Author: Markus Triska
A linear programming problem consists of a set of (linear) constraints, a number of variables and a linear objective function. 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. Consider having 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.
The library(simplex) module provides the following
predicates:
:- use_module(library(simplex)).
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).
radiation(S) :-
gen_state(S0),
post_constraints(S0, S1),
minimize([0.4*x1, 0.5*x2], S1, S).
An example query:
?- radiation(S), variable_value(S, x1, Val1),
variable_value(S, x2, Val2).
Val1 = 15 rdiv 2
Val2 = 9 rdiv 2 ;
knapsack_constrain(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).
knapsack(S) :-
knapsack_constrain(S0),
maximize([7*x(1), 4*x(2)], S0, 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_constrain(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.
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)]).
coins(S) :-
gen_state(S0),
coins(S0, S).
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