It looks like you are using an old version of SAS. That dynamic data set name functionality was introduced in SAS/OR 14.1 in SAS 9.4M3 in 2015:
Data Set Input/Output :: SAS/OR(R) 14.1 User's Guide: Mathematical Programming
Here's a workaround that uses PROC APPEND inside a SUBMIT block:
/* call MILP solver to find top few solutions for SelectTreatment */
num numSolsWanted = 5;
num numSolsFound init 0;
set SOLS = 1..numSolsFound;
set TREATMENTS_s {SOLS};
con ExcludeSolution {s in SOLS}:
sum {t in TREATMENTS_s[s]} SelectTreatment[t] <= card(TREATMENTS_s[s]) - 1;
for {1..numSolsWanted} do;
put numSolsFound=;
solve;
numSolsFound = numSolsFound + 1;
TREATMENTS_s[numSolsFound] = {t in TREATMENTS: SelectTreatment[t].sol > 0.5};
create data want_method from [method] sol=numSolsFound SelectMethod;
create data want_treatment from [treatment] sol=numSolsFound SelectTreatment;
create data want_dt_bloc from [dt bloc] sol=numSolsFound IsGood Score;
submit;
proc append base=want_method_all data=want_method;
run;
proc append base=want_treatment_all data=want_treatment;
run;
proc append base=want_dt_bloc_all data=want_dt_bloc;
run;
endsubmit;
end;
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You might be interested in PROC PSMATCH, which uses a linear assignment solver under the hood for optimal matching:
https://go.documentation.sas.com/doc/en/pgmsascdc/v_065/statug/statug_psmatch_toc.htm
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You can impose an explicit "objective cut" constraint:
con NonnegativeCost:
cost >= 0;
But that will probably actually increase the solve time.
A more promising approach is to try different choices for the ALGORITHM= option. If you can share the SOLVE statement you are currently using, as well as the LP solver's iteration log, I might be able to give a more specific recommendation.
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The simplest way to get your code working is to make the following changes.
1. Change the declaration of two sets:
set <str> GUARDS;
set <str> SHIFTS;
2. Replace the READ DATA statement:
read data guard_availability into GUARDS=[Guard];
read data guard_availability into SHIFTS=[Shift];
read data guard_availability into DATES=[Date];
read data guard_availability into [Guard Shift Date]
Available Hours;
3. Change the CREATE DATA statement:
create data schedule from [Guard Shift Date] Assigned=x;
You might find this book of examples useful:
SAS Help Center: SAS/OR® User's Guide: Mathematical Programming Examples
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Sure. But that would lead to many combinations , here is one of combination:
DATA HAVE;
INPUT COURSE $ ROOM_NO $ ROOM_CAPACITY ROOM_LOAD;
DATALINES;
CHEM101 C101 20 7
CHEM101 C102 20 11
CHEM101 C103 10 8
CHEM101 C104 15 10
BIO302 HB102 22 20
BIO302 HB106 24 12
BIO302 HB902 30 29
BIO302 W102 30 31
BIO302 AB102 40 27
;
RUN;
%macro room(course=);
data temp;
set have;
if COURSE="&course.";
run;
proc optmodel;
set <str> ROOM_NO;
num ROOM_CAPACITY{ROOM_NO};
num ROOM_LOAD{ROOM_NO};
num sum_ROOM_LOAD=sum{i in ROOM_NO} ROOM_LOAD[i];
read data temp into ROOM_NO=[ROOM_NO] ROOM_CAPACITY ROOM_LOAD;
var v{ROOM_NO} binary;
min TotalCost = (sum{i in ROOM_NO} ROOM_CAPACITY[i]*v[i]) - sum_ROOM_LOAD ;
con con: sum{i in ROOM_NO} ROOM_CAPACITY[i]*v[i] >= sum_ROOM_LOAD;
solve ;
create data want from [ROOM]={i in ROOM_NO}
ROOM_CAPACITY=ROOM_CAPACITY[i]
ROOM_LOAD=ROOM_LOAD[i]
flag=v[i].sol;
quit;
data want;
length COURSE ROOM $ 80;
set want;
COURSE="&course.";
run;
proc append base=final_want data=want force;run;
%mend;
proc freq data=have noprint ;
table course/out=course;
run;
proc delete data=final_want;run;
data _null_;
set course;
call execute(catt('%room(course=',course,')'));
run;
proc summary data=final_want ;
by course;
var room_load;
output out=sum(drop=_:) sum=sum;
run;
data final_want2;
merge final_want sum;
by course;
if first.course then cum_ROOM_CAPACITY=0;
new_room_load=0;
if flag=1 then do;
cum_ROOM_CAPACITY+ROOM_CAPACITY;
new_room_load=ifn(sum>=cum_ROOM_CAPACITY,ROOM_CAPACITY,ROOM_CAPACITY-cum_ROOM_CAPACITY+sum);
end;
drop sum cum_ROOM_CAPACITY;
run;
proc print data=final_want2 noobs;run;
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Looks like you just need to read the correct data set:
/*read data WORK.SANTANDER_HIST into [date ativo] {j in PERIODS} <hist_data[date,ativo,j]=col(j)>;*/
read data WORK.SANTANDER_HIST_3 into [date ativo] {j in PERIODS} <hist_data[date,ativo,j]=col(j)>;
With this change, the ERROR messages no longer appear.
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RobPratt,
Unfortunately, The OPTBINNING procedure is running under SAS/Viya , I don't have it. My sas is a stand-alone PC version.
And it looks like PROC OPTBINNING need set up some parameters(cutpoints ,woe,lower bound, upper bound, and WOE trend <-- what does it mean? ) firstly .
I need the code like your posted before for binning categroy variable .Like smoothing and flexibility.
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proc optmodel;
num a{1..2}=[2 12];
num b{1..3}=[2 12 24];
num d{i in 1..10}=i;
num e{i in 1..47}=i;
var var_a{1..2} binary;
var var_b{1..3} binary;
var var_c >=2 <=42 ;
var var_d integer;
impvar sum= 4*sum{i in 1..2} a[i]*var_a[i] +
-2*sum{i in 1..3} b[i]*var_b[i] +
-2*var_c +
4*var_d ;
min obj=sum;
con con_total:sum>=1;
con con_a:sum{i in 1..2} var_a[i]=1;
con con_b:sum{i in 1..3} var_b[i]=1;
con con_d:var_c=2+4*var_d;
solve ;
print a var_a ;
print b var_b ;
print var_c ;
print var_d ;
quit;
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