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## A.33 library(simplex): Solve linear programming problems

- author
- Markus Triska

### A.33.1 Introduction

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.

Solving an LP or integer linear program (ILP) with this library typically comprises 4 stages:

- an initial state is generated with gen_state/1
- all relevant constraints are added with constraint/3
- maximize/3 or minimize/3
are used to obtain a
*solved state*that represents an optimum solution - variable_value/3 and objective/2 are used on the solved state to obtain variable values and the objective function at the optimum.

The most frequently used predicates are thus:

**gen_state**(`-State`)- Generates an initial state corresponding to an empty linear program.
**constraint**(`+Constraint, +S0, -S`)- 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. **maximize**(`+Objective, +S0, -S`)- 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`.`\`

arg{`S`} is unified with an internal representation of the solved instance. **minimize**(`+Objective, +S0, -S`)- Analogous to maximize/3.
**variable_value**(`+State, +Variable, -Value`)`Value`is unified with the value obtained for`Variable`.`State`must correspond to a solved instance.**objective**(`+State, -Objective`)- Unifies
`Objective`with the result of the objective function at the obtained extremum.`State`must correspond to a 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.

### A.33.2 Delayed column generation

*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:

**constraint**(`+Name, +Constraint, +S0, -S`)- Like constraint/3,
and attaches the name
`Name`(an atom or compound term) to the new constraint. **shadow_price**(`+State, +Name, -Value`)- Unifies
`Value`with the shadow price corresponding to the linear constraint whose name is`Name`.`State`must correspond to a solved instance. **constraint_add**(`+Name, +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.

An example application of *delayed column generation* to solve a *bin
packing* task is available from:
**metalevel.at/various/colgen/**

### A.33.3 Solving LPs with special structure

The following predicates allow you to solve specific kinds of LPs more efficiently:

**transportation**(`+Supplies, +Demands, +Costs, -Transport`)- Solves a transportation problem.
`Supplies`and`Demands`must be lists of non-negative integers. Their respective sums must be equal.`Costs`is a list of lists representing the cost matrix, where an entry (*i*,*j*) denotes the integer 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*to*j*. **assignment**(`+Cost, -Assignment`)- Solves a linear assignment problem.
`Cost`is a list of lists representing the quadratic cost matrix, where element (i,j) denotes the integer 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 adjacency matrix.

### A.33.4 Examples

We include a few examples for solving LPs with this library.

#### A.33.4.1 Example 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)). 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.

#### A.33.4.2 Example 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.

#### A.33.4.3 Example 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). C1 = 1, C5 = 2, C20 = 5.