A Smoothing Sample Average Approximation

A Smoothing Sample Average Approximation Method for Stochastic Optimization Problems with CVaR Risk Measure Fanwen Meng1 , Jie Sun2 , Mark Goh3

Abstract This paper is concerned with solving single CVaR and mixed CVaR minimization problems. A CHKS-type smoothing sample average approximation (SAA) method is proposed for solving these two problems, which retains the convexity and smoothness of the original problem and is easy to implement. For any fixed smoothing constant ², this method produces a sequence whose cluster points are weak stationary points of the CVaR optimization problems with probability one. This framework of combining smoothing technique and SAA scheme can be extended to other smoothing functions as well. Practical numerical examples arising from logistics management are presented to show the usefulness of this method.

Key words. Conditional Value-at-Risk, Sample Average Approximation, Smoothing Method, Stochastic Optimization

AMS subject classifications. 91B28, 90C90, 62P05

1

Introduction

It is very important in risk management to choose a proper risk measure, as exemplified in the governmental regulations such as Basel Accord II (2006), which uses Value at Risk (VaR) as 1

The corresponding author. The Logistic Institute - Asia Pacific, National University of Singapore. Email: [email protected]. 2 School of Business and Risk Management Institute, National University of Singapore. Email: [email protected]. 3 School of Business and The Logistics Institute - Asia Pacific, National University of Singapore and University of South Australia. Email: [email protected] and [email protected].

1

a preferred risk measure [8]. Given a confidence level α ∈ (0, 1) and a loss function f (x, y˜) : IRn × IRm → IR, where x is the decision variable and y˜ represents the uncertain factors defined on a probability space (Ω, F, P), the VaR of the random variable f (x, y˜) is defined as the left α-quantile of f , namely VaRα (x) = min{u | P(f (x, y˜) ≤ u) ≥ α}.

(1.1)

Here P(·) stands for the probability. However, as a function of the decision variable, VaR is generally nonconvex and computationally nontractable which makes the resulting VaR optimization problems hard to solve. Due to this and other reasons, a new risk measure, called conditional VaR (CVaR) has been studied extensively in recent literature. For x ∈ IRn , let F (x, ·) denote the distribution of the random variable z˜ = f (x, y˜). For the given confidence level α ∈ (0, 1), CVaR is defined as follows [3, 19]: CVaRα (x) = Eα−tail [˜ z ], where the α-tail cumulative distribution function of z˜ is in the form of  0, if u < VaRα (x),    Fα (x, u) = P(˜ z ≤ u) =    F (x, u) − α , if u ≥ VaRα (x). 1−α

(1.2)

(1.3)

While CVaR is conceptually defined as the expectation of f (x, y˜) in the conditional distribution of its upper α-tail, a more operationally convenient definition by Rockafellar and Uryasev [19] is as follows. CVaRα (x) := min{η(x, u, α) | u ∈ IR},

(1.4)

1 E[f (x, y˜) − u]+ , 1−α

(1.5)

where η(x, u, α) := u +

where the superscript plus denotes the plus function [t]+ := max{0, t}, E denotes the mathematical expectation. It has been shown that CVaR is the best convex approximation to VaR and has lots of nice properties, which makes it widely acceptable in risk management [2, 4, 7, 9, 10, 12, 14, 18, 19, 20]. In this paper, we are interested in studying CVaR-related minmization problems. The mathematical model of these problems can be cast in the following form min CVaRα (x) s.t. x ∈ X ,

(1.6)

where X stands for the feasible region of the problem, which itself may be defined by certain CVaR constraints, but for the time being, we only assume that X is convex and closed. It is known [20] that (1.6) is equivalent to the following stochastic program ¾ ½ 1 + (1.7) E[f (x, y˜) − u] min u+ 1−α (x,u)∈X ×IR 2

in the sense that these two problems achieve the same minimum value and the x-component of the solution to (1.7) is a solution to (1.6). Another very interesting measure of risk, called the mixed CVaR, is defined as λ1 CVaRα1 (x) + · · · + λJ CVaRαJ (x),

(1.8)

