About RobPratt

RobPratt · ‎06-20-2021

The new constraint involves XIT and so must appear after that variable is declared.

RobPratt · ‎06-20-2021

To maximize the number of targets that are covered by at least three cameras, replace Psi, the original objective, and the XIT_CON2 and XIT_CON3 constraints with the following: /* maximize number of triply covered targets */ var IsTriplyCovered {TARGETS} binary; max NumTriplyCovered = sum {t in TARGETS} IsTriplyCovered[t]; con TriplyCoveredCon {t in TARGETS}: XIT[t] >= 3 * IsTriplyCovered[t]; In SAS Viya, you can alternatively use an indicator constraint as follows: con TriplyCoveredCon {t in TARGETS}: IsTriplyCovered[t] = 1 implies XIT[t] >= 3;

RobPratt · ‎06-06-2021

Here's an approach that linearizes the quadratic objective so that you can use the MILP solver: num k = 3; var Psi {TARGETS} integer <= k; set VALUES = 0..k; var IsValue {TARGETS, VALUES} binary; con OneValue {t in TARGETS}: sum {v in VALUES} IsValue[t,v] = 1; con LinkIsValuePsi {t in TARGETS}: sum {v in VALUES} v*IsValue[t,v] = Psi[t]; minimize OptCoverage = sum{t in TARGETS, v in VALUES} (k-v)^2 * IsValue[t,v] -sum{station in STATIONS, orientation in ORIENTATIONS} Chi[station,orientation];

RobPratt · ‎06-06-2021

Can you please also provide the input data sets?

RobPratt · ‎06-02-2021

Please see also this documentation example, for which the code is available in the same github repository: https://github.com/sassoftware/optimization-examples/blob/main/mpex/mpex22.sas

RobPratt · ‎05-22-2021

OK, good, you can also strengthen the formulation by multiplying the upper bound by build[i]. If you do that, the assignFlow <= build constraints are no longer necessary, and it might speed things up if you remove them.

RobPratt · ‎05-21-2021

OK, I get the same objective values as you, but notice that you cannot tell which facilities are built just by looking at the Flow variables. If you add this statement after each solve, you will see that all 184 facilities are built: put (card({i in RDC: Build[i].sol > 0.5}))=; Also notice that the first objective value matches the sum of all build costs.

RobPratt · ‎05-21-2021

I am not able to replicate your results with the data you sent. Instead, I get all 184 facilities built in both solves. Did you maybe run with rdcMINZip3Cap = 0? In any case, you have the wrong objective cut: con ObjectiveCut{<i,j> in ARCS}: assignFlow[i, j] <= min_build; It should instead be: con ObjectiveCut: green_rdc_min_build <= min_build;

RobPratt · ‎05-21-2021

It sounds like you are not imposing the objective cut in the second solve. Are you able to share your full code and data? If you don't want to share publicly, you can send it to me privately.

RobPratt · ‎05-21-2021

Was the build cost of the 176 facilities the same as for the 36 facilities? If so, what don't you like about the second solution? If it builds too many facilities, you might want to include another objective cut to convey that information in the optimization model.

RobPratt · ‎05-21-2021

You almost had it. Just add Flow to the end: create data green_rdc_sol from [i j]={i in RDC, j in ZIP3: Flow[i,j].sol > 0.5} Flow; Regarding SAVE MPS + PROC OPTMILP, that approach of course works but is inefficient because you are adding two extra writes and reads for each solve. I recommend instead using PROC OPTMODEL for both problem generation and the solver calls. If you strongly prefer calling the solver from PROC OPTMILP, you might consider using a SUBMIT/ENDSUBMIT block so you don't need to exit PROC OPTMODEL.

RobPratt · ‎05-21-2021

It sounds like you want to use primary and secondary objectives like this: /* primary objective */ min green_rdc_min_build = sum{i in RDC} cost[i] * Build[i]; solve; num min_build; min_build = green_rdc_min_build; con ObjectiveCut: green_rdc_min_build <= min_build; /* secondary objective */ min green_rdc_min_dist = sum{<i,j> in ARCS} distance[i,j] * assignFlow[i,j]; solve with milp / primalin; Note the use of the PRIMALIN option to warm start the second solver call with the first solution.

