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 6.5.
Table 6.5: Cost and Nutrition Values
Bread |
Milk |
Cheese |
Potato |
Fish |
Yogurt |
|
---|---|---|---|---|---|---|
Cost |
2.0 |
3.5 |
8.0 |
1.5 |
11.0 |
1.0 |
Protein, g |
4.0 |
8.0 |
7.0 |
1.3 |
8.0 |
9.2 |
Fat, g |
1.0 |
5.0 |
9.0 |
0.1 |
7.0 |
1.0 |
Carbohydrates, g |
15.0 |
11.7 |
0.4 |
22.6 |
0.0 |
17.0 |
Calories |
90 |
120 |
106 |
97 |
130 |
180 |
The following SAS code creates the data set fooddata
of Table 6.5:
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.
Output 6.1.1: Optimal Solution to the Diet Problem
Problem Summary | |
---|---|
Objective Sense | Minimization |
Objective Function | f |
Objective Type | Linear |
Number of Variables | 6 |
Bounded Above | 0 |
Bounded Below | 5 |
Bounded Below and Above | 1 |
Free | 0 |
Fixed | 0 |
Number of Constraints | 4 |
Linear LE (<=) | 1 |
Linear EQ (=) | 0 |
Linear GE (>=) | 3 |
Linear Range | 0 |
Constraint Coefficients | 23 |
Performance Information | |
---|---|
Execution Mode | On Client |
Number of Threads | 1 |
Solution Summary | |
---|---|
Solver | LP |
Algorithm | Dual Simplex |
Objective Function | f |
Solution Status | Optimal |
Objective Value | 12.081337881 |
Iterations | 7 |
Primal Infeasibility | 8.881784E-16 |
Dual Infeasibility | 0 |
Bound Infeasibility | 0 |
[1] | diet |
---|---|
Bread | 0.000000 |
Cheese | 0.449499 |
Fish | 0.500000 |
Milk | 0.053599 |
Potato | 1.865168 |
Yogurt | 0.000000 |