P where αi ∈ (0, 1) denote the probability levels and λi > 0 represent weights with Ji=1 λi = 1, i = 1, . . . , J. Clearly, the single CVaR is a special mixed CVaR with J = 1. Again, the mixed CVaR minimization problem ½ ¾ min λ1 CVaRα1 (x) + · · · + λJ CVaRαJ (x) . (1.9) x∈X

is shown [12] to be equivalent to the following problem µ ¶ µ ¶ 1 1 + + E[f (x, y˜) − u1 ] + · · · + λJ u J + E[f (x, y˜) − uJ ] min λ1 u1 + 1 − α1 1 − αJ s. t. (x, u1 , · · · , uJ ) ∈ X × IR × · · · × IR. (1.10) Note that in problems (1.7) and (1.10), if the expectations can be evaluated analytically, then these two problems can be regarded as standard nonlinear programming problems. However, it might not be easy to evaluate or compute the underlying expectations in (1.7) and (1.10). We therefore consider a Monte Carlo simulation based method, called the sample average approximation (SAA) method. See [13, 23, 24, 25, 27]. The basic idea of the method is to generate an independent identically distributed (i.i.d.) samples of y˜ and then approximate the expected value with the sample average. Consequently, the SAA program is a deterministic problem. However, the resulting SAA program may be still nonsmooth due to the nonsmoothness of the plus function in the objective functions of (1.7) and (1.10). To deal with the nonsmoothness of the SAA programs, as well as to facilitate Newton-type methods in solving the SAA programs, we propose a CHKS-smoothing technique in this paper and investigate in what sense the smoothed problem could approximate the original problem. We assume that f (x, y˜) is continuously differentiable and convex in x for any realizations of y˜. In this way, our smoothing SAA method will preserve the convexity and smoothness in the SAA programs. Smoothing technique has been used in solving stochastic optimization problems by Lin, Chen, and Fukushima [11], but it appears to be new to consider them under the framework of SAA methods and CVaR constraints. It will be seen later in this paper that the method can be extended without essential difficulty to other smoothing techniques and nonsmooth loss functions. To show the usefulness of this approach, we provide numerical examples which come from logistics management and present computational results. The rest of this paper is organized as follows. Section 2 presents some basic notions and discusses some properties of CVaR and the SAA programs. Section 3 introduces a smoothing SAA method and analyzes its convergence. Two examples in logistics with CVaR risk measure in combination with their numerical solutions are presented in Section 4. 3

2

Preliminaries

In this section, we first recall some basic notions which will be used in the subsequent analysis. We then discuss the optimality conditions of the CVaR and mixed CVaR optimization problems.

2.1

Compact Set Mappings and Generalized Jacobians

Let k · k denote the Euclidean norm of a vector or a compact set of vectors. When M is a compact set of vectors, we denote the norm of M by kMk := maxM ∈M kM k. For two compact sets C and D, the deviation from C to D, or the excess of C over D, is defined by D(C, D) := sup d(x, D) x∈C

where d(x, D) denotes the distance from point x to set D defined by d(x, D) := inf x0 ∈D kx − x0 k. l

Let A(·, y˜) : V → 2IR be a random compact set-valued mapping, where V ⊂ IRl is a compact set of IRl and y˜ : Ω → Ξ ⊂ IRm is a random vector. A selection of the random set A(v, y˜(ω)) is a random vector A(v, y˜(ω)), which means A(v, y˜(ω)) is measurable. The expectation of A(v, y˜(ω)), denoted by E[A(v, y˜(ω))], is defined as the collection of E[A(v, y˜(ω))] where A(v, y˜(ω)) is a selection. Note that such selections exist, see Artstein and Vitale [1] and references therein. We need a general assumption for our discussion. Assumption 1 E[f (x, y˜) − u]+ is finite for any (x, u) ∈ IRn × IR. Assumption 1 ensures that the objective functions in CVaR minimization (1.4) and mixed CVaR minimization are well defined. Note that if there exists a measurable function κ(˜ y ) such that E[κ(˜ y )] < ∞ and |f (x, y˜)| ≤ κ(˜ y ) for all x ∈ X and y˜ ∈ Ξ, then Assumption 1 holds. For a locally Lipschitz continuous function Φ : Θ ⊆ U → W , where Θ is open, let DΦ denote the set where Φ is differentiable. Then, the B-subdifferential of Φ at u0 ∈ Θ, denoted by ∂B Φ(u0 ), is the set of V such that V = lim JΦ(uk ), k→∞

(2.11)

where uk ∈ DΦ converges to u0 . Hence, Clarke’s generalized Jacobian of Φ at u0 is the convex hull of ∂B Φ(u0 ) [6], i.e., ∂Φ(u0 ) = conv{∂B Φ(u0 )}.

