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RobPratt · ‎12-17-2014

By the way, you also need to initialize PRICE to missing if you want to be able to check its value everywhere: num PRICE {RGN_I, MTH_I, CNTRY_I, CLS_I, MKT_I, MKTTYPE_I} init .;

RobPratt · ‎12-17-2014

Here is one way to express the desired constraints by using two sets of constraints with logical conditions: con y_CON1 {r in RGN_I, m in MTH_I, p in CNTRY_I, c in CLS_I, k in MKT_I, t in MKTTYPE_I: PRICE[r,m,p,c,k,t] = .}: y[r,m,p,c,k,t] = 0; con y_CON2 {r in RGN_I, m in MTH_I, p in CNTRY_I, c in CLS_I, k in MKT_I, t in MKTTYPE_I: PRICE[r,m,p,c,k,t] ne .}: PRICE[r,m,p,c,k,t]*0.9 <= y[r,m,p,c,k,t] <= PRICE[r,m,p,c,k,t]*1.1; And here is an alternative approach that uses one set of range constraints with two IF-THEN-ELSE expressions: con y_CON {r in RGN_I, m in MTH_I, p in CNTRY_I, c in CLS_I, k in MKT_I, t in MKTTYPE_I}: (if PRICE[r,m,p,c,k,t] = . then 0 else PRICE[r,m,p,c,k,t]*0.9) <= y[r,m,p,c,k,t] <= (if PRICE[r,m,p,c,k,t] = . then 0 else PRICE[r,m,p,c,k,t]*1.1); A third approach is to use the FIX statement when PRICE is missing: for {r in RGN_I, m in MTH_I, p in CNTRY_I, c in CLS_I, k in MKT_I, t in MKTTYPE_I: PRICE[r,m,p,c,k,t] = .} fix y[r,m,p,c,k,t] = 0; con y_CON {r in RGN_I, m in MTH_I, p in CNTRY_I, c in CLS_I, k in MKT_I, t in MKTTYPE_I: PRICE[r,m,p,c,k,t] ne .}: PRICE[r,m,p,c,k,t]*0.9 <= y[r,m,p,c,k,t] <= PRICE[r,m,p,c,k,t]*1.1;

RobPratt · ‎12-15-2014

Glad to help. Please mark this question as Answered.

RobPratt · ‎12-14-2014

You can use PROC OPTMODEL in SAS/OR to formulate the problem and call either the MILP solver or possibly the CLP solver: SAS/OR(R) 13.2 User's Guide: Mathematical Programming Besides the v and w decision variables, you need a binary decision variable x[j,k] that indicates whether observation j fails constraint k. You want to fix x[j,k] = 0 for all observations j with FLAG = 0 and all constraints k. If you are using the MILP solver, you can enforce this relationship by introducing big-M constraints (assuming DC also depends on j): /* if x[j,k] = 1 then v + sum {i in ISET } w * DC[i,j] >= 0 */ -v - sum {i in ISET } w * DC[i,j] <= M1[j,k] * (1 - x[j,k]) /* if x[j,k] = 0 then v + sum {i in ISET } w * DC[i,j] < 0 */ v + sum {i in ISET } w * DC[i,j] + epsilon <= M2[j,k] * x[j,k] where M1[j,k] and M2[j,k] are upper bounds on the left hand sides, and epsilon is a small positive number, say 1e-6, that represents how far away from 0 you consider "negative" to be. If you are using the CLP solver, you can enforce the relationship with a single set of REIFY constraints, without big-M or epsilon: REIFY(x[j,k], v + sum {i in ISET } w * DC[i,j] >= 0) But note that the CLP solver requires all variables to be integer, and I don't know whether your v and w have that restriction. There is a bit of ambiguity in the problem statement. Does "fail these constraints" mean fail all constraints or fail at least one constraint? In either case, you need another binary variable y that indicates whether observation j fails, and your objective is to maximize sum {j in JSET} y . If y = 1 means fail all, then you want y <= x[j,k] for all k. If y = 1 means fail at least one, then you want y <= sum {k in KSET} x[j,k].

RobPratt · ‎11-18-2014

Please post your code, along with the data if possible.

RobPratt · ‎11-08-2014

Your latest version is not equivalent to the other two. Instead, it should be: con test{t in ASC}:sum{u in UIN:<t,u> in Map}y>=0; Also, the "Constraint is empty" message is only a NOTE and not an ERROR. If you use the EXPAND statement, you will see that the constraint is 0 >= 0 because the sum is empty. If you really want to avoid generating these constraints, you can use a logical condition: con test{t in ASC: card(slice(<t,*>,Map)) > 0}:sum{u in slice(<t,*>,Map)}y>=0;

RobPratt · ‎11-07-2014

You can write this more compactly by using an "implicit" slice: con test{t in ASC}: sum{<(t),u> in Map} y >= 0; See this doc example for more discussion: SAS/OR(R) 13.2 User's Guide: Mathematical Programming Examples

