A portfolio optimization problem in Markowitz style looks like the following

\[\begin{aligned} \min && f(w) &= \frac12 w^T\Sigma w\\ \textrm{subject to} && w^Tr &= R \\ && w^T e &= 1 \\ && w & \succeq b_L \\ && w & \preceq b_U \end{aligned}\]

which the last two are to bound the weight of each asset in the portfolio. This is a nicely formulated optimization problem and one way to analytically solve it is to use Lagrange multipliers.

Shadow price

Assume we do not have the last two inequality constraints, the Lagrangian for the above problem would be

\[L(w,\lambda) = \frac12w^T\Sigma w - \lambda_1(w^Tr-R) - \lambda_2(w^Te-1)\]

The Lagrangian has the property that for the optimal solution \(w^*\) to the original problem, \(L(w^*,\lambda) = f(w^*)\), namely the Lagrangian function attained the same value as the objective function. This is trivial as we know that the optimizer must satisfy the equality constraints and hence the two extra terms in the Lagrangian always reduced to zero.

Mathematically, we can also make the Lagrangian having two Lagrange multiplier terms added to the objective function instead of subtraction. In doing so, we reversed our solution to \(\lambda_1\) and \(\lambda_2\) above. But if we consider that

\[\frac{\partial L(w^*,\lambda)}{\partial R} = \lambda_1\]

we see that it bears a physical meaning for subtraction, i.e., \(\lambda_1\) indicates how much it changes to the objective function if we marginally increased the boundary value \(R\) on the constraint. In this particular equality constraint, we are pushing the expected portfolio return \(R\) to a higher level and \(\lambda_1\) is the amount of variance increased. Hence the Lagrange multiplier \(\lambda_1\) is called the shadow price for the return \(R\).

Inequality constraints and activeness

A similar Lagrangian can be created when there are inequality constraints, but their Lagrange multiplier is no longer arbitrary:

\[L(w, \lambda, \theta, \phi) =\frac12 w^T\Sigma w -\lambda_1(w^Tr-R) - \lambda_2(w^Te-1) -\theta^T(w-b_L) + \phi^T(w-b_U)\]

The way to think of what sign should a Lagrange multiplier carry is to consider the dual. As we are doing a minimization here, the dual is a maximization problem, namely,

\[g(\lambda,\theta,\phi) = \inf_w L(w,\lambda,\theta,\phi)\]

and according to the max-min inequality we have the weak-duality property

\[\sup_{\lambda,\theta,\phi}\inf_w L(w,\lambda,\theta,\phi) \le \inf_w \sup_{\lambda,\theta,\phi}L(w,\lambda,\theta,\phi)\]

and the equality holds if we have strong duality. The RHS is the solution to the optimization problem and the LHS is the dual problem. Therefore the dual must be less than the optimal solution in the original problem

\[g(\lambda,\theta,\phi) \le \inf_w\sup_{\lambda,\theta,\phi} L(w,\lambda,\theta,\phi)\]

If we consider the Lagrange multipliers associated with inequality constraints \(\theta\) and \(\phi\) to be positive (there is no restriction for equality constraints), we must augment the objective function into \(L(w,\lambda,\theta,\phi)\) with negative values. Hence for \(w-b_L\succeq 0\), we augment it with \(-\theta(w-b_L)\), and for \(w-b_U\preceq 0\), we augment it with \(+\phi(w-b_U)\).

Why we need it in this way? Let us denote the feasible domain as \(\mathcal{D}\) and the optimal solution to the problem as \(w^*\in\mathcal{D}\). An inequality constraint at \(w^*\) shall either has the equality holds (which we call the constraint is active) or not (inactive). An constraint is inactive iff its removal does not change the optimal solution. The boundary of \(\mathcal{D}\) is defined by the constraints as if they are active (which the equality constraints can be assumed always active).

