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RobPratt · ‎02-12-2015

It will help me understand better if you write the desired logical relationship among V, R, A, B, C, and RV as a proposition using IF, THEN, AND, OR, and NOT, without the indexing.

RobPratt · ‎02-12-2015

The proposition if AND {j in 1..n} (P = 1) then Q = 1, where P and Q are binary variables, can be enforced by the linear constraint sum {j in 1..n} P - n + 1 <= Q. Assuming X, W, and Y are binary variables in your problem, you can express the desired proposition in this form by introducing additional binary variables (and big-M constraints) to represent each of the following three propositions: sum{<nod,(n)> in PickUpNode, p in PickSanity:p=nod and nod=1} U <= Ui Ui <= Uj Uj <= Udn Do you have bounds for U?

RobPratt · ‎02-03-2015

As Gergely points out, the financial functions are supported in PROC OPTMODEL. And the documentation contains such an example: SAS/OR(R) 13.2 User's Guide: Mathematical Programming Examples I recommend one of the following remedies: 1. trying different initial values for the variables used in the function 2. using the MULTISTART option in the SOLVE WITH NLP statement: SAS/OR(R) 13.2 User's Guide: Mathematical Programming 3. using the NOINITVAR option in the PROC OPTMODEL statement: SAS/OR(R) 13.2 User's Guide: Mathematical Programming

RobPratt · ‎01-30-2015

Assuming MS >= 0, here are three ways, in increasing order of compactness: con Mycon {i in PickUpNode}: (if i = 'Cary, NC' then 1 else 0) <= sum{<s,d> in RouteTable: s=i} MS[s,d] <= 1; con Mycon {i in PickUpNode}: (if i = 'Cary, NC' then 1 else 0) <= sum{<(i),d> in RouteTable} MS[i,d] <= 1; con Mycon {i in PickUpNode}: (i = 'Cary, NC') <= sum{<(i),d> in RouteTable} MS[i,d] <= 1; You might find these examples useful: SAS/OR(R) 13.2 User's Guide: Mathematical Programming Examples

RobPratt · ‎01-30-2015

Glad to help.

RobPratt · ‎01-30-2015

Seems like you must have declared a numeric parameter Units indexed over 4-tuples somewhere else. I think you probably just need to modify your READ DATA statements as follows, because your OPTMODEL parameter names are not the same as your data set variable names: read data ex1.PUN into PickUpNode=[I VehicleID VehicleSize] PUunits=Units; read data ex1.DUN into DropOffNode=[J VehicleID VehicleSize] revenue DUunits=Units;

RobPratt · ‎01-15-2015

Suppose the depot is node 0 and your solution satisfies x[2,5] = 1, x[5,8] = 1, and x[8,2] = 1. Then this solution contains a subtour of length 3. Disallowing x[i,j] with i = j eliminates only subtours of length 1 but allows longer subtours. So you need constraints to eliminate them. The pictures in the first (TSP) example should help solidify the concept. You want a single connected tour that contains each node exactly once. But the first several iterations of the algorithm yield a collection of disjoint subtours that together contain each node exactly once. The subtour elimination constraints prevent such subtours from arising, and these constraints are added to the problem dynamically. Once you get a solution that is a single tour, you terminate the algorithm.

RobPratt · ‎01-15-2015

You might find the following three documentation examples useful because they all demonstrate subtour elimination: SAS/OR(R) 13.2 User's Guide: Mathematical Programming SAS/OR(R) 13.2 User's Guide: Mathematical Programming Examples SAS/OR(R) 13.2 User's Guide: Mathematical Programming Examples

RobPratt · ‎01-14-2015

Glad to help. Please mark this question Answered.

RobPratt · ‎01-14-2015

Here's one way: data want(keep=ID vol:); set have; array vol[&Vol]; vol1 = frt_val; do j = 2 to &Vol-1; vol = mid_val; end; vol&Vol = bck_val; run;

RobPratt · ‎01-13-2015

First create a SAS data set as follows: data indata; input k i; datalines; 1 1 1 2 1 5 2 3 2 7 3 4 3 6 3 8 3 9 ; In PROC OPTMODEL, you can read these pairs like this: set <num,num> KI; read data indata into KI=[k i]; One approach to express the desired constraint is to use an implicit slice: set KSET = setof {<k,i> in KI} k; con Mycon {k in KSET}: sum {<(k),i> in KI, j in Pages} BlockSize * BlockAssign[i,j] <= MaxDivPrntArea ; If you insist on using DivBlocks, here is an alternative approach that uses the SETOF operator again: set DivBlocks {k in KSET} = setof {<(k),i> in KI} i; con Mycon {k in KSET}: sum {i in DivBlocks , j in Pages} BlockSize * BlockAssign[i,j] <= MaxDivPrntArea ; A third approach passes through the KI set only once to construct both KSET and DivBlocks: set KSET init {}; set DivBlocks {KSET} init {}; for {<k,i> in KI} do; KSET = KSET union {k}; DivBlocks = DivBlocks union {i}; end; con Mycon {k in KSET}: sum {i in DivBlocks , j in Pages} BlockSize * BlockAssign[i,j] <= MaxDivPrntArea ; The doc link referenced in my previous reply contains numerous relevant examples. Search for "sets indexed by other sets" in the pdf.

RobPratt · ‎01-12-2015

The index set for MaxDivPrntArea should match (or be a subset of) the index set for your constraint. More explicitly, suppose you have a SAS data set called mydata that contains k and MaxDivPrntArea. Then you need something like this: set KSET; num MaxDivPrntArea {KSET}; read data mydata into KSET= MaxDivPrntArea; con Mycon {k in KSET}: sum {i in DivBlocks , j in Pages} BlockSize * BlockAssign[i,j] <= MaxDivPrntArea ; Or are you having trouble creating the mydata SAS data set itself? In any case, you might find it useful to read some of the examples here: SAS/OR(R) 13.2 User's Guide: Mathematical Programming Examples

RobPratt · ‎12-21-2014

Instructions can be found here: https://communities.sas.com/message/242631#242631

RobPratt · ‎12-17-2014

The logical condition has to appear last, after all index-set-items: SAS/OR(R) 13.2 User's Guide: Mathematical Programming Here are two ways to accomplish what you want: sum{<o, d> in ARCS, p in PRODUCTS: o in SET_A} sum{<o, d> in ARCS: o in SET_A} sum{p in PRODUCTS}

RobPratt · ‎12-17-2014

You will probably see better performance if you omit the z_CON constraints (which seem to make the problem infeasible) and rewrite the objective function in a way that leads to a sparse Hessian: *min MSE = sum{r in RGN_I} (z - VALUE_TGT )**2; var Error {RGN_I}; con Error_CON {r in RGN_I}: Error = z - VALUE_TGT ; min MSE = sum{r in RGN_I} Error **2;

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