About RobPratt

RobPratt

Can you please share your data set?

RobPratt

For the code I suggested, the following suffices: num FTR {PROD_IDs}; read data SECOND_FTR_PRODS into [PROD_ID] FTR;

RobPratt

con Constraint1 {<PROD_ID, slot, day> in PROD_SLOT_DAY, <PROD_ID2, slot2, (day)> in PROD_SLOT_DAY: FTR[PROD_ID] = FTR[PROD_ID2] and slot2 = slot+1}: PROD_ASSIGN[PROD_ID, slot, day] + PROD_ASSIGN[PROD_ID2, slot2, day] <= 1; con Constraint2 {<PROD_ID, slot, day> in PROD_SLOT_DAY, <PROD_ID2, (slot), day2> in PROD_SLOT_DAY: FTR[PROD_ID] = FTR[PROD_ID2] and dayToNum[day2] = dayToNum[day]+1}: PROD_ASSIGN[PROD_ID, slot, day] + PROD_ASSIGN[PROD_ID2, slot, day2] <= 1;

RobPratt

I am not able to replicate the behavior you are seeing. Please attach the full log from a fresh session.

RobPratt

What does the following yield? proc print data=casuser.vrpdata(where=(ModelWeekNumber=3)); run;

RobPratt

From your log: ModelWeekNumber=3 NOTE: There were 1 rows read from table 'VRPDATA' in caslib 'CASUSER(SRHY-Exp)'. I recommend checking your vrpdata table, which seems to have only one observation with ModelWeekNumber=3.

RobPratt

The simplest way is to use the runOptmodel action with the groupBy parameter. The idea is to write the OPTMODEL code as if you are solving for only one BY group and then make the following modifications. Call PROC CAS instead of PROC OPTMODEL: /*proc optmodel;*/ proc cas noqueue; source pgm; put _BY_LINE_; Use NodeID instead of _N_ as the key column (always a good idea to avoid _N_ with CAS because the data order is unpredictable): read data vrpdata into NODES =[NodeID] ModelNode ModelLat ModelLong Volume Mode; After the end of your OPTMODEL code, call the runOptmodel action: /*quit;*/ endsource; action optimization.runOptmodel / code=pgm groupBy='ModelWeekNumber'; quit; The resulting solution_data CAS table will automatically include any BY variables, so you can use a BY statement in the PROC SGPLOT call to plot one BY group at a time: proc sgplot data=casuser.solution_data noautolegend; by ModelWeekNumber; scatter x=x1 y=y1 / datalabel=i; vector x=x2 y=y2 / xorigin=x1 yorigin=y1 group=k noarrowheads; xaxis display=none; yaxis display=none; run; It is worth noting that you can optionally get a summary table with one row per BY group as follows: action optimization.runOptmodel / code=pgm groupBy='ModelWeekNumber' statusOut='statusOut'; The resulting statusOut table contains information about the solver call for each BY group.

RobPratt

Glad to help. Yes, your understanding is correct (assuming N=6). You can also use the EXPAND statement to help verify what is happening. For example: expand DesiredConstraint1; Here are the first several lines of the resulting output: Constraint DesiredConstraint1[AAA,9,DAY_1,11,DAY_2]: PROD_ASSIGN[AAA,9,DAY_1] + PROD_ASSIGN[AAA,11,DAY_2] <= 1 Constraint DesiredConstraint1[AAA,9,DAY_1,11,DAY_3]: PROD_ASSIGN[AAA,9,DAY_1] + PROD_ASSIGN[AAA,11,DAY_3] <= 1 Constraint DesiredConstraint1[AAA,9,DAY_1,12,DAY_2]: PROD_ASSIGN[AAA,9,DAY_1] + PROD_ASSIGN[AAA,12,DAY_2] <= 1 Constraint DesiredConstraint1[AAA,9,DAY_1,12,DAY_3]: PROD_ASSIGN[AAA,9,DAY_1] + PROD_ASSIGN[AAA,12,DAY_3] <= 1 Constraint DesiredConstraint1[AAA,9,DAY_1,9,DAY_3]: PROD_ASSIGN[AAA,9,DAY_1] + PROD_ASSIGN[AAA,9,DAY_3] <= 1 Constraint DesiredConstraint1[AAA,9,DAY_1,9,DAY_2]: PROD_ASSIGN[AAA,9,DAY_1] + PROD_ASSIGN[AAA,9,DAY_2] <= 1 Constraint DesiredConstraint1[AAA,9,DAY_1,10,DAY_3]: PROD_ASSIGN[AAA,9,DAY_1] + PROD_ASSIGN[AAA,10,DAY_3] <= 1 Constraint DesiredConstraint1[AAA,9,DAY_1,10,DAY_2]: PROD_ASSIGN[AAA,9,DAY_1] + PROD_ASSIGN[AAA,10,DAY_2] <= 1

RobPratt

create data want from [item] x UseItem;

