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617 lines
23 KiB
Python
617 lines
23 KiB
Python
5 years ago
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"""
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Simplex method for solving linear programming problems
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"""
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import numpy as np
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from warnings import warn
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from .optimize import OptimizeResult, OptimizeWarning, _check_unknown_options
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from ._linprog_util import _postsolve
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def _pivot_col(T, tol=1.0E-12, bland=False):
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"""
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Given a linear programming simplex tableau, determine the column
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of the variable to enter the basis.
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Parameters
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----------
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T : 2D array
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A 2D array representing the simplex tableau, T, corresponding to the
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linear programming problem. It should have the form:
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[[A[0, 0], A[0, 1], ..., A[0, n_total], b[0]],
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[A[1, 0], A[1, 1], ..., A[1, n_total], b[1]],
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.
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.
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.
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[A[m, 0], A[m, 1], ..., A[m, n_total], b[m]],
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[c[0], c[1], ..., c[n_total], 0]]
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for a Phase 2 problem, or the form:
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[[A[0, 0], A[0, 1], ..., A[0, n_total], b[0]],
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[A[1, 0], A[1, 1], ..., A[1, n_total], b[1]],
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.
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.
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.
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[A[m, 0], A[m, 1], ..., A[m, n_total], b[m]],
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[c[0], c[1], ..., c[n_total], 0],
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[c'[0], c'[1], ..., c'[n_total], 0]]
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for a Phase 1 problem (a problem in which a basic feasible solution is
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sought prior to maximizing the actual objective. ``T`` is modified in
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place by ``_solve_simplex``.
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tol : float
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Elements in the objective row larger than -tol will not be considered
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for pivoting. Nominally this value is zero, but numerical issues
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cause a tolerance about zero to be necessary.
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bland : bool
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If True, use Bland's rule for selection of the column (select the
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first column with a negative coefficient in the objective row,
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regardless of magnitude).
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Returns
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-------
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status: bool
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True if a suitable pivot column was found, otherwise False.
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A return of False indicates that the linear programming simplex
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algorithm is complete.
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col: int
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The index of the column of the pivot element.
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If status is False, col will be returned as nan.
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"""
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ma = np.ma.masked_where(T[-1, :-1] >= -tol, T[-1, :-1], copy=False)
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if ma.count() == 0:
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return False, np.nan
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if bland:
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return True, np.nonzero(ma.mask == False)[0][0]
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return True, np.ma.nonzero(ma == ma.min())[0][0]
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def _pivot_row(T, basis, pivcol, phase, tol=1.0E-12, bland=False):
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"""
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Given a linear programming simplex tableau, determine the row for the
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pivot operation.
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Parameters
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----------
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T : 2D array
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A 2D array representing the simplex tableau, T, corresponding to the
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linear programming problem. It should have the form:
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[[A[0, 0], A[0, 1], ..., A[0, n_total], b[0]],
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[A[1, 0], A[1, 1], ..., A[1, n_total], b[1]],
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.
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.
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.
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[A[m, 0], A[m, 1], ..., A[m, n_total], b[m]],
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[c[0], c[1], ..., c[n_total], 0]]
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for a Phase 2 problem, or the form:
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[[A[0, 0], A[0, 1], ..., A[0, n_total], b[0]],
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[A[1, 0], A[1, 1], ..., A[1, n_total], b[1]],
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.
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.
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.
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[A[m, 0], A[m, 1], ..., A[m, n_total], b[m]],
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[c[0], c[1], ..., c[n_total], 0],
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[c'[0], c'[1], ..., c'[n_total], 0]]
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for a Phase 1 problem (a Problem in which a basic feasible solution is
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sought prior to maximizing the actual objective. ``T`` is modified in
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place by ``_solve_simplex``.
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basis : array
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A list of the current basic variables.
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pivcol : int
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The index of the pivot column.
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phase : int
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The phase of the simplex algorithm (1 or 2).
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tol : float
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Elements in the pivot column smaller than tol will not be considered
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for pivoting. Nominally this value is zero, but numerical issues
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cause a tolerance about zero to be necessary.