2.2

Optimality Conditions

Let z := (x, u). With a little abuse of notations, we write the random vector y˜ simply as y, 1 η(z, α) := η(x, u, α) = u + 1−α E[f (x, y) − u]+ , and Z := X × IR. Recall that minimizing CVaR 4

over X is equivalent to solving the following problem: min η(z, α) := E[g(z, α, y)],

(2.12)

z∈Z

1 where g(z, α, y) := u + 1−α [f (x, y) − u]+ . Note that (2.12) is a convex problem since Z is a convex set and η(·, α) is a convex function. Let NZ (¯ z ) denote the normal cone of Z at z¯. Then, according to [17, Theorem 23.8], the optimality condition of problem (2.12) can be written as

0 ∈ ∂z E[g(z, α, y)] + NZ (z).

(2.13)

Note that, due to the convexity of the problem under consideration, we have ∂z E[g(z, α, y)] = E[∂z g(z, α, y)]. It is known that (2.13) can be further relaxed as: Ã ! 0 1 0∈ + E[A(z, y)] + NZ (z), (2.14) 1−α 1 where

( Ã A(z, y) :=

ν

∂x f (x, y) −1

!

) : ν ∈ ∂ (max{0, t}) , t = f (x, y) − u . n+1

It is not hard to show that the set-valued mapping A(·, y) : Z → 2IR is upper semicontinuous, which is an important property in analyzing the convergence of the stationary points. We say z ∈ Z a weak stationary point of (2.12) if it satisfies (2.14) and z ∈ Z a stationary point of (2.12) if it satisfies (2.13). Obviously, if a point is a stationary point of (2.12), this point must be a weak stationary point but not vice versa. Similarly, we consider the optimality conditions for the mixed CVaR minimization problem. We still use the notation z to denote (x, u1 , . . . , uJ ) and Z to denote X × IR × · · · × IR. Recall that (1.9) is equivalent to µ ¶ ¶ µ 1 1 E[f (x, y) − uJ ]+ min λ1 u1 + E[f (x, y) − u1 ]+ + · · · + λJ uJ + 1 − α1 1 − αJ s. t. x ∈ X , u1 , · · · , uJ ∈ IR. (2.15) We say z ∈ Z a weak stationary point of problem (2.15) if it satisfies à ! 0 λJ λ1 E[A1 (z, y)] + · · · + E[AJ (z, y)] + NZ (z), 0∈ + 1 − α1 1 − αJ λ

(2.16)

) ! ∂ f (x, y) x : ν ∈ ∂ (max{0, t}) , t = f (x, y) − ui , where λ = (λ1 , . . . , λJ )T , Ai (z, y) := ν βi βi is a vector of all zero entries except its i-th entry being −1, i.e., βi = (0, . . . , −1, . . . , 0)T ∈ IRJ . We say z ∈ Z a stationary point of problem (2.15) if it satisfies ( Ã

0 ∈ λ1 E[∂z g1 (x, u1 , α1 , y)] + · · · + λJ E[∂z gJ (x, uJ , αJ , y)] + NZ (z), where gi (x, ui , αi , y) := ui +

1 1−αi [f (x, y)

− ui ]+ , i = 1, . . . , J. 5

(2.17)

2.3

The SAA Counterparts

We now consider the SAA method for solving CVaR minimization problem. For the single CVaR case, let {y 1 , y 2 , · · · , y N } be an independent identically distributed (i.i.d) sample of y. The SAA program of (2.12) is as follows: ) ( N 1 X g(z, α, y i ) | z ∈ Z . (2.18) min gN (z, α) := N i=1

Clearly, (2.18) is a deterministic optimization problem. In order to solve the original problem (2.12), the SAA method solves a sequence of (2.18) and let the sample size N increase. Note that, in the analysis, α ∈ (0, 1) is taken as a fixed scalar. Hence, unless otherwise specified, the underlying derivatives are taken with respect to z. A generalized Karush-Kuhn-Tucker (GKKT) condition for (2.18) can be stated as 0 ∈ ∂z gN (z, α) + NZ (z). We say z is a weak GKKT point of (2.18) if it satisfies à ! N X 0 1 A(z, y i ) + NZ (z). (2.19) 0∈ + N (1 − α) 1 i=1