RobPratt · ‎05-10-2021

The following does what you want: data zData; input bucket Z_score; datalines; 1 -2.38219471237423 2 -0.642830996719877 3 -0.190964264784506 4 0.379313913718413 ; %let minYear = 2015; %let maxYear = 2018; data actualData; input bucket actual&minYear-actual&maxYear; datalines; 1 0.011488175674031 0.006578316410400 0.005081508424610 0.011271589592748 2 0.323791550923876 0.223834610249151 0.180403588330841 0.312637786039641 3 0.500800479192011 0.373187463052586 0.384055738171937 0.439063500503939 4 0.693121340842155 0.592898832157161 0.571297020124314 0.733773269523783 ; proc optmodel; set BUCKETS; num z_score {BUCKETS}; read data zData into BUCKETS=[bucket] z_score; set YEARS = &minYear...&maxYear; num actual {BUCKETS, YEARS}; read data actualData into [bucket] {y in YEARS} <actual[bucket,y]=col('actual'||y)>; var Index {YEARS}; var Predicted {BUCKETS, YEARS}; var Error {BUCKETS, YEARS}; min SSE = sum {b in BUCKETS, y in YEARS} Error[b,y]^2; con PredictedCon {b in BUCKETS, y in YEARS}: Predicted[b,y] = CDF('NORMAL', Index[y]+z_score[b]); con ErrorCon {b in BUCKETS, y in YEARS}: Error[b,y] = Predicted[b,y] - actual[b,y]; solve; print actual; print Predicted; print {y in YEARS} (sum {b in BUCKETS} Error[b,y]^2) Index; quit; The SAS System The OPTMODEL Procedure Solution Summary Solver NLP Algorithm Interior Point Direct Objective Function SSE Solution Status Optimal Objective Value 0.0048361802 Optimality Error 7.917181E-10 Infeasibility 3.991606E-10 Iterations 2 Presolve Time 0.00 Solution Time 0.01 actual 2015 2016 2017 2018 1 0.0114882 0.0065783 0.0050815 0.0112716 2 0.3237916 0.2238346 0.1804036 0.3126378 3 0.5008005 0.3731875 0.3840557 0.4390635 4 0.6931213 0.5928988 0.5712970 0.7337733 Predicted 2015 2016 2017 2018 1 0.0135008 0.0059413 0.0052249 0.0123153 2 0.3184164 0.2188015 0.2057626 0.3058023 3 0.4919160 0.3728249 0.3559350 0.4777037 4 0.7088449 0.5971192 0.5796242 0.6965001 [1] Index 2015 0.000359105 0.17070 2016 0.000043681 -0.13342 2017 0.001503220 -0.17838 2018 0.002930174 0.13505

RobPratt · ‎04-21-2021

Watch this Ask the Expert session to learn to diagnose and understand why an optimization problem is infeasible and how to make it feasible. Watch the webinar Join Rob Pratt to understand the ways to diagnose infeasibility. Perfect for SAS/OR® (SAS®9) and SAS Optimization (SAS® Viya®) users, especially those who use PROC OPTMODEL or the runOptmodel action. During this webinar you will learn: How to use the “IIS=” option. How to use the network solver. How to use the output of one solver as input for another solver. How to modify an infeasible problem as little as possible to make it feasible. The questions from the Q&A segment held at the end of the webinar are listed below, and the slides from the webinar are attached. How do you use the LINEARIZE option for the example shown? See on-demand webinar for a demo of this. Starts at minute 40:30. Are there other ways of diagnosing infeasibility? Yes. Suppose you’re solving a business problem, you’ve modeled it, and the solver is telling you the problem is infeasible, but you think it’s feasible. If you can construct what should be a feasible solution, even a trivial one, then you can pass that information to the solver by fixing the decision variables to the values in that solution that you manually constructed. When you fix the variables and then call the solver, the solver will tell you which constraint is violated. This is another simple way to diagnose infeasibility if you know a feasible solution. It could point to a data error, a modeling error, or a coding error. Recommended Resources SAS Optimization Support SAS/OR Support Mathematical Optimization, Discrete-Event Simulation, and OR Community Operations Research: Optimize, Simulate, Understand Blogs SAS Global Forum 2021 Want more tips? Be sure to subscribe to the Ask the Expert board to receive follow up Q&A, slides and recordings from other SAS Ask the Expert webinars.

RobPratt · ‎04-20-2021

That works, or you can use 4 integer variables and 1 linear constraint: proc optmodel; num numLists = 4; num target = 5; var X {1..numLists} >= 0 <= target integer; con SumToTarget: sum {i in 1..numLists} X[i] = target; solve with clp / findallsolns; create data want from [combination set]={s in 1.._NSOL_, i in 1..numLists} count=X[i].sol[s]; quit; proc sort data=want; by combination set; run;

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