data data1;
input person $ next $;
datalines;
A B
B C
;
data data2;
input Rank Person1 $ Money;
datalines;
1 A 100
1 A 120
2 A 120
1 B 110
1 B 120
2 B 120
;
proc optmodel;
set <str> PEOPLE;
str next {PEOPLE};
read data data1 into PEOPLE=[person] next;
set<num, str, num> rank_per_mny;
read data data2 into rank_per_mny = [Rank Person1 Money];
put (setof {<r,p,m> in rank_per_mny, <(r),next[p],m2> in rank_per_mny} <r,p,m,m2>)=;
quit;
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Glad to help.
Yes, you can relax the new variable, and this idea is mentioned in the doc example:
"In both formulations, the Product variable can be relaxed to be nonnegative instead of binary. The integrality of Assign , together with the various Product_def* constraints, automatically implies integrality of Product . For real-world problems, you should try both ways to determine which alternative performs better in specific cases."
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You are asking about a product of binary variables, not a product of sets. See this doc example that illustrates how to linearize this product.
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You have not specified any integer variables. Use the INTEGER option in the VAR statement and then the .RELAX suffix to relax x[2]. Also, you need not enter the zero coefficients explicitly. In fact, you can avoid those last three constraints by instead using the .UB variable suffix:
proc optmodel;
var x{1..3}>=0 integer;
x[2].relax = 1;
max z=900*x[1]+1500*x[2]+1000*x[3];
con c1: 50000*x[1]+12000*x[2]+8000*x[3]<=250000;
x[1].ub = 4;
x[2].ub = 15;
x[3].ub = 20;
solve with milp/presolver=none;
print x;
quit;
If you change the objective coefficient for x[1] from 900 to 9000, you do get an optimal objective value of 42,250.
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The colon : operator (which you can think of as "such that") allows you to select a subset based on a logical expression:
http://go.documentation.sas.com/#!/?docsetId=ormpug&docsetVersion=14.2&docsetTarget=ormpug_optmodel_details63.htm&locale=en
The following expressions are all equivalent and count the number of customers whose product is A or B:
sum {c in customers} (product[c] IN {'A', 'B'})
sum {c in customers} (if product[c] IN {'A', 'B'} then 1 else 0)
sum {c in customers} (if product[c] IN {'A', 'B'} then 1)
sum {c in customers: product[c] IN {'A', 'B'}} 1
card({c in customers: product[c] IN {'A', 'B'})
See also my reply here, which is similar:
https://communities.sas.com/t5/Mathematical-Optimization/Optimize-Traveling-Distance/td-p/308817/page/2
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The CREATE DATA statement has a LENGTH= option:
http://go.documentation.sas.com/#!/?docsetId=ormpug&docsetVersion=14.2&docsetTarget=ormpug_optmodel_syntax11.htm&locale=en#ormpug.optmodel.npxcreatestmt
Correct syntax for your sum is:
sum {c in customers: product[c] IN {'A', 'B'}}
For new users of PROC OPTMODEL, I recommend this book of examples.
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Did you try this out? The following valid cuts seem to help by strengthening the master problem:
/* additional valid cuts */
/* smallest positive distance from p */
num spd {p in ORDERS} = min {w in LOCATIONS_p[p] diff {p}} distance[p,w];
/* if On[p] = 0 then Eta[p] >= spd[p] */
con OnOrPositiveDistance {p in ORDERS}:
Eta[p] >= spd[p] * (1 - On[p]);
problem Master include
On Eta
MasterObjective
Cover
max_LOCATIONs_on
OptimalityCut OnOrPositiveDistance;
For scenario 3, the optimal objective value is between 142.8M and 144.8M.
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I have a couple of suggestions for improvement.
1. An arbitrarily large value like 100000 for M can cause numerical difficulties for optimization solvers. In this case, the supplycons constraint implies that X[d,c] <= supply[d], so you can instead use the following tighter constraint:
con conserv {d in DC, c in City}:X[d,c]-supply[d]*Y[d]<=0;
2. You can in fact avoid the conserv constraint altogether if you tighten the supplycons constraint as follows:
con supplycons {d in DC}:sum{c in City} X[d,c]<=supply[d]*Y[d];
The resulting model (without conserv and with the new supplycons) is tighter, smaller, and more numerically stable. For your small instance, even the original model solves quickly. But these best practices can make a big difference for larger instances.
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Assuming D is bounded both above and below, you can linearize this as follows:
var X binary;
con C1:
Ad <= D;
/* if X = 1 then D <= 0 else redundant */
con C2:
D <= D.ub * (1 - X);
/* if X = 1 then D <= Ad else redundant */
con C3:
D - Ad <= (D.ub - D.lb) * (1 - X);
/* if X = 0 then D >= 0 else redundant */
con C4:
D >= D.lb * X;
/* if X = 0 then Ad >= 0 else redundant */
con C5:
Ad >= D.lb * X;
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You are missing some angle brackets and parentheses. Try this:
con sbcls_dsp{<grp,dpt,cls,sbcls> in itm_hrchy}:
sum{<(grp),(dpt),(cls),(sbcls),ssn,itm,dsp,prc> in dsp_prc_choice}
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