Similarly, for the mixed CVaR minimization case, let {y 1 , y 2 , · · · , y N } be an i.i.d sample of y. The SAA program of (2.15) is as follows: Ã Ã ! ! N N X 1 1 X min gN (z, α1 , . . . , αJ ) := λ1 g1 (z, α1 , y i ) + · · · + λJ gJ (z, αJ , y i ) . (2.20) z∈Z N N i=1

i=1

We say z is a weak GKKT point of (2.20) if it satisfies à ! N N X X 0 λ1 λJ i 0∈ + A1 (z, y ) + · · · + AJ (z, y i ) + NZ (z). (1 − α1 )N (1 − αJ )N λ i=1 i=1

(2.21)

With the notions of stationary points and by the strong law of large numbers, one can derive the convergence of the GKKT points of SAA method, which can be found in [12]. In next section, we shall apply the smoothing method in combination with SAA approach to the original CVaR minimization problems under consideration.

3

A Smoothing SAA Method

In this section, we use some smoothing techniques to overcome computational difficulties arising from the nonsmoothness of problems (2.12) and (2.15). For simplicity in analysis, we mainly investigate the convergence of the smoothing SAA method in the case when f (·, y) is continuously differentiable and the distribution of random variable y is continuous. At the end of this section, we briefly discuss the situations when the loss function is nondifferentiable and y follows 6

the discrete distribution, respectively. Since the arguments on the mixed CVaR minimization problem are similar to those for single CVaR problem, we shall concentrate on the single case and omit the details of the mixed case.

3.1

The Case of Single CVaR

Note that the nonsmoothness of the objective η in (2.12) is essentially due to that of the plus function [·]+ . The following are three well-known smoothing functions for the plus function [t]+ , see [15] for instance. • The neural network smoothing function: Φ(ε, t) = t + ε ln(1 + e−t/ε ); • The Chen-Harker-Kanzow-Smale (CHKS) smoothing function: √ 4ε2 + t2 + t ; Φ(ε, t) = 2 • The uniform smoothing function:   t        1 Φ(ε, t) = (t + ε/2)2  2ε        0

(3.22)

(3.23)

if t ≥ ε/2 if − ε/2 < t < ε/2

(3.24)

if t ≤ −ε/2.

Here ε > 0 is the smoothing parameter. Using the smoothing function Φ of [·]+ , we then define the following function η˜(ε, z, α) := u +

1 E[Φ(ε, f (x, y) − u)]. 1−α

(3.25)

Then, η˜ can be rewritten as η˜(ε, z, α) = E[ψ(ε, z, α, y)], where ψ(ε, z, α, y) := u +

1 Φ(ε, f (x, y) − u). 1−α

Note that η˜ is convex in z if f (x, y) is convex in x. Hence, one can employ the well-developed smoothing Newton methods for solving the resulting optimization problem. For further discussions on smoothing functions and smoothing techniques, see [15] and the references therein. In the following, we consider the smoothed counterpart using the CHKS function. We have the following results.

7

Proposition 3.1 Given ε > 0 and α ∈ (0, 1). Suppose that: (a) f (x, y) is twice continuously differentiable in x w.p.1; (b) E[∇x f (x, y)] is well defined and ∇x f (x, y) is dominated by an integrable function κ2 (y). Then, η˜(ε, z, α) is continuously differentiable in z and ∇z η˜(ε, z, α) = E[∇z ψ(ε, z, α, y)]. Further, suppose that (c) E[∇2xx f (x, y)] and E[∇x f (x, y)∇x f (x, y)T ] are well defined and ∇2xx f (x, y) and ∇x f (x, y)∇x f (x, y)T are dominated by an integrable function κ3 (y). Then, η˜(ε, z, α) is twice continuously differentiable in z and ∇2zz η˜(ε, z, α) = E[∇2zz ψ(ε, z, α, y)]. 1 (f (x, y)−u) and h2 (ε, z, α, y) := Proof. Let h1 (ε, z, α, y) := u+ 2(1−α) Then, ψ(ε, z, α, y) = h1 (ε, z, α, y) + h2 (ε, z, α, y), and

1 2(1−α)

p

4ε2 + (f (x, y) − u)2 .

η˜(ε, z, α) = E[h1 (ε, z, α, y)] + E[h2 (ε, z, α, y)].

