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""" 

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 _format_A_constraints(A, n_x, sparse_lhs=False): 

"""Format the left hand side of the constraints to a 2D array 

 

Parameters 

---------- 

A : 2D array 

2D array such that ``A @ x`` gives the values of the upper-bound 

(in)equality constraints at ``x``. 

n_x : int 

The number of variables in the linear programming problem. 

sparse_lhs : bool 

Whether either of `A_ub` or `A_eq` are sparse. If true return a 

coo_matrix instead of a numpy array. 

 

Returns 

------- 

np.ndarray or sparse.coo_matrix 

2D array such that ``A @ x`` gives the values of the upper-bound 

(in)equality constraints at ``x``. 

 

""" 

if sparse_lhs: 

return sps.coo_matrix( 

(0, n_x) if A is None else A, dtype=float, copy=True 

) 

elif A is None: 

return np.zeros((0, n_x), dtype=float) 

else: 

return np.array(A, dtype=float, copy=True) 

 

 

def _format_b_constraints(b): 

"""Format the upper bounds of the constraints to a 1D array 

 

Parameters 

---------- 

b : 1D array 

1D array of values representing the upper-bound of each (in)equality 

constraint (row) in ``A``. 

 

Returns 

------- 

1D np.array 

1D array of values representing the upper-bound of each (in)equality 

constraint (row) in ``A``. 

 

""" 

if b is None: 

return np.array([], dtype=float) 

b = np.array(b, dtype=float, copy=True).squeeze() 

return b if b.size != 1 else b.reshape((-1)) 

 

 

def _clean_inputs(c, A_ub=None, b_ub=None, A_eq=None, b_eq=None, bounds=None, 

x0=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. 

x0 : 1D array, optional 

Starting values of the independent variables, which will be refined by 

the optimization algorithm. 

 

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). 

x0 : 1D array, optional 

Starting values of the independent variables, which will be refined by 

the optimization algorithm. 

""" 

if c is None: 

raise TypeError 

 

try: 

c = np.array(c, dtype=np.float, copy=True).squeeze() 

except ValueError: 

raise TypeError( 

"Invalid input for linprog: c must be a 1D array of numerical " 

"coefficients") 

else: 

# If c is a single value, convert it to a 1D array. 

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 must be a 1D array and 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") 

 

sparse_lhs = sps.issparse(A_eq) or sps.issparse(A_ub) 

try: 

A_ub = _format_A_constraints(A_ub, n_x, sparse_lhs=sparse_lhs) 

except ValueError: 

raise TypeError( 

"Invalid input for linprog: A_ub must be a 2D array " 

"of numerical values") 

else: 

n_ub = A_ub.shape[0] 

if len(A_ub.shape) != 2 or A_ub.shape[1] != n_x: 

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") 

 

try: 

b_ub = _format_b_constraints(b_ub) 

except ValueError: 

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") 

else: 

if b_ub.shape != (n_ub,): 

raise ValueError( 

"Invalid input for linprog: b_ub must be a 1D array; b_ub " 

"must not have more than one non-singleton dimension and " 

"the number of rows in A_ub must equal 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") 

 

try: 

A_eq = _format_A_constraints(A_eq, n_x, sparse_lhs=sparse_lhs) 

except ValueError: 

raise TypeError( 

"Invalid input for linprog: A_eq must be a 2D array " 

"of numerical values") 

else: 

n_eq = A_eq.shape[0] 

if len(A_eq.shape) != 2 or A_eq.shape[1] != n_x: 

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") 

 

try: 

b_eq = _format_b_constraints(b_eq) 

except ValueError: 

raise TypeError( 

"Invalid input for linprog: b_eq must be a 1D array of " 

"numerical values, each representing the upper bound of an " 

"inequality constraint (row) in A_eq") 

else: 

if b_eq.shape != (n_eq,): 

raise ValueError( 

"Invalid input for linprog: b_eq must be a 1D array; b_eq " 

