0 |
| Primal Infeasibility | 0 |
| Dual Infeasibility | 0 |
| Bound Infeasibility | 0 |
| [1] | [2] | flow |
|---|---|---|
| 1 | 4 | 10 |
| 2 | 1 | 0 |
| 2 | 3 | 10 |
| 2 | 6 | 10 |
| 3 | 4 | 5 |
| 3 | 5 | 5 |
| 4 | 7 | 10 |
| 5 | 6 | 0 |
| 5 | 7 | 5 |
| 6 | 8 | 10 |
| 7 | 8 | 0 |
The optimal solution is represented graphically in Output 6.5.3.
Output 6.5.3: Minimum Cost Network Flow Problem: Optimal Solution

The iteration log is displayed in Output 6.5.4.
Output 6.5.4: Log: Solution Progress
| NOTE: There were 11 observations read from the data set WORK.ARCDATA. |
| NOTE: There were 8 observations read from the data set WORK.NODEDATA. |
| NOTE: Problem generation will use 4 threads. |
| NOTE: The problem has 11 variables (0 free, 0 fixed). |
| NOTE: The problem has 8 linear constraints (0 LE, 8 EQ, 0 GE, 0 range). |
| NOTE: The problem has 22 linear constraint coefficients. |
| NOTE: The problem has 0 nonlinear constraints (0 LE, 0 EQ, 0 GE, 0 range). |
| NOTE: The PRESOLVER value NONE is applied because a pure network has been found. |
| 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 20.0000000 109.0000000 0.00 |
| 2 0 20.0000000 109.0000000 0.00 |
| 3 5.0000000 15.0000000 104.0000000 0.00 |
| 4 5.0000000 15.0000000 103.0000000 0.00 |
| 5 75.0000000 15.0000000 103.0000000 0.00 |
| 6 75.0000000 15.0000000 99.0000000 0.00 |
| 7 130.0000000 10.0000000 96.0000000 0.00 |
| 8 270.0000000 0 0 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 PROCEDURE OPTMODEL printed pages 40-41. |
| STATUS=OK ALGORITHM=NS SOLUTION_STATUS=OPTIMAL OBJECTIVE=270 |
| PRIMAL_INFEASIBILITY=0 DUAL_INFEASIBILITY=0 BOUND_INFEASIBILITY=0 ITERATIONS=8 |
| ITERATIONS2=0 PRESOLVE_TIME=0.00 SOLUTION_TIME=0.00 |
Now, suppose there is a budget on the flow that comes out of arc 2: the total arc cost of flow that comes out of arc 2 cannot exceed 50. You can use the following call to PROC OPTMODEL to find the minimum-cost flow:
proc optmodel;
set <str> NODES;
num supply_demand {NODES};
set <str,str> ARCS;
num arcLower {ARCS};
num arcUpper {ARCS};
num arcCost {ARCS};
read data arcdata into ARCS=[_tail_ _head_]
arcLower=_lo_ arcUpper=_capac_ arcCost=_cost_;
read data nodedata into NODES=[_node_] supply_demand=_sd_;
var flow {<i,j> in ARCS} >= arcLower[i,j] <= arcUpper[i,j];
min obj = sum {<i,j> in ARCS} arcCost[i,j] * flow[i,j];
con balance {i in NODES}:
sum {<(i),j> in ARCS} flow[i,j] - sum {<j,(i)> in ARCS} flow[j,i]
= supply_demand[i];
con budgetOn2:
sum {<i,j> in ARCS: i='2'} arcCost[i,j] * flow[i,j] <= 50;
solve with lp / algorithm=ns scale=none logfreq=1;
print flow;
quit;
%put &_OROPTMODEL_;
The optimal solution is displayed in Output 6.5.5.
