Difference between revisions of "Mixed-Integer linear programming"

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If you do this, your nodes will have the following solution order:
If you do this, your nodes will have the following solution order:


Node 0 \((z=82.8)\)
Node 0 \((z=82.80)\)
Node 1 \((z=80.67)\)
 


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Revision as of 17:10, 6 April 2015

Class date(s): 25 March 2015
Download video: Link [787 M]

Download video: Link [905 M]

Download video: Link [884 M]

Resources

Scroll down, if necessary, to see the resources.

Date Class number Topic Slides/handouts for class Video file References and Notes
25 March 11B
  • Representation of integer variables
  • Types of problems that can be solved with integer variables

Handout from class

Video

See the GAMS codes below

30 March 12A

More problems that can be solved with integer variables

  • Knapsack problem
  • Mutual exclusivity constraints
  • Dependence constraints
  • Allocation problems

Handout from class

Video

See the GAMS codes below

01 April 12B

The branch and bound procedure to solve integer problems

Handout from class

Sequence of the branch and bound steps

Video

See code below to solve the original problem, and then the relaxed problem.

See another example problem below (with solution).

06 April 13A

Working towards understanding schedule problems in engineering

  • Gantt chart approach
  • What the search variables mean
  • What objective function choices exist
  • Setting up the integer (disjunctive) variables constraints

Handout from class

See code below to simple scheduling problem.

08 April 13B

More problems that can be solved with integer variables:

  • Scheduling problems
  • Choice of locations
  • Travelling salesmen problem


Solving a basic ILP

free variable        income    "total income";
positive variables   x1, x2;
binary variable      delta     "use ingredient x3 or not at all";

EQUATIONS
obj    "maximize income",
blend  "blending constraint";
obj..   income =E= 18*x1 - 3*x2 - 9*(20*delta);
blend.. 2*x1 + x2 + 7*(20*delta) =L= 150;
x1.up = 25;
x2.up = 30;

model recipe /all/;
SOLVE recipe using MIP maximizing income;

Solving the knapsack problem

free variable        value "total value";
sets                 j     "item j" /1*5/;
binary variables     x(j)  "whether to include item in the knapsack";

parameter   v(j) "value of object j"
/1   4,
 2   2,
 3   10,
 4   1,
 5   2/;

parameter   w(j) "weight of object j"
/1  12,
 2   1,
 3   4,
 4   1,
 5   2/;

EQUATIONS
obj     "maximize value",
weight  "weight constraint";

obj..    value =e= sum(j, v(j)*x(j));
weight.. sum(j, w(j)*x(j)) =L= 15;

model knapsack /all/;
solve knapsack using mip maximizing value;

Solving the knapsack problem for selecting amount projects, with constraints

free variable        value "total value";
sets                 j     "item j" /1*5/;
binary variables     x(j)  "whether to include item in the project";

parameter   v(j) "value of object j"
/1   50,
 2   72,
 3   25,
 4   41,
 5   17/;

parameter   w(j) "cost of object j"
/1   8,
 2   21,
 3   15,
 4   10,
 5   7/;

EQUATIONS
obj     "maximize value",
weight  "weight constraint",
me1v2   "1 and 2 are mutually exclusive",
me1v3v5 "1, 3 and 5 are mutually exclusive",
d4v3    "4 depends on 3";

obj..    value =e= sum(j, v(j)*x(j));
weight.. sum(j, w(j)*x(j)) =L= 35;

me1v2..    x('1') + x('2') =L= 1;
me1v3v5..  x('1') + x('3') + x('5') =L= 1;
d4v3..     x('4') =L= x('3');

model projects /all/;
solve projects using mip maximizing value;

Binary problem relaxation

This is the problem we looked at in class for the generators.

Notice that the only differences are:

  • the binary variables become positive variables, thepr
  • the problem is solved as an LP not an MIP
  • the constraints for .LO and .UP are added, where required, depending on the partial solution solved
Original problem solved in class Relaxed problem solved in class
free variable        cost "total cost";
sets                 j     "item j" /1*4/;
binary variables     x(j)  "whether to use generator";

parameter   c(j) "cost of using generator j"
/1   7,
 2   12,
 3   5,
 4   14/;

parameter   p(j) "power of generator j"
/1   300,
 2   600,
 3   500,
 4   1600/;

EQUATIONS
obj    "cost",
power_constraint;

obj..      cost =e= sum(j, c(j)*x(j));
power_constraint..  sum(j, p(j)*x(j)) =G= 700;

MODEL purchase /all/;
SOLVE purchase using MIP minimizing cost;
free variable        cost "total cost";
sets                 j     "item j" /1*4/;
positive variables   x(j)  "whether to use generator";

parameter   c(j) "cost of using generator j"
/1   7,
 2   12,
 3   5,
 4   14/;

parameter   p(j) "power of generator j"
/1   300,
 2   600,
 3   500,
 4   1600/;

EQUATIONS
obj    "cost",
power_constraint;

obj..      cost =e= sum(j, c(j)*x(j));
power_constraint..  sum(j, p(j)*x(j)) =G= 700;

* Solve the problem for the partial solution: (#, 0, #, 1)
x.LO('1')=0;
x.UP('1')=1;

x.LO('2')=0;
x.UP('2')=0;

x.LO('3')=0;
x.UP('3')=1;

x.LO('4')=1;
x.UP('4')=1;

MODEL purchase /all/;
SOLVE purchase using LP minimizing cost;

Example problem to practice

The objective is to minimize the objective function value, \(z\). The table or partial solutions is provided below. The only constraints are that all the 3 search variables must be binary (0 or 1). The "relaxed" refers to the relaxed problem solution.

4G3-2015-MIP-Practice-problem.png

Use the depth-first search method, and when branching, choose the branch with \(x_i = 0\) before the \(x_i=1\) branch.

If you do this, your nodes will have the following solution order:

Node 0 \((z=82.80)\) Node 1 \((z=80.67)\)