A linear programming problem or simply linear program (LP) consists of:
The goal is to assign values to the variables so as to maximize (or minimize) the value of the objective function while satisfying all constraints.
Many optimization problems can be modeled in this way. As one basic
example, consider a knapsack with fixed capacity C, and a number of
items with sizes s(i) and values v(i). The
goal is to put as many items as possible in the knapsack (not exceeding
its capacity) while maximizing the sum of their values.
As another example, suppose you are given a set of coins with certain values, and you are to find the minimum number of coins such that their values sum up to a fixed amount. Instances of these problems are solved below.
Solving an LP or integer linear program (ILP) with this library typically comprises 4 stages:
The most frequently used predicates are thus:
Left Op C,
where Left is a list of Coefficient*Variable
terms (variables in the context of linear programs can be atoms or
compound terms) and C is a non-negative numeric constant. The
list represents the sum of its elements. Op can be =, =<
or >=. The coefficient 1 can be omitted. An
integrality constraint is of the form integral(Variable)
and constrains Variable to an integral value.Coefficient*Variable terms that represents the sum of its
elements, with respect to the linear program corresponding to state
S0. \arg{S} is unified with an
internal representation of the solved instance.All numeric quantities are converted to rationals via rationalize/1, and rational arithmetic is used throughout solving linear programs. In the current implementation, all variables are implicitly constrained to be non-negative. This may change in future versions, and non-negativity constraints should therefore be stated explicitly.
Delayed column generation means that more constraint columns are added to an existing LP. The following predicates are frequently used when this method is applied:
Coefficient*Variable terms.
The terms are added to the left-hand side of the constraint named Name. S
is unified with the resulting state.An example application of delayed column generation to solve a bin packing task is available from: metalevel.at/various/colgen/
The following predicates allow you to solve specific kinds of LPs more efficiently:
We include a few examples for solving LPs with this library.
This is the "radiation therapy" example, taken from Introduction to Operations Research by Hillier and Lieberman.
Prolog DCG notation is used to implicitly thread the state through posting the constraints:
:- use_module(library(simplex)).
radiation(S) :-
gen_state(S0),
post_constraints(S0, S1),
minimize([0.4*x1, 0.5*x2], S1, S).
post_constraints -->
constraint([0.3*x1, 0.1*x2] =< 2.7),
constraint([0.5*x1, 0.5*x2] = 6),
constraint([0.6*x1, 0.4*x2] >= 6),
constraint([x1] >= 0),
constraint([x2] >= 0).
An example query:
?- radiation(S), variable_value(S, x1, Val1),
variable_value(S, x2, Val2).
Val1 = 15 rdiv 2,
Val2 = 9 rdiv 2.
Here is an instance of the knapsack problem described above, where
C = 8, and we have two types of items: One item with value
7 and size 6, and 2 items each having size 4 and value 4. We introduce
two variables, x(1) and x(2) that denote how
many items to take of each type.
:- use_module(library(simplex)).
knapsack(S) :-
knapsack_constraints(S0),
maximize([7*x(1), 4*x(2)], S0, S).
knapsack_constraints(S) :-
gen_state(S0),
constraint([6*x(1), 4*x(2)] =< 8, S0, S1),
constraint([x(1)] =< 1, S1, S2),
constraint([x(2)] =< 2, S2, S).
An example query yields:
?- knapsack(S), variable_value(S, x(1), X1),
variable_value(S, x(2), X2).
X1 = 1
X2 = 1 rdiv 2.
That is, we are to take the one item of the first type, and half of one of the items of the other type to maximize the total value of items in the knapsack.
If items can not be split, integrality constraints have to be imposed:
knapsack_integral(S) :-
knapsack_constraints(S0),
constraint(integral(x(1)), S0, S1),
constraint(integral(x(2)), S1, S2),
maximize([7*x(1), 4*x(2)], S2, S).
Now the result is different:
?- knapsack_integral(S), variable_value(S, x(1), X1),
variable_value(S, x(2), X2).
X1 = 0
X2 = 2
That is, we are to take only the two items of the second type. Notice in particular that always choosing the remaining item with best performance (ratio of value to size) that still fits in the knapsack does not necessarily yield an optimal solution in the presence of integrality constraints.
We are given:
The task is to find a minimal number of these coins that
amount to 111 units in total. We introduce variables c(1), c(5)
and c(20) denoting how many coins to take of the respective
type:
:- use_module(library(simplex)).
coins(S) :-
gen_state(S0),
coins(S0, S).
coins -->
constraint([c(1), 5*c(5), 20*c(20)] = 111),
constraint([c(1)] =< 3),
constraint([c(5)] =< 20),
constraint([c(20)] =< 10),
constraint([c(1)] >= 0),
constraint([c(5)] >= 0),
constraint([c(20)] >= 0),
constraint(integral(c(1))),
constraint(integral(c(5))),
constraint(integral(c(20))),
minimize([c(1), c(5), c(20)]).
An example query:
?- coins(S), variable_value(S, c(1), C1),
variable_value(S, c(5), C5),
variable_value(S, c(20), C20).
C1 = 1,
C5 = 2,
C20 = 5.