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Python

"""
Method agnostic utility functions for linear progamming
"""
import numpy as np
import scipy.sparse as sps
from warnings import warn
from .optimize import OptimizeWarning
from scipy.optimize._remove_redundancy import (
_remove_redundancy, _remove_redundancy_sparse, _remove_redundancy_dense
)
def _check_sparse_inputs(options, A_ub, A_eq):
"""
Check the provided ``A_ub`` and ``A_eq`` matrices conform to the specified
optional sparsity variables.
Parameters
----------
A_ub : 2D array, optional
2D array such that ``A_ub @ x`` gives the values of the upper-bound
inequality constraints at ``x``.
A_eq : 2D array, optional
2D array such that ``A_eq @ x`` gives the values of the equality
constraints at ``x``.
options : dict
A dictionary of solver options. All methods accept the following
generic options:
maxiter : int
Maximum number of iterations to perform.
disp : bool
Set to True to print convergence messages.
For method-specific options, see :func:`show_options('linprog')`.
Returns
-------
A_ub : 2D array, optional
2D array such that ``A_ub @ x`` gives the values of the upper-bound
inequality constraints at ``x``.
A_eq : 2D array, optional
2D array such that ``A_eq @ x`` gives the values of the equality
constraints at ``x``.
options : dict
A dictionary of solver options. All methods accept the following
generic options:
maxiter : int
Maximum number of iterations to perform.
disp : bool
Set to True to print convergence messages.
For method-specific options, see :func:`show_options('linprog')`.
"""
# This is an undocumented option for unit testing sparse presolve
_sparse_presolve = options.pop('_sparse_presolve', False)
if _sparse_presolve and A_eq is not None:
A_eq = sps.coo_matrix(A_eq)
if _sparse_presolve and A_ub is not None:
A_ub = sps.coo_matrix(A_ub)
sparse = options.get('sparse', False)
if not sparse and (sps.issparse(A_eq) or sps.issparse(A_ub)):
options['sparse'] = True
warn("Sparse constraint matrix detected; setting 'sparse':True.",
OptimizeWarning)
return options, A_ub, A_eq
def _clean_inputs(c, A_ub=None, b_ub=None, A_eq=None, b_eq=None, bounds=None):
"""
Given user inputs for a linear programming problem, return the
objective vector, upper bound constraints, equality constraints,
and simple bounds in a preferred format.
Parameters
----------
c : 1D array
Coefficients of the linear objective function to be minimized.
A_ub : 2D array, optional
2D array such that ``A_ub @ x`` gives the values of the upper-bound
inequality constraints at ``x``.
b_ub : 1D array, optional
1D array of values representing the upper-bound of each inequality
constraint (row) in ``A_ub``.
A_eq : 2D array, optional
2D array such that ``A_eq @ x`` gives the values of the equality
constraints at ``x``.
b_eq : 1D array, optional
1D array of values representing the RHS of each equality constraint
(row) in ``A_eq``.
bounds : sequence, optional
``(min, max)`` pairs for each element in ``x``, defining
the bounds on that parameter. Use None for one of ``min`` or
``max`` when there is no bound in that direction. By default
bounds are ``(0, None)`` (non-negative).
If a sequence containing a single tuple is provided, then ``min`` and
``max`` will be applied to all variables in the problem.
Returns
-------
c : 1D array
Coefficients of the linear objective function to be minimized.
A_ub : 2D array, optional
2D array such that ``A_ub @ x`` gives the values of the upper-bound
inequality constraints at ``x``.
b_ub : 1D array, optional
1D array of values representing the upper-bound of each inequality
constraint (row) in ``A_ub``.
A_eq : 2D array, optional
2D array such that ``A_eq @ x`` gives the values of the equality
constraints at ``x``.
b_eq : 1D array, optional
1D array of values representing the RHS of each equality constraint
(row) in ``A_eq``.
bounds : sequence of tuples
``(min, max)`` pairs for each element in ``x``, defining
the bounds on that parameter. Use None for each of ``min`` or
``max`` when there is no bound in that direction. By default
bounds are ``(0, None)`` (non-negative).
