@mkeintz Yes, I included both objectives, with one of them commented out.   Yes, you can optimize multiple objectives sequentially, with a primary objective, secondary objective, tertiary objective, and so on.  Here is one such example: SAS Help Center: Distribution 1: Which Factories and Depots to Supply Which Customers ... View more

Did you get any error message in the log?  I suspect that your SAS version does not support indicator constraints (IMPLIES), in which case you should instead use the second (commented out) version of the RespectCapacity constraint. ... View more

@mlogan Yes, my solution already does that.  HB106 has NEW_ROOM_LOAD = 0.  And Room_Status = 'Available' if and only if NEW_ROOM_LOAD = 0. ... View more

Here's one way to solve the problem with PROC OPTMODEL: proc optmodel; /* declare parameters and read input data set */ set <str,str> COURSE_ROOM; num capacity {COURSE_ROOM}; num load {COURSE_ROOM}; read data HAVE into COURSE_ROOM=[course room_no] capacity=room_capacity load=room_load; /* calculate demand per course */ set <str> COURSES init {}; num demand {COURSES} init 0; for {<c,r> in COURSE_ROOM} do; COURSES = COURSES union {c}; demand[c] = demand[c] + load[c,r]; end; /* declare decision variables */ var NewRoomLoad {<c,r> in COURSE_ROOM} >= 0 <= capacity[c,r]; var IsAvailable {COURSE_ROOM} binary; /* declare objective */ /* max NumAvailable = sum {<c,r> in COURSE_ROOM} IsAvailable[c,r];*/ max AvailableCapacity = sum {<c,r> in COURSE_ROOM} capacity[c,r] * IsAvailable[c,r]; /* declare constraints */ con RespectCapacity {<c,r> in COURSE_ROOM}: IsAvailable[c,r] = 1 implies NewRoomLoad[c,r] = 0; /* con RespectCapacity {<c,r> in COURSE_ROOM}:*/ /* NewRoomLoad[c,r] <= NewRoomLoad[c,r].ub * (1 - IsAvailable[c,r]);*/ con SatisfyDemand {c in COURSES}: sum {<(c),r> in COURSE_ROOM} NewRoomLoad[c,r] = demand[c]; /* call MILP solver */ solve; /* create output data set */ create data WANT from [COURSE ROOM_NO]={<c,r> in COURSE_ROOM} ROOM_CAPACITY=capacity ROOM_LOAD=load NEW_ROOM_LOAD=NewRoomLoad Room_Status=(if IsAvailable[c,r] > 0.5 then 'Available' else ''); quit; The first MAX statement (commented out) maximizes the number of available rooms.  The second MAX statement maximizes the available capacity.  For your sample data, both objectives yield the same available rooms: COURSE ROOM_NO ROOM_CAPACITY ROOM_LOAD NEW_ROOM_LOAD Room_Status CHEM101 C101 20 7 20   CHEM101 C102 20 11 16   CHEM101 C103 10 8 0 Available CHEM101 C104 15 10 0 Available BIO302 HB102 22 20 19   BIO302 HB106 24 12 0 Available BIO302 HB902 30 29 30   BIO302 W102 30 31 30   BIO302 AB102 40 27 40     If you are using a SAS version that does not support indicator constraints (IMPLIES), you can use the second (commented out) RespectCapacity constraint instead.   It might also be worth considering a secondary objective like minimizing the number of reassigned students.  Also, is there any concern about the reassignments introducing schedule conflicts where the same student is enrolled in different courses that meet at the same time? ... View more

If you want hist_asset to depend on DATES, COMMS, and PERIODS, you must first populate all three index sets before you populate the parameter.  You already populated COMMS and PERIODS, so you just need DATES: set DATES; read data WORK.santander_hist_3 into DATES=[date]; num hist_asset {DATES, COMMS, PERIODS}; read data WORK.santander_hist_3 into [date ativo] {j in PERIODS} <hist_asset[date, ativo, j]=col(j)>; print hist_asset;   For the IMPVAR, you have the indices of hist_asset out of order: /*impvar amount_date_comm_period {date in DATES}=sum{comm in COMMS,period in PERIODS} amount[comm,period]*hist_asset[comm,date,period];*/ impvar amount_date_comm_period {date in DATES}=sum{comm in COMMS,period in PERIODS} amount[comm,period]*hist_asset[date,comm,period]; ... View more

