A new global algorithm for factor-risk-constrained mean-variance portfolio selection

Abstract

We consider the factor-risk-constrained mean-variance portfolio-selection (MVPS) problem that allows managers to construct portfolios with desired factor-risk characteristics. Its optimization model is a non-convex quadratically constrained quadratic program that is known to be NP-hard. In this paper, we investigate the new global algorithm for factor-risk-constrained MVPS problem based on the successive convex optimization (SCO) method and the semi-definite relaxation (SDR) with a second-order cone (SOC) constraint. We first develop an SCO algorithm and show that it converges to a KKT point of the problem. We then develop a new global algorithm for factor-risk-constrained MVPS, which integrates the SCO method, the SDR with an SOC constraint, the branch-and-bound framework and the adaptive branch-and-cut rule for factor-related variables, to find a globally optimal solution to the underlying problem within a pre-specified $\epsilon$-tolerance. We establish the global convergence of the proposed algorithm and its complexity. Preliminary numerical results demonstrate the effectiveness of the proposed algorithm in finding a globally optimal solution to medium- and large-scale instances of factor-risk-constrained MVPS.

Appendix: proof of prosition 2

In this appendix, we present the proof of Prosition 2, which needs the following two lemmas.

Lemma 8

Let $\{x^k \}$ and $\{\xi ^k \}$ be the infinite sequences generated by Algorithm 2. Then,

$$\begin{aligned} g^+(x^{k})-g^+(x^{k+1}) \ge \delta _{\min }\Vert \xi ^{k+1}-\xi ^k\Vert _2^2 \end{aligned}$$

for all k, where $\delta _{\min }=\min \{\delta _j,j\in J\}$.

Proof

We first rewrite problem (27) as the following

$$\begin{aligned} \min \limits _{x\in {{{\mathcal {D}}}}} \tau _k\max \left\{ \max _{j\in J}\left\{ {\tilde{g}}_j(x;\xi ^k)\right\} ,0\right\} . \end{aligned}$$

(40)

Note that two problems (27) and (40) are equivalent in the sense that they have the same optimal value and the same solution set. From Step 1, $(x^{k+1},\nu ^{k+1})$ is the optimal solution of problem (27). Thus, $x^{k+1}$ is the optimal solution of problem (40), and

$$\begin{aligned} \nu ^{k+1}=\max \left\{ \max _{j\in J}\left\{ {\tilde{g}}_j(x^{k+1};\xi ^k)\right\} ,0\right\} . \end{aligned}$$

(41)

Note that $x^{k}$ is a feasible solution of problem (40) and $\tau _k>0$. Thus,

$$\begin{aligned} \nu ^{k+1} \le \max \left\{ \max _{j\in J}\left\{ {\tilde{g}}_j(x^{k};\xi ^k)\right\} ,0\right\} . \end{aligned}$$

(42)

By noting that

$$\begin{aligned} {\tilde{g}}_j(x;\xi ^k)=g_j(x)+\delta _j\Vert Mx-\xi ^k\Vert ^2_2, \end{aligned}$$

(43)

and that $\xi ^k=Mx^k$ for all k, we yield ${\tilde{g}}_j(x^k;\xi ^k)=g_j(x^k)$ for all k. It then follows from (42) that

$$\begin{aligned} \nu ^{k+1}\le g^+(x^k),\quad \forall k. \end{aligned}$$

(44)

On the other hand, note that $\xi ^{k+1}=Mx^{k+1}$, we have from (41) that $v^{k+1}\ge 0$ and

$$\begin{aligned} \nu ^{k+1}\ge & {} {\tilde{g}}_j(x^{k+1};\xi ^k)=g_j(x^{k+1})+\delta _j\Vert \xi ^k -\xi ^{k+1}\Vert ^2_2\\\ge & {} g_j(x^{k+1})+\delta _{\min }\Vert \xi ^k-\xi ^{k+1}\Vert ^2_2,\quad j\in J, \end{aligned}$$

where $\delta _{\min }=\min \{\delta _j,j\in J\}$. The above inequality yields

$$\begin{aligned} \nu ^{k+1}-\delta _{\min }\Vert \xi ^k-\xi ^{k+1}\Vert ^2_2 \ge \max _{j\in J}\left\{ g_j(x^{k+1})\right\} =g(x^{k+1})>0, \end{aligned}$$

which further implies

$$\begin{aligned} \nu ^{k+1}\ge g^+(x^{k+1})+\delta _{\min }\Vert \xi ^k-\xi ^{k+1}\Vert ^2_2. \end{aligned}$$