"""
try:
if c is None:
raise TypeError
try:
c = np.asarray(c, dtype=float).copy().squeeze()
except BaseException: # typically a ValueError and shouldn't be, IMO
raise TypeError
if c.size == 1:
c = c.reshape((-1))
n_x = len(c)
if n_x == 0 or len(c.shape) != 1:
raise ValueError(
"Invalid input for linprog: c should be a 1D array; it must "
"not have more than one non-singleton dimension")
if not(np.isfinite(c).all()):
raise ValueError(
"Invalid input for linprog: c must not contain values "
"inf, nan, or None")
except TypeError:
raise TypeError(
"Invalid input for linprog: c must be a 1D array of numerical "
"coefficients")
try:
try:
if sps.issparse(A_eq) or sps.issparse(A_ub):
A_ub = sps.coo_matrix(
(0, n_x), dtype=float) if A_ub is None else sps.coo_matrix(
A_ub, dtype=float).copy()
else:
A_ub = np.zeros(
(0, n_x), dtype=float) if A_ub is None else np.asarray(
A_ub, dtype=float).copy()
except BaseException:
raise TypeError
n_ub = A_ub.shape[0]
if len(A_ub.shape) != 2 or A_ub.shape[1] != len(c):
raise ValueError(
"Invalid input for linprog: A_ub must have exactly two "
"dimensions, and the number of columns in A_ub must be "
"equal to the size of c ")
if (sps.issparse(A_ub) and not np.isfinite(A_ub.data).all()
or not sps.issparse(A_ub) and not np.isfinite(A_ub).all()):
raise ValueError(
"Invalid input for linprog: A_ub must not contain values "
"inf, nan, or None")
except TypeError:
raise TypeError(
"Invalid input for linprog: A_ub must be a numerical 2D array "
"with each row representing an upper bound inequality constraint")
try:
try:
b_ub = np.array(
[], dtype=float) if b_ub is None else np.asarray(
b_ub, dtype=float).copy().squeeze()
except BaseException:
raise TypeError
if b_ub.size == 1:
b_ub = b_ub.reshape((-1))
if len(b_ub.shape) != 1:
raise ValueError(
"Invalid input for linprog: b_ub should be a 1D array; it "
"must not have more than one non-singleton dimension")
if len(b_ub) != n_ub:
raise ValueError(
"Invalid input for linprog: The number of rows in A_ub must "
"be equal to the number of values in b_ub")
if not(np.isfinite(b_ub).all()):
raise ValueError(
"Invalid input for linprog: b_ub must not contain values "
"inf, nan, or None")
except TypeError:
raise TypeError(
"Invalid input for linprog: b_ub must be a 1D array of "
"numerical values, each representing the upper bound of an "
"inequality constraint (row) in A_ub")
try:
try:
if sps.issparse(A_eq) or sps.issparse(A_ub):
A_eq = sps.coo_matrix(
(0, n_x), dtype=float) if A_eq is None else sps.coo_matrix(
A_eq, dtype=float).copy()
else:
A_eq = np.zeros(
(0, n_x), dtype=float) if A_eq is None else np.asarray(
A_eq, dtype=float).copy()
except BaseException:
raise TypeError
n_eq = A_eq.shape[0]
if len(A_eq.shape) != 2 or A_eq.shape[1] != len(c):
raise ValueError(
"Invalid input for linprog: A_eq must have exactly two "
"dimensions, and the number of columns in A_eq must be "
"equal to the size of c ")
if (sps.issparse(A_eq) and not np.isfinite(A_eq.data).all()
or not sps.issparse(A_eq) and not np.isfinite(A_eq).all()):
raise ValueError(
"Invalid input for linprog: A_eq must not contain values "
"inf, nan, or None")
except TypeError:
raise TypeError(
"Invalid input for linprog: A_eq must be a 2D array with each "
"row representing an equality constraint")
try:
try:
b_eq = np.array(
[], dtype=float) if b_eq is None else np.asarray(
b_eq, dtype=float).copy().squeeze()
except BaseException:
raise TypeError
if b_eq.size == 1:
b_eq = b_eq.reshape((-1))
if len(b_eq.shape) != 1:
raise ValueError(
"Invalid input for linprog: b_eq should be a 1D array; it "
"must not have more than one non-singleton dimension")
if len(b_eq) != n_eq:
raise ValueError(
"Invalid input for linprog: the number of rows in A_eq "
"must be equal to the number of values in b_eq")
if not(np.isfinite(b_eq).all()):
raise ValueError(
"Invalid input for linprog: b_eq must not contain values "
"inf, nan, or None")
except TypeError:
raise TypeError(
"Invalid input for linprog: b_eq must be a 1D array of "
"numerical values, each representing the right hand side of an "
"equality constraints (row) in A_eq")
# "If a sequence containing a single tuple is provided, then min and max
# will be applied to all variables in the problem."
# linprog doesn't treat this right: it didn't accept a list with one tuple
# in it
try:
if isinstance(bounds, str):
raise TypeError
if bounds is None or len(bounds) == 0:
bounds = [(0, None)] * n_x
elif len(bounds) == 1:
b = bounds[0]
if len(b) != 2:
raise ValueError(
"Invalid input for linprog: exactly one lower bound and "
"one upper bound must be specified for each element of x")
bounds = [b] * n_x
elif len(bounds) == n_x:
try:
len(bounds[0])
except BaseException:
bounds = [(bounds[0], bounds[1])] * n_x
for i, b in enumerate(bounds):
if len(b) != 2:
raise ValueError(
"Invalid input for linprog, bound " +
str(i) +
" " +
str(b) +
": exactly one lower bound and one upper bound must "
"be specified for each element of x")
elif (len(bounds) == 2 and np.isreal(bounds[0])
and np.isreal(bounds[1])):
bounds = [(bounds[0], bounds[1])] * n_x
else:
raise ValueError(
"Invalid input for linprog: exactly one lower bound and one "
"upper bound must be specified for each element of x")
clean_bounds = [] # also creates a copy so user's object isn't changed
for i, b in enumerate(bounds):
if b[0] is not None and b[1] is not None and b[0] > b[1]:
raise ValueError(
"Invalid input for linprog, bound " +
str(i) +
" " +
str(b) +
": a lower bound must be less than or equal to the "
"corresponding upper bound")
if b[0] == np.inf:
raise ValueError(
"Invalid input for linprog, bound " +
str(i) +
" " +
str(b) +
": infinity is not a valid lower bound")
if b[1] == -np.inf:
raise ValueError(
"Invalid input for linprog, bound " +
str(i) +
" " +
str(b) +
": negative infinity is not a valid upper bound")
lb = float(b[0]) if b[0] is not None and b[0] != -np.inf else None
ub = float(b[1]) if b[1] is not None and b[1] != np.inf else None
clean_bounds.append((lb, ub))
bounds = clean_bounds
except ValueError as e:
if "could not convert string to float" in e.args[0]:
raise TypeError
else:
raise e
except TypeError as e:
print(e)
raise TypeError(
"Invalid input for linprog: bounds must be a sequence of "
"(min,max) pairs, each defining bounds on an element of x ")
return c, A_ub, b_ub, A_eq, b_eq, bounds
def _presolve(c, A_ub, b_ub, A_eq, b_eq, bounds, rr, tol=1e-9):
"""
Given inputs for a linear programming problem in preferred format,
presolve the problem: identify trivial infeasibilities, redundancies,
and unboundedness, tighten bounds where possible, and eliminate fixed
variables.