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bland : bool
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If True, use Bland's rule for selection of the row (if more than one
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row can be used, choose the one with the lowest variable index).
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Returns
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-------
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status: bool
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True if a suitable pivot row was found, otherwise False. A return
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of False indicates that the linear programming problem is unbounded.
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row: int
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The index of the row of the pivot element. If status is False, row
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will be returned as nan.
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"""
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if phase == 1:
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k = 2
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else:
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k = 1
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ma = np.ma.masked_where(T[:-k, pivcol] <= tol, T[:-k, pivcol], copy=False)
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if ma.count() == 0:
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return False, np.nan
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mb = np.ma.masked_where(T[:-k, pivcol] <= tol, T[:-k, -1], copy=False)
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q = mb / ma
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min_rows = np.ma.nonzero(q == q.min())[0]
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if bland:
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return True, min_rows[np.argmin(np.take(basis, min_rows))]
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return True, min_rows[0]
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def _apply_pivot(T, basis, pivrow, pivcol, tol=1e-12):
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"""
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Pivot the simplex tableau inplace on the element given by (pivrow, pivol).
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The entering variable corresponds to the column given by pivcol forcing
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the variable basis[pivrow] to leave the basis.
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Parameters
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----------
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T : 2D array
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A 2D array representing the simplex tableau, T, corresponding to the
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linear programming problem. It should have the form:
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[[A[0, 0], A[0, 1], ..., A[0, n_total], b[0]],
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[A[1, 0], A[1, 1], ..., A[1, n_total], b[1]],
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.
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.
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.
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[A[m, 0], A[m, 1], ..., A[m, n_total], b[m]],
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[c[0], c[1], ..., c[n_total], 0]]
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for a Phase 2 problem, or the form:
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[[A[0, 0], A[0, 1], ..., A[0, n_total], b[0]],
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[A[1, 0], A[1, 1], ..., A[1, n_total], b[1]],
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.
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.
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.
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[A[m, 0], A[m, 1], ..., A[m, n_total], b[m]],
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[c[0], c[1], ..., c[n_total], 0],
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[c'[0], c'[1], ..., c'[n_total], 0]]
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for a Phase 1 problem (a problem in which a basic feasible solution is
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sought prior to maximizing the actual objective. ``T`` is modified in
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place by ``_solve_simplex``.
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basis : 1D array
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An array of the indices of the basic variables, such that basis[i]
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contains the column corresponding to the basic variable for row i.
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Basis is modified in place by _apply_pivot.
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pivrow : int
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Row index of the pivot.
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pivcol : int
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Column index of the pivot.
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"""
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basis[pivrow] = pivcol
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pivval = T[pivrow, pivcol]
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T[pivrow] = T[pivrow] / pivval
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for irow in range(T.shape[0]):
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if irow != pivrow:
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T[irow] = T[irow] - T[pivrow] * T[irow, pivcol]
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# The selected pivot should never lead to a pivot value less than the tol.
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if np.isclose(pivval, tol, atol=0, rtol=1e4):
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message = (
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"The pivot operation produces a pivot value of:{0: .1e}, "
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"which is only slightly greater than the specified "
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"tolerance{1: .1e}. This may lead to issues regarding the "
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"numerical stability of the simplex method. "
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"Removing redundant constraints, changing the pivot strategy "
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"via Bland's rule or increasing the tolerance may "
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"help reduce the issue.".format(pivval, tol))
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warn(message, OptimizeWarning)
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def _solve_simplex(T, n, basis, maxiter=1000, phase=2, status=0, message='',
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callback=None, tol=1.0E-12, nit0=0, bland=False, _T_o=None):
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"""
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Solve a linear programming problem in "standard form" using the Simplex
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Method. Linear Programming is intended to solve the following problem form:
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Minimize::
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c @ x
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Subject to::
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A @ x == b
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x >= 0
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Parameters
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----------
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T : 2D array
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A 2D array representing the simplex tableau, T, corresponding to the
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linear programming problem. It should have the form:
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[[A[0, 0], A[0, 1], ..., A[0, n_total], b[0]],
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[A[1, 0], A[1, 1], ..., A[1, n_total], b[1]],
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.
