If you have SAS Viya, you can use the LINEARIZE option to automatically linearize the problem:
solve linearize;
Problem Summary
Objective Sense
Maximization
Objective Function
bts
Objective Type
Nonlinear
Number of Variables
15
Bounded Above
0
Bounded Below
0
Bounded Below and Above
15
Free
0
Fixed
0
Binary
0
Integer
3
Number of Constraints
5
Linear LE (<=)
5
Linear EQ (=)
0
Linear GE (>=)
0
Linear Range
0
Constraint Coefficients
15
Solution Summary
Solver
MILP
Algorithm
Branch and Cut
Objective Function
bts
Solution Status
Optimal
Objective Value
292
Relative Gap
0
Absolute Gap
0
Primal Infeasibility
0
Bound Infeasibility
0
Integer Infeasibility
0
Best Bound
292
Nodes
1
Solutions Found
3
Iterations
10
Presolve Time
0.03
Solution Time
0.04
p1g
p2g
p3g
p4g
p5g
p1s
p2s
p3s
p4s
p5s
p1t
p2t
p3t
p4t
p5t
1.25
1.25
1.25
1.25
6
2.5
2.5
2.5
2.5
0
2.25
2.25
2.25
2.25
0
But notice that some variables are fractional. That is because the INTEGER keyword applies only to three of the variables in your VAR statements. What you intended was instead:
var p1g<=6>=0 integer, p2g<=6>=0 integer, p3g<=6>=0 integer, p4g<=6>=0 integer, p5g<=6>=0 integer;
var p1s<=6>=0 integer, p2s<=6>=0 integer, p3s<=6>=0 integer, p4s<=6>=0 integer, p5s<=6>=0 integer;
var p1t<=6>=0 integer, p2t<=6>=0 integer, p3t<=6>=0 integer, p4t<=6>=0 integer, p5t<=6>=0 integer;
And this yields integer values for all variables:
p1g
p2g
p3g
p4g
p5g
p1s
p2s
p3s
p4s
p5s
p1t
p2t
p3t
p4t
p5t
6
2
0
0
3
0
4
0
6
0
0
0
6
0
3
A more compact and less error-prone approach is to use index sets:
proc optmodel;
set PROVIDERS = 1..5;
set LOCATIONS = /g s t/;
var NumDays {PROVIDERS, LOCATIONS} >= 0 <= 6 integer;
/* declare objective */
num officeDemand {LOCATIONS} = [80, 70, 90];
impvar OfficeVisit = sum {l in LOCATIONS} min(7*sum {p in PROVIDERS} NumDays[p,l], officeDemand[l]);
num videoDemand {LOCATIONS} = [25, 31, 27];
impvar VideoVisit = sum {l in LOCATIONS} min(3*sum {p in PROVIDERS} NumDays[p,l], videoDemand[l]);
/*declare constraints*/
con ProviderLimit {p in PROVIDERS}: sum {l in LOCATIONS} NumDays[p,l] <= 6;
max bts = OfficeVisit + VideoVisit;
solve linearize;
print NumDays;
quit;
NumDays
g
s
t
1
6
0
0
2
2
4
0
3
0
0
6
4
0
6
0
5
3
0
3
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