(3.26)

So, to show the desired results, it suffices to consider the two expectation items in (3.26) respectively. By [24, Proposition 3.2] and assumption, it is clear that E[h1 (ε, z, α, y)] is twice continuously differentiable with ∇z E[h1 (ε, z, α, y)] = E[∇z h1 (ε, z, α, y)],

∇2zz E[h1 (ε, z, α, y)] = E[∇2zz h1 (ε, z, α, y)]. (3.27)

We now study the differentiable properties of the second term in (3.26). Observe that à ! ∇x f (x, y) f (x, y) − u p ∇z h2 (ε, z, α, y) = −1 2(1 − α) 4ε2 + (f (x, y) − u)2 and ! à 2 f (x, y) 0 ∇ xx ∇2zz h2 (ε, z, α, y) = 0 0 à ∇x f (x, y)∇x f (x, y)T 4ε2 p + −∇x f (x, y)T 2(1 − α) (4ε2 + (f (x, y) − u)2 )3 f (x, y) − u p 2(1 − α) 4ε2 + (f (x, y) − u)2

! −∇x f (x, y) . 1

Again, by [24, Proposition 3.2] together with the assumption, E[h2 (ε, z, α, y)] is twice continuously differentiable with ∇z E[h2 (ε, z, α, y)] = E[∇z h2 (ε, z, α, y)],

∇2zz E[h2 (ε, z, α, y)] = E[∇2zz h2 (ε, z, α, y)]. (3.28)

Thus, it follows from (3.27) and (3.28) that ∇z E[h(ε, z, α, y)] = ∇z E[h1 (ε, z, α, y)] + ∇z E[h2 (ε, z, α, y)] = E[∇z h1 (ε, z, α, y)] + E[∇z h2 (ε, z, α, y)] = E[∇z h(ε, z, α, y)]

8

and ∇2zz E[h(ε, z, α, y)] = ∇2zz E[h1 (ε, z, α, y)] + ∇2zz E[h2 (ε, z, α, y)] = E[∇2zz h1 (ε, z, α, y)] + E[∇2zz h2 (ε, z, α, y)] = E[∇2zz h(ε, z, α, y)], which lead to the desired results immediately. Next, we study the following smoothed problem of (2.12). min E[ψ(ε, z, α, y)] z∈Z

(3.29)

Based on the above discussion, η˜ is twice continuously differentiable in ε and z. Suppose further that f (·, y) is convex for a.e. y ∈ Ω. It is not difficult to show that the smoothed function η˜(ε, ·, α) is convex in z. Therefore, (3.29) is a smoothed convex problem. Proposition 3.2 Given α ∈ (0, 1) and ε > 0. Suppose that: (a) X is convex; (b) f (x, y) is continuously differentiable in x w.p.1, E[∇x f (x, y)] is well defined, and ∇x f (x, y) is dominated by an integrable function; (c) f (·, y) is convex for a.e. y ∈ Ω. Then, a point z¯ ∈ Z is an optimal solution of (3.29) if and only if the following condition holds: 0 ∈ E [∇z ψ(ε, z¯, α, y)] + NZ (¯ z ) = E [H(ε, z¯, α, y)] + NZ (¯ z ),

(3.30)

where p    H(ε, z, α, y) :=   

 4ε2 + (f (x, y) − u)2 + f (x, y) − u p ∇x f (x, y) 2(1 − α) 4ε2 + (f (x, y) − u)2   . p  2 2 4ε + (f (x, y) − u) + f (x, y) − u  p 1− 2(1 − α) 4ε2 + (f (x, y) − u)2

(3.31)

Proof. With the help of Proposition 3.1, it is not hard to derive the desired result. We omit the details for brevity. Let {y 1 , y 2 , · · · , y N } be an i.i.d sample of y. Then, the SAA program of the smoothed problem (3.29) is as follows: ( ) N 1 X min ψN (ε, z, α) := ψ(ε, z, α, y i ) | z ∈ Z (3.32) N i=1

So, the GKKT conditions of (3.32) can be written as 0∈

N 1 X H(ε, z, α, y i ) + NZ (z). N i=1

9

(3.33)

