About RobPratt

RobPratt · ‎10-30-2023

Here's a way to enforce #2, because the top state is at most 30% if and only if all states are at most 30%. By using the SETOF operator, you can avoid separately reading all the states from a data set. set STATES = setof {s in studentSet} state[s]; con stateCon {st in STATES}: sum{s in studentSet: state[s] = st} score[s] * Inclusion[s] <= 0.3 * sum{s in studentSet} score[s] * Inclusion[s]; Alternatively, you can reuse totalScore: set STATES = setof {s in studentSet} state[s]; con stateCon {st in STATES}: sum{s in studentSet: state[s] = st} score[s] * Inclusion[s] <= 0.3 * totalScore;

RobPratt · ‎10-30-2023

num ATsol {row}; num AT_Volsol {row}; cofor {i in row} do; id = i; solve; /* Save each individual solution */ ATsol[id] = AT.sol; AT_Volsol[id] = AT_vol.sol; end; create data solutionAT from [i] ATsol; create data solutionATVol from [i] AT_Volsol;

RobPratt · ‎10-02-2023

For modern logistics and supply chain management, efficiently solving Vehicle Routing Problems (VRPs) has become an essential task for organizations to ensure timely deliveries, minimize cost, reduce environmental impact and enhance customer satisfaction. In a typical VRP, a set of customers must be assigned to a fleet of vehicles. The vehicles start and end at a common depot and visit all customers once. There are different variants of VRPs, such as Capacitated Vehicle Routing Problem (CVRP) and Vehicle Routing Problem with Profits (VRPP). This presentation first introduces the fundamental aspects of VRPs and their applications. Next, it presents several novel approaches that efficiently solve VRPs with time windows and other constraints. Those new approaches were motivated by real-world use cases and are available in SAS® Optimization.

RobPratt · ‎09-29-2023

Please share your input and output if possible.

RobPratt · ‎09-01-2023

You have parentheses where you should instead use curly braces. Here is correct syntax: con Mycon9: sum {i in OBS: Campaign_Line_Item[i] in {'Walmart DSP|Display|Cross Device|Standard Audience|In Market Christmas|47490|7974163' ,'Walmart DSP|Display|Cross Device|Custom Audience|Heavy Spenders Home Decor|47490|8003807' ,'Walmart DSP|Display|Cross Device|Propensity Audience|Holiday Decor|47490|8005360' ,'Walmart DSP|Display|Cross Device|Propensity Audience|Ornaments|47490|7974160' ,'Walmart DSP|Display|Cross Device|Custom Audience|Nov&DecShoppers|47490|7974178' ,'Walmart|Display|Cross Device|Contextual Targeting|Holiday Decor|CY23|47490|7974064'}} Optimized_Budget[i] <= 140000; I recommend instead defining the set separately to make the code easier for human consumption: set CAMPAIGNS = {'Walmart DSP|Display|Cross Device|Standard Audience|In Market Christmas|47490|7974163' ,'Walmart DSP|Display|Cross Device|Custom Audience|Heavy Spenders Home Decor|47490|8003807' ,'Walmart DSP|Display|Cross Device|Propensity Audience|Holiday Decor|47490|8005360' ,'Walmart DSP|Display|Cross Device|Propensity Audience|Ornaments|47490|7974160' ,'Walmart DSP|Display|Cross Device|Custom Audience|Nov&DecShoppers|47490|7974178' ,'Walmart|Display|Cross Device|Contextual Targeting|Holiday Decor|CY23|47490|7974064'}; con Mycon9: sum {i in OBS: Campaign_Line_Item[i] in CAMPAIGNS} Optimized_Budget[i] <= 140000;

RobPratt · ‎09-01-2023

If you want to find a minimum number of lists to cover all the names, that is the set cover problem. If the lists have different prices and you want to minimize the total cost, that is the weighted set cover problem, mentioned on the same page. If you want to maximize the number of names covered subject to a budget on the number or total cost of the lists, that is the maximum coverage problem, also known as the partial set cover problem. You can solve all of these variants by calling the MILP solver in PROC OPTMODEL. You would declare a binary decision variable for each list and a constraint for each name. For the partial set cover problem, you would declare an additional binary decision variable for each name to indicate whether it is covered.

RobPratt · ‎08-31-2023

Yes, as long as 50000 is <= Input_Budget[j] for campaign j.

