Example 6.18 Migration to OPTMODEL: Production, Inventory, Distribution

The following example shows how to solve Example 6.4 using PROC OPTMODEL. The input data sets are the same as in that example.

The following PROC OPTMODEL code read the data sets, build the linear programming model, solve the model, and output the optimal solution to SAS data sets called ARC1 and NODE2:

proc optmodel;
   set <str> NODES;
   num _supdem_ {NODES} init 0;
   read data node0 into NODES=[_node_] _supdem_;

   set <str,str> ARCS;
   num _lo_ {ARCS} init 0;
   num _capac_ {ARCS} init .;
   num _cost_ {ARCS};
   num diagonal {ARCS};
   num factory {ARCS};
   str key_id {ARCS};
   str mth_made {ARCS};
   str _name_ {ARCS};
   read data arc0 nomiss into ARCS=[_tail_ _head_] _lo_ _capac_ _cost_
      diagonal factory key_id mth_made _name_;
   NODES = NODES union (union {<i,j> in ARCS} {i,j});

   var Flow {<i,j> in ARCS} >= _lo_[i,j];
   for {<i,j> in ARCS: _capac_[i,j] ne .} Flow[i,j].ub = _capac_[i,j];
   min obj = sum {<i,j> in ARCS} _cost_[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] = _supdem_[i];

   num infinity = constant('BIG');
   num excess = sum {i in NODES} _supdem_[i];
   if (excess > 0) then do;
      /* change equality constraint to le constraint for supply nodes */
      for {i in NODES: _supdem_[i] > 0} balance[i].lb = -infinity;
   end;
   else if (excess < 0) then do;
      /* change equality constraint to ge constraint for demand nodes */
      for {i in NODES: _supdem_[i] < 0} balance[i].ub = infinity;
   end;

   solve;

   num _supply_ {<i,j> in ARCS} =
      (if _supdem_[i] ne 0 then _supdem_[i] else .);
   num _demand_ {<i,j> in ARCS} =
      (if _supdem_[j] ne 0 then -_supdem_[j] else .);
   num _fcost_ {<i,j> in ARCS} = _cost_[i,j] * Flow[i,j].sol;

   create data arc1 from [_tail_ _head_]
      _cost_ _capac_ _lo_ _name_ _supply_ _demand_ _flow_=Flow _fcost_
      _rcost_ =
         (if Flow[_tail_,_head_].rc ne 0 then Flow[_tail_,_head_].rc else .)
      _status_ = Flow.status diagonal factory key_id mth_made;
   create data node2 from [_node_]
      _supdem_ = (if _supdem_[_node_] ne 0 then _supdem_[_node_] else .)
      _dual_ = balance.dual;
quit;

The statements use both single-dimensional (NODES) and multiple-dimensional (ARCS) index sets, which are populated from the corresponding data set variables in the READ DATA statements. The _SUPDEM_, _LO_, and _CAPAC_ parameters are given initial values, and the NOMISS option in the READ DATA statement tells OPTMODEL to read only the nonmissing values from the input data set. The balance constraint is initially declared as an equality, but depending on the total supply or demand, the sense of this constraint is changed to $\le $ or $\ge $ by relaxing the constraint’s lower or upper bound, respectively. The ARC1 output data set contains most of the same information as in Example 6.4, including reduced cost, basis status, and dual values. The _ANUMB_ and _TNUMB_ values do not apply here.

The PROC PRINT statements are similar to Example 6.4.

options ls=80 ps=54;
proc print data=arc1 heading=h width=min;
   var _tail_ _head_ _cost_ _capac_ _lo_ _name_
   _supply_ _demand_ _flow_ _fcost_;
   sum _fcost_;
run;
proc print data=arc1 heading=h width=min;
   var _rcost_ _status_ diagonal factory key_id mth_made;
run;
proc print data=node2;
run;

The output data sets contain the same optimal solution as Output 6.4.2, Output 6.4.3, and Output 6.4.4. The log is displayed in Output 6.18.1.

Output 6.18.1: OPTMODEL Log

NOTE: There were 8 observations read from the data set WORK.NODE0.              
NOTE: There were 64 observations read from the data set WORK.ARC0.              
NOTE: Problem generation will use 4 threads.                                    
NOTE: The problem has 64 variables (0 free, 0 fixed).                           
NOTE: The problem has 20 linear constraints (4 LE, 16 EQ, 0 GE, 0 range).       
NOTE: The problem has 128 linear constraint coefficients.                       
NOTE: The problem has 0 nonlinear constraints (0 LE, 0 EQ, 0 GE, 0 range).      
NOTE: The OPTMODEL presolver is disabled for linear problems.                   
NOTE: The LP presolver value AUTOMATIC is applied.                              
NOTE: The LP presolver removed 0 variables and 0 constraints.                   
NOTE: The LP presolver removed 0 constraint coefficients.                       
NOTE: The presolved problem has 64 variables, 20 constraints, and 128           
      constraint coefficients.                                                  
NOTE: The LP solver is called.                                                  
NOTE: The Dual Simplex algorithm is used.                                       
                           Objective                                            
      Phase Iteration        Value         Time                                 
       D 1          1    0.000000E+00         0                                 
       D 2          2   -4.020320E+06         0                                 
       D 2         32   -1.281110E+06         0                                 
NOTE: Optimal.                                                                  
NOTE: Objective = -1281110.35.                                                  
NOTE: The Dual Simplex solve time is 0.00 seconds.                              
NOTE: The data set WORK.ARC1 has 64 observations and 16 variables.              
NOTE: The data set WORK.NODE2 has 20 observations and 3 variables.