About RobPratt

RobPratt · ‎07-21-2013

min z = rosenbrock(x);

RobPratt · ‎07-19-2013

FCMP enablement in PROC OPTMODEL is not documented or supported but has been available as a hidden feature for several releases. Official support for this functionality is on the roadmap for a coming release.

RobPratt · ‎07-16-2013

Here's one way to impose the "min volume for conflict group 5" constraint: num cust_ranking {i in cust} = max {<(i),j> in CUST_camp} ranking[i,j]; print cust_ranking; con cust_conflict_5: sum {<i,j> in CUST_CAMP: cust_ranking >= 100000 and conflict_group[i,j] = 5} assign[i,j] >= 20000; You can also store the high-ranking customers in a set and use that set instead: set CUST_HIGHRANK = {i in cust: cust_ranking >= 100000}; con cust_conflict_5: sum {<i,j> in CUST_CAMP: i in CUST_HIGHRANK and conflict_group[i,j] = 5} assign[i,j] >= 20000; You can mimic this logic to get the other constraints.

RobPratt · ‎07-04-2013

You might try making portmean an explicit variable instead, to reduce the density of the expression for portvar. Replace the impvar portmean statement with: var portmean; con portmean_con: portmean = (1/card(DATES)) * (sum {t in DATES} (sum {i in STOCKS } ((1/card(STOCKS )) + (1/card(STOCKS )) * sum {k in CHARACTERISTICS} Theta * char[i,t,k]) * ret[i,t]));

RobPratt · ‎07-03-2013

If I understand correctly, the following should do what you want: set CONFLICT_GROUPS {i in CUST} = setof {<(i),j> in CUST_CAMP} conflict_group[i,j]; con cust_conflict {i in cust, g in CONFLICT_GROUPS}: sum {<(i),j> in CUST_CAMP: conflict_group[i,j] = g} assign[i,j] <= 1; expand cust_conflict;

RobPratt · ‎12-06-2012

You are correct that PROC OPTMODEL does not support such matrix syntax. In particular, there is no INV() function in PROC OPTMODEL, and matrix products are expressed using sums of products, with explicit indices, as you have indicated. There are currently no plans to support IML-like syntax, but thank you for your suggestion. Rob Pratt

RobPratt · ‎10-31-2012

As of SAS 9.3 released in July 2011, the NLPU, NLPC, SQP, and IPNLP solvers are no longer under active development. You should instead use the NLP solver (SOLVE WITH NLP), which accesses two algorithms: INTERIORPOINT (default) and ACTIVESET. For more details, see: http://support.sas.com/documentation/cdl/en/ormpug/63975/HTML/default/viewer.htm#ormpug_nlpsolver_sect005.htm Also, FCMP support for PROC OPTMODEL is undocumented and not yet fully functional.

RobPratt · ‎07-27-2012

For your "cardinality-constrained" assignment problem, you can modify both sets of = 1 constraints to <= 1 and then include a new constraint that forces the desired number of assignments: con cardinality: sum{i in NSET, j in NSET} x[i,j] = 10; In that case, you should also declare x as binary and use the MILP solver.

RobPratt · ‎07-24-2012

For a given pair of states, your problem can be formulated as a linear assignment problem, where the cost of assigning record i from one state to record j from the other state is the absolute difference of the property values. See the following usage note for how to solve a linear assignment problem using the LP solver in SAS/OR: http://support.sas.com/kb/35/026.html In SAS/OR 12.1, which will be released next month, there is also a linear assignment problem algorithm available in the new OPTNET procedure. Note that without your threshold constraint, if there are n records for each of the two states, then just sorting the values for each state and matching them up in that order is optimal. This property does not hold in general, but for your absolute difference cost, it is easy to show that any other matching will have higher cost. (If two values are out of order, interchanging them will improve the cost.)

RobPratt · ‎06-29-2012

Add this line before the SOLVE statement: for {j in 1..n: and {i in w} (m[i,j] = 0)} fix wt = 0;

RobPratt · ‎06-11-2012

Yes, remove w from wt declaration: var wt {1..n} >= 0 <= 1 init 1; Replace wt[i,j] with wt everywhere. Also, omit weighting constraint. Finally: con swt_con: swt = 1; Or you can omit impvar swt and do this: con swt_con: sum {j in 1..n} wt = 1;

RobPratt · ‎06-09-2012

Here is how to discretize a bounded variable y: proc optmodel; var y >= 0 <= 3; num num_steps = 10; set STEPS = 0..num_steps-1; num a {s in STEPS} = (s - y.lb) / (y.ub - y.lb); var yb {STEPS} binary; con assign: sum {s in STEPS} yb = 1; con link: sum {s in STEPS} a * yb = y; quit;

RobPratt · ‎06-09-2012

And here's how you can do the sums if you declare wt and sp as in my earlier reply: impvar wt_sum {i in w} = sum {j in 1..n} wt[i,j]; impvar sp_sum {i in w} = sum {j in 1..n} sp[i,j];

RobPratt · ‎06-09-2012

Here's an alternative approach that uses fewer PROC OPTMODEL statements: %let n = 3; proc optmodel; num n = &n; set w; num weeks{w}; num dep{w}; num m{w,1..n}; read data M into w = [_n_] dep {j in 1..n} <m[_N_,j]=col('m'||j)>; var wt {w,1..n} >= 0 <= 1 init 1; impvar sp {i in w, j in 1..n} = m[i,j] * wt[i,j]; print sp; quit;

RobPratt · ‎05-28-2012

If you multiply both sides by y, you get x + y z = c y. If y and z are binary variables, you can replace the product y z with a new binary variable w and add some constraints. If only one of y and z is binary and the other is bounded, you can linearize the product in a similar manner. If neither of y and z is binary, you can discretize one of them and apply the same techniques. For details, please see Chapters 10 and 24 in the new book of mathematical programming examples available as part of the SAS/OR documentation: https://support.sas.com/pubscat/bookdetails.jsp?pc=64980 (click on View This Title Online for a free pdf version).

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