An Assignment Problem (lp12)

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/*          S A S   S A M P L E   L I B R A R Y                 */
/*                                                              */
/*    NAME: LP12                                                */
/*   TITLE: An Assignment Problem (lp12)                        */
/* PRODUCT: OR                                                  */
/*  SYSTEM: ALL                                                 */
/*    KEYS: LP                                                  */
/*   PROCS: LP                                                  */
/*    DATA:                                                     */
/*                                                              */
/* SUPPORT:                             UPDATE:                 */
/*     REF:                                                     */
/*    MISC:                                                     */
/*                                                              */
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title 'An Assignment Problem';

data object;
   input machine customer
         grade1 grade2 grade3 grade4 grade5 grade6;
   datalines;
1 1 102 140 105  105  125  148
1 2 115 133 118  118  143  166
1 3  70 108  83   83   88   86
1 4  79 117  87   87  107  105
1 5  77 115  90   90  105  148
2 1 123 150 125  124  154   .
2 2 130 157 132  131  166   .
2 3 103 130 115  114  129   .
2 4 101 128 108  107  137   .
2 5 118 145 130  129  154   .
3 1  83  .   .    97  122  147
3 2 119  .   .   133  163  180
3 3  67  .   .    91  101  101
3 4  85  .   .   104  129  129
3 5  90  .   .   114  134  179
4 1 108 121  79   .   112  132
4 2 121 132  92   .   130  150
4 3  78  91  59   .    77   72
4 4 100 113  76   .   109  104
4 5  96 109  77   .   105  145
;


data demand;
   input customer
         grade1 grade2 grade3 grade4 grade5 grade6;
   datalines;
1 100 100 150  150  175  250
2 300 125 300  275  310  325
3 400   0 400  500  340    0
4 250   0 750  750    0    0
5   0 600 300    0  210  360
;


data resource;
   input machine
         grade1 grade2 grade3 grade4 grade5 grade6 avail;
   datalines;
1 .250 .275 .300  .350  .310  .295  744
2 .300 .300 .305  .315  .320  .     244
3 .350 .    .     .320  .315  .300  790
4 .280 .275 .260  .     .250  .295  672
;


/*  build the linear programming model */

data model;
   array grade{6} grade1-grade6;
   length _type_ $ 8 _row_ $ 8 _col_ $ 8;
   keep _type_ _row_ _col_ _coef_;

   ncust=5;
   nmach=4;
   ngrade=6;


/* generate the objective function */

_type_='MAX';
_row_='OBJ';
do k=1 to nmach;
   do i=1 to ncust;
      link readobj;     /* read the objective coefficient data */
      do j=1 to ngrade;
         if grade{j}^=. then do;
            _col_='X'||put(i,1.)||put(j,1.)||put(k,1.);
            _coef_=grade{j};
            output;
         end;
      end;
   end;
end;

/* generate the demand constraints */

do i=1 to ncust;
   link readdmd;        /* read the demand data */
   do j=1 to ngrade;
      if grade{j}^=. then do;
         _type_='EQ';
         _row_='DEMAND'||put(i,1.)||put(j,1.);
         _col_='_RHS_';
         _coef_=grade{j};
         output;
         _type_=' ';
         do k=1 to nmach;
            _col_='X'||put(i,1.)||put(j,1.)||put(k,1.);
            _coef_=1.0;
            output;
         end;
      end;
   end;
end;

/* generate the machine constraints */

do k=1 to nmach;
   link readres;       /* read the machine data */
   _type_='LE';
   _row_='MACHINE'||put(k,1.);
   _col_='_RHS_';
   _coef_=avail;
   output;
   _type_=' ';
   do i=1 to ncust;
      do j=1 to ngrade;
         if grade{j}^=. then do;
            _col_='X'||put(i,1.)||put(j,1.)||put(k,1.);
            _coef_=grade{j};
            output;
            end;
         end;
      end;
   end;

readobj: set object;
return;
readdmd: set demand;
return;
readres: set resource;
return;
run;

/* solve the linear program */

proc lp data=model sparsedata noprint primalout=primal;
run;

   /* report the solution */

   data solution;
      set primal;
      keep customer grade machine amount;
      if substr(_var_,1,1)='X' then do;
        if _value_^=0 then do;
          customer = substr(_var_,2,1);
          grade    = substr(_var_,3,1);
          machine  = substr(_var_,4,1);
          amount   = _value_;
          output;
        end;
     end;
   run;


proc tabulate data=solution;
   class customer grade machine;
   var   amount;
   table (machine*customer), (grade*amount);
run;