RobPratt

Here's one way to enforce your first desired constraint: num dayToNum {DayNumbers}; num count init 0; for {d in DayNumbers} do; count = count + 1; dayToNum[d] = count; end; print dayToNum; num N = 2; con DesiredConstraint1 {<PROD_ID, slot, day> in PROD_SLOT_DAY, <(PROD_ID), slot2, day2> in PROD_SLOT_DAY: dayToNum[day2] in dayToNum[day]+1..dayToNum[day]+N}: PROD_ASSIGN[PROD_ID, slot, day] + PROD_ASSIGN[PROD_ID, slot2, day2] <= 1; But notice that this conflicts with your fixed assignments of AAA to days 6, 7, and 8, so the resulting problem is infeasible. For your second desired constraint, you can use multiple objectives as follows: max NumPreferences = sum {<PROD_ID, slot, day> in PREF_PROD_SHOWCD inter PROD_SLOT_DAY} PROD_ASSIGN[PROD_ID, slot, day]; solve obj NumPreferences; con ObjectiveCut: NumPreferences >= NumPreferences.sol; solve obj TotalPreferenceWeight with milp / primalin; The idea is to first maximize the number of preferences that can be satisfied, then impose a constraint that enforces that maximum, and then maximize your original objective function. I have used the PRIMALIN option to warm start the second solve with the solution from the first solve.

RobPratt

Here are two ways, both using the MILP solver through PROC OPTMODEL: /* automated linearization in SAS Viya */ proc optmodel; set ITEMS; num x {ITEMS}; read data tab1 into ITEMS=[_N_] x; var UseItem {ITEMS} binary; min Error = abs(sum {j in ITEMS} x[j]*UseItem[j] - &target); solve linearize; print x UseItem; quit; /* manual linearization */ proc optmodel; set ITEMS; num x {ITEMS}; read data tab1 into ITEMS=[_N_] x; var UseItem {ITEMS} binary; var Over >= 0; var Under >= 0; min Error = Over + Under; con sum {j in ITEMS} x[j]*UseItem[j] - Over + Under = &target; solve; print x UseItem; quit; Both yield optimal objective value 0 with the following solution: [1] x UseItem 1 22.57 0 2 62.43 1 3 56.01 1 4 52.71 1 5 24.44 1 6 20.34 1 7 16.00 0 8 14.73 1 9 17.52 1 10 20.15 1 11 22.76 1 12 18.59 1 13 30.14 1 14 14.63 0 15 23.13 0 16 11.00 1 17 26.39 0 18 149.18 1

RobPratt

The following examples illustrate how to do what you want: 42332 - BY group processing with PROC OPTMODEL and the runOptmodel action (sas.com) SAS Help Center: Efficiency Analysis: How to Use Data Envelopment Analysis to Compare Efficiencies of Garages

RobPratt

In the optimization model, vehicles start and end at the depot. But in the implementation of the resulting optimal solution, the vehicles do not actually leave the depot (that is, do not actually traverse the zero-distance dummy arcs out of the depot). You can just omit those arcs from the output data set by specifying "i ne depot" as follows, yielding for each vehicle a path that starts at a customer node and ends at the depot: create data solution_data from [i j k]= {<i,j> in LOGICAL_ARCS, k in VEHICLES: UseArc[i,j,k].sol > 0.5 and i ne depot} StartNode=ModelNode[i] StartCity=ModelNode[i] x1=ModelLat[i] y1=ModelLong[i] EndNode=ModelNode[j] EndCity=ModelNode[j] x2=ModelLat[j] y2=ModelLong[j] flow=Flow[i,j,k] dist=dist[i,j] function='line' drawspace='datavalue'; For a similar problem transformation regarding the traveling salesman problem, see the "A BETTER LOWER BOUND FROM TSP" section in my SAS Global Forum 2014 paper The Traveling Baseball Fan Problem and the OPTMODEL Procedure.

RobPratt

The optimization model requires vehicles to start and end at the depot. To solve your problem variant where the vehicles can start anywhere but must end at the depot, it suffices to change the arc costs to 0 for all arcs out of the depot. In the resulting optimal solution, you can then drop those dummy arcs out of the depot. This is a known optimization modeling trick to convert your problem to a standard form just by changing the data.

RobPratt

Here's how to increment the flow instead of decrementing: /* the amount of supply picked up by vehicle k at node i must equal Volume[i] if UseNode[i,k] = 1; otherwise, it must equal 0 */ con FlowBalance {i in NODES diff {depot}, k in VEHICLES}: sum {<(i),j> in LOGICAL_ARCS} Flow[i,j,k] - sum {<j,(i)> in LOGICAL_ARCS} Flow[j,i,k] = Volume[i] * UseNode[i,k]; /* each vehicle is empty when it leaves the depot */ for {<(depot),j> in LOGICAL_ARCS, k in VEHICLES} fix Flow[depot,j,k] = 0; Here's how to start at any node and always end at the depot, by imposing cost 0 out of the depot: num dist {<i,j> in ARCS} = (if i ne depot then GEODIST(ModelLat[i],ModelLong[i],ModelLat[j],ModelLong[j],'M') else 0); /* if i=depot, this means that we want to treat the first leg from depot as 0 so that the truck goes from depot as home base free */

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