Parameters
----------
c : 1D array
Coefficients of the linear objective function to be minimized.
A_ub : 2D array, optional
2D array such that ``A_ub @ x`` gives the values of the upper-bound
inequality constraints at ``x``.
b_ub : 1D array, optional
1D array of values representing the upper-bound of each inequality
constraint (row) in ``A_ub``.
A_eq : 2D array, optional
2D array such that ``A_eq @ x`` gives the values of the equality
constraints at ``x``.
b_eq : 1D array, optional
1D array of values representing the RHS of each equality constraint
(row) in ``A_eq``.
bounds : sequence of tuples
``(min, max)`` pairs for each element in ``x``, defining
the bounds on that parameter. Use None for each of ``min`` or
``max`` when there is no bound in that direction.
rr : bool
If ``True`` attempts to eliminate any redundant rows in ``A_eq``.
Set False if ``A_eq`` is known to be of full row rank, or if you are
looking for a potential speedup (at the expense of reliability).
tol : float
The tolerance which determines when a solution is "close enough" to
zero in Phase 1 to be considered a basic feasible solution or close
enough to positive to serve as an optimal solution.
Returns
-------
c : 1D array
Coefficients of the linear objective function to be minimized.
c0 : 1D array
Constant term in objective function due to fixed (and eliminated)
variables.
A_ub : 2D array, optional
2D array such that ``A_ub @ x`` gives the values of the upper-bound
inequality constraints at ``x``.
b_ub : 1D array, optional
1D array of values representing the upper-bound of each inequality
constraint (row) in ``A_ub``.
A_eq : 2D array, optional
2D array such that ``A_eq @ x`` gives the values of the equality
constraints at ``x``.
b_eq : 1D array, optional
1D array of values representing the RHS of each equality constraint
(row) in ``A_eq``.
bounds : sequence of tuples
``(min, max)`` pairs for each element in ``x``, defining
the bounds on that parameter. Use None for each of ``min`` or
``max`` when there is no bound in that direction. Bounds have been
tightened where possible.
x : 1D array
Solution vector (when the solution is trivial and can be determined
in presolve)
undo: list of tuples
(index, value) pairs that record the original index and fixed value
for each variable removed from the problem
complete: bool
Whether the solution is complete (solved or determined to be infeasible
or unbounded in presolve)
status : int
An integer representing the exit status of the optimization::
0 : Optimization terminated successfully
1 : Iteration limit reached
2 : Problem appears to be infeasible
3 : Problem appears to be unbounded
4 : Serious numerical difficulties encountered
message : str
A string descriptor of the exit status of the optimization.
References
----------
.. [5] Andersen, Erling D. "Finding all linearly dependent rows in
large-scale linear programming." Optimization Methods and Software
6.3 (1995): 219-227.
.. [8] Andersen, Erling D., and Knud D. Andersen. "Presolving in linear
programming." Mathematical Programming 71.2 (1995): 221-245.
"""
# ideas from Reference [5] by Andersen and Andersen
# however, unlike the reference, this is performed before converting
# problem to standard form
# There are a few advantages:
# * artificial variables have not been added, so matrices are smaller
# * bounds have not been converted to constraints yet. (It is better to
# do that after presolve because presolve may adjust the simple bounds.)
# There are many improvements that can be made, namely:
# * implement remaining checks from [5]
# * loop presolve until no additional changes are made
# * implement additional efficiency improvements in redundancy removal [2]
undo = [] # record of variables eliminated from problem
# constant term in cost function may be added if variables are eliminated
c0 = 0
complete = False # complete is True if detected infeasible/unbounded
x = np.zeros(c.shape) # this is solution vector if completed in presolve
status = 0 # all OK unless determined otherwise
message = ""
# Standard form for bounds (from _clean_inputs) is list of tuples
# but numpy array is more convenient here
# In retrospect, numpy array should have been the standard
bounds = np.array(bounds)
lb = bounds[:, 0]
ub = bounds[:, 1]
lb[np.equal(lb, None)] = -np.inf
ub[np.equal(ub, None)] = np.inf
bounds = bounds.astype(float)
lb = lb.astype(float)
ub = ub.astype(float)
m_eq, n = A_eq.shape
m_ub, n = A_ub.shape
if (sps.issparse(A_eq)):
A_eq = A_eq.tolil()
A_ub = A_ub.tolil()
def where(A):
return A.nonzero()
vstack = sps.vstack
else:
where = np.where
vstack = np.vstack
# zero row in equality constraints
zero_row = np.array(np.sum(A_eq != 0, axis=1) == 0).flatten()
if np.any(zero_row):
if np.any(
np.logical_and(
zero_row,
np.abs(b_eq) > tol)): # test_zero_row_1
# infeasible if RHS is not zero
status = 2
message = ("The problem is (trivially) infeasible due to a row "
"of zeros in the equality constraint matrix with a "
"nonzero corresponding constraint value.")
complete = True
return (c, c0, A_ub, b_ub, A_eq, b_eq, bounds,
x, undo, complete, status, message)
else: # test_zero_row_2
# if RHS is zero, we can eliminate this equation entirely
A_eq = A_eq[np.logical_not(zero_row), :]
b_eq = b_eq[np.logical_not(zero_row)]
# zero row in inequality constraints
zero_row = np.array(np.sum(A_ub != 0, axis=1) == 0).flatten()
if np.any(zero_row):
if np.any(np.logical_and(zero_row, b_ub < -tol)): # test_zero_row_1
# infeasible if RHS is less than zero (because LHS is zero)
status = 2
message = ("The problem is (trivially) infeasible due to a row "
"of zeros in the equality constraint matrix with a "
"nonzero corresponding constraint value.")
