The OPTMILP Procedure

Example 6.1 Simple Integer Linear Program

This example illustrates a model in an MPS-format data table. This data table is passed to PROC OPTMILP, and a solution is found.

Consider a scenario where you have a container with a set of limiting attributes (volume V and weight W) and a set I of items that you want to pack. Each item type i has a certain value p Subscript i, a volume v Subscript i, and a weight w Subscript i. You must choose at most four items of each type so that the total value is maximized and all the chosen items fit into the container. Let x Subscript i be the number of items of type i to be included in the container. This model can be formulated as the following integer linear program:

StartLayout 1st Row 1st Column max 2nd Column sigma-summation Underscript i element-of upper I Endscripts p Subscript i Baseline x Subscript i 3rd Column Blank 2nd Row 1st Column normal s period normal t period 2nd Column sigma-summation Underscript i element-of upper I Endscripts v Subscript i Baseline x Subscript i 3rd Column less-than-or-equal-to 4th Column upper V 5th Column Blank 6th Column left-parenthesis normal v normal o normal l normal u normal m normal e normal bar normal c normal o normal n right-parenthesis 3rd Row 1st Column Blank 2nd Column sigma-summation Underscript i element-of upper I Endscripts w Subscript i Baseline x Subscript i 3rd Column less-than-or-equal-to 4th Column upper W 5th Column Blank 6th Column left-parenthesis normal w normal e normal i normal g normal h normal t normal bar normal c normal o normal n right-parenthesis 4th Row 1st Column Blank 2nd Column x Subscript i 3rd Column less-than-or-equal-to 4th Column 4 5th Column for-all i element-of upper I 5th Row 1st Column Blank 2nd Column x Subscript i Baseline element-of double-struck upper Z Superscript plus 3rd Column Blank 4th Column Blank 5th Column for-all i element-of upper I EndLayout

Constraint (volume_con) enforces the volume capacity limit, while constraint (weight_con) enforces the weight capacity limit. An instance of this problem can be saved in an MPS-format data table by using the following code:

data mylib.ex1data;
   input _id_ field1 $ field2 $ field3 $ field4 field5 $ field6;
   datalines;
1  NAME       .             ex1data           .     .                 .
2  ROWS       .             .                 .     .                 .
3  MAX        z             .                 .     .                 .
4  L          volume_con    .                 .     .                 .
5  L          weight_con    .                 .     .                 .
6  COLUMNS    .             .                 .     .                 .
7  .          .MRK0         'MARKER'          .     'INTORG'          .
8  .          x[1]          z                 1     volume_con       10
9  .          x[1]          weight_con       12     .                 .
10 .          x[2]          z                 2     volume_con      300
11 .          x[2]          weight_con       15     .                 .
12 .          x[3]          z                 3     volume_con      250
13 .          x[3]          weight_con       72     .                 .
14 .          x[4]          z                 4     volume_con      610
15 .          x[4]          weight_con      100     .                 .
16 .          x[5]          z                 5     volume_con      500
17 .          x[5]          weight_con      223     .                 .
18 .          x[6]          z                 6     volume_con      120
19 .          x[6]          weight_con       16     .                 .
20 .          x[7]          z                 7     volume_con       45
21 .          x[7]          weight_con       73     .                 .
22 .          x[8]          z                 8     volume_con      100
23 .          x[8]          weight_con       12     .                 .
24 .          x[9]          z                 9     volume_con      200
25 .          x[9]          weight_con      200     .                 .
26 .          x[10]         z                10     volume_con       61
27 .          x[10]         weight_con      110     .                 .
28 .          .MRK1         'MARKER'          .     'INTEND'          .
29 RHS        .             .                 .     .                 .
30 .          .RHS.         volume_con     1000     .                 .
31 .          .RHS.         weight_con      500     .                 .
32 BOUNDS     .             .                 .     .                 .
33 UP         .BOUNDS.      x[1]              4     .                 .
34 UP         .BOUNDS.      x[2]              4     .                 .
35 UP         .BOUNDS.      x[3]              4     .                 .
36 UP         .BOUNDS.      x[4]              4     .                 .
37 UP         .BOUNDS.      x[5]              4     .                 .
38 UP         .BOUNDS.      x[6]              4     .                 .
39 UP         .BOUNDS.      x[7]              4     .                 .
40 UP         .BOUNDS.      x[8]              4     .                 .
41 UP         .BOUNDS.      x[9]              4     .                 .
42 UP         .BOUNDS.      x[10]             4     .                 .
43 ENDATA     .             .                 .     .                 .
;

