Optimization Action Set
Simple Integer Linear Program
This section contains PROC CAS code.
Note: Input data must be accessible in your CAS session, either as a CAS table or as a transient-scope table. A CAS table has a two-level name: the first level is your CAS engine libref, and the second level is the table name. You refer to this table in the CAS procedure by specifying only the second level. For more information about two-level names, see Chapter 3, Shared Concepts (SAS Optimization: Mathematical Optimization Procedures). A transient-scope table is called directly from the action and exists in memory for the duration of the action. For more information about accessing data, see SAS Viya: System Programming Guide. For more information about PROC CAS and programming in CASL, see SAS Cloud Analytic Services: CASL Programmer’s Guide and SAS Cloud Analytic Services: CASL Reference.
This example illustrates a model in an MPS-format data table. This data table is created in CAS, and the solveMilp action solves the corresponding mixed integer linear program.
Consider a scenario where you have a container with a set of limiting attributes (volume V and weight W), and you have a set I of items that you want to pack. Each item type i has a certain value 
, a volume 
, and a weight 
. 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 
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:
The constraint (volume_con) enforces the volume capacity limit, and the constraint (weight_con) enforces the weight capacity limit. The following DATA step shows an instance of this problem in an MPS-format data table. The DATA step assumes that your CAS engine libref is named mycas, but you can substitute any appropriately defined CAS engine libref.
data mycas.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 . . . . .
;
Alternatively, you can use the upload action on a CSV file that has the following content:
_id_ ,field1 ,field2 ,field3 ,field4 ,field5 ,field6 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 appear in field4. The coefficients of (volume_con) appear in field6. The coefficients 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 correct order even if the table is stored on separate machines in CAS.
You can solve this problem by using the following statements to call the solveMilp action:
proc cas;
loadactionset "optimization";
action optimization.solveMilp result=r status=s /
data = {name = "ex1data"}
primalOut = {name = "ex1soln" replace = true};
run;
print r.ProblemSummary; run;
print r.SolutionSummary; run;
action table.fetch / table = "ex1soln"; run;
quit;
The progress of the solver is shown in Output 2.13.1.
Output 2.13.1: Simple Integer Linear Program solveMilp Log
| NOTE: Active Session now MYSESS. |
| NOTE: Added action set 'optimization'. |
| 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. |
The problem summary and solution summary are shown in Output 2.13.2 and Output 2.13.3.
Output 2.13.2: Problem Summary
| Problem Summary | |
|---|---|
| Problem Name | ex1data |
| Objective Sense | Maximization |
| Objective Function | z |
| RHS | .RHS. |
| Number of Variables | 10 |
| Bounded Above | 0 |
| Bounded Below | 0 |
| Bounded Above and Below | 10 |
| Free | 0 |
| Fixed | 0 |
| Binary | 0 |
| Integer | 10 |
| Number of Constraints | 2 |
| LE (<=) | 2 |
| EQ (=) | 0 |
| GE (>=) | 0 |
| Range | 0 |
| Constraint Coefficients | 20 |
Output 2.13.3: Solution Summary
| Solution Summary | |
|---|---|
| Solver | MILP |
| Algorithm | Branch and Cut |
| Objective Function | z |
| Solution Status | Optimal |
| Objective Value | 87 |
| Relative Gap | 0 |
| Absolute Gap | 0 |
| Primal Infeasibility | 0 |
| Bound Infeasibility | 0 |
| Integer Infeasibility | 0 |
| Best Bound | 87 |
| Nodes | 1 |
| Solutions Found | 5 |
| Iterations | 14 |
| Presolve Time | 0.00 |
| Solution Time | 0.05 |
The data table ex1soln displayed in Output 2.13.4 shows the optimal solution that is found.
Output 2.13.4: Simple Integer Linear Program Solution
| Selected Rows from Table EX1SOLN | |||||||||
|---|---|---|---|---|---|---|---|---|---|
| _Index_ | Objective Function ID | RHS ID | Variable Name | Variable Type | Objective Coefficient | Lower Bound | Upper Bound | Variable Value | Solution |
| 1 | z | .RHS. | x[1] | I | 1 | 0 | 4 | 0 | 1 |
| 2 | z | .RHS. | x[2] | I | 2 | 0 | 4 | 0 | 1 |
| 3 | z | .RHS. | x[3] | I | 3 | 0 | 4 | 0 | 1 |
| 4 | z | .RHS. | x[4] | I | 4 | 0 | 4 | 0 | 1 |
| 5 | z | .RHS. | x[5] | I | 5 | 0 | 4 | 0 | 1 |
| 6 | z | .RHS. | x[6] | I | 6 | 0 | 4 | 3 | 1 |
| 7 | z | .RHS. | x[7] | I | 7 | 0 | 4 | 1 | 1 |
| 8 | z | .RHS. | x[8] | I | 8 | 0 | 4 | 4 | 1 |
| 9 | z | .RHS. | x[9] | I | 9 | 0 | 4 | 0 | 1 |
| 10 | z | .RHS. | x[10] | I | 10 | 0 | 4 | 3 | 1 |
The optimal solution is 
, and 
, with a total value of 87. From this solution, you can compute the total volume used, which is 988 (
); the total weight used is 499 (
).
Simple Integer Linear Program
This section contains Lua code for the analysis in the CASL version of this example, which contains details about the results.
Note: In order to run this code, the data that are described in the CASL version need to be accessible to the CAS server. One way to do this is to convert the ex1data data to the comma-separated-value (CSV) file ex1data.csv and then use the following code to load the CSV file into CAS:
s:loadtable{casLib="casuser", path="ex1data.csv"}
For more information about coding in Lua, see Getting Started with SAS Viya for Lua and SAS Viya: System Programming Guide.
The following code solves the simple integer linear program stored in the data table ex1data:
s:optimization_solveMilp{
data = {name = "ex1data"},
primalOut = {name = "ex1soln", replace = true}}
Simple Integer Linear Program
This section contains Python code for the analysis in the CASL version of this example, which contains details about the results.
Note: In order to run this code, the data that are described in the CASL version need to be accessible to the CAS server. One way to do this is to convert the ex1data data to the comma-separated-value (CSV) file ex1data.csv and then use the following code to load the CSV file into CAS:
s.upload_file('ex1data.csv')
For more information about coding in Python, see Getting Started with SAS Viya for Python and SAS Viya: System Programming Guide.
The following code solves the simple integer linear program stored in the data table ex1data:
s.optimization.solveMilp(
data = {"name": "ex1data"},
primalOut = {"name": "ex1soln", "replace": True})
Simple Integer Linear Program
This section contains R code for the analysis in the CASL version of this example, which contains details about the results.
Note: In order to run this code, the data that are described in the CASL version need to be accessible to the CAS server. One way to do this is to convert the ex1data data to the comma-separated-value (CSV) file ex1data.csv and then use the following code to load the CSV file into CAS:
m <- s$upload("ex1data.csv", casOut=list(name="ex1data"))
For more information about coding in R, see Getting Started with SAS Viya for R and SAS Viya: System Programming Guide.
The following code solves the simple integer linear program stored in the data table ex1data:
cas.optimization.solveMilp(s,
data = "ex1data",
primalOut = list(name="ex1soln", replace=TRUE))