About RobPratt

RobPratt · ‎01-10-2023

Can you please show what you tried?

RobPratt · ‎12-16-2022

When you are ready to migrate, you might find this example helpful: PROC NLP: Rewriting NLP Models for PROC OPTMODEL - 9.3 (sas.com)

RobPratt · ‎11-03-2022

Here's one approach that uses the MILP solver in PROC OPTMODEL: proc optmodel; /* declare parameters and read data */ set OBJECTS; str group {OBJECTS}; read data have1 into OBJECTS=[object] group; set <str,num> GROUP_PLACE; num n {GROUP_PLACE}; read data have2 into GROUP_PLACE=[group place] n; set OBJECT_GROUP_PLACE = {o in OBJECTS, <g,p> in GROUP_PLACE: group[o] = g}; /* X[o,g,p] = 1 if object o is assigned to group g and place p; 0 otherwise */ var X {OBJECT_GROUP_PLACE} binary; /* assign each object to exactly one group-place */ con AssignOnce {o in OBJECTS}: sum {<(o),g,p> in OBJECT_GROUP_PLACE} X[o,g,p] = 1; /* assign correct number of objects to each group-place */ con Cardinality {<g,p> in GROUP_PLACE}: sum {<o,(g),(p)> in OBJECT_GROUP_PLACE} X[o,g,p] = n[g,p]; /* call MILP solver with no objective */ solve noobj; /* create output data set */ create data want from [object group place]={<o,g,p> in OBJECT_GROUP_PLACE: X[o,g,p].sol > 0.5}; quit; If you had some measure of desirability of placing object o in group g and place p, you could instead maximize the total desirability rather than assigning arbitrarily.

RobPratt · ‎09-26-2022

You have not supplied everything to be able to run your code, but the NLP solver does not support integer variables. (Edit: I see now that you have changed your code to use LSO instead of NLP. The LSO solver, now called the black-box solver, supports integer variables but does not guarantee optimality. Better to use MILP here.) Fortunately, you can easily linearize the nonlinear constraint as follows: con overmax: 100 * sum {rownum in sset} over[rownum]*x[rownum] <= &over_target. * sum {rownum in sset} qtd[rownum]*x[rownum]; Then you can call the MILP solver: solve with milp; Or just: solve; Alternatively, if you specify the LINEARIZE option (available in SAS Viya only) without changing your original model, OPTMODEL will do this linearization for you automatically: solve linearize;

RobPratt · ‎09-21-2022

The final four constraint declarations have the effect of setting many variables to 0. The presolver will remove these explicit constraints and the corresponding variables, but a much more efficient approach is to modify the index sets to avoid creating the variables in the first place. Please see this documentation example: SAS Help Center: Sparse Modeling Making similar changes in your code will reduce the memory needed to store the data and generate the problem, thereby freeing up more memory to be used by the solver. Please let me know if this does not resolve your out-of-memory issue.

RobPratt · ‎08-11-2022

Yes, that looks correct if i is a facility and k is a customer. This constraint enforces Assign2[i,k] = 1 if and only if Assign2[i,’377’] = 1.

RobPratt · ‎08-10-2022

If you have binary decision variables X[i,j] that indicate whether customer i is assigned to facility j, the desired constraints are X[2,j] = X[3,j] and X[2,j] = X[4,j] for all j.

RobPratt · ‎08-09-2022

It looks like you are using an older version that does not support comparison of tuples. I recommend upgrading, but until then here are three workarounds: = (if <i,t> in {<SOURCE,0>} then 1 else if <i,t> in {<SINK,n+1>} then -1 else 0); = (if i = SOURCE and t = 0 then 1 else if i = SINK and t = n+1 then -1 else 0); = (if i = SOURCE then 1 else if i = SINK then -1 else 0);

