Mixed-Integer linear programming

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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

01 April 12B

The branch and bound procedure to solve integer problems

Handout from class

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

06 April 13A

More problems that can be solved with integer variables

  • Allocation problems
  • Set covering problems
  • Choice of locations
  • Travelling salesmen problem
  • Scheduling 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

Here is an example of an unusual case, where the binary problem (original) has the same optimum as the relaxed problem (LP). In this case the LP solution happens to be at values of 0 and 1 for the search variables, so this relaxed problem has the same optimum as the binary problem that it relaxed.

Original problem Relaxed problem
free variable        cost "total cost";
sets                 j     "item j" /1*4/;
binary variables     x(j)  "whether to include item in the purchase";

parameter   c(j) "cost of object j"
/1   3,
 2   4,
 3   6,
 4   14/;

EQUATIONS
obj    "cost",
LP_constraint,
IP_constraint,
NLP_constraint;

obj..    cost =e= sum(j, c(j)*x(j));
LP_constraint..  sum(j, x(j))    =G= 1;
IP_constraint..  x('2') + x('4') =G= 1;
NLP_constraint.. x('3') + x('4') =G= 1;

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 include item in the purchase";

parameter   c(j) "cost of object j"
/1   3,
 2   4,
 3   6,
 4   14/;

EQUATIONS
obj    "cost",
LP_constraint,
IP_constraint,
NLP_constraint;

obj..    cost =e= sum(j, c(j)*x(j));
LP_constraint..  sum(j, x(j))    =G= 1;
IP_constraint..  x('2') + x('4') =G= 1;
NLP_constraint.. x('3') + x('4') =G= 1;

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

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;