Exemplo n.º 1
0
// replaceBland uses the Bland rule to find the indices to swap if the minimum
// move is 0. The indices to be swapped are replace and minIdx (following the
// nomenclature in the main routine).
func replaceBland(A mat64.Matrix, ab *mat64.Dense, xb []float64, basicIdxs, nonBasicIdx []int, r, move []float64) (replace, minIdx int, err error) {
	// Use the traditional bland rule, except don't replace a constraint which
	// causes the new ab to be singular.
	for i, v := range r {
		if v > -blandNegTol {
			continue
		}
		minIdx = i
		err = computeMove(move, minIdx, A, ab, xb, nonBasicIdx)
		if err != nil {
			// Either unbounded or something went wrong.
			return -1, -1, err
		}
		replace = floats.MinIdx(move)
		if math.Abs(move[replace]) > blandZeroTol {
			// Large enough that it shouldn't be a problem
			return replace, minIdx, nil
		}
		// Find a zero index where replacement is non-singular.
		biCopy := make([]int, len(basicIdxs))
		for replace, v := range move {
			if v > blandZeroTol {
				continue
			}
			copy(biCopy, basicIdxs)
			biCopy[replace] = nonBasicIdx[minIdx]
			abTmp := extractColumns(A, biCopy)
			// If the condition number is reasonable, use this index.
			if mat64.Cond(abTmp, 1) < 1e16 {
				return replace, minIdx, nil
			}
		}
	}
	return -1, -1, ErrBland
}
Exemplo n.º 2
0
func simplex(initialBasic []int, c []float64, A mat64.Matrix, b []float64, tol float64) (float64, []float64, []int, error) {
	err := verifyInputs(initialBasic, c, A, b)
	if err != nil {
		if err == ErrUnbounded {
			return math.Inf(-1), nil, nil, ErrUnbounded
		}
		return math.NaN(), nil, nil, err
	}
	m, n := A.Dims()

	// There is at least one optimal solution to the LP which is at the intersection
	// to a set of constraint boundaries. For a standard form LP with m variables
	// and n equality constraints, at least m-n elements of x must equal zero
	// at optimality. The Simplex algorithm solves the standard-form LP by starting
	// at an initial constraint vertex and successively moving to adjacent constraint
	// vertices. At every vertex, the set of non-zero x values is the "basic
	// feasible solution". The list of non-zero x's are maintained in basicIdxs,
	// the respective columns of A are in ab, and the actual non-zero values of
	// x are in xb.
	//
	// The LP is equality constrained such that A * x = b. This can be expanded
	// to
	//  ab * xb + an * xn = b
	// where ab are the columns of a in the basic set, and an are all of the
	// other columns. Since each element of xn is zero by definition, this means
	// that for all feasible solutions xb = ab^-1 * b.
	//
	// Before the simplex algorithm can start, an initial feasible solution must
	// be found. If initialBasic is non-nil a feasible solution has been supplied.
	// Otherwise the "Phase I" problem must be solved to find an initial feasible
	// solution.

	var basicIdxs []int // The indices of the non-zero x values.
	var ab *mat64.Dense // The subset of columns of A listed in basicIdxs.
	var xb []float64    // The non-zero elements of x. xb = ab^-1 b

	if initialBasic != nil {
		// InitialBasic supplied. Panic if incorrect length or infeasible.
		if len(initialBasic) != m {
			panic("lp: incorrect number of initial vectors")
		}
		ab = extractColumns(A, initialBasic)
		xb, err = initializeFromBasic(ab, b)
		if err != nil {
			panic(err)
		}
		basicIdxs = make([]int, len(initialBasic))
		copy(basicIdxs, initialBasic)
	} else {
		// No inital basis supplied. Solve the PhaseI problem.
		basicIdxs, ab, xb, err = findInitialBasic(A, b)
		if err != nil {
			return math.NaN(), nil, nil, err
		}
	}

	// basicIdxs contains the indexes for an initial feasible solution,
	// ab contains the extracted columns of A, and xb contains the feasible
	// solution. All x not in the basic set are 0 by construction.