Glad to help, and no need for embarrassment.  Please start a new thread if you have additional questions. ... View more

The OPTBINNING procedure solves this problem: https://go.documentation.sas.com/doc/en/pgmsascdc/v_060/casml/casml_optbinning_toc.htm   As the Overview section mentions, a shortest path approach is used under the hood, but I cannot share the details.   Here are a couple of related posts, each of which has links to others: https://communities.sas.com/t5/Mathematical-Optimization/Trying-to-use-PROC-OPTMODEL-for-monotonic-supervised-optimal/td-p/553822 https://communities.sas.com/t5/SAS-Programming/Binning-categorize-continuous-var-into-categories/td-p/936719 ... View more

Glad to help.  Please start a new thread for your new question. ... View more

The Infeasibility of 2 means that some constraints are violated.  You might need to change some solver options, like increasing POPSIZE= or NABSFCONV=.   Here is code to call the black-box solver for different numGroups and return the best solution found:   proc optmodel printlevel=0; set OBS; str purpose {OBS}; str good_bad {OBS}; read data have into OBS=[_N_] purpose good_bad; set <str> PURPOSES init {}; num bad {PURPOSES} init 0; num good {PURPOSES} init 0; str pThis, gbThis; for {i in OBS} do; pThis = purpose[i]; gbThis = good_bad[i]; PURPOSES = PURPOSES union {pThis}; if gbThis = 'bad' then bad[pThis] = bad[pThis] + 1; else if gbThis = 'good' then good[pThis] = good[pThis] + 1; end; print bad good; num total_n_bad = sum {p in PURPOSES} bad[p]; num total_n_good = sum {p in PURPOSES} good[p]; put total_n_bad=; put total_n_good=; /* num numGroups = 2;*/ num numGroups; set GROUPS = 1..numGroups; var IsPurposeGroup {PURPOSES, GROUPS} binary; impvar N_bad {g in GROUPS} = sum {p in PURPOSES} bad[p] * IsPurposeGroup[p,g]; impvar N_good {g in GROUPS} = sum {p in PURPOSES} good[p] * IsPurposeGroup[p,g]; impvar Bad_dist {g in GROUPS} = N_bad[g] / total_n_bad; impvar Good_dist {g in GROUPS} = N_good[g] / total_n_good; impvar Woe {g in GROUPS} = (Bad_dist[g] - Good_dist[g]) * log(Bad_dist[g]/Good_dist[g]); max IV = sum {g in GROUPS} Woe[g]; con OneGroupPerPurpose {p in PURPOSES}: sum {g in GROUPS} IsPurposeGroup[p,g] = 1; con Bad_dist_lb {g in GROUPS}: Bad_Dist[g] >= 0.05; con Good_dist_lb {g in GROUPS}: Good_Dist[g] >= 0.05; /* for {p in {'1','2','3'}} fix IsPurposeGroup[p,1] = 1;*/ /* for {p in {'8','X'}} fix IsPurposeGroup[p,2] = 1;*/ /* solve with blackbox;*/ num bestIV init -1; num bestNumGroups init .; num assignedGroup {PURPOSES}; do numGroups = 2..card(PURPOSES); put numGroups=; solve with blackbox; if _solution_status_ ne 'FAILED' and bestIV < IV then do; print N_bad N_good Bad_dist Good_dist Woe; bestIV = IV; bestNumGroups = numGroups; for {p in PURPOSES} do; for {g in GROUPS: IsPurposeGroup[p,g].sol > 0.5} do; assignedGroup[p] = g; leave; end; end; end; put bestIV= bestNumGroups=; end; print bestIV bestNumGroups; print assignedGroup; quit;     For the German credit data, the best solution found has numGroups = 6, but the objective value is slightly worse than from your GA: [1] N_bad N_good Bad_dist Good_dist Woe 1 89 145 0.29667 0.207143 0.0321570 2 18 94 0.06000 0.134286 0.0598464 3 31 43 0.10333 0.061429 0.0217940 4 42 77 0.14000 0.110000 0.0072349 5 58 123 0.19333 0.175714 0.0016836 6 62 218 0.20667 0.311429 0.0429590 bestIV bestNumGroups 0.16567 6 [1] assignedGroup 0 1 1 2 2 5 3 6 4 3 5 4 6 3 8 2 9 4 X 3   The black-box solver is not guaranteed to find a globally optimal (or even a feasible) solution.  Now that I understand the problem that you want to solve, I have an idea to find a globally optimal solution by using the MILP solver instead and will share that later. ... View more