complete = True
return (c, c0, A_ub, b_ub, A_eq, b_eq, bounds,
x, undo, complete, status, message)
else: # test_zero_row_2
# if LHS is >= 0, we can eliminate this constraint entirely
A_ub = A_ub[np.logical_not(zero_row), :]
b_ub = b_ub[np.logical_not(zero_row)]
# zero column in (both) constraints
# this indicates that a variable isn't constrained and can be removed
A = vstack((A_eq, A_ub))
if A.shape[0] > 0:
zero_col = np.array(np.sum(A != 0, axis=0) == 0).flatten()
# variable will be at upper or lower bound, depending on objective
x[np.logical_and(zero_col, c < 0)] = ub[
np.logical_and(zero_col, c < 0)]
x[np.logical_and(zero_col, c > 0)] = lb[
np.logical_and(zero_col, c > 0)]
if np.any(np.isinf(x)): # if an unconstrained variable has no bound
status = 3
message = ("If feasible, the problem is (trivially) unbounded "
"due to a zero column in the constraint matrices. If "
"you wish to check whether the problem is infeasible, "
"turn presolve off.")
complete = True
return (c, c0, A_ub, b_ub, A_eq, b_eq, bounds,
x, undo, complete, status, message)
# variables will equal upper/lower bounds will be removed later
lb[np.logical_and(zero_col, c < 0)] = ub[
np.logical_and(zero_col, c < 0)]
ub[np.logical_and(zero_col, c > 0)] = lb[
np.logical_and(zero_col, c > 0)]
# row singleton in equality constraints
# this fixes a variable and removes the constraint
singleton_row = np.array(np.sum(A_eq != 0, axis=1) == 1).flatten()
rows = where(singleton_row)[0]
cols = where(A_eq[rows, :])[1]
if len(rows) > 0:
for row, col in zip(rows, cols):
val = b_eq[row] / A_eq[row, col]
if not lb[col] - tol <= val <= ub[col] + tol:
# infeasible if fixed value is not within bounds
status = 2
message = ("The problem is (trivially) infeasible because a "
"singleton row in the equality constraints is "
"inconsistent with the bounds.")
complete = True
return (c, c0, A_ub, b_ub, A_eq, b_eq, bounds,
x, undo, complete, status, message)
else:
# sets upper and lower bounds at that fixed value - variable
# will be removed later
lb[col] = val
ub[col] = val
A_eq = A_eq[np.logical_not(singleton_row), :]
b_eq = b_eq[np.logical_not(singleton_row)]
# row singleton in inequality constraints
# this indicates a simple bound and the constraint can be removed
# simple bounds may be adjusted here
# After all of the simple bound information is combined here, get_Abc will
# turn the simple bounds into constraints
singleton_row = np.array(np.sum(A_ub != 0, axis=1) == 1).flatten()
cols = where(A_ub[singleton_row, :])[1]
rows = where(singleton_row)[0]
if len(rows) > 0:
for row, col in zip(rows, cols):
val = b_ub[row] / A_ub[row, col]
if A_ub[row, col] > 0: # upper bound
if val < lb[col] - tol: # infeasible
complete = True
elif val < ub[col]: # new upper bound
ub[col] = val
else: # lower bound
if val > ub[col] + tol: # infeasible
complete = True
elif val > lb[col]: # new lower bound
lb[col] = val
if complete:
status = 2
message = ("The problem is (trivially) infeasible because a "
"singleton row in the upper bound constraints is "
"inconsistent with the bounds.")
return (c, c0, A_ub, b_ub, A_eq, b_eq, bounds,
x, undo, complete, status, message)
A_ub = A_ub[np.logical_not(singleton_row), :]
b_ub = b_ub[np.logical_not(singleton_row)]
# identical bounds indicate that variable can be removed
i_f = np.abs(lb - ub) < tol # indices of "fixed" variables
i_nf = np.logical_not(i_f) # indices of "not fixed" variables
# test_bounds_equal_but_infeasible
if np.all(i_f): # if bounds define solution, check for consistency
residual = b_eq - A_eq.dot(lb)
slack = b_ub - A_ub.dot(lb)
if ((A_ub.size > 0 and np.any(slack < 0)) or
(A_eq.size > 0 and not np.allclose(residual, 0))):
status = 2
message = ("The problem is (trivially) infeasible because the "
"bounds fix all variables to values inconsistent with "
"the constraints")
complete = True
return (c, c0, A_ub, b_ub, A_eq, b_eq, bounds,
x, undo, complete, status, message)
ub_mod = ub
lb_mod = lb
if np.any(i_f):
c0 += c[i_f].dot(lb[i_f])
b_eq = b_eq - A_eq[:, i_f].dot(lb[i_f])
b_ub = b_ub - A_ub[:, i_f].dot(lb[i_f])
c = c[i_nf]
x = x[i_nf]
A_eq = A_eq[:, i_nf]
A_ub = A_ub[:, i_nf]
# record of variables to be added back in
undo = [np.nonzero(i_f)[0], lb[i_f]]
# don't remove these entries from bounds; they'll be used later.
# but we _also_ need a version of the bounds with these removed
lb_mod = lb[i_nf]
ub_mod = ub[i_nf]
# no constraints indicates that problem is trivial
if A_eq.size == 0 and A_ub.size == 0:
b_eq = np.array([])
b_ub = np.array([])
# test_empty_constraint_1
if c.size == 0:
status = 0
message = ("The solution was determined in presolve as there are "
"no non-trivial constraints.")