(45)

It then follows from (44) and (45) that for all k, one has

$$\begin{aligned} g^+(x^{k})-g^+(x^{k+1}) \ge \delta _{\min }\Vert \xi ^{k+1}-\xi ^k\Vert _2^2. \end{aligned}$$

The proof of the lemma is finished. $\square$

Lemma 9

Let the sequence $\{\tau _k\}$ be generated by Algorithm 2. Suppose that Assumption 1 holds at any $x\in {{{\mathcal {D}}}}$ with $g(x)\ge 0$. Then, there exists some index $k_0$ such that $\tau _k= \tau _{k_0}$ for all $k\ge k_0$.

Proof

Suppose, by contradiction, that $\tau _k\rightarrow +\infty$ as $k\rightarrow \infty$. Then (31) implies that $\tau _k<\gamma _k=\min \{\Vert \xi ^{k+1}-\xi ^k\Vert ^{-1},\Vert \eta ^{k+1}\Vert _1+\delta \}$ happens an infinite number of times. It then further follows that there exists a subsequence of iteration indices $\{k_i\}$, such that

$$\begin{aligned} \lim _{i\rightarrow \infty } \Vert \xi ^{k_i+1}-\xi ^{k_i}\Vert =0,\quad \quad \lim _{i\rightarrow \infty }\Vert \eta ^{k_i+1}\Vert _1=+\infty . \end{aligned}$$

(46)

Taking a further subsequence, if necessary, we can assume that $\{x^{k_i+1}\}\rightarrow {\bar{x}}$ as $i\rightarrow \infty$. By Step 1, since $(x^{k_i+1},\nu ^{k_i+1})$ is the optimal solution of convex problem (27) with $k=k_i$ and $(\eta ^{k_i+1},\mu ^{k_i+1})$ is the associated Lagrange multipliers, we have from the necessary optimality condition (29) that

$$\begin{aligned} \left\{ \begin{array}{l} -\sum _{j\in J}\eta _j^{k_i+1}\nabla {\tilde{g}}_j(x^{k_i+1};\xi ^{k_i}) \in {{{\mathcal {N}}}}_{{{{\mathcal {D}}}}}(x^{k_i+1}),\\ \eta _j^{k_i+1} \left[ {\tilde{g}}_j(x^{k_i+1};\xi ^{k_i})-\nu ^{k_i+1}\right] =0, ~j \in J,\\ \nu ^{k+1}\ge 0,~~{\tilde{g}}_j(x^{k+1};\xi ^{k})\le \nu ^{k+1} ,~~j\in J. \end{array}\right. \end{aligned}$$

(47)

Since $\xi ^{k_i+1}=Mx^{k_i+1}$ by Step 1, we have

$$\begin{aligned} {\tilde{g}}_j(x^{k_i+1};\xi ^{k_i})=g_j(x^{k_i+1})+\delta _j\Vert \xi ^{k_i+1}-\xi ^{k_i}\Vert ^2, ~j\in J. \end{aligned}$$

(48)

If $g({\bar{x}})<0$, then when $k_i$ is sufficiently large, one has

$$\begin{aligned} {\tilde{g}}_j(x^{k_i+1};\xi ^{k_i})-\nu ^{k_i+1}<0,\quad j\in J. \end{aligned}$$

Thus, by (47), $\eta _j^{k_i+1}=0$, $j\in J$ and then $\Vert \eta ^{k_i+1}\Vert _1=0$, which contradicts $\Vert \eta ^{k_i+1}\Vert _1=+\infty$ as $i\rightarrow \infty$. Therefore, $g({\bar{x}})\ge 0$. Let

$$\begin{aligned} \lim _{i\rightarrow \infty }\frac{\eta ^{k_i+1}}{\Vert \eta ^{k_i+1}\Vert _1}={\bar{\eta }}. \end{aligned}$$