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.
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.
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[A[m, 0], A[m, 1], ..., A[m, n_total], b[m]],
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[c[0], c[1], ..., c[n_total], 0]]
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for a Phase 2 problem, or the form:
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[[A[0, 0], A[0, 1], ..., A[0, n_total], b[0]],
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[A[1, 0], A[1, 1], ..., A[1, n_total], b[1]],
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.
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.
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.
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[A[m, 0], A[m, 1], ..., A[m, n_total], b[m]],
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[c[0], c[1], ..., c[n_total], 0],
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[c'[0], c'[1], ..., c'[n_total], 0]]
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for a Phase 1 problem (a problem in which a basic feasible solution is
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sought prior to maximizing the actual objective. ``T`` is modified in
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place by ``_solve_simplex``.
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n : int
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The number of true variables in the problem.
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basis : 1D array
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An array of the indices of the basic variables, such that basis[i]
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contains the column corresponding to the basic variable for row i.
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Basis is modified in place by _solve_simplex
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maxiter : int
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The maximum number of iterations to perform before aborting the
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optimization.
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phase : int
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The phase of the optimization being executed. In phase 1 a basic
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feasible solution is sought and the T has an additional row
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representing an alternate objective function.
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callback : callable, optional (simplex only)
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If a callback function is provided, it will be called within each
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iteration of the simplex algorithm. The callback must require a
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`scipy.optimize.OptimizeResult` consisting of the following fields:
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x : 1D array
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The independent variable vector which optimizes the linear
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programming problem.
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fun : float
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Value of the objective function.
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success : bool
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True if the algorithm succeeded in finding an optimal solution.
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slack : 1D array
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The values of the slack variables. Each slack variable
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corresponds to an inequality constraint. If the slack is zero,
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the corresponding constraint is active.
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con : 1D array
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The (nominally zero) residuals of the equality constraints,
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that is, ``b - A_eq @ x``
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phase : int
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The phase of the optimization being executed. In phase 1 a basic
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feasible solution is sought and the T has an additional row
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representing an alternate objective function.
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status : int
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An integer representing the exit status of the optimization::
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0 : Optimization terminated successfully
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1 : Iteration limit reached
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2 : Problem appears to be infeasible
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3 : Problem appears to be unbounded
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4 : Serious numerical difficulties encountered
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nit : int
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The number of iterations performed.
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message : str
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A string descriptor of the exit status of the optimization.
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tol : float
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The tolerance which determines when a solution is "close enough" to
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zero in Phase 1 to be considered a basic feasible solution or close
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enough to positive to serve as an optimal solution.
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nit0 : int
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The initial iteration number used to keep an accurate iteration total
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in a two-phase problem.
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bland : bool
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If True, choose pivots using Bland's rule [3]_. In problems which
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fail to converge due to cycling, using Bland's rule can provide
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convergence at the expense of a less optimal path about the simplex.
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Returns
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-------
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nit : int
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The number of iterations. Used to keep an accurate iteration total
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in the two-phase problem.
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status : int
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An integer representing the exit status of the optimization::
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0 : Optimization terminated successfully
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1 : Iteration limit reached
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2 : Problem appears to be infeasible
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3 : Problem appears to be unbounded
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4 : Serious numerical difficulties encountered
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"""
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nit = nit0
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complete = False
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if phase == 1:
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m = T.shape[0]-2
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elif phase == 2:
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m = T.shape[0]-1
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else:
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raise ValueError("Argument 'phase' to _solve_simplex must be 1 or 2")
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if phase == 2:
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# Check if any artificial variables are still in the basis.
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# If yes, check if any coefficients from this row and a column
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# corresponding to one of the non-artificial variable is non-zero.
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# If found, pivot at this term. If not, start phase 2.
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# Do this for all artificial variables in the basis.
