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

Sorry, this thread is getting hard to follow. What do you want to integrate into which PROC OPTMODEL code? ... View more

Here are a few ways to implement BY-group processing: https://support.sas.com/kb/42/332.html ... View more

Here's one way to enforce that x is in {2,6,10,...,42}: var x; set MYSET = 2..42 by 4; var y {MYSET} binary; con Mycon1: sum {i in MYSET} y[i] = 1; con Mycon2: sum {i in MYSET} i*y[i] = x; Here's another way that uses fewer variables and constraints: var x >= 2 <= 42; var y integer; con Mycon: x = 2 + 4*y; See also https://or.stackexchange.com/questions/6545/how-to-linearize-membership-in-a-finite-set ... View more

Thanks for posting this reply.  +1   A slight improvement is that for variables d and e with consecutive integer domains, you can avoid num d, num e, con_d, and con_e by using integer variables: var var_d >= 1 <= 10 integer; var var_e >= 1 <= 47 integer; ... View more

The following PROC OPTMODEL code solves an ordinary least squares regression:   proc optmodel; /* declare sets and input parameters */ set OBS; num y {OBS}; num a {OBS}; num b {OBS}; num c {OBS}; num d {OBS}; num e {OBS}; /* read input data set */ read data input into OBS=[_N_] y a b c d e; /* declare decision variables */ var Beta {0..5}; var Error {OBS}; impvar Estimate {i in OBS} = Beta[0] + Beta[1]*a[i] + Beta[2]*b[i] + Beta[3]*c[i] + Beta[4]*d[i] + Beta[5]*e[i]; /* declare objective to minimize sum of squared errors */ min SSE = sum {i in OBS} Error[i]^2; /* declare constraints */ con ErrorCon {i in OBS}: Error[i] = Estimate[i] - y[i]; /* call quadratic programming (QP) solver */ solve; /* print regression coefficients */ print Beta; /* create output data sets */ create data BetaData from [j] Beta; create data EstimateData from [i] a b c d e y Estimate Error; quit; The resulting regression coefficients are:   [1] Beta 0 416.9191 1 19.3854 2 18.5124 3 -13.0827 4 -22.7077 5 -1.4499   Those are the same values that are returned by the following PROC REG call: proc reg data=input; model y = a b c d e; quit; For the additional constraints, do you want to impose bounds on Beta[j]? Or do you instead want to solve an optimization problem where a, b, c, d, and e are the (bounded) decision variables and you want to minimize the predicted value of y?  That is, with the fixed values of Beta[j] found earlier, do you want to minimize the linear function Beta[0] + Beta[1]*a + Beta[2]*b + Beta[3]*c + Beta[4]*d + Beta[5]*e subject to some constraints on a, b, c, d, and e? ... View more