elif (np.any(np.logical_and(c < 0, ub_mod == np.inf)) or
np.any(np.logical_and(c > 0, lb_mod == -np.inf))):
# test_no_constraints()
# test_unbounded_no_nontrivial_constraints_1
# test_unbounded_no_nontrivial_constraints_2
status = 3
message = ("The problem is (trivially) unbounded "
"because there are no non-trivial constraints and "
"a) at least one decision variable is unbounded "
"above and its corresponding cost is negative, or "
"b) at least one decision variable is unbounded below "
"and its corresponding cost is positive. ")
else: # test_empty_constraint_2
status = 0
message = ("The solution was determined in presolve as there are "
"no non-trivial constraints.")
complete = True
x[c < 0] = ub_mod[c < 0]
x[c > 0] = lb_mod[c > 0]
# where c is zero, set x to a finite bound or zero
x_zero_c = ub_mod[c == 0]
x_zero_c[np.isinf(x_zero_c)] = ub_mod[c == 0][np.isinf(x_zero_c)]
x_zero_c[np.isinf(x_zero_c)] = 0
x[c == 0] = x_zero_c
# if this is not the last step of presolve, should convert bounds back
# to array and return here
# *sigh* - convert bounds back to their standard form (list of tuples)
# again, in retrospect, numpy array would be standard form
lb[np.equal(lb, -np.inf)] = None
ub[np.equal(ub, np.inf)] = None
bounds = np.hstack((lb[:, np.newaxis], ub[:, np.newaxis]))
bounds = bounds.tolist()
for i, row in enumerate(bounds):
for j, col in enumerate(row):
if str(
col) == "nan": # comparing col to float("nan") and
# np.nan doesn't work. should use np.isnan
bounds[i][j] = None
# remove redundant (linearly dependent) rows from equality constraints
n_rows_A = A_eq.shape[0]
redundancy_warning = ("A_eq does not appear to be of full row rank. To "
"improve performance, check the problem formulation "
"for redundant equality constraints.")
if (sps.issparse(A_eq)):
if rr and A_eq.size > 0: # TODO: Fast sparse rank check?
A_eq, b_eq, status, message = _remove_redundancy_sparse(A_eq, b_eq)
if A_eq.shape[0] < n_rows_A:
warn(redundancy_warning, OptimizeWarning)
if status != 0:
complete = True
return (c, c0, A_ub, b_ub, A_eq, b_eq, bounds,
x, undo, complete, status, message)
# This is a wild guess for which redundancy removal algorithm will be
# faster. More testing would be good.
small_nullspace = 5
if rr and A_eq.size > 0:
try: # TODO: instead use results of first SVD in _remove_redundancy
rank = np.linalg.matrix_rank(A_eq)
except Exception: # oh well, we'll have to go with _remove_redundancy_dense
rank = 0
if rr and A_eq.size > 0 and rank < A_eq.shape[0]:
warn(redundancy_warning, OptimizeWarning)
dim_row_nullspace = A_eq.shape[0]-rank
if dim_row_nullspace <= small_nullspace:
A_eq, b_eq, status, message = _remove_redundancy(A_eq, b_eq)
if dim_row_nullspace > small_nullspace or status == 4:
A_eq, b_eq, status, message = _remove_redundancy_dense(A_eq, b_eq)
if A_eq.shape[0] < rank:
message = ("Due to numerical issues, redundant equality "
"constraints could not be removed automatically. "
"Try providing your constraint matrices as sparse "
"matrices to activate sparse presolve, try turning "
"off redundancy removal, or try turning off presolve "
"altogether.")
status = 4
if status != 0:
complete = True
return (c, c0, A_ub, b_ub, A_eq, b_eq, bounds,
x, undo, complete, status, message)
def _parse_linprog(c, A_ub, b_ub, A_eq, b_eq, bounds, options):
"""
Parse the provided linear programming problem
``_parse_linprog`` employs two main steps ``_check_sparse_inputs`` and
``_clean_inputs``. ``_check_sparse_inputs`` checks for sparsity in the
provided constraints (``A_ub`` and ``A_eq) and if these match the provided
sparsity optional values.
``_clean inputs`` checks of the provided inputs. If no violations are
identified the objective vector, upper bound constraints, equality
constraints, and simple bounds are returned in the expected format.
Parameters
----------
c : 1D array
Coefficients of the linear objective function to be minimized.
A_ub : 2D array, optional
2D array such that ``A_ub @ x`` gives the values of the upper-bound
inequality constraints at ``x``.
b_ub : 1D array, optional
1D array of values representing the upper-bound of each inequality
constraint (row) in ``A_ub``.
A_eq : 2D array, optional
2D array such that ``A_eq @ x`` gives the values of the equality
constraints at ``x``.
b_eq : 1D array, optional
1D array of values representing the RHS of each equality constraint
(row) in ``A_eq``.
bounds : sequence
``(min, max)`` pairs for each element in ``x``, defining
the bounds on that parameter. Use None for one of ``min`` or
``max`` when there is no bound in that direction. By default
bounds are ``(0, None)`` (non-negative). If a sequence containing a
single tuple is provided, then ``min`` and ``max`` will be applied to
all variables in the problem.
options : dict
A dictionary of solver options. All methods accept the following
generic options:
maxiter : int
Maximum number of iterations to perform.
disp : bool
Set to True to print convergence messages.
For method-specific options, see :func:`show_options('linprog')`.
Returns
-------
c : 1D array
Coefficients of the linear objective function to be minimized.
A_ub : 2D array, optional
2D array such that ``A_ub @ x`` gives the values of the upper-bound
inequality constraints at ``x``.
b_ub : 1D array, optional
1D array of values representing the upper-bound of each inequality
constraint (row) in ``A_ub``.
