Mixed-Integer linear programming

From Optimization for Chemical Engineering: 4G3
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Class date(s): 25 March 2015
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Resources

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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

30 March 12B

More problems that can be solved with integer variables

  • Allocation problems
  • Set covering problems
  • Choice of locations

Solutions methods of problems with integer variables

  • Introducing the branch and bound procedure


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;