In the COLUMNS section of this data table, the name of the objective is z, and the objective coefficients p Subscript i appear in field4. The coefficients v Subscript i of (volume_con) appear in field6. The coefficients w Subscript i of (weight_con) appear in field4. In the RHS section, the bounds V and W appear in field4. The _id_ column ensures that the data can be read in the right order even if the table is stored on separate machines in CAS. This problem can be solved by using the following statements to call the OPTMILP procedure:

proc optmilp data=mylib.ex1data primalout=mylib.ex1soln;
run;

The progress of the solver is shown in Output 6.1.1.

Output 6.1.1: Simple Integer Linear Program PROC OPTMILP Log

NOTE: The problem ex1data has 10 variables (0 binary, 10 integer, 0 free, 0     
      fixed).                                                                   
NOTE: The problem has 2 constraints (2 LE, 0 EQ, 0 GE, 0 range).                
NOTE: The problem has 20 constraint coefficients.                               
NOTE: The initial MILP heuristics are applied.                                  
NOTE: The MILP presolver value AUTOMATIC is applied.                            
NOTE: The MILP presolver removed 2 variables and 0 constraints.                 
NOTE: The MILP presolver removed 4 constraint coefficients.                     
NOTE: The MILP presolver modified 0 constraint coefficients.                    
NOTE: The presolved problem has 8 variables, 2 constraints, and 16 constraint   
      coefficients.                                                             
NOTE: The MILP solver is called.                                                
NOTE: The parallel Branch and Cut algorithm is used.                            
NOTE: The Branch and Cut algorithm is using up to 80 threads.                   
          Node   Active   Sols    BestInteger      BestBound      Gap    Time   
             0        1      4     85.0000000    158.0000000   46.20%       0   
             0        1      4     85.0000000     88.0955497    3.51%       0   
             0        1      4     85.0000000     87.4545455    2.81%       0   
NOTE: The MILP presolver is applied again.                                      
             0        1      5     87.0000000     87.4545455    0.52%       0   
NOTE: Optimal.                                                                  
NOTE: Objective = 87.                                                           
NOTE: The Cloud Analytic Services server processed the request in 0.517744      
      seconds.                                                                  
NOTE: The data set MYLIB.EX1SOLN has 10 observations and 9 variables.           


The data table ex1soln is shown in Output 6.1.2.

Output 6.1.2: Simple Integer Linear Program Solution

Example 1 Solution Data

Objective
Function ID
RHS IDVariable NameVariable
Type
Objective
Coefficient
Lower
Bound
Upper
Bound
Variable
Value
Solution
z.RHS.x[1]I10401
z.RHS.x[2]I20401
z.RHS.x[3]I30401
z.RHS.x[4]I40401
z.RHS.x[5]I50401
z.RHS.x[6]I60431
z.RHS.x[7]I70411
z.RHS.x[8]I80441
z.RHS.x[9]I90401
z.RHS.x[10]I100431


The optimal solution is x 6 equals 3 comma x 7 equals 1 comma x 8 equals 4, and x 10 equals 3, with a total value of 87. From this solution, you can compute the total volume used, which is 988 (less-than-or-equal-to upper V equals 1000); the total weight used is 499 (less-than-or-equal-to upper W equals 500). The problem summary and solution summary are shown in Output 6.1.3.

Output 6.1.3: Simple Integer Linear Program Summary

 

The OPTMILP Procedure

Problem Summary
Problem Nameex1data
Objective SenseMaximization
Objective Functionz
RHS.RHS.
  
Number of Variables10
Bounded Above0
Bounded Below0
Bounded Above and Below10
Free0
Fixed0
Binary0
Integer10
  
Number of Constraints2
LE (<=)2
EQ (=)0
GE (>=)0
Range0
  
Constraint Coefficients20

Solution Summary
SolverMILP
AlgorithmBranch and Cut
Objective Functionz
Solution StatusOptimal
Objective Value87
  
Relative Gap0
Absolute Gap0
Primal Infeasibility0
Bound Infeasibility0
Integer Infeasibility0
  
Best Bound87
Nodes1
Solutions Found5
Iterations11
  
Presolve Time0.00
Solution Time0.01


Last updated: September 09, 2026