The OPTLP Procedure

Example 10.3 The Diet Problem

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 10.6.

Table 10.6: 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 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 use the following SAS code to create the MPS-format input data set:

data ex3;
   input field1 $ field2 $ field3 $ field4 field5 $ field6;
   datalines;
NAME        .          EX3      .     .         .
ROWS        .          .        .     .         .
 N          diet       .        .     .         .
 G          calories   .        .     .         .
 L          protein    .        .     .         . 
 G          fat        .        .     .         .
 G          carbs      .        .     .         .
COLUMNS     .          .        .     .         .
.           br         diet     2     calories  90
.           br         protein  4     fat       1
.           br         carbs    15    .         .
.           mi         diet     3.5   calories  120
.           mi         protein  8     fat       5
.           mi         carbs    11.7  .         .
.           ch         diet     8     calories  106
.           ch         protein  7     fat       9
.           ch         carbs    .4    .         .
.           po         diet     1.5   calories  97
.           po         protein  1.3   fat       .1
.           po         carbs    22.6  .         .
.           fi         diet     11    calories  130
.           fi         protein  8     fat       7
.           fi         carbs    0     .         .
.           yo         diet     1     calories  180
.           yo         protein  9.2   fat       1
.           yo         carbs    17    .         .
RHS         .          .        .     .         .
.           .          calories 300   protein   10
.           .          fat      8     carbs     10
BOUNDS      .          .        .     .         .
UP          .          mi       1     .         .
LO          .          fi       .5    .         .
ENDATA      .          .        .     .         .
;

You can solve the diet problem by using PROC OPTLP as follows:

proc optlp data=ex3
   presolver = none
   algorithm = ps
   primalout = ex3pout
   dualout   = ex3dout
   logfreq   = 1;
run;

The solution summary and the optimal primal solution are displayed in Output 10.3.1.

Output 10.3.1: Diet Problem: Solution Summary and Optimal Primal Solution

Solution Summary

Obs	Label1	cValue1	nValue1
1	Solver	LP	.
2	Algorithm	Primal Simplex	.
3	Objective Function	diet	.
4	Solution Status	Optimal	.
5	Objective Value	12.081337881	12.081338
6			.
7	Primal Infeasibility	0	0
8	Dual Infeasibility	0	0
9	Bound Infeasibility	0	0
10			.
11	Iterations	9	9.000000
12	Presolve Time	0.00	0
13	Solution Time	0.03	0.031200

Primal Solution

Obs	Objective Function ID	Variable Name	Variable Type	Objective Coefficient	Lower Bound	Upper Bound	Variable Value	Variable Status	Reduced Cost
1	diet	br	N	2.0	0.0	1.7977E308	0.00000	L	1.19066
2	diet	mi	D	3.5	0.0	1	0.05360	B	0.00000
3	diet	ch	N	8.0	0.0	1.7977E308	0.44950	B	0.00000
4	diet	po	N	1.5	0.0	1.7977E308	1.86517	B	0.00000
5	diet	fi	O	11.0	0.5	1.7977E308	0.50000	L	5.15641
6	diet	yo	N	1.0	0.0	1.7977E308	0.00000	L	1.10849

The cost of the optimal diet is 12.08 units.