	// nonBasicIdx is the set of nonbasic variables.
	nonBasicIdx := make([]int, 0, n-m)
	inBasic := make(map[int]struct{})
	for _, v := range basicIdxs {
		inBasic[v] = struct{}{}
	}
	for i := 0; i < n; i++ {
		_, ok := inBasic[i]
		if !ok {
			nonBasicIdx = append(nonBasicIdx, i)
		}
	}

	// cb is the subset of c for the basic variables. an and cn
	// are the equivalents to ab and cb but for the nonbasic variables.
	cb := make([]float64, len(basicIdxs))
	for i, idx := range basicIdxs {
		cb[i] = c[idx]
	}
	cn := make([]float64, len(nonBasicIdx))
	for i, idx := range nonBasicIdx {
		cn[i] = c[idx]
	}
	an := extractColumns(A, nonBasicIdx)

	bVec := mat64.NewVector(len(b), b)
	cbVec := mat64.NewVector(len(cb), cb)

	// Temporary data needed each iteration. (Described later)
	r := make([]float64, n-m)
	move := make([]float64, m)

	// Solve the linear program starting from the initial feasible set. This is
	// the "Phase 2" problem.
	//
	// Algorithm:
	// 1) Compute the "reduced costs" for the non-basic variables. The reduced
	// costs are the lagrange multipliers of the constraints.
	// 	 r = cn - an^T * ab^-T * cb
	// 2) If all of the reduced costs are positive, no improvement is possible,
	// and the solution is optimal (xn can only increase because of
	// non-negativity constraints). Otherwise, the solution can be improved and
	// one element will be exchanged in the basic set.
	// 3) Choose the x_n with the most negative value of r. Call this value xe.
	// This variable will be swapped into the basic set.
	// 4) Increase xe until the next constraint boundary is met. This will happen
	// when the first element in xb becomes 0. The distance xe can increase before
	// a given element in xb becomes negative can be found from
	//	xb = Ab^-1 b - Ab^-1 An xn
	//     = Ab^-1 b - Ab^-1 Ae xe
	//     = bhat + d x_e
	//  xe = bhat_i / - d_i
	// where Ae is the column of A corresponding to xe.
	// The constraining basic index is the first index for which this is true,
	// so remove the element which is min_i (bhat_i / -d_i), assuming d_i is negative.
	// If no d_i is less than 0, then the problem is unbounded.
	// 5) If the new xe is 0 (that is, bhat_i == 0), then this location is at
	// the intersection of several constraints. Use the Bland rule instead
	// of the rule in step 4 to avoid cycling.
	for {
		// Compute reduced costs -- r = cn - an^T ab^-T cb
		var tmp mat64.Vector
		err = tmp.SolveVec(ab.T(), cbVec)
		if err != nil {
			break
		}
		data := make([]float64, n-m)
		tmp2 := mat64.NewVector(n-m, data)
		tmp2.MulVec(an.T(), &tmp)
		floats.SubTo(r, cn, data)

		// Replace the most negative element in the simplex. If there are no
		// negative entries then the optimal solution has been found.
		minIdx := floats.MinIdx(r)
		if r[minIdx] >= -tol {
			break
		}

		for i, v := range r {
			if math.Abs(v) < rRoundTol {
				r[i] = 0
			}
		}

		// Compute the moving distance.
		err = computeMove(move, minIdx, A, ab, xb, nonBasicIdx)
		if err != nil {
			if err == ErrUnbounded {
				return math.Inf(-1), nil, nil, ErrUnbounded
			}
			break
		}

		// Replace the basic index along the tightest constraint.
		replace := floats.MinIdx(move)
		if move[replace] <= 0 {
			replace, minIdx, err = replaceBland(A, ab, xb, basicIdxs, nonBasicIdx, r, move)
			if err != nil {
				if err == ErrUnbounded {
					return math.Inf(-1), nil, nil, ErrUnbounded
				}
				break
			}
		}

		// Replace the constrained basicIdx with the newIdx.
		basicIdxs[replace], nonBasicIdx[minIdx] = nonBasicIdx[minIdx], basicIdxs[replace]
		cb[replace], cn[minIdx] = cn[minIdx], cb[replace]
		tmpCol1 := mat64.Col(nil, replace, ab)
		tmpCol2 := mat64.Col(nil, minIdx, an)
		ab.SetCol(replace, tmpCol2)
		an.SetCol(minIdx, tmpCol1)

