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
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
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
library(simplex) module provides the following
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.
This is the "radiation therapy" example, taken from "Introduction to
Operations Research" by Hillier and Lieberman. DCG notation is used to
implicitly thread the state through posting the constraints:
Cost is a list of lists representing the quadratic cost
matrix, where element (i,j) denotes the cost of assigning entity i
to entity j. An assignment with minimal cost is computed and
unified with Assignment as a list of lists, representing an
Adds a linear or integrality constraint to the linear program
corresponding to state S0. A linear constraint is of the form
"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.
+Constraint, +S0, -S)
and attaches the name Name (an atom or compound term) to the
+Left, +S0, -S)
Left is a list of Coefficient*Variable terms. The terms are
added to the left-hand side of the constraint named
Name. S is unified with the resulting state.
Generates an initial state corresponding to an empty linear program.
Maximizes the objective function, stated as a list of
"Coefficient*Variable" terms that represents the sum of its elements,
with respect to the linear program corresponding to state S0. S
is unified with an internal representation of the solved instance.
Analogous to maximize/3.
Unifies Objective with the result of the objective function
at the obtained extremum. State must correspond to a solved
Unifies Value with the shadow price corresponding to the
linear constraint whose name is Name. State must
correspond to a solved instance.
+Demands, +Costs, -Transport)
Supplies and Demands are both lists of positive
numbers. Their respective sums must be equal. Costs is a list
of lists representing the cost matrix, where an entry (i,j) denotes the
cost of transporting one unit from i to j. A
transportation plan having minimum cost is computed and unified with Transport
in the form of a list of lists that represents the transportation
matrix, where element (i,j) denotes how many units to ship from i
Value is unified with the value obtained for
Variable. State must correspond to a solved
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).
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 ;
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.
constraint([6*x(1), 4*x(2)] =< 8, S0, S1),
constraint([x(1)] =< 1, S1, S2),
constraint([x(2)] =< 2, S2, S).
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
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
We are given 3 coins each worth 1, 20 coins each worth 5, and 10 coins
each worth 20 units of money. The task is to find a minimal number of
these coins that amount to 111 units of money. We introduce variables
c(1), c(5) and c(20) denoting how many coins to take of the respective
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),
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