The solution \(w^*\) is a point on this boundary. As we are studying a minimization problem, \(f(w^*)\) is increasing into \(\mathcal{D}\) and decreasing away from \(\mathcal{D}\). Similarly, if \(w-b_L\succeq 0\) is active, \(w^*-b_L=0\) and it is increasing into \(\mathcal{D}\) and decreasing away from it (and similarly for \(w-b_U\preceq 0\)). In summary, we have

  \(w^*+\delta\in\mathcal{D}\) \(w^*+\delta \notin\mathcal{D}\)
\(f(w^*)\) \(f(w^*+\delta)\ge f(w^*)\) \(f(w^*+\delta)\le f(w^*)\)
\(w^*-b_L = 0\) \(w^*+\delta-b_L\succeq 0\) \(w^*+\delta-b_L\preceq 0\)
\(w^*-b_U = 0\) \(w^*+\delta-b_U\preceq 0\) \(w^*+\delta-b_U\succeq 0\)

and we need to make \(L(w^*+\delta,\lambda,\theta,\phi)\ge L(w^*,\lambda,\theta,\phi)\) for \(w^*+\delta\in\mathcal{D}\) so that we can find the optimal solution \(w^* = \arg\min L(w,\lambda,\theta,\phi)\) as a saddle point.

Karush-Kuhn-Tucker conditions

The KKT conditions state that

  1. \(\nabla L(w^*,\lambda,\theta,\phi)=0\) at the optimal solution \(w^*\)
  2. Primal constraints are satisfied for \(w^*\)
  3. Dual constraints \(\theta\ge 0\) and \(\phi\ge 0\) are satisfied, i.e. the Lagrange multipliers for inequality constraints are non-negative
  4. Complementary slackness: \(\theta\odot(w^*-b_L)=0\) and \(\phi\odot(w^*-b_U)=0\), i.e., the Lagrange multiplier will be zero if the corresponding inequality constraint is inactive

Solution

We can use the KKT conditions to solve for the above optimization problem. Since

\[L(w, \lambda, \theta, \phi) =\frac12 w^T\Sigma w -\lambda_1(w^Tr-R) - \lambda_2(w^Te-1) -\theta^T(w-b_L) + \phi^T(w-b_U)\]

The first condition states that

\[\nabla_w L(w, \lambda, \theta, \phi) = \Sigma w -\lambda_1r - \lambda_2 e -\theta + \phi = 0\]

the second condition states that

\[\begin{aligned} w^Tr - R = -\nabla_{\lambda_1} L(w, \lambda, \theta, \phi) &=0 \\ w^Te - 1 = -\nabla_{\lambda_2} L(w, \lambda, \theta, \phi) &=0 \\ w - b_L & \succeq 0 \\ w - b_U & \preceq 0 \end{aligned}\]

the third condition states that

\[\theta \ge 0;\qquad\phi \ge 0\]

and the fourth condition states that

\[\theta\odot(w-b_L)=0;\qquad\phi\odot(w-b_U)=0.\]

Assume \(w\) is a vector of \(n\) elements, we have \(3n+2\) unknowns (\(w,\theta,\phi\) are \(n\)-vectors and \(\lambda\) is 2-vector), \(n+2+0+2n=3n+2\) equalities from the four conditions, and \(0+2n+2n+0=4n\) inequalities. This should sufficient to provide a solution, but note that the equations from fourth condition are nonlinear as it includes \(\theta\odot w\) and \(\phi\odot w\) terms. To make it a system of linear equations, we can consider various combination of activeness of inequality constraints to simplify it. It would be tremendously easier if none of the inequality constraints are active (e.g., when \(b_L=-\infty\) and \(b_U=\infty\), which for sure \(\theta=\phi=\mathbf{0}\) based on the complementart slackness), in this case we have

\[\begin{aligned} \Sigma w - \lambda_1r-\lambda_2e &=0 \\ w &= \Sigma^{-1}(\lambda_1r+\lambda_2e) \\ &= \lambda_1\Sigma^{-1}r+\lambda_2\Sigma^{-1}e \end{aligned}\]

substitute:

\[\begin{aligned} w^Tr - R &= \lambda_1 r^T\Sigma^{-1}r + \lambda_2 e^T\Sigma^{-1}r - R = 0 \\ w^Te - 1 &= \lambda_1r^T\Sigma^{-1}e+\lambda_2e^T\Sigma^{-1}e - 1 = 0 \end{aligned}\]

therefore

\[\begin{aligned} \begin{bmatrix}r^T\Sigma^{-1}e & r^T\Sigma^{-1}e\\ r^T\Sigma^{-1}e & e^T\Sigma^{-1}e\end{bmatrix} \begin{bmatrix}\lambda_1\\ \lambda_2\end{bmatrix} &= \begin{bmatrix}R\\ 1\end{bmatrix} \\ \begin{bmatrix}\lambda_1\\ \lambda_2\end{bmatrix} &= \begin{bmatrix}r^T\Sigma^{-1}e & r^T\Sigma^{-1}e\\ r^T\Sigma^{-1}e & e^T\Sigma^{-1}e\end{bmatrix}^{-1}\begin{bmatrix}R\\ 1\end{bmatrix} \end{aligned}\]

and substitute back to above for \(w^*\). But the solution under this condition must not violate the second conditions, namely, \(b_L \preceq w^* \preceq b_U\). In fact we can also solve for both \(w\) and \(\lambda\) together in a matrix form equation of

\[\begin{aligned} \Sigma w -\lambda_1 r - \lambda_2 e &= 0 \\ w^Tr - R &=0 \\ w^Te - 1 &=0 \\ \implies\quad \begin{bmatrix}\Sigma & r & e\\ r^T & 0 & 0\\ e^T & 0 & 0\end{bmatrix} \begin{bmatrix}w\\ -\lambda_1\\ -\lambda_2\end{bmatrix} &= \begin{bmatrix}0\\ R\\ 1\end{bmatrix} \\ \therefore\quad \begin{bmatrix}w\\ -\lambda_1\\ -\lambda_2\end{bmatrix} &= \begin{bmatrix}\Sigma & r & e\\ r^T & 0 & 0\\ e^T & 0 & 0\end{bmatrix}^{-1}\begin{bmatrix}0\\ R\\ 1\end{bmatrix}. \end{aligned}\]

But the essence of using Karush-Kuhn-Tucker conditions to solve an optimization problem with inequality constraints is to make it combinatorial. Assume \(b_L\prec b_U\) and in some reasonable finite range (e.g. \(b_L=\mathbf{0}\) and \(b_U=e\)), to solve this we need to test all combinations of activeness of inequality constraints. In above, we have \(2n\) inequalities from the second KKT condition and there are \(2^{2n}\) combinations of activeness. When an inequality is active, its equality holds and the corresponding Lagrange multiplier can be non-zero. Hence a new set of linear equations are created and we can solve for \(w\) and other Lagrange multipliers, but we need to validate the solution not violating the KKT conditions, especially that of \(b_L \preceq w \preceq b_U\), and check the objective function. For example, if all inequality constraints are active, the optimization problem has its solution presented as

\[\begin{aligned} \begin{bmatrix}\Sigma & r & e & I & I\\ r^T & 0 & 0 & 0 & 0\\ e^T & 0 & 0 & 0 & 0 \\I & 0 & 0 & 0 & 0\\ I & 0 & 0 & 0 & 0\end{bmatrix} \begin{bmatrix}w\\ -\lambda_1\\ -\lambda_2\\ -\theta\\ \phi\end{bmatrix} &= \begin{bmatrix}0\\ R\\ 1\\ b_L\\ b_U\end{bmatrix} \\ \therefore\quad \begin{bmatrix}w\\ -\lambda_1\\ -\lambda_2\\ -\theta\\ \phi\end{bmatrix} &= \begin{bmatrix}\Sigma & r & e & I & I\\ r^T & 0 & 0 & 0 & 0\\ e^T & 0 & 0 & 0 & 0 \\I & 0 & 0 & 0 & 0\\ I & 0 & 0 & 0 & 0\end{bmatrix}^{-1} \begin{bmatrix}0\\ R\\ 1\\ b_L\\ b_U\end{bmatrix} \end{aligned}\]

and if some constraints are inactive, some of the rows and columns above shall be removed. After checking all combinations of activeness, the best solution based on the objective function are selected.