A_eq : 2D array, optional
2D array such that ``A_eq @ x`` gives the values of the equality
constraints at ``x``.
b_eq : 1D array, optional
1D array of values representing the RHS of each equality constraint
(row) in ``A_eq``.
bounds : sequence, optional
``(min, max)`` pairs for each element in ``x``, defining
the bounds on that parameter. Use None for one of ``min`` or
``max`` when there is no bound in that direction. By default
bounds are ``(0, None)`` (non-negative).
If a sequence containing a single tuple is provided, then ``min`` and
``max`` will be applied to all variables in the problem.
options : dict, optional
A dictionary of solver options. All methods accept the following
generic options:
maxiter : int
Maximum number of iterations to perform.
disp : bool
Set to True to print convergence messages.
For method-specific options, see :func:`show_options('linprog')`.
"""
if options is None:
options = {}
solver_options = {k: v for k, v in options.items()}
solver_options, A_ub, A_eq = _check_sparse_inputs(solver_options, A_ub, A_eq)
# Convert lists to numpy arrays, etc...
c, A_ub, b_ub, A_eq, b_eq, bounds = _clean_inputs(
c, A_ub, b_ub, A_eq, b_eq, bounds)
return c, A_ub, b_ub, A_eq, b_eq, bounds, solver_options
def _get_Abc(c, c0=0, A_ub=None, b_ub=None, A_eq=None, b_eq=None, bounds=None,
undo=[]):
"""
Given a linear programming problem of the form:
Minimize::
c @ x
Subject to::
A_ub @ x <= b_ub
A_eq @ x == b_eq
lb <= x <= ub
where ``lb = 0`` and ``ub = None`` unless set in ``bounds``.
Return the problem in standard form:
Minimize::
c @ x
Subject to::
A @ x == b
x >= 0
by adding slack variables and making variable substitutions as necessary.
Parameters
----------
c : 1D array
Coefficients of the linear objective function to be minimized.
Components corresponding with fixed variables have been eliminated.
c0 : float
Constant term in objective function due to fixed (and eliminated)
variables.
A_ub : 2D array, optional
2D array such that ``A_ub @ x`` gives the values of the upper-bound
inequality constraints at ``x``.
b_ub : 1D array, optional
1D array of values representing the upper-bound of each inequality
constraint (row) in ``A_ub``.
A_eq : 2D array, optional
2D array such that ``A_eq @ x`` gives the values of the equality
constraints at ``x``.
b_eq : 1D array, optional
1D array of values representing the RHS of each equality constraint
(row) in ``A_eq``.
bounds : sequence of tuples
``(min, max)`` pairs for each element in ``x``, defining
the bounds on that parameter. Use None for each of ``min`` or
``max`` when there is no bound in that direction. Bounds have been
tightened where possible.
undo: list of tuples
(`index`, `value`) pairs that record the original index and fixed value
for each variable removed from the problem
Returns
-------
A : 2D array
2D array such that ``A`` @ ``x``, gives the values of the equality
constraints at ``x``.
b : 1D array
1D array of values representing the RHS of each equality constraint
(row) in A (for standard form problem).
c : 1D array
Coefficients of the linear objective function to be minimized (for
standard form problem).
c0 : float
Constant term in objective function due to fixed (and eliminated)
variables.
References
----------
.. [9] Bertsimas, Dimitris, and J. Tsitsiklis. "Introduction to linear
programming." Athena Scientific 1 (1997): 997.
"""
if sps.issparse(A_eq):
sparse = True
A_eq = sps.lil_matrix(A_eq)
A_ub = sps.lil_matrix(A_ub)
def hstack(blocks):
return sps.hstack(blocks, format="lil")
def vstack(blocks):
return sps.vstack(blocks, format="lil")
zeros = sps.lil_matrix
eye = sps.eye
else:
sparse = False
hstack = np.hstack
vstack = np.vstack
zeros = np.zeros
eye = np.eye
fixed_x = set()
if len(undo) > 0:
# these are indices of variables removed from the problem
# however, their bounds are still part of the bounds list
fixed_x = set(undo[0])
# they are needed elsewhere, but not here
bounds = [bounds[i] for i in range(len(bounds)) if i not in fixed_x]
# in retrospect, the standard form of bounds should have been an n x 2
# array. maybe change it someday.
# modify problem such that all variables have only non-negativity bounds
bounds = np.array(bounds)
lbs = bounds[:, 0]
ubs = bounds[:, 1]
m_ub, n_ub = A_ub.shape
lb_none = np.equal(lbs, None)
ub_none = np.equal(ubs, None)
lb_some = np.logical_not(lb_none)
ub_some = np.logical_not(ub_none)
