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Published byKiara Ibbotson Modified about 1 year ago

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SCIP Optimization Suite

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Three main software –zimpl: Compiler of ZIMPL modeling language –soplex: LP solver (implementation of Simplex) –scip: An advanced implementation of B&B to solve ILP All these are available in single packages –SCIP optimization suite –Source code –zimpl, soplex, scip are standalone applications –Binary (Linux & Windows) –Single executable scip application that is linked by zimpl & soplex 2

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How Does It Work? As a programming library –It has API, you can call the functions in C, C++ As a standalone solver –Develop a model in zimpl language –Compile/Translate your model: zimpl model.zpl –Solve it LP problems: soplex model.lp ILP, MIP problems: scip -f model.lp –scip by itself calls zimpl if the input file is not.lp scip -f model.zpl 3

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What we need Parameters Variables Sets Objective function Constraints 5

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Sets Set of numbers Set of strings Set of tuples 6

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Set operations 7

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{,,,, …} {1, 6, 7, 8, 9} 8

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Indexed Sets The arrays of ZIMPL Each element has its own index –A set indexes another set Refer to i-th element by S[i] Example set I := {1, 2, 4}; set A[I] := {10}, {20}, {30,40,50}; set B[ in I] := {10 * i}; 9

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Parameters set A := {1,2,3}; param B := 10; param C[A] := 10, 20, 30; param D[A] := 100 default 0; param E := min A; param F := max in A : C[i]; 10

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These operations are used in zimpl models to generate the numerical model. Most operations are applicable only on parameters, cannot be used for variables, because they are not linear 11

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Variables “real”, “binary”, or “integer” –Default is “real” var x1; var x2 integer; var x3 binary; set A := {1,2,3}; var x4[A] real; var x5[A * A] integer >=0 <= 10; 12

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Objective “maximize” or “minimize” var x1; var x2; var x3; maximize obj1: x1 + x2 + x3; minimize obj2: 2*x1 + 3*x2; 13

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Objective set A := {1,2,3}; param B[A] := 10, 20, 30; var X[A]; maximize obj1: sum in A: X[i]; minimize obj2: sum in A: B[i] * X[i]; 14

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Constraint subto name: constraint –“ ”, “==“ –There is not “>” and “<“ subto c1: x1 <= 10; subto c2: x1 + x2 <= 20; subto c3: x1 + x2 + x3 == 100; 15

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Constraint set A := {1,2,3}; param B[A] := 10, 20, 30; var X[A]; subto c1: forall in A: X[i] <= B[i]; 16

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Expressions forall expression –forall in S: x[i] <= b[i]; sum expression –sum in S: x[i]; if expression –forall in S: x[i] <= if (i mod 3 == 0) then A[i] else B[i] end; 17

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Example 18

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set I := {1,2,3}; set J := {1,2}; param c[I] := 1, 20, 300; param A[J * I] := 1, 2, 3, 30, 20, 10; param b[J] := 20, 200; var X[I]; maximize obj: sum in I: c[i] * X[i]; subto const: forall in J: sum in I: A[j,i] * X[i] <= b[j]; 19

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Realistic Problems A model for the problem Multiple instances –Each instance has its own data Separation between model and data –Create a general model –Read data from file 20

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Reading Set and Parameters from file “read filename as template” set A := {read "a.txt" as " "}; a.txt 1 2 3 4 5 21

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set A := {read "a.txt" as " "}; param B[A] := read "b.txt" as " 2s"; a.txtb.txt 11 aa 23 bb 32 cc 22

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set I := {read "I.txt" as " "}; set J := {read "J.txt" as " "}; param c[I] := read "c.txt" as " 2n"; param A[J * I] := read "A.txt" as " 3n"; param b[J] := read "b.txt" as " 2n"; var X[I]; maximize obj: sum in I: c[i] * X[i]; subto const: forall in J: sum in I: A[j,i] * X[i] <= b[j]; 24

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Integrality Complexity: An Example Consider the LP problem –I = 1000 –J = 100 If variables are “real” – Solution time = 0.21 sec. – Objective = 1727.05 If variables are “integer” –Solution time = 22.85 sec. –Objective = 1724 26

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