Output 6.5.5: Network Simplex Solver: Primal Solution Output
| Problem Summary | |
|---|---|
| Objective Sense | Minimization |
| Objective Function | obj |
| Objective Type | Linear |
| Number of Variables | 11 |
| Bounded Above | 0 |
| Bounded Below | 0 |
| Bounded Below and Above | 11 |
| Free | 0 |
| Fixed | 0 |
| Number of Constraints | 9 |
| Linear LE (<=) | 1 |
| Linear EQ (=) | 8 |
| Linear GE (>=) | 0 |
| Linear Range | 0 |
| Constraint Coefficients | 24 |
| Performance Information | |
|---|---|
| Execution Mode | On Client |
| Number of Threads | 1 |
| Solution Summary | |
|---|---|
| Solver | LP |
| Algorithm | Network Simplex |
| Objective Function | obj |
| Solution Status | Optimal |
| Objective Value | 274 |
| Iterations | 5 |
| Iterations2 | 5 |
| Primal Infeasibility | 8.881784E-16 |
| Dual Infeasibility | 0 |
| Bound Infeasibility | 0 |
| [1] | [2] | flow |
|---|---|---|
| 1 | 4 | 12 |
| 2 | 1 | 2 |
| 2 | 3 | 10 |
| 2 | 6 | 8 |
| 3 | 4 | 3 |
| 3 | 5 | 7 |
| 4 | 7 | 10 |
| 5 | 6 | 2 |
| 5 | 7 | 5 |
| 6 | 8 | 10 |
| 7 | 8 | 0 |
The optimal solution is represented graphically in Output 6.5.6.
Output 6.5.6: Minimum Cost Network Flow Problem: Optimal Solution (with Budget Constraint)

The iteration log is displayed in Output 6.5.7. Note that the network simplex solver extracts a subnetwork in this case.
Output 6.5.7: Log: Solution Progress
| NOTE: There were 11 observations read from the data set WORK.ARCDATA. |
| NOTE: There were 8 observations read from the data set WORK.NODEDATA. |
| NOTE: Problem generation will use 4 threads. |
| NOTE: The problem has 11 variables (0 free, 0 fixed). |
| NOTE: The problem has 9 linear constraints (1 LE, 8 EQ, 0 GE, 0 range). |
| NOTE: The problem has 24 linear constraint coefficients. |
| NOTE: The problem has 0 nonlinear constraints (0 LE, 0 EQ, 0 GE, 0 range). |
| NOTE: The LP presolver value AUTOMATIC is applied. |
| NOTE: The LP presolver removed 6 variables and 7 constraints. |
| NOTE: The LP presolver removed 15 constraint coefficients. |
| NOTE: The presolved problem has 5 variables, 2 constraints, and 9 constraint |
| coefficients. |
| NOTE: The LP solver is called. |
| NOTE: The Network Simplex algorithm is used. |
| NOTE: The network has 1 rows (50.00%), 5 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 |
| 1 259.9800000 5.0200000 34.0000000 0.00 |
| 2 264.9900000 0.0100000 31.0000000 0.00 |
| 3 265.0300000 0 2.0000000 0.00 |
| 4 255.0300000 0 0 0.00 |
| 5 270.0000000 0 0 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 Dual Simplex algorithm is used. |
| Objective Entering Leaving |
| Phase Iteration Value Time Variable Variable |
| D 2 1 2.700000e+02 0 flow['5','6'] |
| budgetOn2 (S) |
| D 2 2 2.700000e+02 0 flow['3','4'] |
| flow['2','3'] |
| D 2 3 2.740000e+02 0 |
| P 2 4 2.740000e+02 0 |
| P 2 5 2.740000e+02 0 |
| NOTE: Optimal. |
| NOTE: Objective = 274. |
| NOTE: The Simplex solve time is 0.00 seconds. |
| NOTE: The PROCEDURE OPTMODEL printed pages 47-48. |
| STATUS=OK ALGORITHM=NS SOLUTION_STATUS=OPTIMAL OBJECTIVE=274 |
| PRIMAL_INFEASIBILITY=8.881784E-16 DUAL_INFEASIBILITY=0 BOUND_INFEASIBILITY=0 |
| ITERATIONS=5 ITERATIONS2=5 PRESOLVE_TIME=0.00 SOLUTION_TIME=0.00 |