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# Ref: "An Introduction to Linear Programming and Game Theory"
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# by Paul R. Thie, Gerard E. Keough, 3rd Ed,
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# Chapter 3.7 Redundant Systems (pag 102)
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for pivrow in [row for row in range(basis.size)
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if basis[row] > T.shape[1] - 2]:
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non_zero_row = [col for col in range(T.shape[1] - 1)
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if abs(T[pivrow, col]) > tol]
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if len(non_zero_row) > 0:
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pivcol = non_zero_row[0]
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_apply_pivot(T, basis, pivrow, pivcol)
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nit += 1
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if len(basis[:m]) == 0:
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solution = np.zeros(T.shape[1] - 1, dtype=np.float64)
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else:
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solution = np.zeros(max(T.shape[1] - 1, max(basis[:m]) + 1),
|
||
|
|
dtype=np.float64)
|
||
|
|
|
||
|
|
while not complete:
|
||
|
|
# Find the pivot column
|
||
|
|
pivcol_found, pivcol = _pivot_col(T, tol, bland)
|
||
|
|
if not pivcol_found:
|
||
|
|
pivcol = np.nan
|
||
|
|
pivrow = np.nan
|
||
|
|
status = 0
|
||
|
|
complete = True
|
||
|
|
else:
|
||
|
|
# Find the pivot row
|
||
|
|
pivrow_found, pivrow = _pivot_row(T, basis, pivcol, phase, tol, bland)
|
||
|
|
if not pivrow_found:
|
||
|
|
status = 3
|
||
|
|
complete = True
|
||
|
|
|
||
|
|
if callback is not None:
|
||
|
|
solution[basis[:n]] = T[:n, -1]
|
||
|
|
x = solution[:m]
|
||
|
|
c, A_ub, b_ub, A_eq, b_eq, bounds, undo = _T_o
|
||
|
|
x, fun, slack, con, _, _ = _postsolve(
|
||
|
|
x, c, A_ub, b_ub, A_eq, b_eq, bounds, undo=undo, tol=tol
|
||
|
|
)
|
||
|
|
res = OptimizeResult({
|
||
|
|
'x': x,
|
||
|
|
'fun': fun,
|
||
|
|
'slack': slack,
|
||
|
|
'con': con,
|
||
|
|
'status': status,
|
||
|
|
'message': message,
|
||
|
|
'nit': nit,
|
||
|
|
'success': status == 0 and complete,
|
||
|
|
'phase': phase,
|
||
|
|
'complete': complete,
|
||
|
|
})
|
||
|
|
callback(res)
|
||
|
|
|
||
|
|
if not complete:
|
||
|
|
if nit >= maxiter:
|
||
|
|
# Iteration limit exceeded
|
||
|
|
status = 1
|
||
|
|
complete = True
|
||
|
|
else:
|
||
|
|
_apply_pivot(T, basis, pivrow, pivcol)
|
||
|
|
nit += 1
|
||
|
|
return nit, status
|
||
|
|
|
||
|
|
|
||
|
|
def _linprog_simplex(c, c0, A, b, maxiter=1000, disp=False, callback=None,
|
||
|
|
tol=1.0E-12, bland=False, _T_o=None, **unknown_options):
|
||
|
|
"""
|
||
|
|
Minimize a linear objective function subject to linear equality and
|
||
|
|
non-negativity constraints using the two phase simplex method.
|
||
|
|
Linear programming is intended to solve problems of the following form:
|
||
|
|
|
||
|
|
Minimize::
|
||
|
|
|
||
|
|
c @ x
|
||
|
|
|
||
|
|
Subject to::
|
||
|
|
|
||
|
|
A @ x == b
|
||
|
|
x >= 0
|
||
|
|
|
||
|
|
Parameters
|
||
|
|
----------
|
||
|
|
c : 1D array
|
||
|
|
Coefficients of the linear objective function to be minimized.
|
||
|
|
c0 : float
|
||
|
|
Constant term in objective function due to fixed (and eliminated)
|
||
|
|
variables. (Purely for display.)