We say z is a stationary point of the smoothed problem (3.29) if it satisfies (3.30) and say z is a stationary point of the smoothed SAA problem (3.32) if it satisfies (3.33). The following is the main result of this section. Theorem 3.1 Let α ∈ (0, 1). Suppose all conditions in Proposition 3.2 are satisfied. (i) For ε 6= 0 fixed, let {zN (ε)} be a sequence of stationary points that satisfies (3.33) and z ∗ (ε) be an accumulation point. Assume that there exists a compact set Z containing a neighborhood of z ∗ (ε) such that k∇x f (x, y)k ≤ κ4 (y) for any z(= (x, u)) ∈ Z, where p p E[ (1 − 2α)2 + κ4 (y)2 + (1 + κ4 (y)2 ] < ∞. Then, w.p.1. z ∗ (ε) is a stationary point of (3.29). (ii) Let ε = εN with εN → 0 as N → ∞. Let {z(εN )} be a sequence of stationary points that satisfies (3.33) and z ∗ be an accumulation point. Assume that there exists a compact set Z containing a neighborhood of z ∗ such that k∇x f (x, y)k ≤ κ4 (y) for any z ∈ Z, where p p E[ (1 − 2α)2 + κ4 (y)2 + (1 + κ4 (y)2 ] < ∞. Then, w.p.1. z ∗ is a weak stationary point of (2.12), namely, it satisfies (2.14). Proof. Part (i). Evidently, H(ε, ·, α, y) is continuous on Z for every y ∈ Ω. So, according to (3.31), and by assumption, for any z ∈ Z, kH(ε, ·, α, y)k ≤

i hp p 1 (1 − 2α)2 + κ4 (y)2 + 1 + κ4 (y)2 , 2(1 − α)

(3.34)

in which the right hand side is an integral random function. Without loss of generality, assume that zN (ε) → z ∗ (ε) as N → ∞. Since zN (ε) is a stationary point of problem (3.32), we have 1 PN H(ε, zN (ε), α, y i ) + NZ (zN (ε)). Then, we have 0∈ N i=1 0∈

N N 1 X 1 X H(ε, z ∗ (ε), α, y i ) + H(ε, zN (ε), α, y i ) − E[H(ε, zN (ε), α, y)] N N i=1

+E[H(ε, zN (ε), α, y)] −

1 N

i=1 N X

H(ε, z ∗ (ε), α, y i ) + NZ (zN (ε)).

i=1

By the strong law of large numbers, it yields that N 1 X lim H(ε, z ∗ (ε), α, y i ) = E[H(ε, z ∗ (ε), α, y)], w.p.1. N →∞ N i=1

10

Thus, by the Lebesgue dominated convergence theorem again and the strong law of large numbers, we have # " N 1 X lim E[H(ε, zN (ε), α, y)] − H(ε, z ∗ (ε), α, y i ) N →∞ N i=1 · ¸ N 1 X = E lim H(ε, zN (ε), α, y) − lim H(ε, z ∗ (ε), α, y i ) N →∞ N →∞ N i=1

= E[H(ε, z ∗ (ε), α, y)] − E[H(ε, z ∗ (ε), α, y)] = 0. In addition, note that for N sufficiently large, ° ° N ° °1 X ° ° H(ε, zN (ε), α, y i ) − E[H(ε, zN (ε), α, y)]° ° ° °N i=1

∞

° ° N °1 X ° ° ° ≤ max ° H(ε, z, α, y i ) − E[H(ε, z, α, y)]° , ° z∈Z ° N i=1

∞

L∞ -norm

where k · k∞ denotes the usual (maximum norm) of a vector. By [21], the last term tends to zero w.p.1 as N → ∞. Thereby, ° ° N °1 X ° ° ° H(ε, zN (ε), α, y i ) − E[H(ε, zN (ε), α, y)]° → 0 ° ° °N i=1

∞

as N → ∞. Thus, 0 ∈ E[H(ε, z ∗ (ε), α, y)] + NZ (z ∗ (ε)) w.p.1. This shows part (i). Part (ii). Note that H(ε, ·, α, y) is dominated by an integral random function on Z as given in (3.34). Without loss of generality, we assume that z(εN ) → z ∗ as N → ∞. To show z ∗ is a weak stationary point of (2.12), we only need to show ¯ ∗ , y)] + NZ (z ∗ ), 0 ∈ E[A(z where

(3.35)