RobPratt · ‎08-09-2022

Here are the requested modifications to your code, together with optional PUT, PRINT, and EXPAND statements to help see what is going on: data DYINC_M; input FRIDX TOIDX INC; datalines; 1 2 -0.23548 1 3 -0.14027 1 4 -0.22685 2 1 0.014942 2 3 -0.02719 2 4 -0.11378 3 1 0.076665 3 2 -0.06068 3 4 -0.05206 4 1 0.059285 4 2 -0.07806 4 3 0.017148 ; data HISINCAVG_M; input DYIDX INC; datalines; 1 -0.072240154 2 -0.075171085 3 -0.077492213 ; PROC OPTMODEL; num n = 4; SET TS = 1..n; SET IDX = 1..n; NUM SOURCE = 0; NUM SINK = 1 + MAX {I in IDX} I; SET NODES = (IDX cross TS) UNION {<SOURCE,0>, <SINK,n+1>}; SET ARCS = setof {i in IDX} <SOURCE,i,0> union setof {i in IDX, j in IDX diff {i}, t in TS} <i,j,t> union setof {i in IDX} <i,SINK,n> ; put NODES=; put ARCS=; NUM INCS {IDX, IDX}; NUM INCAVG {TS}; READ DATA DYINC_M INTO [FRIDX TOIDX] INCS=INC; print INCS; READ DATA HISINCAVG_M INTO [DYIDX] INCAVG=INC; print INCAVG; VAR USEARC {ARCS} BINARY; NUM COST {<i,j,t> in ARCS} = if t in 1..n-1 then (INCS[I,J] - INCAVG[T])**2; print cost; MIN TOTALCOST = SUM{<i,j,t> in ARCS} COST[i,j,t]*USEARC[i,j,t]; con FlowBalance {<i,t> in NODES}: sum {<(i),j,(t)> in ARCS} UseArc[i,j,t] - sum {<j,(i),(t)-1> in ARCS} UseArc[j,i,t-1] = (if <i,t> = <SOURCE,0> then 1 else if <i,t> = <SINK,n+1> then -1 else 0); con VisitOnce {j in IDX}: sum {<i,(j),t> in ARCS} UseArc[i,j,t] = 1; expand; solve; print {<i,j,t> in ARCS: UseArc[i,j,t].sol > 0.5} cost; QUIT; The resulting optimal solution uses the following arcs: [1] [2] [3] COST 0 1 0 0.00000000 1 3 1 0.00462806 2 5 4 0.00000000 3 4 2 0.00053412 4 2 3 0.00000032 The corresponding path is 1->3->4->2.

RobPratt · ‎08-08-2022

The standard TSP model does not allow time-dependent costs. Instead, you can solve your problem as a side-constrained shortest path problem in a directed acyclic network with 18 nodes: (0,0): this is a dummy source node (i,t) for i in 1..4 and t in 1..4 (5,5): this is a dummy sink node The arcs are from (i,t) to (j,t+1), where i in 0..4, j in 1..5, and t in 0..4, and you can compute the arc costs according to your formula. The decision variables are UseArc[i,j,t], and you need a flow balance constraint for each node, as well as a side constraint sum {<i,(j),t> in ARCS} UseArc[i,j,t] >= 1 for j in 1..4. This approach is a simpler version of the one illustrated in the "Network Formulation" section of https://support.sas.com/resources/papers/proceedings14/SAS101-2014.pdf.

RobPratt · ‎08-08-2022

The standard approach to this variant is to introduce a dummy node with zero-cost edges to the other nodes, as shown in the “A Better Lower Bound from TSP” section here: https://support.sas.com/resources/papers/proceedings14/SAS101-2014.pdf In the resulting tour, the two neighbors of the dummy node are the starting and ending nodes.