|
||
|
|
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 right hand side of each equality
|
||
|
|
constraint (row) in ``A``.
|
||
|
|
callback : callable, optional (simplex only)
|
||
|
|
If a callback function is provided, it will be called within each
|
||
|
|
iteration of the simplex algorithm. The callback must require a
|
||
|
|
`scipy.optimize.OptimizeResult` consisting of the following fields:
|
||
|
|
|
||
|
|
x : 1D array
|
||
|
|
The independent variable vector which optimizes the linear
|
||
|
|
programming problem.
|
||
|
|
fun : float
|
||
|
|
Value of the objective function.
|
||
|
|
success : bool
|
||
|
|
True if the algorithm succeeded in finding an optimal solution.
|
||
|
|
slack : 1D array
|
||
|
|
The values of the slack variables. Each slack variable
|
||
|
|
corresponds to an inequality constraint. If the slack is zero,
|
||
|
|
the corresponding constraint is active.
|
||
|
|
con : 1D array
|
||
|
|
The (nominally zero) residuals of the equality constraints, that
|
||
|
|
is, ``b - A_eq @ x``
|
||
|
|
phase : int
|
||
|
|
The phase of the optimization being executed. In phase 1 a basic
|
||
|
|
feasible solution is sought and the T has an additional row
|
||
|
|
representing an alternate objective function.
|
||
|
|
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
|
||
|
|
|
||
|
|
nit : int
|
||
|
|
The number of iterations performed.
|
||
|
|
message : str
|
||
|
|
A string descriptor of the exit status of the optimization.
|
||
|
|
Options
|
||
|
|
-------
|
||
|
|
maxiter : int
|
||
|
|
The maximum number of iterations to perform.
|
||
|
|
disp : bool
|
||
|
|
If True, print exit status message to sys.stdout
|
||
|
|
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.
|
||
|
|
bland : bool
|
||
|
|
If True, use Bland's anti-cycling rule [3]_ to choose pivots to
|
||
|
|
prevent cycling. If False, choose pivots which should lead to a
|
||
|
|
converged solution more quickly. The latter method is subject to
|
||
|
|
cycling (non-convergence) in rare instances.
|
||
|
|
|
||
|
|
Returns
|
||
|
|
-------
|
||
|
|
x : 1D array
|
||
|
|
Solution vector.
|
||
|
|
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.
|
||
|
|
iteration : int
|
||
|
|
The number of iterations taken to solve the problem.
|
||
|
|
|
||
|
|
References
|
||
|
|
----------
|
||
|
|
.. [1] Dantzig, George B., Linear programming and extensions. Rand
|
||
|
|
Corporation Research Study Princeton Univ. Press, Princeton, NJ,
|
||
|
|
1963
|
||
|
|
.. [2] Hillier, S.H. and Lieberman, G.J. (1995), "Introduction to
|
||
|
|
Mathematical Programming", McGraw-Hill, Chapter 4.
|
||
|
|
.. [3] Bland, Robert G. New finite pivoting rules for the simplex method.
|
||
|
|
Mathematics of Operations Research (2), 1977: pp. 103-107.
|
||
|
|
|
||
|
|
|
||
|
|
Notes
|
||
|
|
-----
|
||
|
|
The expected problem formulation differs between the top level ``linprog``
|
||
|
|
module and the method specific solvers. The method specific solvers expect a
|
||
|
|
problem in standard form:
|
||
|
|
|
||
|
|
Minimize::
|
||
|
|
|
||
|
|
c @ x
|
||
|
|
|
||
|
|
Subject to::
|
||
|
|
|
||
|
|
A @ x == b
|
||
|
|
x >= 0
|
||
|
|
|
||
|
|
Whereas the top level ``linprog`` module expects a problem of 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``.
|
||
|
|
|
||
|
|
The original problem contains equality, upper-bound and variable constraints
|
||
|
|
whereas the method specific solver requires equality constraints and
|
||
|
|
variable non-negativity.