RobPratt · ‎11-07-2014

data indata; input a b c f; datalines; 1 5 4 0.2 1 6 5 0.3 2 5 3 0.21 2 6 4 0.31 3 5 2 0.42 3 6 3 0.81 4 5 1 0.92 4 6 2 0.11 ; proc optmodel; set OBS; set PARAMS = {'a','b','c'}; num data {OBS, PARAMS}; num f {OBS}; read data indata into OBS=[_N_] {j in PARAMS} <data[_N_,j]=col(j)> f; print data f; set VALUES {j in PARAMS} = setof {i in OBS} data[i,j]; var X {j in PARAMS, v in VALUES } >= 0 <= 10 init 1; impvar Error {i in OBS} = prod {j in PARAMS} X[j,data[i,j]] - f; min SSE = sum {i in OBS} Error^2; solve with NLP / ms; num prod {i in OBS} = prod {j in PARAMS} X[j,data[i,j]].sol; print X; print prod f Error; quit;

RobPratt · ‎11-06-2014

You might find these two examples helpful: 42332 - BY group processing with PROC OPTMODEL SAS/OR(R) 13.2 User's Guide: Mathematical Programming Examples If not, please post what you have tried so far.

RobPratt · ‎10-30-2014

You can try PROC OPTLSO: SAS/OR(R) 13.2 User's Guide: Local Search Optimization But if your problem can be linearized, you might see better performance by using the MILP solver instead. We have used over two dozen approaches to tackle various MINLP problems. Can you share any details for your problem?

RobPratt · ‎10-02-2014

You can use PROC OPTMODEL to formulate the problem and solve it with the mixed integer linear programming (MILP) solver. You need to introduce one binary variable per player, with the interpretation that the variable equals 1 if and only if that player is selected. Each of the four conditions corresponds to a linear constraint over these variables. And the objective function to be maximized is also a linear function of these variables.

RobPratt · ‎08-20-2014

The simplest approach here is to use the DROP statement (SAS/OR(R) 13.2 User's Guide: Mathematical Programming) to drop the first constraint: drop constr1; More generally, you can use the PROBLEM and USE PROBLEM statements to handle multiple problems within one PROC OPTMODEL session: SAS/OR(R) 13.2 User's Guide: Mathematical Programming

RobPratt · ‎07-29-2014

This seems to solve fine in SAS/OR 13.1 (the latest production release, from December 2013). The binary variables take integer values in the resulting solution, and summing up the _OBJCOEF_ * _VALUE_ from the PrimalSolution data set yields the same objective value that is shown in the log. What behavior are you seeing instead? NOTE: The problem PROBLEM has 8042 variables (2555 binary, 0 integer, 0 free, 0 fixed). NOTE: The problem has 3686 constraints (2202 LE, 1472 EQ, 12 GE, 0 range). NOTE: The problem has 17188 constraint coefficients. NOTE: The MILP presolver value MODERATE is applied. NOTE: The MILP presolver removed 0 variables and 24 constraints. NOTE: The MILP presolver removed 24 constraint coefficients. NOTE: The MILP presolver modified 0 constraint coefficients. NOTE: The presolved problem has 8042 variables, 3662 constraints, and 17164 constraint coefficients. NOTE: The MILP solver is called. Node Active Sols BestInteger BestBound Gap Time 0 1 0 . 303907325 . 0 0 1 0 . 303907325 . 0 0 1 0 . 303907325 . 0 NOTE: The MILP solver added 1 cuts with 3 cut coefficients at the root. 27 10 2 6096111710 303907326 1905.91% 4 52 23 3 5570543121 303907327 1732.97% 5 100 25 3 5570543121 303907327 1732.97% 5 200 53 3 5570543121 303907327 1732.97% 6 220 59 4 303907696 303907327 0.00% 7 NOTE: Optimal within relative gap. NOTE: Objective = 303907695.65. NOTE: The data set WORK.PRIMALSOLUTION has 8042 observations and 8 variables. NOTE: The data set WORK.DUALSOLUTION has 3686 observations and 8 variables. NOTE: PROCEDURE OPTMILP used (Total process time): real time 30.95 seconds cpu time 7.48 seconds

RobPratt · ‎07-25-2014

Yes, binary variables are supported in the objective function for the MILP solver. Can you please share your code, sparsedata data set, or MPS data set? By the way, OPTMODEL is the recommended procedure for formulating and solving mathematical optimization problems.

RobPratt · ‎07-13-2014

Here's one way to use PROC OPTMODEL to solve this optimization problem with one decision variable: proc optmodel; set ISET; set JSET = 1..5; num one {ISET, JSET}; num three {ISET, JSET}; read data one into ISET=[_N_] {j in JSET} <one[_N_,j]=col('t'||j)>; read data three into [_N_] {j in JSET} <three[_N_,j]=col('t'||j)>; var Index; impvar Two {i in ISET, j in JSET} = probit(one[i,j] - Index); min TotalError = sum {i in ISET, j in JSET} ((three[i,j]-Two[i,j])/three[i,j])^2; solve; print Two; print Index; quit; But you need to clean up the data before you can use PROBIT.

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