The NETFLOW Procedure

Missing S Supply and Missing D Demand Values

In some models, you may want a node to be either a supply or demand node but you want the node to supply or demand the optimal number of flow units. To indicate that a node is such a supply node, use a missing S value in the SUPPLY list variable in the ARCDATA= data set or the SUPDEM list variable in the NODEDATA= data set. To indicate that a node is such a demand node, use a missing D value in the DEMAND list variable in the ARCDATA= data set or the SUPDEM list variable in the NODEDATA= data set.

Suppose the oil example in the section "Introductory Example" is changed so that crude oil can be obtained from either the Middle East or U.S.A. in any amounts. You should specify that the node "middle east" is a supply node, but you do not want to stipulate that it supplies 100 units, as before. The node "u.s.a." should also remain a supply node, but you do not want to stipulate that it supplies 80 units. You must specify that these nodes have missing S supply capabilities.

  
   title  'Oil Industry Example'; 
   title3 'Crude Oil can come from anywhere'; 
   data miss_s; 
      missing S; 
      input   _node_&$15. _sd_; 
      datalines; 
   middle east         S 
   u.s.a.              S 
   servstn1 gas      -95 
   servstn1 diesel   -30 
   servstn2 gas      -40 
   servstn2 diesel   -15 
   ;
 

The following PROC NETFLOW run uses the same ARCDATA= and CONDATA= data sets used in the section "Introductory Example".

  
   proc netflow 
      nodedata=miss_s       /* the supply (missing S) and */ 
                            /* demand data                */ 
      arcdata=arcd1         /* the arc descriptions       */ 
      condata=cond1         /* the side constraints       */ 
      conout=solution;      /* the solution data set      */ 
      run; 
   print some_arcs('middle east' 'u.s.a.',_all_)/short; 
  
   proc print; 
      sum _fcost_; 
      run;
 

The following messages appear on the SAS log:

  
   NOTE: Number of nodes= 14 . 
   NOTE: Number of supply nodes= 2 . 
   NOTE: Of these, 2 have unspecified (.S) supply capability. 
   NOTE: Number of demand nodes= 4 . 
   NOTE: Total supply= 0 , total demand= 180 . 
   NOTE: Number of arcs= 18 . 
   NOTE: Number of iterations performed (neglecting any 
         constraints)= 9 . 
   NOTE: Of these, 0 were degenerate. 
   NOTE: Optimum (neglecting any constraints) found. 
   NOTE: Minimal total cost= 50040 . 
   NOTE: Number of <= side constraints= 0 . 
   NOTE: Number of == side constraints= 2 . 
   NOTE: Number of >= side constraints= 2 . 
   NOTE: Number of arc and nonarc variable side 
         constraint coefficients= 8 . 
   NOTE: Number of iterations, optimizing with 
         constraints= 3 . 
   NOTE: Of these, 0 were degenerate. 
   NOTE: Optimum reached. 
   NOTE: Minimal total cost= 50075 .
 

The PRINT statement reports the arcs directed away from the supply nodes, shown in Figure 5.14. The amount of crude obtained from the Middle East and U.S.A. is 30 and 150 units, respectively.

Oil Industry Example   Crude Oil can come from anywhere The NETFLOW Procedure _N_ _from_ _to_ _cost_ _capac_ _lo_ _name_ _FLOW_ 1 middle east refinery 1 63 95 20 m_e_ref1 20 2 u.s.a. refinery 1 55 99999999 0   125 3 middle east refinery 2 81 80 10 m_e_ref2 10 4 u.s.a. refinery 2 49 99999999 0   25

Figure 5.14: Print Statement, Oil Example, Missing S Supplies

The CONOUT= data set is shown in Figure 5.15.