Here's a straightforward approach that uses the black-box solver:   data have; input good_bad $ purpose $; datalines; good 1 bad 1 good 2 good 2 bad 2 good 3 good 8 good 8 bad X bad X ; proc optmodel; set OBS; str purpose {OBS}; str good_bad {OBS}; read data have into OBS=[_N_] purpose good_bad; set <str> PURPOSES init {}; num bad {PURPOSES} init 0; num good {PURPOSES} init 0; str pThis, gbThis; for {i in OBS} do; pThis = purpose[i]; gbThis = good_bad[i]; PURPOSES = PURPOSES union {pThis}; if gbThis = 'bad' then bad[pThis] = bad[pThis] + 1; else if gbThis = 'good' then good[pThis] = good[pThis] + 1; end; print bad good; num total_n_bad = sum {p in PURPOSES} bad[p]; num total_n_good = sum {p in PURPOSES} good[p]; put total_n_bad=; put total_n_good=; num numGroups = 2; set GROUPS = 1..numGroups; var IsPurposeGroup {PURPOSES, GROUPS} binary; impvar N_bad {g in GROUPS} = sum {p in PURPOSES} bad[p] * IsPurposeGroup[p,g]; impvar N_good {g in GROUPS} = sum {p in PURPOSES} good[p] * IsPurposeGroup[p,g]; impvar Bad_dist {g in GROUPS} = N_bad[g] / total_n_bad; impvar Good_dist {g in GROUPS} = N_good[g] / total_n_good; impvar Woe {g in GROUPS} = (Bad_dist[g] - Good_dist[g]) * log(Bad_dist[g]/Good_dist[g]); max IV = sum {g in GROUPS} Woe[g]; con OneGroupPerPurpose {p in PURPOSES}: sum {g in GROUPS} IsPurposeGroup[p,g] = 1; con Bad_dist_lb {g in GROUPS}: Bad_Dist[g] >= 0.05; con Good_dist_lb {g in GROUPS}: Good_Dist[g] >= 0.05; /* for {p in {'1','2','3'}} fix IsPurposeGroup[p,1] = 1;*/ /* for {p in {'8','X'}} fix IsPurposeGroup[p,2] = 1;*/ solve with blackbox; print N_bad N_good Bad_dist Good_dist Woe; num assignedGroup {PURPOSES}; for {p in PURPOSES} do; for {g in GROUPS: IsPurposeGroup[p,g].sol > 0.5} do; assignedGroup[p] = g; leave; end; end; print assignedGroup; quit;   On my machine, this yields a maximum IV of 1.5796959506.   Uncomment the FIX statements to recover your sample solution with IV = 0.1155245301.   Please verify whether this solves your problem for a fixed value of numGroups.  Then I can show you how to find the best numGroups. ... View more

There are a few issues here: A COFOR loop does not achieve any parallelism unless the body of the loop contains at least one SOLVE statement. If n_good[g] = 0, then good_dist[g] = 0, yielding a division by zero in the argument of the LOG function and hence a missing value for woe[g]. Your objective function IV depends only on numeric parameters and not on decision variables. It looks like you are trying to solve a separate optimization problem for each group.  If so, I recommend first writing the code to solve only one group.  If not, please provide an algebraic description of the optimization problem you want to solve. ... View more