Implementation

The function below shows how the above optimization can be solved numerically. It try out all combinations of activeness and find the solution using the matrix equation described above. The best solution is then returned.

import numpy as np

def markowitz(ret, cov, r, lb=np.nan, ub=np.nan):
    """Markowitz minimizer with bounds constraints for a specified portfolio return

    Args:
        ret: A vector of N asset returns
        cov: NxN matrix of covariance of asset returns
        r (float): portfolio return to achieve
        lb, ub (float or vector): lowerbound and upperbound for the portfolio weights,
              if float, all weights are subject the same bound
    Returns:
		A (N+2) vector of portfolio weights and the Lagrange multipliers or
		None if no solution can be found
    """
    # Sanitation
    ret = np.array(ret).squeeze()
    cov = np.array(cov).squeeze()
    r = float(r)
    N = len(ret)
    if ret.shape != (N,):
        raise ValueError("Asset returns `ret` should be a vector")
    if cov.shape != (N,N):
        raise ValueError("Covariance matrix `cov` should be in shape ({},{}) to match the return vector".format(N,N))
    if isinstance(lb, (float,int)):
        lb = np.ones(N) * lb
    if isinstance(ub, (float,int)):
        ub = np.ones(N) * ub
    lb = lb.squeeze()
    ub = ub.squeeze()
    if lb.shape != (N,):
        raise ValueError("Lowerbound `lb` should be in shape (%d,) to match the return vector" % N)
    if ub.shape != (N,):
        raise ValueError("Upperbound `ub` should be in shape (%d,) to match the return vector" % N)
    if (lb > ub).any():
        raise ValueError("Lowerbound must no greater than upperbound")
    # Construct matrices as templates for the equation AX=B
    A = np.zeros((N+2+N+N,N+2+N+N))
    A[:N, :N] = cov
    A[:N, N] = A[N, :N] = ret
    A[:N, N+1] = A[N+1, :N] = np.ones(N)
    A[:N, N+2:N+N+2] = A[N+2:N+N+2, :N] = A[:N, N+N+2:] = A[N+N+2:, :N] = np.eye(N)
    b = np.zeros((N+2+N+N,1))
    b[N:N+2, 0] = [r, 1]
    b[N+2:N+N+2, 0] = lb
    b[N+N+2:, 0] = ub
    # Try all activeness combinations and track the best result to minimize objective
    bitmaps = 2**(2*N)
    best_obj = np.inf
    best_vector = None
    for bitmap in range(bitmaps):
        # constraints 0 to N-1 are for lowerbound and N to 2N-1 are for upperbound
        # row/column N+2+i corresponds to the constraint i
        inactive = [N+2+i for i in range(2*N) if bitmap & (2**i)]
        active = [N+2+i for i in range(2*N) if i not in inactive]
        # verify no conflicting active constraints
        if any(N+i in active for i in active):
            continue # conflicting activeness found, skip this one
        # Delete some rows and columns from the template for this activeness combination
        A_ = np.delete(np.delete(A, inactive, axis=0), inactive, axis=1)
        b_ = np.delete(b, inactive, axis=0)
        # Solve and check using matrix algebra
        try:
            x_ = (np.linalg.inv(A_) @ b_).squeeze()
            w = x_[:N]
            if (w < lb).any() or (w > ub).any():
                continue # solution not in feasible domain, try next one
            obj_val = w @ cov @ w # compute the covariance, i.e., objective function * 2
            if obj_val < best_obj:
                # Lower variance found, save the solution vector
                best_obj = obj_val
                x = np.zeros(N+2+N+N)
                x[:N+2] = x_[:N+2]    # w and negative lambda 
                x[active] = x_[N+2:]  # negative theta and phi
                x[N:N+2+N] *= -1      # lambda and theta are negated
                best_vector = x
        except np.linalg.LinAlgError:
            pass # no solution found for this combination
    return best_vector