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"""
Simplex method for solving linear programming problems
"""
import numpy as np
from warnings import warn
from .optimize import OptimizeResult, OptimizeWarning, _check_unknown_options
from ._linprog_util import _postsolve
def _pivot_col(T, tol=1.0E-12, bland=False):
"""
Given a linear programming simplex tableau, determine the column
of the variable to enter the basis.
Parameters
----------
T : 2D array
A 2D array representing the simplex tableau, T, corresponding to the
linear programming problem. It should have the form:
[[A[0, 0], A[0, 1], ..., A[0, n_total], b[0]],
[A[1, 0], A[1, 1], ..., A[1, n_total], b[1]],
.
.
.
[A[m, 0], A[m, 1], ..., A[m, n_total], b[m]],
[c[0], c[1], ..., c[n_total], 0]]
for a Phase 2 problem, or the form:
[[A[0, 0], A[0, 1], ..., A[0, n_total], b[0]],
[A[1, 0], A[1, 1], ..., A[1, n_total], b[1]],
.
.
.
[A[m, 0], A[m, 1], ..., A[m, n_total], b[m]],
[c[0], c[1], ..., c[n_total], 0],
[c'[0], c'[1], ..., c'[n_total], 0]]
for a Phase 1 problem (a problem in which a basic feasible solution is
sought prior to maximizing the actual objective. ``T`` is modified in
place by ``_solve_simplex``.
tol : float
Elements in the objective row larger than -tol will not be considered
for pivoting. Nominally this value is zero, but numerical issues
cause a tolerance about zero to be necessary.
bland : bool
If True, use Bland's rule for selection of the column (select the
first column with a negative coefficient in the objective row,
regardless of magnitude).
Returns
-------
status: bool
True if a suitable pivot column was found, otherwise False.
A return of False indicates that the linear programming simplex
algorithm is complete.
col: int
The index of the column of the pivot element.
If status is False, col will be returned as nan.
"""
ma = np.ma.masked_where(T[-1, :-1] >= -tol, T[-1, :-1], copy=False)
if ma.count() == 0:
return False, np.nan
if bland:
return True, np.nonzero(ma.mask == False)[0][0]
return True, np.ma.nonzero(ma == ma.min())[0][0]
def _pivot_row(T, basis, pivcol, phase, tol=1.0E-12, bland=False):
"""
Given a linear programming simplex tableau, determine the row for the
pivot operation.
Parameters
----------
T : 2D array
A 2D array representing the simplex tableau, T, corresponding to the
linear programming problem. It should have the form:
[[A[0, 0], A[0, 1], ..., A[0, n_total], b[0]],
[A[1, 0], A[1, 1], ..., A[1, n_total], b[1]],
.
.
.
[A[m, 0], A[m, 1], ..., A[m, n_total], b[m]],
[c[0], c[1], ..., c[n_total], 0]]
for a Phase 2 problem, or the form:
[[A[0, 0], A[0, 1], ..., A[0, n_total], b[0]],
[A[1, 0], A[1, 1], ..., A[1, n_total], b[1]],
.
.
.
[A[m, 0], A[m, 1], ..., A[m, n_total], b[m]],
[c[0], c[1], ..., c[n_total], 0],
[c'[0], c'[1], ..., c'[n_total], 0]]
for a Phase 1 problem (a Problem in which a basic feasible solution is
sought prior to maximizing the actual objective. ``T`` is modified in
place by ``_solve_simplex``.
basis : array
A list of the current basic variables.
pivcol : int
The index of the pivot column.
phase : int
The phase of the simplex algorithm (1 or 2).
tol : float
Elements in the pivot column smaller than tol will not be considered
for pivoting. Nominally this value is zero, but numerical issues
cause a tolerance about zero to be necessary.
bland : bool
If True, use Bland's rule for selection of the row (if more than one
row can be used, choose the one with the lowest variable index).
Returns
-------
status: bool
True if a suitable pivot row was found, otherwise False. A return
of False indicates that the linear programming problem is unbounded.
row: int
The index of the row of the pivot element. If status is False, row
will be returned as nan.
"""
if phase == 1:
k = 2
else:
k = 1
ma = np.ma.masked_where(T[:-k, pivcol] <= tol, T[:-k, pivcol], copy=False)
if ma.count() == 0:
return False, np.nan
mb = np.ma.masked_where(T[:-k, pivcol] <= tol, T[:-k, -1], copy=False)
q = mb / ma
min_rows = np.ma.nonzero(q == q.min())[0]
if bland:
return True, min_rows[np.argmin(np.take(basis, min_rows))]
return True, min_rows[0]
def _apply_pivot(T, basis, pivrow, pivcol, tol=1e-12):
"""
Pivot the simplex tableau inplace on the element given by (pivrow, pivol).
The entering variable corresponds to the column given by pivcol forcing
the variable basis[pivrow] to leave the basis.
Parameters
----------
T : 2D array
A 2D array representing the simplex tableau, T, corresponding to the
linear programming problem. It should have the form:
[[A[0, 0], A[0, 1], ..., A[0, n_total], b[0]],
[A[1, 0], A[1, 1], ..., A[1, n_total], b[1]],
.
