This example uses PROC OPTMODEL features to simplify the construction of a mathematically formulated model. The model is based on the example “An Assignment Problem” in Chapter 5: The LP Procedure in SAS/OR 12.3 User's Guide: Mathematical Programming Legacy Procedures. A single invocation of PROC OPTMODEL replaces several steps in the PROC LP statements.
The model assigns production of various grades of cloth to a set of machines in order to maximize profit while meeting customer demand. Each machine has different capacities to produce the various grades of cloth. (See the PROC LP example “An Assignment Problem” for more details.) The mathematical formulation, where represents the amount of cloth of grade to produce on machine for customer , follows:
The OBJECT
, DEMAND
, and RESOURCE
data sets are the same as in the PROC LP example. A new data set, GRADE
, is added to help separate the data from the model.
title 'An Assignment Problem'; data grade(drop=i); do i = 1 to 6; grade = 'grade'||put(i,1.); output; end; run;
data object; input machine customer grade1 grade2 grade3 grade4 grade5 grade6; datalines; 1 1 102 140 105 105 125 148 1 2 115 133 118 118 143 166 1 3 70 108 83 83 88 86 1 4 79 117 87 87 107 105 1 5 77 115 90 90 105 148 2 1 123 150 125 124 154 . 2 2 130 157 132 131 166 . 2 3 103 130 115 114 129 . 2 4 101 128 108 107 137 . 2 5 118 145 130 129 154 . 3 1 83 . . 97 122 147 3 2 119 . . 133 163 180 3 3 67 . . 91 101 101 3 4 85 . . 104 129 129 3 5 90 . . 114 134 179 4 1 108 121 79 . 112 132 4 2 121 132 92 . 130 150 4 3 78 91 59 . 77 72 4 4 100 113 76 . 109 104 4 5 96 109 77 . 105 145 ;
data demand; input customer grade1 grade2 grade3 grade4 grade5 grade6; datalines; 1 100 100 150 150 175 250 2 300 125 300 275 310 325 3 400 0 400 500 340 0 4 250 0 750 750 0 0 5 0 600 300 0 210 360 ;
data resource; input machine grade1 grade2 grade3 grade4 grade5 grade6 avail; datalines; 1 .250 .275 .300 .350 .310 .295 744 2 .300 .300 .305 .315 .320 . 244 3 .350 . . .320 .315 .300 790 4 .280 .275 .260 . .250 .295 672 ;
The following PROC OPTMODEL statements read the data sets, build the linear programming model, solve the model, and output
the optimal solution to a SAS data set called SOLUTION
:
proc optmodel; /* declare index sets */ set CUSTOMERS; set <str> GRADES; set MACHINES; /* declare parameters */ num return {CUSTOMERS, GRADES, MACHINES} init 0; num demand {CUSTOMERS, GRADES}; num cost {GRADES, MACHINES} init 0; num avail {MACHINES}; /* read the set of grades */ read data grade into GRADES=[grade]; /* read the set of customers and their demands */ read data demand into CUSTOMERS=[customer] {j in GRADES} <demand[customer,j]=col(j)>; /* read the set of machines, costs, and availability */ read data resource nomiss into MACHINES=[machine] {j in GRADES} <cost[j,machine]=col(j)> avail; /* read objective data */ read data object nomiss into [machine customer] {j in GRADES} <return[customer,j,machine]=col(j)>; /* declare the model */ var AmountProduced {CUSTOMERS, GRADES, MACHINES} >= 0; max TotalReturn = sum {i in CUSTOMERS, j in GRADES, k in MACHINES} return[i,j,k] * AmountProduced[i,j,k]; con req_demand {i in CUSTOMERS, j in GRADES}: sum {k in MACHINES} AmountProduced[i,j,k] = demand[i,j]; con req_avail {k in MACHINES}: sum {i in CUSTOMERS, j in GRADES} cost[j,k] * AmountProduced[i,j,k] <= avail[k]; /* call the solver and save the results */ solve; create data solution from [customer grade machine] = {i in CUSTOMERS, j in GRADES, k in MACHINES: AmountProduced[i,j,k].sol ne 0} amount=AmountProduced; /* print optimal solution */ print AmountProduced; quit;
The statements use both numeric (NUM) and character (STR) index sets, which are populated from the corresponding data set
variables in the READ DATA statements. The OPTMODEL parameters can be either single-dimensional (AVAIL) or multiple-dimensional
(COST, DEMAND, RETURN). The RETURN and COST parameters are given initial values of 0, and the NOMISS option in the READ DATA
statement tells PROC OPTMODEL to read only the nonmissing values from the input data sets. The model declaration is nearly
identical to the mathematical formulation. The logical condition AmountProduced[i,j,k].sol ne 0
in the CREATE DATA statement ensures that only the nonzero parts of the solution appear in the SOLUTION
data set. In the PROC LP example, the creation of this data set required postprocessing of the PROC LP output data set.
The solver produces the following problem summary and solution summary:
Output 5.3.1: LP Solver Result
An Assignment Problem |
Problem Summary | |
---|---|
Objective Sense | Maximization |
Objective Function | TotalReturn |
Objective Type | Linear |
Number of Variables | 120 |
Bounded Above | 0 |
Bounded Below | 120 |
Bounded Below and Above | 0 |
Free | 0 |
Fixed | 0 |
Number of Constraints | 34 |
Linear LE (<=) | 4 |
Linear EQ (=) | 30 |
Linear GE (>=) | 0 |
Linear Range | 0 |
Constraint Coefficients | 220 |
Solution Summary | |
---|---|
Solver | LP |
Algorithm | Dual Simplex |
Objective Function | TotalReturn |
Solution Status | Optimal |
Objective Value | 871426.03763 |
Primal Infeasibility | 0 |
Dual Infeasibility | 0 |
Bound Infeasibility | 0 |
Iterations | 60 |
Presolve Time | 0.00 |
Solution Time | 0.00 |
The SOLUTION
data set can be processed by PROC TABULATE as follows to create a compact representation of the solution:
proc tabulate data=solution; class customer grade machine; var amount; table (machine*customer), (grade*amount=''*sum=''); run;
These statements produce the table shown in Output 5.3.2.
Output 5.3.2: An Assignment Problem
An Assignment Problem |
grade | |||||||
---|---|---|---|---|---|---|---|
grade1 | grade2 | grade3 | grade4 | grade5 | grade6 | ||
machine | customer | . | 100.00 | 150.00 | 150.00 | 175.00 | 250.00 |
1 | 1 | ||||||
2 | . | . | 300.00 | . | . | . | |
3 | . | . | 256.72 | 210.31 | . | . | |
4 | . | . | 750.00 | . | . | . | |
5 | . | 92.27 | . | . | . | . | |
2 | 3 | . | . | 143.28 | . | 340.00 | . |
5 | . | . | 300.00 | . | . | . | |
3 | 2 | . | . | . | 275.00 | 310.00 | 325.00 |
3 | . | . | . | 289.69 | . | . | |
4 | . | . | . | 750.00 | . | . | |
5 | . | . | . | . | 210.00 | 360.00 | |
4 | 1 | 100.00 | . | . | . | . | . |
2 | 300.00 | 125.00 | . | . | . | . | |
3 | 400.00 | . | . | . | . | . | |
4 | 250.00 | . | . | . | . | . | |
5 | . | 507.73 | . | . | . | . |