Then,

$$\begin{aligned} {\bar{\eta }}_j=0~\mathrm{if}~j\not \in J({\bar{x}})~ \mathrm{and} \sum _{j\in J({\bar{x}})} {\bar{\eta }}_j=1. \end{aligned}$$

By the definition of ${{{\mathcal {N}}}}_{{{{\mathcal {D}}}}}(x^{k_i+1})$, the first inclusion of (47) yields

$$\begin{aligned} -\left( \sum _{j\in J}\eta _j^{k_i+1}\nabla {\tilde{g}}_j(x^{k_i+1};\xi ^{k_i})\right) ^T(x-x^{k_i+1})\le 0,~\forall x\in {{{\mathcal {D}}}}. \end{aligned}$$

(49)

Note that since $\xi ^{k+1}=Mx^{k+1}$ by Step 1, we obtain

$$\begin{aligned} \nabla {\tilde{g}}_j(x^{k+1};\xi ^k)=\nabla g_j(x^{k+1}) +2\delta _j(\xi ^{k+1}-\xi ^k), ~j\in J, \end{aligned}$$

which, by (46) and $\{x^{k_i+1}\}\rightarrow {\bar{x}}$ as $i\rightarrow \infty$, implies

$$\begin{aligned} \lim _{i\rightarrow \infty } \nabla {\tilde{g}}_j(x^{k_i+1};\xi ^{k_i})=\nabla g_j({\bar{x}}),~j\in J. \end{aligned}$$

(50)

By dividing the inequality (49) by $\Vert \eta ^{k_i+1}\Vert _1$ and letting $i\rightarrow \infty$, we obtain

$$\begin{aligned} -\left( \sum _{j\in J({\bar{x}})}{\bar{\eta }}_j\nabla g_j({\bar{x}}) \right) ^T(x-{\bar{x}})\le 0,~\forall x\in {{{\mathcal {D}}}}, \end{aligned}$$

which in turn implies that

$$\begin{aligned} 0\in \sum \limits _{j\in J({\bar{x}})}{\bar{\eta }}_j\nabla g_j({\bar{x}})+{{{\mathcal {N}}}}_{{{{\mathcal {D}}}}}({\bar{x}}). \end{aligned}$$

This, by $\sum _{j\in J({\bar{x}})}{\bar{\eta }}_j=1$, contradicts Condition (9) holds at ${\bar{x}}$ with $g({\bar{x}})\ge 0$. $\square$

Proof of Prosition 2

By Lemma 8, $\lim _{k\rightarrow \infty } \Vert \xi ^{k+1}-\xi ^{k}\Vert =0$. Then from (30), $\gamma _k=\Vert \eta ^{k+1}\Vert _1+\delta$ when k is sufficiently large. By Lemma 9, there exists some index $k_0$ such that $\tau _k= \tau _{k_0}$ for all $k\ge k_0$. Hence, from (31), $\tau _k\ge \Vert \eta ^{k+1}\Vert _1+\delta$ for all k sufficiently large. In view of the second relation of (29), it follows that $\mu ^{k+1}> 0$, thus, $\nu ^{k+1} = 0$ when k is sufficiently large. From the last relation of (29), we have that ${\tilde{g}}_j(x^{k+1};\xi ^{k})\le 0$ for $j\in J$ when k is sufficiently large. Note that since $\xi ^{k+1}=Mx^{k+1}$ by Step 1, from (43) we have that ${\tilde{g}}_j(x^{k+1};\xi ^{k})=g_j(x^{k+1})+\delta _j\Vert \xi ^{k+1}-\xi ^{k}\Vert ^2$. Since $\lim _{k\rightarrow \infty } \Vert \xi ^{k+1}-\xi ^{k}\Vert =0$, we obtain $\lim _{k\rightarrow \infty }{\tilde{g}}_j(x^{k+1};\xi ^{k})=g_j({\bar{x}})$. Consequently, $g_j({\bar{x}})\le 0$ and so ${\bar{x}}\in {{\mathcal {F}}}$. $\square$