RobPratt · ‎08-31-2023

OK, I think you want the following: con Mycon1: sum {i in OBS: Media_Tactics[i]='Offsite Display Walmart Network'} Optimized_Budget[i] <= 350000; con Mycon2 {j in CAMPAIGNS}: sum {i in OBS: Campaign_Line_Item[i]=j} Optimized_Budget[i] <= Input_Budget[j];

RobPratt · ‎08-31-2023

No, that syntax will yield errors, and the logical condition compares a sum of decision variables to a string, so it would never be true. Do you maybe want the following? con Mycon1: sum {i in OBS: Media_Tactics[i]='Offsite Display Walmart Network'} Optimized_Budget[i] <= 200000; If not, what is the business rule that you are trying to enforce?

RobPratt · ‎08-31-2023

Not sure I follow. Do you mean that you want to use a logical condition with respect to the input data to determine whether to impose a single constraint?

RobPratt · ‎08-30-2023

The inputs for PROC OPTMILP are decision variables, bounds, integrality restrictions, linear constraints, and a linear objective. By default, the MILP solver that is invoked by PROC OPTMILP finds one optimal solution, which is an assignment of values to decision variables such that the variable bounds, integrality restrictions, and linear constraints are satisfied and the objective value is the best possible. To retrieve the resulting solution, you can use the PRIMALOUT= option to name an output data table that will contain one observation per decision variable. The _VAR_ column contains the name of each variable, and the _VALUE_ column contains the value of that variable in the solution returned by the MILP solver. For binary variables, the _TYPE_ column will be B, and the _VALUE_ will be "close to" 0 or 1 (within the integer tolerance, whose default is 1e-5). If this explanation does not provide sufficient clarity, maybe it would help if you can supply your specific input and output that you want to interpret.

RobPratt · ‎08-24-2023

Your Mycon1 looks like an attempt to do that: con Mycon1 {i in OBS}: sum {j in OBS} Optimized_Budget[j] <= &overall_input_budget * (1+&allowed_budget_inc) * (1-&value_add_pct); This family of constraints is indexed by i, but nothing in the constraint refers to i, so you end up with the exact same constraint over and over again. Instead, I think you intended to declare a single constraint: con Mycon1: sum {j in OBS} Optimized_Budget[j] <= &overall_input_budget * (1+&allowed_budget_inc) * (1-&value_add_pct);

RobPratt · ‎08-24-2023

It is hard to tell for sure from just looking at your log, but here is what I suspect is going on. You declared this constraint: con Mycon7 {i in OBS}: Optimized_Budget[i]<=if Media_Tactics[i]='Offsite Display Walmart Network' then 2000000; For the IF-THEN/ELSE expression, there is a default of ELSE 0, so what you declared is equivalent to this: con Mycon7 {i in OBS}: Optimized_Budget[i]<=if Media_Tactics[i]='Offsite Display Walmart Network' then 2000000 else 0; Because Optimized_Budget[i] is declared with >= 0 and this constraint forces that variable to be <= 0 if Media_Tactics[i] is not 'Offsite Display Walmart Network', the effect is that Optimized_Budget[i] = 0 in that case. I think you instead meant for the constraint to be enforced only if the condition is satisfied, which you can do as follows: con Mycon7 {i in OBS: Media_Tactics[i]='Offsite Display Walmart Network'}: Optimized_Budget[i] <= 2000000; And for a constraint with only one variable, it is more efficient to instead omit the constraint and just change the variable bounds: for {i in OBS: Media_Tactics[i]='Offsite Display Walmart Network'} Optimized_Budget[i].ub = 2000000; If I guessed wrong about your intent, I recommend trying the suggestion in the log to use the IIS= option to help diagnose infeasibility: solve with lp / iis=1; expand / iis;

RobPratt · ‎08-04-2023

You cannot fix an implicit variable, and you can fix a variable only to a constant. Instead you can modify the IMPVAR statement as follows: /* impvar VolBasedCost {i in ISN} = Volume_ISN[i] * sum {m in MOL} volRate[m] * IsMol[i,m];*/ impvar VolBasedCost {i in ISN} = sum {m in MOL} max(Volume_ISN[i] * volRate[m], volRate_Min[m]) * IsMol[i,m]; Because of the OneMol constraint, each i will have IsMol[i,m] = 1 for exactly one m. The new IMPVAR declaration will enforce VolBasedCost[i] to be the larger of Volume_ISN[i] * volRate[m] and volRate_Min[m] for that m.

RobPratt · ‎07-28-2023

I already replied to Santha in response to a private message, but for the benefit of the rest of the community, here was my recommendation: The introduction of Proportion into the optimization model seems to complicate things. I recommend going back to the model that you said worked smoothly last Saturday (https://communities.sas.com/t5/Mathematical-Optimization/AssetOpt-OutputViews-FullModel/td-p/885576) and computing Proportion after the fact if you need it for reporting.

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