Obs _from_ _to_ _cost_ _capac_ _lo_ _name_ _SUPPLY_ _DEMAND_ _FLOW_ _FCOST_ _RCOST_ _ANUMB_ _TNUMB_ _STATUS_ 1 refinery 1 r1 200 175 50 thruput1 . . 145.00 29000.00 . 7 2 KEY_ARC BASIC 2 refinery 2 r2 220 100 35 thruput2 . . 35.00 7700.00 17 8 3 LOWERBD NONBASIC 3 r1 ref1 diesel 0 75 0   . . 36.25 0.00 . 10 5 KEY_ARC BASIC 4 r1 ref1 gas 0 140 0 r1_gas . . 108.75 0.00 . 9 5 KEY_ARC BASIC 5 r2 ref2 diesel 0 75 0   . . 8.75 0.00 . 12 6 KEY_ARC BASIC 6 r2 ref2 gas 0 100 0 r2_gas . . 26.25 0.00 . 11 6 KEY_ARC BASIC 7 middle east refinery 1 63 95 20 m_e_ref1 S . 20.00 1260.00 8 2 1 LOWERBD NONBASIC 8 u.s.a. refinery 1 55 99999999 0   S . 125.00 6875.00 . 3 4 KEY_ARC BASIC 9 middle east refinery 2 81 80 10 m_e_ref2 S . 10.00 810.00 32 4 1 LOWERBD NONBASIC 10 u.s.a. refinery 2 49 99999999 0   S . 25.00 1225.00 . 5 4 KEY_ARC BASIC 11 ref1 diesel servstn1 diesel 18 99999999 0   . 30 30.00 540.00 . 17 8 KEY_ARC BASIC 12 ref2 diesel servstn1 diesel 36 99999999 0   . 30 0.00 0.00 12 18 10 LOWERBD NONBASIC 13 ref1 gas servstn1 gas 15 70 0   . 95 68.75 1031.25 . 13 7 KEY_ARC BASIC 14 ref2 gas servstn1 gas 17 35 5   . 95 26.25 446.25 . 14 9 NONKEY ARC BASIC 15 ref1 diesel servstn2 diesel 17 99999999 0   . 15 6.25 106.25 . 19 8 KEY_ARC BASIC 16 ref2 diesel servstn2 diesel 23 99999999 0   . 15 8.75 201.25 . 20 10 KEY_ARC BASIC 17 ref1 gas servstn2 gas 22 60 0   . 40 40.00 880.00 . 15 7 KEY_ARC BASIC 18 ref2 gas servstn2 gas 31 99999999 0   . 40 0.00 0.00 7 16 9 LOWERBD NONBASIC                     50075.00

Figure 5.15: Missing S SUPDEM Values in NODEDATA

The optimal supplies of nodes "middle east" and "u.s.a." are 30 and 150 units, respectively. For this example, the same optimal solution is obtained if these nodes had supplies less than these values (each supplies 1 unit, for example) and the THRUNET option was specified in the PROC NETFLOW statement. With the THRUNET option active, when total supply exceeds total demand, the specified nonmissing demand values are the lowest number of flow units that must be absorbed by the corresponding node. This is demonstrated in the following PROC NETFLOW run. The missing S is most useful when nodes are to supply optimal numbers of flow units and it turns out that for some nodes, the optimal supply is 0.

  
   data miss_s_x; 
      missing S; 
      input   _node_&$15. _sd_; 
      datalines; 
   middle east         1 
   u.s.a.              1 
   servstn1 gas      -95 
   servstn1 diesel   -30 
   servstn2 gas      -40 
   servstn2 diesel   -15 
   ; 
  
   proc netflow 
      thrunet 
      nodedata=miss_s_x     /* No supply (missing S)      */ 
      arcdata=arcd1         /* the arc descriptions       */ 
      condata=cond1         /* the side constraints       */ 
      conout=solution;      /* the solution data set      */ 
      run; 
   print some_arcs('middle east' 'u.s.a.',_all_)/short; 
  
   proc print; 
      sum _fcost_; 
      run;
 

The following messages appear on the SAS log. Note that the Total supply= 2, not 0 as in the last run.

  
   NOTE: Number of nodes= 14 . 
   NOTE: Number of supply nodes= 2 . 
   NOTE: Number of demand nodes= 4 . 
   NOTE: Total supply= 2 , total demand= 180 . 
   NOTE: Number of arcs= 18 . 
   NOTE: Number of iterations performed (neglecting any 
         constraints)= 13 . 
   NOTE: Of these, 0 were degenerate. 
   NOTE: Optimum (neglecting any constraints) found. 
   NOTE: Minimal total cost= 50040 . 
   NOTE: Number of <= side constraints= 0 . 
   NOTE: Number of == side constraints= 2 . 
   NOTE: Number of >= side constraints= 2 . 
   NOTE: Number of arc and nonarc variable side 
         constraint coefficients= 8 . 
   NOTE: Number of iterations, optimizing with 
         constraints= 3 . 
   NOTE: Of these, 0 were degenerate. 
   NOTE: Optimum reached. 
   NOTE: Minimal total cost= 50075 .
 

The PRINT statement and the CONDATA= data set are very similar; the supplies of the supply nodes are 1, not missing S. Otherwise, the solutions are identical.

If total supply exceeds total demand, any missing S values are ignored. If total demand exceeds total supply, any missing D values are ignored.