		// Compute the new xb.
		xbVec := mat64.NewVector(len(xb), xb)
		err = xbVec.SolveVec(ab, bVec)
		if err != nil {
			break
		}
	}
	// Found the optimum successfully or died trying. The basic variables get
	// their values, and the non-basic variables are all zero.
	opt := floats.Dot(cb, xb)
	xopt := make([]float64, n)
	for i, v := range basicIdxs {
		xopt[v] = xb[i]
	}
	return opt, xopt, basicIdxs, err
}
Exemplo n.º 3
0
// findInitialBasic finds an initial basic solution, and returns the basic
// indices, ab, and xb.
func findInitialBasic(A mat64.Matrix, b []float64) ([]int, *mat64.Dense, []float64, error) {
	m, n := A.Dims()
	basicIdxs := findLinearlyIndependent(A)
	if len(basicIdxs) != m {
		return nil, nil, nil, ErrSingular
	}

	// It may be that this linearly independent basis is also a feasible set. If
	// so, the Phase I problem can be avoided.
	ab := extractColumns(A, basicIdxs)
	xb, err := initializeFromBasic(ab, b)
	if err == nil {
		return basicIdxs, ab, xb, nil
	}

	// This set was not feasible. Instead the "Phase I" problem must be solved
	// to find an initial feasible set of basis.
	//
	// Method: Construct an LP whose optimal solution is a feasible solution
	// to the original LP.
	// 1) Introduce an artificial variable x_{n+1}.
	// 2) Let x_j be the most negative element of x_b (largest constraint violation).
	// 3) Add the artificial variable to A with:
	//      a_{n+1} = b - \sum_{i in basicIdxs} a_i + a_j
	//    swap j with n+1 in the basicIdxs.
	// 4) Define a new LP:
	//   minimize  x_{n+1}
	//   subject to [A A_{n+1}][x_1 ... x_{n+1}] = b
	//          x, x_{n+1} >= 0
	// 5) Solve this LP. If x_{n+1} != 0, then the problem is infeasible, otherwise
	// the found basis can be used as an initial basis for phase II.
	//
	// The extra column in Step 3 is defined such that the vector of 1s is an
	// initial feasible solution.

	// Find the largest constraint violator.
	// Compute a_{n+1} = b - \sum{i in basicIdxs}a_i + a_j. j is in basicIDx, so
	// instead just subtract the basicIdx columns that are not minIDx.
	minIdx := floats.MinIdx(xb)
	aX1 := make([]float64, m)
	copy(aX1, b)
	col := make([]float64, m)
	for i, v := range basicIdxs {
		if i == minIdx {
			continue
		}
		mat64.Col(col, v, A)
		floats.Sub(aX1, col)
	}

	// Construct the new LP.
	// aNew = [A, a_{n+1}]
	// bNew = b
	// cNew = 1 for x_{n+1}
	aNew := mat64.NewDense(m, n+1, nil)
	aNew.Copy(A)
	aNew.SetCol(n, aX1)
	basicIdxs[minIdx] = n // swap minIdx with n in the basic set.
	c := make([]float64, n+1)
	c[n] = 1

	// Solve the Phase 2 linear program.
	_, xOpt, newBasic, err := simplex(basicIdxs, c, aNew, b, 1e-10)
	if err != nil {
		return nil, nil, nil, errors.New(fmt.Sprintf("lp: error finding feasible basis: %s", err))
	}

	// If n+1 is part of the solution basis then the problem is infeasible. If
	// not, then the problem is feasible and newBasic is an initial feasible
	// solution.
	if math.Abs(xOpt[n]) > phaseIZeroTol {
		return nil, nil, nil, ErrInfeasible
	}

	// The value is zero. First, see if it's not in the basis (feasible solution).
	basicIdx := -1
	basicMap := make(map[int]struct{})
	for i, v := range newBasic {
		if v == n {
			basicIdx = i
		}
		basicMap[v] = struct{}{}
		xb[i] = xOpt[v]
	}
	if basicIdx == -1 {
		// Not in the basis.
		ab = extractColumns(A, newBasic)
		return newBasic, ab, xb, nil
	}

	// The value is zero, but it's in the basis. See if swapping in another column
	// finds a feasible solution.
	for i := range xOpt {
		if _, inBasic := basicMap[i]; inBasic {
			continue
		}
		newBasic[basicIdx] = i
		ab := extractColumns(A, newBasic)
		xb, err := initializeFromBasic(ab, b)
		if err == nil {
			return newBasic, ab, xb, nil
		}
	}
	return nil, nil, nil, ErrInfeasible
}