The following works fine for me. Maybe you have not declared your index set Slope1 properly. (You will need to change my declarations if some of your data set variables are strings instead of numbers.) data Slope1; input Physician Product Linear_piece Slope; datalines; 1 2 3 4 5 6 7 8 ; proc optmodel; set <num,num,num> Slope1; num slope {Slope1}; read data Slope1 into Slope1 = [Physician Product Linear_piece] Slope; /* Slope[Physician, Product, Linear_piece] = col("Slope");*/ print slope; quit;
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Since PROC MODEL is part of SAS/ETS and not SAS/OR, you might have better luck in the Forecasting and Econometrics community: https://communities.sas.com/community/support-communities/sas_forecasting
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I'll be glad to help, but I still need more details. What does your constraint that involves expmatrix look like? It looks like you have overloaded z as the name your objective function and as the third argument of expmatrix.
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FCMP enablement has actually been hidden for several releases and is officially supported in 13.1. You might try using it with your earlier release and see if it just works. Using SUBMIT instead would require an indirect approach with multiple solver calls. And I would need more explicit details about your intended constraint.
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Now that PROC OPTMODEL supports FCMP functions and routines, I think the examples from this post will help you: https://communities.sas.com/message/146656#146656 Note that expmatrix is one of the routines provided by PROC FCMP. The SUBMIT block is useful, but in this case I think FCMP support is the more appropriate OPTMODEL feature for your needs.
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I'm not quite sure what you mean by embedding a constraint. The job shop example shown by Ed uses SUBMIT to call PROC CLP, and the resulting output provides information that can be used to add valid constraints to the MILP model. Can you provide some more detail about your use case?
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Here's an example that uses a SUBMIT block to call a DATA step and PROC GPLOT to plot intermediate solutions during a classical row-generation approach to the TSP: SAS/OR(R) 13.1 User's Guide: Mathematical Programming If that example isn't helpful, maybe you can describe what you had in mind.
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Your "link" constraints enforce only one side of the implication (if diet > 0 then IsUsed = 1) and the solution does respect that. If you also want the converse (if IsUsed =1 then diet > 0), then you need another set of constraints. You also need to decide how small of a value for diet is considered to be positive. For example, would you accept diet = 1e-6 as positive? Suppose that your tolerance is called epsilon. Then the following constraints should do what you want: con link2 {al in FOOD}: diet[al] >= epsilon * IsUsed[al];
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For each triple (j1,j2,j3) where j1 < j2 < j3, the constraints I suggested exclude x[j1] = 1, x[j2] = 0, x[j3] = 1, even if j1, j2, and j3 are not consecutive. Your example 1 0 0 1 1 is excluded by the constraint indexed by (j1,j2,j3) = (1,2,4). (It is also excluded by a few other triples.)
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Here's a way to do it by using PROC OPTMODEL in SAS/OR:
proc optmodel;
num n = 5;
var x {1..n} binary;
var y binary;
con ZeroOrThree:
sum {j in 1..n} x[j] = 3*y;
/* exclude (1,...,0,...,1) */
con Consecutive {j1 in 1..n-2, j2 in j1+1..n-1, j3 in j2+1..n}:
x[j1] + (1 - x[j2]) + x[j3] <= 2;
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The first two equations are true. The third equation requires cost['gooda'] and cost['goodb'] to be defined. You could assign those costs to be 0, and then the third equation would be true. Alternatively, you could modify the sum in the third equation in any of the following three ways: sum{p in gooditem inter products} cost sum{p in gooditem: p in products} cost sum{p in gooditem} (if p in products then cost ) Each of these approaches would yield 3.
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You would just need a second SET and READ DATA for the other data set: set<string> subproducts_set; read data subproducts_table into subproducts_set=[subproducts];
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Here's one way to do it with the MILP solver in PROC OPTMODEL, by minimizing the absolute error between the sum and the target: data indata; input wages; datalines; 14611 250125 2000135 850 502525 55458 248788 2211 11252458 2251215 ; %let target = 11520255; proc optmodel; set RECORDS; num wages {RECORDS}; read data indata into RECORDS=[_N_] wages; var UseRecord {RECORDS} binary; var Surplus >= 0; var Slack >= 0; min Error = Surplus + Slack; con Error_con: sum {i in RECORDS} wages * UseRecord - Surplus + Slack = ⌖ solve; print wages UseRecord; quit;
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