# if preprocessing is on, lb == ub can't happen
# if preprocessing is off, then it would be best to convert that
# to an equality constraint, but it's tricky to make the other
# required modifications from inside here.
# unbounded below: substitute xi = -xi' (unbounded above)
l_nolb_someub = np.logical_and(lb_none, ub_some)
i_nolb = np.nonzero(l_nolb_someub)[0]
lbs[l_nolb_someub], ubs[l_nolb_someub] = (
-ubs[l_nolb_someub], lbs[l_nolb_someub])
lb_none = np.equal(lbs, None)
ub_none = np.equal(ubs, None)
lb_some = np.logical_not(lb_none)
ub_some = np.logical_not(ub_none)
c[i_nolb] *= -1
if len(i_nolb) > 0:
if A_ub.shape[0] > 0: # sometimes needed for sparse arrays... weird
A_ub[:, i_nolb] *= -1
if A_eq.shape[0] > 0:
A_eq[:, i_nolb] *= -1
# upper bound: add inequality constraint
i_newub = np.nonzero(ub_some)[0]
ub_newub = ubs[ub_some]
n_bounds = np.count_nonzero(ub_some)
A_ub = vstack((A_ub, zeros((n_bounds, A_ub.shape[1]))))
b_ub = np.concatenate((b_ub, np.zeros(n_bounds)))
A_ub[range(m_ub, A_ub.shape[0]), i_newub] = 1
b_ub[m_ub:] = ub_newub
A1 = vstack((A_ub, A_eq))
b = np.concatenate((b_ub, b_eq))
c = np.concatenate((c, np.zeros((A_ub.shape[0],))))
# unbounded: substitute xi = xi+ + xi-
l_free = np.logical_and(lb_none, ub_none)
i_free = np.nonzero(l_free)[0]
n_free = len(i_free)
A1 = hstack((A1, zeros((A1.shape[0], n_free))))
c = np.concatenate((c, np.zeros(n_free)))
A1[:, range(n_ub, A1.shape[1])] = -A1[:, i_free]
c[np.arange(n_ub, A1.shape[1])] = -c[i_free]
# add slack variables
A2 = vstack([eye(A_ub.shape[0]), zeros((A_eq.shape[0], A_ub.shape[0]))])
A = hstack([A1, A2])
# lower bound: substitute xi = xi' + lb
# now there is a constant term in objective
i_shift = np.nonzero(lb_some)[0]
lb_shift = lbs[lb_some].astype(float)
c0 += np.sum(lb_shift * c[i_shift])
if sparse:
b = b.reshape(-1, 1)
A = A.tocsc()
b -= (A[:, i_shift] * sps.diags(lb_shift)).sum(axis=1)
b = b.ravel()
else:
b -= (A[:, i_shift] * lb_shift).sum(axis=1)
return A, b, c, c0
def _display_summary(message, status, fun, iteration):
"""
Print the termination summary of the linear program
Parameters
----------
message : str
A string descriptor of the exit status of the optimization.
status : int
An integer representing the exit status of the optimization::
0 : Optimization terminated successfully
1 : Iteration limit reached
2 : Problem appears to be infeasible
3 : Problem appears to be unbounded
4 : Serious numerical difficulties encountered
fun : float
Value of the objective function.
iteration : iteration
The number of iterations performed.
"""
print(message)
if status in (0, 1):
print(" Current function value: {0: <12.6f}".format(fun))
print(" Iterations: {0:d}".format(iteration))
def _postsolve(x, c, A_ub=None, b_ub=None, A_eq=None, b_eq=None, bounds=None,
complete=False, undo=[], tol=1e-8):
"""
Given solution x to presolved, standard form linear program x, add
fixed variables back into the problem and undo the variable substitutions
to get solution to original linear program. Also, calculate the objective
function value, slack in original upper bound constraints, and residuals
in original equality constraints.
Parameters
----------
x : 1D array
Solution vector to the standard-form problem.
c : 1D array
Original coefficients of the linear objective function to be minimized.
A_ub : 2D array, optional
2D array such that ``A_ub @ x`` gives the values of the upper-bound
inequality constraints at ``x``.
b_ub : 1D array, optional
1D array of values representing the upper-bound of each inequality
constraint (row) in ``A_ub``.
A_eq : 2D array, optional
2D array such that ``A_eq @ x`` gives the values of the equality
constraints at ``x``.
b_eq : 1D array, optional
1D array of values representing the RHS of each equality constraint
(row) in ``A_eq``.
bounds : sequence of tuples
Bounds, as modified in presolve
complete : bool
Whether the solution is was determined in presolve (``True`` if so)
undo: list of tuples
(`index`, `value`) pairs that record the original index and fixed value
for each variable removed from the problem
tol : float
Termination tolerance; see [1]_ Section 4.5.
Returns
-------
x : 1D array
Solution vector to original linear programming problem
fun: float
optimal objective value for original problem
slack : 1D array
The (non-negative) slack in the upper bound constraints, that is,
``b_ub - A_ub @ x``
con : 1D array
The (nominally zero) residuals of the equality constraints, that is,
``b - A_eq @ x``
lb : 1D array
The lower bound constraints on the original variables
ub: 1D array
The upper bound constraints on the original variables
"""
# note that all the inputs are the ORIGINAL, unmodified versions
# no rows, columns have been removed
# the only exception is bounds; it has been modified
# we need these modified values to undo the variable substitutions
# in retrospect, perhaps this could have been simplified if the "undo"
# variable also contained information for undoing variable substitutions
n_x = len(c)
# we don't have to undo variable substitutions for fixed variables that
# were removed from the problem
no_adjust = set()
# if there were variables removed from the problem, add them back into the
# solution vector
if len(undo) > 0:
no_adjust = set(undo[0])
x = x.tolist()
for i, val in zip(undo[0], undo[1]):
x.insert(i, val)
x = np.array(x)
# now undo variable substitutions
# if "complete", problem was solved in presolve; don't do anything here
if not complete and bounds is not None: # bounds are never none, probably
n_unbounded = 0
for i, b in enumerate(bounds):
if i in no_adjust:
continue
lb, ub = b
if lb is None and ub is None:
n_unbounded += 1
x[i] = x[i] - x[n_x + n_unbounded - 1]
else:
if lb is None:
x[i] = ub - x[i]
else:
x[i] += lb
n_x = len(c)
x = x[:n_x] # all the rest of the variables were artificial
fun = x.dot(c)
slack = b_ub - A_ub.dot(x) # report slack for ORIGINAL UB constraints
# report residuals of ORIGINAL EQ constraints
con = b_eq - A_eq.dot(x)
# Patch for bug #8664. Detecting this sort of issue earlier
# (via abnormalities in the indicators) would be better.
bounds = np.array(bounds) # again, this should have been the standard form
lb = bounds[:, 0]
ub = bounds[:, 1]
lb[np.equal(lb, None)] = -np.inf
ub[np.equal(ub, None)] = np.inf
return x, fun, slack, con, lb, ub
def _check_result(x, fun, status, slack, con, lb, ub, tol, message):
"""
Check the validity of the provided solution.