 ν   Ã !   ∇ f (x, y) x ¯ y) := 0 + 1 A(z, y) =  1 − α  A(z, : ν ∈ ∂(max{0, t}), t = f (x, y) − u . ν   1−α 1 1− 1−α Note that for z ∈ Z and y ∈ Ω,    1  ∇ f (x, y)      2(1 − α) x  if f (x, y) − u = 0;  ,   1   1 −   2(1 − α)           1 H(0, z, α, y) := lim H(ε, z, α, y) =  1 − α ∇x f (x, y)  ε→0+  if f (x, y) − u > 0;  ,  1   1−    1−α      Ã !     0   , if f (x, y) − u < 0.  1 11

According to the assumption, it follows that for z ∈ Z ½ ¾ p 1 p 1 2 2 2 2 kH(0, z, α, y)k ≤ max 1, k∇x f (x, y)k + (1 − 2α) , k∇x f (x, y)k + α 1−α 2(1 − α) ½ ¾ 1 p 1 p ≤ max 1, κ4 (y)2 + (1 − 2α)2 , κ4 (y)2 + 1 1−α (1 − α) ½ h i¾ p p 1 2 2 2 ≤ max 1, κ4 (y) + (1 − 2α) + κ4 (y) + 1 , 1−α hence, H(0, ·, α, y) is dominated by an integral random function on Z, thereby, E[H(0, z, α, y)] ¯ y) for any z ∈ Z is well-defined for z ∈ Z. Additionally, it is easy to see that H(0, z, α, y) ⊂ A(z, ¯ y)] for z ∈ Z. Thus, to show (3.35) and y ∈ Ω, which leads to that E[H(0, z, α, y)] ⊂ E[A(z, holds, it suffices to prove that 0 ∈ E[H(0, z ∗ , α, y)] + NZ (z ∗ ). On the other hand, let ε0 > 0 be a fixed small number. Evidently, H(·, ·, α, y) : [0, ε0 ) × Z → IRn+1 is a continuous vector-valued mapping for every y ∈ Ω. In addition, based on the above discussion and part (i), it follows that H(·, ·, α, y) is dominated by an integral random function for (ε, z) ∈ [0, ε0 ) × Z. Since z(εN ) is a stationary point of problem (3.32), we have 0∈

N 1 X H(εN , z(εN ), α, y i ) + NZ (z(εN )). N

(3.36)

i=1

Then, N N 1 X 1 X H(0, z ∗ , α, y i ) + H(εN , z(εN ), α, y i ) − E[H(εN , z(εN ), α, y)] N N

0 ∈

i=1

i=1

N 1 X H(0, z ∗ , α, y i ) + NZ (z(εN )). +E[H(εN , z(εN ), α, y)] − N i=1

By the strong law of large numbers and Lebesgue dominated convergence theorem, for the first, fourth, and fifth terms on the right hand side of the above equation, we have N 1 X H(0, z ∗ , α, y i ) = E[H(0, z ∗ , α, y)], lim N →∞ N

w.p.1.

(3.37)

i=1

" lim

N →∞

# N 1 X ∗ i E[H(εN , z(εN ), α, y)] − H(0, z , α, y ) = E[ lim H(εN , z(εN ), α, y)] − N →∞ N

1 N →∞ N lim

i=1

N X

H(0, z ∗ , α, y i )] = E[H(0, z ∗ , α, y)] − E[H(0, z ∗ , α, y)] = 0,

i=1

12

w.p.1.

(3.38)

1 PN H(εN , z(εN ), α, y i ) − E[H(εN , z(εN ), α, y)]k. Note that Finally, we estimate the term k N i=1 for N large enough, we have (εN , z(εN )) ∈ [0, ε0 ] × Z. Then k

N 1 X H(εN , z(εN ), α, y i ) − E[H(εN , z(εN ), α, y)]k∞ N i=1

N 1 X ≤ max k H(ε, z, α, y i ) − E[H(ε, z, α, y)]k∞ . N (ε,z)∈[0,ε0 ]×Z i=1

1 PN Again, by [21], limN →∞ max(ε,z)∈[0,ε0 ]×Z k H(ε, z, α, y i ) − E[H(ε, z, α, y)]k∞ = 0, w.p.1, N i=1 thus, lim k

N →∞

N 1 X H(εN , z(εN ), α, y i ) − E[H(εN , z(εN ), α, y)]k = 0, N

w.p.1

(3.39)

i=1

Therefore, by (3.36) together with (3.37), (3.38), and (3.39), w.p.1 z ∗ satisfies 0 ∈ E[H(0, z ∗ , α, y)] + NZ (z ∗ ), which implies that w.p.1 z ∗ satisfies (3.35), that is, z ∗ is a weak stationary point of (2.12). This completes the proof. Note that in the above analysis, we do not specify the feasible region X . In fact, X can sometimes be defined by finitely many equality and/or inequality constraint functions. In this case, one can derive the optimality condition using the Lagrangian function as discussed in classical nonlinear programming. In terms of numerical computation, we can apply some welldeveloped Newton methods on nonsmooth equations or nonsmooth optimization problems and let the smoothing parameter ε tend to zero, see [15, 16], for instance.