.
.
[A[m, 0], A[m, 1], ..., A[m, n_total], b[m]],
[c[0], c[1], ..., c[n_total], 0]]
for a Phase 2 problem, or the form:
[[A[0, 0], A[0, 1], ..., A[0, n_total], b[0]],
[A[1, 0], A[1, 1], ..., A[1, n_total], b[1]],
.
.
.
[A[m, 0], A[m, 1], ..., A[m, n_total], b[m]],
[c[0], c[1], ..., c[n_total], 0],
[c'[0], c'[1], ..., c'[n_total], 0]]
for a Phase 1 problem (a problem in which a basic feasible solution is
sought prior to maximizing the actual objective. ``T`` is modified in
place by ``_solve_simplex``.
basis : 1D array
An array of the indices of the basic variables, such that basis[i]
contains the column corresponding to the basic variable for row i.
Basis is modified in place by _apply_pivot.
pivrow : int
Row index of the pivot.
pivcol : int
Column index of the pivot.
"""
basis[pivrow] = pivcol
pivval = T[pivrow, pivcol]
T[pivrow] = T[pivrow] / pivval
for irow in range(T.shape[0]):
if irow != pivrow:
T[irow] = T[irow] - T[pivrow] * T[irow, pivcol]
# The selected pivot should never lead to a pivot value less than the tol.
if np.isclose(pivval, tol, atol=0, rtol=1e4):
message = (
"The pivot operation produces a pivot value of:{0: .1e}, "
"which is only slightly greater than the specified "
"tolerance{1: .1e}. This may lead to issues regarding the "
"numerical stability of the simplex method. "
"Removing redundant constraints, changing the pivot strategy "
"via Bland's rule or increasing the tolerance may "
"help reduce the issue.".format(pivval, tol))
warn(message, OptimizeWarning)
def _solve_simplex(T, n, basis, maxiter=1000, phase=2, status=0, message='',
callback=None, tol=1.0E-12, nit0=0, bland=False, _T_o=None):
"""
Solve a linear programming problem in "standard form" using the Simplex
Method. Linear Programming is intended to solve the following problem form:
Minimize::
c @ x
Subject to::
A @ x == b
x >= 0
Parameters
----------
T : 2D array
A 2D array representing the simplex tableau, T, corresponding to the
linear programming problem. It should have the form:
[[A[0, 0], A[0, 1], ..., A[0, n_total], b[0]],
[A[1, 0], A[1, 1], ..., A[1, n_total], b[1]],
.
.
.
[A[m, 0], A[m, 1], ..., A[m, n_total], b[m]],
[c[0], c[1], ..., c[n_total], 0]]
for a Phase 2 problem, or the form:
[[A[0, 0], A[0, 1], ..., A[0, n_total], b[0]],
[A[1, 0], A[1, 1], ..., A[1, n_total], b[1]],
.
.
.
[A[m, 0], A[m, 1], ..., A[m, n_total], b[m]],
[c[0], c[1], ..., c[n_total], 0],
[c'[0], c'[1], ..., c'[n_total], 0]]
for a Phase 1 problem (a problem in which a basic feasible solution is
sought prior to maximizing the actual objective. ``T`` is modified in
place by ``_solve_simplex``.
n : int
The number of true variables in the problem.
basis : 1D array
An array of the indices of the basic variables, such that basis[i]
contains the column corresponding to the basic variable for row i.
Basis is modified in place by _solve_simplex
maxiter : int
The maximum number of iterations to perform before aborting the
optimization.
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.
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.
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.
nit0 : int
The initial iteration number used to keep an accurate iteration total
in a two-phase problem.
bland : bool
If True, choose pivots using Bland's rule [3]_. In problems which
fail to converge due to cycling, using Bland's rule can provide
convergence at the expense of a less optimal path about the simplex.
Returns
-------
nit : int
The number of iterations. Used to keep an accurate iteration total
in the two-phase 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
"""
nit = nit0
complete = False
if phase == 1:
m = T.shape[0]-2
elif phase == 2:
m = T.shape[0]-1
else:
raise ValueError("Argument 'phase' to _solve_simplex must be 1 or 2")
if phase == 2:
# Check if any artificial variables are still in the basis.
# If yes, check if any coefficients from this row and a column
# corresponding to one of the non-artificial variable is non-zero.
# If found, pivot at this term. If not, start phase 2.
# Do this for all artificial variables in the basis.
# Ref: "An Introduction to Linear Programming and Game Theory"
# by Paul R. Thie, Gerard E. Keough, 3rd Ed,
# Chapter 3.7 Redundant Systems (pag 102)
for pivrow in [row for row in range(basis.size)
if basis[row] > T.shape[1] - 2]:
non_zero_row = [col for col in range(T.shape[1] - 1)
if abs(T[pivrow, col]) > tol]
if len(non_zero_row) > 0:
pivcol = non_zero_row[0]
_apply_pivot(T, basis, pivrow, pivcol)
nit += 1
if len(basis[:m]) == 0:
solution = np.zeros(T.shape[1] - 1, dtype=np.float64)
else:
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)