A valid (optimal) solution satisfies all bounds, all slack variables are
negative and all equality constraint residuals are strictly non-zero.
Further, the lower-bounds, upper-bounds, slack and residuals contain
no nan values.
Parameters
----------
x : 1D array
Solution vector to original linear programming problem
fun: float
optimal objective value for original problem
status : int
An integer representing the exit status of the optimization::
0 : Optimization terminated successfully
1 : Iteration limit reached
2 : Problem appears to be infeasible
3 : Problem appears to be unbounded
4 : Serious numerical difficulties encountered
slack : 1D array
The (non-negative) slack in the upper bound constraints, that is,
``b_ub - A_ub @ x``
con : 1D array
The (nominally zero) residuals of the equality constraints, that is,
``b - A_eq @ x``
lb : 1D array
The lower bound constraints on the original variables
ub: 1D array
The upper bound constraints on the original variables
message : str
A string descriptor of the exit status of the optimization.
tol : float
Termination tolerance; see [1]_ Section 4.5.
Returns
-------
status : int
An integer representing the exit status of the optimization::
0 : Optimization terminated successfully
1 : Iteration limit reached
2 : Problem appears to be infeasible
3 : Problem appears to be unbounded
4 : Serious numerical difficulties encountered
message : str
A string descriptor of the exit status of the optimization.
"""
# Somewhat arbitrary, but status 5 is very unusual
tol = np.sqrt(tol) * 10
contains_nans = (
np.isnan(x).any()
or np.isnan(fun)
or np.isnan(slack).any()
or np.isnan(con).any()
)
if contains_nans:
is_feasible = False
else:
invalid_bounds = (x < lb - tol).any() or (x > ub + tol).any()
invalid_slack = status != 3 and (slack < -tol).any()
invalid_con = status != 3 and (np.abs(con) > tol).any()
is_feasible = not (invalid_bounds or invalid_slack or invalid_con)
if status == 0 and not is_feasible:
status = 4
message = ("The solution does not satisfy the constraints, yet "
"no errors were raised and there is no certificate of "
"infeasibility or unboundedness. This is known to occur "
"if the `presolve` option is False and the problem is "
"infeasible. If you encounter this under different "
"circumstances, please submit a bug report. Otherwise, "
"please enable presolve.")
elif status == 0 and contains_nans:
status = 4
message = ("Numerical difficulties were encountered but no errors "
"were raised. This is known to occur if the 'presolve' "
"option is False, 'sparse' is True, and A_eq includes "
"redundant rows. If you encounter this under different "
"circumstances, please submit a bug report. Otherwise, "
"remove linearly dependent equations from your equality "
"constraints or enable presolve.")
elif status == 2 and is_feasible:
# Occurs if the simplex method exits after phase one with a very
# nearly basic feasible solution. Postsolving can make the solution
#basic, however, this solution is NOT optimal
raise ValueError(message)
return status, message
def _postprocess(x, c, A_ub=None, b_ub=None, A_eq=None, b_eq=None, bounds=None,
complete=False, undo=[], status=0, message="", tol=1e-8,
iteration=None, disp=False):
"""
Given solution x to presolved, standard form linear program x, add
fixed variables back into the problem and undo the variable substitutions
to get solution to original linear program. Also, calculate the objective
function value, slack in original upper bound constraints, and residuals
in original equality constraints.
Parameters
----------
x : 1D array
Solution vector to the standard-form problem.
c : 1D array
Original coefficients of the linear objective function to be minimized.
A_ub : 2D array, optional
2D array such that ``A_ub @ x`` gives the values of the upper-bound
inequality constraints at ``x``.
b_ub : 1D array, optional
1D array of values representing the upper-bound of each inequality
constraint (row) in ``A_ub``.
A_eq : 2D array, optional
2D array such that ``A_eq @ x`` gives the values of the equality
constraints at ``x``.
b_eq : 1D array, optional
1D array of values representing the RHS of each equality constraint
(row) in ``A_eq``.
bounds : sequence of tuples
Bounds, as modified in presolve
complete : bool
Whether the solution is was determined in presolve (``True`` if so)
undo: list of tuples
(`index`, `value`) pairs that record the original index and fixed value
for each variable removed from the problem
status : int
An integer representing the exit status of the optimization::
0 : Optimization terminated successfully
1 : Iteration limit reached
2 : Problem appears to be infeasible
3 : Problem appears to be unbounded
4 : Serious numerical difficulties encountered
message : str
A string descriptor of the exit status of the optimization.
tol : float
Termination tolerance; see [1]_ Section 4.5.
Returns
-------
x : 1D array
Solution vector to original linear programming problem
fun: float
optimal objective value for original problem
slack : 1D array
The (non-negative) slack in the upper bound constraints, that is,
``b_ub - A_ub @ x``
con : 1D array
The (nominally zero) residuals of the equality constraints, that is,
``b - A_eq @ x``
status : int
An integer representing the exit status of the optimization::
0 : Optimization terminated successfully
1 : Iteration limit reached
2 : Problem appears to be infeasible
3 : Problem appears to be unbounded
4 : Serious numerical difficulties encountered
message : str
A string descriptor of the exit status of the optimization.
"""
x, fun, slack, con, lb, ub = _postsolve(
x, c, A_ub, b_ub, A_eq, b_eq,
bounds, complete, undo, tol
)
status, message = _check_result(
x, fun, status, slack, con,
lb, ub, tol, message
)
if disp:
_display_summary(message, status, fun, iteration)
return x, fun, slack, con, status, message