RobPratt · ‎07-15-2022

data have; input Date mmddyy10.; datalines; 1/4/2022 1/7/2022 1/8/2022 ; data want(drop=first); set have; retain first; if _N_ = 1 then first = Date; Group2 = 1 + floor((Date-first)/2); run; proc print data=want; format Date mmddyy10.; run; Obs Date Group2 1 01/04/2022 1 2 01/07/2022 2 3 01/08/2022 3

RobPratt · ‎07-12-2022

Besides what @PaigeMiller proposed, here are two alternatives: data want; set have; group1 + 1; group2 = ceil(group1/2); group3 = ceil(group1/3); run; data want; set have; group1 = _N_; group2 = ceil(_N_/2); group3 = ceil(_N_/3); run;

RobPratt · ‎07-07-2022

OK, I think I understand what you want. You cannot separately assign volume and weight for the same item. I recommend introducing a new variable to represent the proportion of the item assigned to a box, along with implicit variables to represent the volume and weight: var BoxesNeeded {ISN,BOX} >= 0 integer; var Proportion {ISN,BOX} >= 0 <= 1; impvar VolumeInsideBox {i in ISN, b in BOX} = Volume_ISN[i] * Proportion[i,b]; impvar WeightInsideBox {i in ISN, b in BOX} = Weight_ISN[i] * Proportion[i,b]; The implicit variables for costs are then as follows: impvar PerVol_Based_Costs = sum {i in ISN, b in BOX: <Org_ISN[i],Des_ISN[i],b> in PerVol_Based_Rate_NoZeroes} PerVol_Based_Rate[Org_ISN[i],Des_ISN[i],b] * VolumeInsideBox[i,b]; impvar PerWeight_Based_Costs = sum {i in ISN, b in BOX: <Org_ISN[i],Des_ISN[i],b> in PerWeight_Based_Rate_NoZeroes} PerWeight_Based_Rate[Org_ISN[i],Des_ISN[i],b] * WeightInsideBox[i,b]; And the constraints are: /* Constraints - START */ for {i in ISN, b in BOX: Is_BoxAvalable_for_a_lane[Org_ISN[i],Des_ISN[i],b] = 0} fix BoxesNeeded[i,b] = 0; con SumProportionToOne {i in ISN}: sum {b in BOX} Proportion[i,b] = 1; con Volume_Constraint {i in ISN, b in BOX}: Volume_Capacity[Org_ISN[i],Des_ISN[i],b] * BoxesNeeded[i,b] >= VolumeInsideBox[i,b]; con Weight_Constraint {i in ISN, b in BOX}: Weight_Capacity[Org_ISN[i],Des_ISN[i],b] * BoxesNeeded[i,b] >= WeightInsideBox[i,b]; con Volume_Constraint_MinThreshold {i in ISN, b in BOX}: Volume_Min[Org_ISN[i],Des_ISN[i],b] * BoxesNeeded[i,b] <= VolumeInsideBox[i,b]; con Weight_Constraint_MinThreshold {i in ISN, b in BOX}: Weight_Min[Org_ISN[i],Des_ISN[i],b] * BoxesNeeded[i,b] <= WeightInsideBox[i,b]; /* Constraints - END */ For your sample data, the resulting optimal solution has an objective value of 500,000 and uses 1x10T and 2x8T: [1] [2] BoxesNeeded Proportion VolumeInsideBox WeightInsideBox SHA_LAX_2 10T 1 0.66667 21.333 10000 SHA_LAX_2 12T 0 0.00000 0.000 0 SHA_LAX_2 20F 0 0.00000 0.000 0 SHA_LAX_2 40F 0 0.00000 0.000 0 SHA_LAX_2 40H 0 0.00000 0.000 0 SHA_LAX_2 45F 0 0.00000 0.000 0 SHA_LAX_2 8T 2 0.33333 10.667 5000 SHA_LAX_2 LCL 0 0.00000 0.000 0 SHA_LAX_2 Level0 0 0.00000 0.000 0 SHA_LAX_2 Level1 0 0.00000 0.000 0 SHA_LAX_2 Level2 0 0.00000 0.000 0 Your proposed solution of 1x10T and 1x12T would have violated the minimum volume threshold of 36 (> 32) for 12T.

RobPratt · ‎07-06-2022

Yes, I think you will need another decision variable because the costs no longer depend on just the number of boxes used. If I understand correctly, you will need to keep track of individual boxes of the same type. But I don't quite understand what your Volume_Constraint_MinThreshold and Weight_Constraint_MinThreshold constraints are intended to enforce. For example, can you please explain what "10T has a min of 2500 kgs" means?

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