Are there researcher-paper scores that you want to maximize or minimize? ... View more

Yes, that is expected behavior.  The Benders decomposition algorithm is a new feature in SAS Viya 2024.05, just released last week. ... View more

Hi Ethan,   I'm glad you enjoyed the presentation.  Attached is a .zip file that contains the three .sas files and one .sasnb file.  Please let me know if you have any further questions.   Thanks, Rob ... View more

You might be interested in this: https://blogs.sas.com/content/subconsciousmusings/2024/04/05/planning-your-solar-eclipse-journey-with-network-analytics/ ... View more

Both of those error messages mean that <'34944801',SI> is not in the index set for v but you are trying to access v['34944801',SI].  The first error message mentions line 103, which you did not show but should correspond to the MIN statement where v is used.  This error arises because the sum is over all <r,c,a> in REMI_CLASS_AOP, which you have not restricted to satisfy your condition on p.  To avoid this error, you can explicitly check membership in REMI_CLASS in the MIN statement: min f = abs(sum {<r,c,a> in REMI_CLASS_AOP: <r,c> in REMI_CLASS} (if char(c,1) = 'M' then 1 else -1)*v[r,c]*p[r,c,a]*X[r,c,a]); To avoid the second error, you can make a similar change to the CREATE DATA statement: create data SolData1 from [r c a]={<r,c,a> in REMI_CLASS_AOP: <r,c> in REMI_CLASS} v[r,c] p X; To restrict v < 100, you can modify the READ DATA statement: read data sas.VData(where=(v < 100)) into REMI_CLASS=[remi class] v; The subsequent INTER SETOF does not then need modification. ... View more

1. This statement should appear after the READ DATA PData and before the SOLVE statement.  The CON statements are just declarations and not programming statements, so it doesn't matter whether you update REMI_CLASS before or after the CON statements. 2. Yes, you can use the dummy index r as part of a logical condition in a constraint. ... View more

Yes, you can do that, but you need to move the final right parenthesis to correctly enforce the logical condition you want: not((q1[r,c,a]-q2[r,c,a])/q2[r,c,a]<0.05) To avoid infeasibility, you would still need to modify REMI_CLASS:   REMI_CLASS = REMI_CLASS inter setof {<r,c,a> in REMI_CLASS_AOP: 9 <= p[r,c,a] < 11.5 and (q1[r,c,a]-q2[r,c,a])/q2[r,c,a]<0.05} <r,c>; The alternative approach of omitting the Pcon and Qcon constraints and instead using the WHERE= option in the READ DATA statement would avoid the dual maintenance of checking these conditions in two different places. ... View more

If q1 and q2 depend on <r,c,a>, I recommend the following: set <str,str,str> REMI_CLASS_AOP; num p {REMI_CLASS_AOP}; num q1 {REMI_CLASS_AOP}; num q2 {REMI_CLASS_AOP}; read data sas.PData(where=((q1-q2)/q2 < 0.05)) into REMI_CLASS_AOP=[remi class aop] p q1 q2; ... View more

If you don't want to modify the input data before calling PROC OPTMODEL, the next best approach is get rid of the disallowed <r,c,a> tuples as soon as possible by not even reading them: read data PData(where=(9 <= p < 11.5)) into REMI_CLASS_AOP=[remi class aop] p; Now you don't need an explicit constraint to force the disallowed variables to 0 because those <r,c,a> tuples no longer even appear in the index set. To make the resulting problem feasible, you can modify the REMI_CLASS set as before:   REMI_CLASS = REMI_CLASS inter (setof {<r,c,a> in REMI_CLASS_AOP} <r,c>); ... View more

I ran with your csv files and got the following log messages: ERROR: The expression for f was not linearized. ERROR: Unable to linearize. The black-box solver might be suitable for this problem. NOTE: Unable to create problem instance due to previous errors. This error arises from the fact that some p values are missing in PDATA.csv.  For example: 35691501 PF   6228 35691502 PD   6227 35691503 PI   6229   You can avoid the error by modifying PDATA.csv or the PData data set or by using a WHERE= clause in the READ DATA statement:   read data PData(where=(p ne .)) into REMI_CLASS_AOP=[remi class aop] p; But then the resulting problem is infeasible because some <r,c> pairs have no corresponding <r,c,a> triple and the OneValue[r,c] constraint becomes 0 = 1.  One way to avoid this infeasibility is to update the REMI_CLASS set after reading PData:     REMI_CLASS = REMI_CLASS inter (setof {<r,c,a> in REMI_CLASS_AOP} <r,c>); After these changes, the SOLVE LINEARIZE statement runs successfully.   ... View more

1. That behavior (nonlinear objective in Problem Summary and MILP solver in Solution Summary) is expected when you use SOLVE LINEARIZE.  The original objective is nonlinear because of ABS, but the automated linearization converts to a linear problem and then calls the MILP solver.   3. Regarding "sometimes doesn't work": Do you get an error message?  Do you get wrong results?  What SAS version are you using?   5. Whether the expectation of getting an optimal objective value close to zero is too optimistic depends on the data.  The solver will find such a solution if one exists.  A good exercise would be to independently verify the feasibility and objective value of the resulting solution returned by the solver. ... View more

1. Yes, you can use an explicit constraint: con Pcon {<r,c,a> in REMI_CLASS_AOP: not(9 < p[r,c,a] < 12)}: X[r,c,a] = 0; But it is more efficient to omit those <r,c,a> tuples.  See this doc example. This is an answer to original question #1.  Sorry, I don't understand your edited question #1. 2. You can use the CREATE DATA statement.  Here are two alternatives:   create data SolData1 from [r c a] v[r,c] p X; create data SolData2 from [r c a]={<r,c,a> in REMI_CLASS_AOP: X[r,c,a].sol > 0.5} v[r,c] p X; 3. What SOLVE statement did you use? 4. That is not a bug.  Those values are within the default integer tolerance of 1e-5. 5. Not sure what you mean.  Can you please clarify?   ... View more

To restrict the p values, you can either modify the input data or use a WHERE clause in the READ DATA statement: read data PData(where=(9 < p < 12)) into REMI_CLASS_AOP=[remi class aop] p; You cannot use the CLP solver here because the variables needed to linearize the absolute value are continuous. ... View more