|
||
|
|
|
||
|
|
``linprog`` module converts the original problem to standard form by
|
||
|
|
converting the simple bounds to upper bound constraints, introducing
|
||
|
|
non-negative slack variables for inequality constraints, and expressing
|
||
|
|
unbounded variables as the difference between two non-negative variables.
|
||
|
|
"""
|
||
|
|
_check_unknown_options(unknown_options)
|
||
|
|
|
||
|
|
status = 0
|
||
|
|
messages = {0: "Optimization terminated successfully.",
|
||
|
|
1: "Iteration limit reached.",
|
||
|
|
2: "Optimization failed. Unable to find a feasible"
|
||
|
|
" starting point.",
|
||
|
|
3: "Optimization failed. The problem appears to be unbounded.",
|
||
|
|
4: "Optimization failed. Singular matrix encountered."}
|
||
|
|
|
||
|
|
n, m = A.shape
|
||
|
|
|
||
|
|
# All constraints must have b >= 0.
|
||
|
|
is_negative_constraint = np.less(b, 0)
|
||
|
|
A[is_negative_constraint] *= -1
|
||
|
|
b[is_negative_constraint] *= -1
|
||
|
|
|
||
|
|
# As all constraints are equality constraints the artificial variables
|
||
|
|
# will also be basic variables.
|
||
|
|
av = np.arange(n) + m
|
||
|
|
basis = av.copy()
|
||
|
|
|
||
|
|
# Format the phase one tableau by adding artificial variables and stacking
|
||
|
|
# the constraints, the objective row and pseudo-objective row.
|
||
|
|
row_constraints = np.hstack((A, np.eye(n), b[:, np.newaxis]))
|
||
|
|
row_objective = np.hstack((c, np.zeros(n), c0))
|
||
|
|
row_pseudo_objective = -row_constraints.sum(axis=0)
|
||
|
|
row_pseudo_objective[av] = 0
|
||
|
|
T = np.vstack((row_constraints, row_objective, row_pseudo_objective))
|
||
|
|
|
||
|
|
nit1, status = _solve_simplex(T, n, basis, phase=1, callback=callback,
|
||
|
|
maxiter=maxiter, tol=tol, bland=bland, _T_o=_T_o)
|
||
|
|
# if pseudo objective is zero, remove the last row from the tableau and
|
||
|
|
# proceed to phase 2
|
||
|
|
if abs(T[-1, -1]) < tol:
|
||
|
|
# Remove the pseudo-objective row from the tableau
|
||
|
|
T = T[:-1, :]
|
||
|
|
# Remove the artificial variable columns from the tableau
|
||
|
|
T = np.delete(T, av, 1)
|
||
|
|
else:
|
||
|
|
# Failure to find a feasible starting point
|
||
|
|
status = 2
|
||
|
|
nit2 = nit1
|
||
|
|
messages[status] = (
|
||
|
|
"Phase 1 of the simplex method failed to find a feasible "
|
||
|
|
"solution. The pseudo-objective function evaluates to {0:.1e} "
|
||
|
|
"which exceeds the required tolerance of {1} for a solution to be "
|
||
|
|
"considered 'close enough' to zero to be a basic solution. "
|
||
|
|
"Consider increasing the tolerance to be greater than {0:.1e}. "
|
||
|
|
"If this tolerance is unacceptably large the problem may be "
|
||
|
|
"infeasible.".format(abs(T[-1, -1]), tol)
|
||
|
|
)
|
||
|
|
|
||
|
|
if status == 0:
|
||
|
|
# Phase 2
|
||
|
|
nit2, status = _solve_simplex(T, n, basis, maxiter=maxiter,
|
||
|
|
phase=2, callback=callback, tol=tol,
|
||
|
|
nit0=nit1, bland=bland, _T_o=_T_o)
|
||
|
|
|
||
|
|
solution = np.zeros(n + m)
|
||
|
|
solution[basis[:n]] = T[:n, -1]
|
||
|
|
x = solution[:m]
|
||
|
|
|
||
|
|
return x, status, messages[status], int(nit2)
|
||
|
|
|