Scryer Prolog documentation

Introduction {#simplex-intro}

A linear programming problem or simply linear program (LP) consists of:

• a set of linear constraints

• a set of variables

• 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. 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.

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.

Example 1 {#simplex-ex-1}

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)).
:- use_module(library(dcgs)).

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).
   S = solved(...), Val1 = 15 rdiv 2, Val2 = 9 rdiv 2.

Example 2 {#simplex-ex-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).
   S = solved(...), X1 = 1 rdiv 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.

Example 3 {#simplex-ex-3}

We are given:

• 3 coins each worth 1 unit

• 20 coins each worth 5 units and

• 10 coins each worth 20 units.

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).
   S = solved(...), C1 = 1 rdiv 1, C5 = 2 rdiv 1, C20 = 5 rdiv 1.

@author Markus Triska