Linear Programmnig (LP)

minimize  c^T x + C
s.t.      Gx <= h
          Ax == b
          lb <= x <= ub
          x_i is integer or continuous variablce

This optimization problem is called Linear Programming (LP) Problem. For example, the following problem is one of the LP.

from flopt import Variable, Problem

# variables
a = Variable(name="a", lowBound=0, upBound=1, cat="Integer")
b = Variable(name="b", lowBound=1, upBound=2, cat="Continuous")
c = Variable(name="c", lowBound=1, upBound=3, cat="Continuous")

# problem
prob = Problem(name="LP")
prob += a + b + c + 2
prob += a + b == 2
prob += b - c <= 3

print(prob)
>>> Name: None
>>>   Type         : Problem
>>>   sense        : minimize
>>>   objective    : a+b+c+2
>>>   #constraints : 2
>>>   #variables   : 3 (Continuous 2, Integer 1)

flopt to LP

We can convert this into Lp form as follows.

from flopt.convert import LpStructure
lp = LpStructure.fromFlopt(prob)

To show the contents of lp,

print(lp.show())
>>> LpStructure
>>> obj  c.T.dot(x) + C
>>> s.t. Gx <= h
>>>      Ax == b
>>>      lb <= x <= ub

>>> #x
>>> 3

>>> c
>>> [1. 1. 1.]

>>> C
>>> 2

>>> G
>>> [[ 0.  1. -1.]]

>>> h
>>> [3.]

>>> A
>>> [[1. 1. 0.]]

>>> b
>>> [2.]

>>> lb
>>> [0. 1. 1.]

>>> ub
>>> [1. 2. 3.]

>>> x
>>> [Variable("a", 0, 1, "Integer", 0) Variable("b", 1, 2, "Continuous", 1.5)
>>>  Variable("c", 1, 3, "Continuous", 2.0)]

Formulation with only equal constraints

You can obtain the formulaton with only eqaual constraints by .toAllEq()

minimize  c^T x + C
s.t.      Ax == b
          lb <= x <= ub
          x_i is integer or continuous variablce

print(lp.toAllEq())
>>> LpStructure
>>>   #x  4
>>>   #c  (4,)
>>>   #C  2
>>>   #G  None  (0-element None %)
>>>   #h  None
>>>   #A  (2, 4)  (0-element 37.500 %)
>>>   #b  (2,)
>>>   #lb 4
>>>   #ub 4

To make the formulation easier to read, we show it in the form of flopt.

print(lp.toAllEq().toFlopt().show())
>>> Name: None
>>>   Type         : Problem
>>>   sense        : minimize
>>>   objective    : a+b+c+2
>>>   #constraints : 2
>>>   #variables   : 4 (Continuous 3, Integer 1)
>>>
>>>   C 0, name None, a+b-2.0 == 0
>>>   C 1, name None, b-c+__s_0-3.0 == 0

__s_0 is a slack variable for an equal constraint.

Formulation with only non-equal constraints

You can obtain the formulaton with only non-eqaual constraints by .toAllNeq()

minimize  c^T x + C
s.t.      Gx <= h
          lb <= x <= ub
          x_i is integer or continuous variablce

print(lp.toAllNeq())
>>> LpStructure
>>>   #x  3
>>>   #c  (3,)
>>>   #C  2
>>>   #G  (3, 3)  (0-element 33.333 %)
>>>   #h  (3,)
>>>   #A  None  (0-element None %)
>>>   #b  None
>>>   #lb 3
>>>   #ub 3

To make the formulation easier to read, we show it in the form of flopt.

print(lp.toAllNeq().toFlopt().show())
>>> Name: None
>>>   Type         : Problem
>>>   sense        : minimize
>>>   objective    : a+b+c+2
>>>   #constraints : 3
>>>   #variables   : 3 (Continuous 2, Integer 1)
>>>
>>>   C 0, name None, b-c-3.0 <= 0
>>>   C 1, name None, a+b-2.0 <= 0
>>>   C 2, name None, -a-b+2.0 <= 0

LP to flopt

# make Lp model
c = [1, 1, 1]
C = 2
A = [[1, 0, 1],
     [1, -1, 0]]
b = [2, 3]
lb = [1, 1, 0]
ub = [2, 3, 1]
types=["Binary", "Continuous", "Continuous"]

from flopt.convert import LpStructure
prob = LpStructure(c, C, A=A, b=b, lb=lb, ub=ub, types=types).toFlopt()

prob.show()
>>> Name: None
>>>   Type         : Problem
>>>   sense        : minimize
>>>   objective    : x_0+x_1+x_2+2
>>>   #constraints : 2
>>>   #variables   : 3 (Continuous 2, Binary 1)
>>>
>>>   C 0, name None, x_0+x_2-2.0 == 0
>>>   C 1, name None, x_0-x_1-3.0 == 0