"must not have more than one non-singleton dimension and " 

"the number of rows in A_eq must equal 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") 

 

# x0 gives a (optional) starting solution to the solver. If x0 is None, 

# skip the checks. Initial solution will be generated automatically. 

if x0 is not None: 

try: 

x0 = np.array(x0, dtype=float, copy=True).squeeze() 

except ValueError: 

raise TypeError( 

"Invalid input for linprog: x0 must be a 1D array of " 

"numerical coefficients") 

if x0.ndim == 0: 

x0 = x0.reshape((-1)) 

if len(x0) == 0 or x0.ndim != 1: 

raise ValueError( 

"Invalid input for linprog: x0 should be a 1D array; it " 

"must not have more than one non-singleton dimension") 

if not x0.size == c.size: 

raise ValueError( 

"Invalid input for linprog: x0 and c should contain the " 

"same number of elements") 

if not np.isfinite(x0).all(): 

raise ValueError( 

"Invalid input for linprog: x0 must not contain values " 

"inf, nan, or None") 

 

# "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, x0 

 

 

def _presolve(c, A_ub, b_ub, A_eq, b_eq, bounds, x0, 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. 

x0 : 1D array, optional 

Starting values of the independent variables, which will be refined by 

the optimization algorithm. 

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) 

x0 : 1D array 

Starting values of the independent variables, which will be refined by 

the optimization algorithm (if solution is not 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, x0, 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, x0, 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, x0, 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, x0, 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, x0, 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, x0, 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] 

# user guess x0 stays separate from presolve solution x 

if x0 is not None: 

x0 = x0[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, x0, 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, x0, undo, complete, status, message) 

 

 

def _parse_linprog(c, A_ub, b_ub, A_eq, b_eq, bounds, options, x0): 

""" 

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')`. 

x0 : 1D array, optional 

Starting values of the independent variables, which will be refined by 

the optimization algorithm. Currently compatible only with the 

'revised simplex' method, and only if x0 is a basic feasible solution 

of 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, 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')`. 

x0 : 1D array, optional 

Starting values of the independent variables, which will be refined by 

the optimization algorithm. Currently compatible only with the 

'revised simplex' method, and only if x0 is a basic feasible solution 

of the problem. 

""" 

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, x0 = _clean_inputs( 

c, A_ub, b_ub, A_eq, b_eq, bounds, x0) 

return c, A_ub, b_ub, A_eq, b_eq, bounds, solver_options, x0 

 

 

def _get_Abc(c, c0=0, A_ub=None, b_ub=None, A_eq=None, b_eq=None, bounds=None, 

x0=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. 

x0 : 1D array 

Starting values of the independent variables, which will be refined by 

the optimization algorithm 

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. 

x0 : 1D array 

Starting values of the independent variables, which will be refined by 

the optimization algorithm 

 

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 x0 is not None: 

x0[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],)))) 

if x0 is not None: 

x0 = np.concatenate((x0, 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))) 

if x0 is not None: 

x0 = np.concatenate((x0, 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] 

if x0 is not None: 

i_free_neg = x0[i_free] < 0 

x0[np.arange(n_ub, A1.shape[1])[i_free_neg]] = -x0[i_free[i_free_neg]] 

x0[i_free[i_free_neg]] = 0 

 

# 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) 

if x0 is not None: 

x0[i_shift] -= lb_shift 

 

return A, b, c, c0, x0 

 

 

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, copy=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 

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) 

copy = True 

if copy: 

x = np.array(x, copy=True) 

 

# 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 within the " 

"required tolerance of " + "{:.2E}".format(tol) + ", 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. This can also occur due to the limited " 

"accuracy of the `interior-point` method. Check whether " 

"the slack and constraint residuals are acceptable; " 

"if not, consider enabling presolve, reducing option " 

"`tol`, and/or using method `revised simplex`. " 

"If you encounter this message under different " 

"circumstances, please submit a bug report.") 

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