The Linear Programming Solver

Consider the problem of diet optimization. There are six different foods: bread, milk, cheese, potato, fish, and yogurt. The cost and nutrition values per unit are displayed in Table 6.6.

The following SAS code creates the data set fooddata of Table 6.6:

data fooddata;
   infile datalines;
   input name $  cost  prot  fat  carb  cal;
   datalines;                      
Bread   2     4     1    15    90
Milk    3.5   8     5    11.7  120
Cheese  8     7     9    0.4   106
Potato  1.5   1.3   0.1  22.6  97
Fish    11    8     7    0     130
Yogurt  1     9.2   1    17    180
;

The objective is to find a minimum-cost diet that contains at least 300 calories, not more than 10 grams of protein, not less than 10 grams of carbohydrates, and not less than 8 grams of fat. In addition, the diet should contain at least 0.5 unit of fish and no more than 1 unit of milk.

You can model the problem and solve it by using PROC OPTMODEL as follows:

proc optmodel;
   /* declare index set */
   set<str> FOOD;

   /* declare variables */
   var diet{FOOD} >= 0;

   /* objective function */
   num cost{FOOD};
   min f=sum{i in FOOD}cost[i]*diet[i];

   /* constraints */
   num prot{FOOD};
   num fat{FOOD};
   num carb{FOOD};
   num cal{FOOD};
   num min_cal, max_prot, min_carb, min_fat;
   con cal_con: sum{i in FOOD}cal[i]*diet[i] >= 300;
   con prot_con: sum{i in FOOD}prot[i]*diet[i] <= 10;
   con carb_con: sum{i in FOOD}carb[i]*diet[i] >= 10;
   con fat_con: sum{i in FOOD}fat[i]*diet[i] >= 8;

   /* read parameters */
   read data fooddata into FOOD=[name] cost prot fat carb cal;

   /* bounds on variables */
   diet['Fish'].lb = 0.5;
   diet['Milk'].ub = 1.0;

   /* solve and print the optimal solution */
   solve with lp/logfreq=1; /* print each iteration to log */
   print diet;

The optimal solution and the optimal objective value are displayed in Output 6.1.1.