SCIP Optimization Suite. Three main software –zimpl: Compiler of ZIMPL modeling language...

SCIP Optimization Suite

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SCIP Optimization Suite

• 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

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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

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max( )

mins.t.

~

i ii

ij i ji

f x cx

a x b j

=

"

å

å

5

What we need

• Parameters • Variables

• Sets

• Objective function

• Constraints

, ,i ij jc a b

ix

,i I j JÎ Î

( )f x

~ij i ji

a x bå

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Sets

• Set of numbers

• Set of strings

• Set of tuples

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Set operations

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{<1, “hi”>, <2, “hi”>, <3, “hi”>, <1, “ha”>, …}

{1, 6, 7, 8, 9}

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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]

• Exampleset I := {1, 2, 4};

set A[I] := <1> {10}, <2> {20}, <4> {30,40,50};

set B[ <i> in I] := {10 * i};

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Parameters

• set A := {1,2,3};

• param B := 10;

• param C[A] := <1> 10, <2> 20, <3> 30;

• param D[A] := <1> 100 default 0;

• param E := min A;

• param F := max <i> in A : C[i];

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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

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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;

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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;

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Objective

• set A := {1,2,3};

• param B[A] := <1> 10, <2> 20, <3> 30;

• var X[A];

• maximize obj1: sum <i> in A: X[i];

• minimize obj2: sum <i> in A: B[i] * X[i];

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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;

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Constraint

• set A := {1,2,3};

• param B[A] := <1> 10, <2> 20, <3> 30;

• var X[A];

• subto c1: forall <i> in A: X[i] <= B[i];

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Expressions

• forall expression– forall <i> in S: x[i] <= b[i];

• sum expression– sum <i> in S: x[i];

• if expression– forall <i> in S: x[i] <= if (i mod 3 == 0) then A[i]

else B[i] end;

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Example

max

s.t.

[1,20,300]

1 2 3 20,

30 20 10 200

Tc x

Ax b

c

A b

£

=

é ù é ùê ú ê ú= =ê ú ê úê ú ê úë û ë û

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set I := {1,2,3};set J := {1,2};

param c[I] := <1> 1, <2> 20, <3> 300;param A[J * I] := <1,1> 1, <1,2> 2, <1,3> 3, <2,1>

30, <2,2> 20, <2,3> 10;param b[J] := <1> 20, <2> 200;

var X[I];

maximize obj: sum <i> in I: c[i] * X[i];

subto const: forall <j> in J:sum <i> in I: A[j,i] * X[i] <= b[j];

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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

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Reading Set and Parameters from file

• “read filename as template”

• set A := {read "a.txt" as "<1n>"};

• a.txt12345

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• set A := {read "a.txt" as "<1n>"};

• param B[A] := read "b.txt" as "<1n> 2s";

• a.txt b.txt

1 1 aa

2 3 bb

3 2 cc

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max

s.t.

Tc x

Ax b£

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set I := {read "I.txt" as "<1n>"};set J := {read "J.txt" as "<1n>"};

param c[I] := read "c.txt" as "<1n> 2n";param A[J * I] := read "A.txt" as "<1n,2n> 3n";param b[J] := read "b.txt" as "<1n> 2n";

var X[I];

maximize obj: sum <i> in I: c[i] * X[i];

subto const: forall <j> in J:sum <i> in I: A[j,i] * X[i] <= b[j];

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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