""" 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