3.2

The Case of nondifferentiable loss functions

In practice, the loss function f (·, y) might not be differentiable. For instance, f (·, y) is piecewise linear/quadratic or even piecewise smooth sometimes. In this case, we resort to some welldeveloped smoothing techniques. One popular way to deal with the nonsmooth functions is the so-called smoothing function technique. Recall that for a locally Lipschitz continuous function ˆ ν) : IRl ×IR → IR is called a smoothing function of ψ if it satisfies the following: ψ : IRl → IR, ψ(s, ˆ 0) = ψ(s); (i) for every s ∈ IRl , ψ(s, (ii) for every s ∈ IRl , ψˆ is locally Lipschitz continuous at (s, 0); (iii) ψˆ is continuously differentiable on IRl × IR \ {0}.

13

where ν ∈ IR is called a smoothing parameter. Note that it is easy to verify the CHKS smoothing function discussed earlier satisfies these three conditions. In practice, we often consider ν > 0, which is driven to zero in computation. For a given nondifferentiable function, one can construct its smoothing counterpart, see [15] for a detailed discussion. Now, let fˆ(·, ν, y) be the smoothing function of f (·, y). We then derive the following problem: min E[ˆ g (z, ν, α, y)]

(3.40)

z∈Z

where gˆ(z, ν, α, y) := u +

1 [fˆ(x, ν, y) − u]+ , 1−α

and ν > 0 is the smoothing parameter. Clearly, problem (3.40) possesses the same setting as discussed previously in this section. Thus, we can apply the same arguments, such as using the CHKS smoothing function, to (3.40), and derive a smoothed SAA program. Similarly, for a (small) fixed positive scalar ν, we can derive the convergence of the corresponding smoothing SAA method. Here we omit the details for brevity.

3.3

The Case of mixed CVaR optimization

Following similar arguments above, the smoothed problem of mixed CVaR optimization problem (2.15) is as follows min{λ1 E[ψ1 (ε, z, α1 , y)] + · · · + λJ E[ψJ (ε, z, αJ , y)]} z∈Z

1 Ψ(ε, f (x, y) − ui ) for i = 1, . . . , J, and Ψ is the CHKS smoothing where ψi (ε, z, αi , y) = ui + 1−α i 1 N function. Let {y , · · · , y } be an i.i.d sample of y, then, the corresponding SAA smoothing program is ( ) N N X 1 X 1 min λ1 ψ1 (ε, z, α1 , y i ) + · · · + λJ ψJ (ε, z, αJ , y i ) z∈Z N N i=1

i=1

Note that under assumptions as Theorem 3.1, the above SAA smoothing program is actually a deterministic smoothing problem. With similar discussions as Section 4.1, we can derive the same convergence results as Theorem 3.1 for the smoothing SAA programs. We omit the details here.

3.4

Discussions on the discrete distribution case

We now consider the case where the random variable y follows a discrete distribution with S finitely many scenarios {y 1 , y 2 , · · · , y S }, each with probability pi , i = 1, · · · , S. Here, we only 14

discuss the mixed CVaR minimization problem (1.10). Note that in this case (1.10) can be written as ! Ã ! Ã S S 1 X 1 X i i + + pi [f (x, y ) − uJ ] pi [f (x, y ) − u1 ] + · · · + λJ u1 + min λ1 u1 + 1 − α1 1 − αJ i=1

i=1

s. t.

(x, u1 , . . . , uJ ) ∈ X × IR × · · · × IR. (3.41)

Since problem (3.41) is nonsmooth, using the CHKS smoothing function, we then derive the corresponding smoothing problem below: Ã ·q ¸! S X 1 i 4ε2 + (f (x, y i ) − u1 )2 + f (x, y ) − u1 + ··· + min λ1 u1 + pi 2(1 − α1 ) i=1 Ã ·q ¸! S X 1 λJ uJ + pi 4ε2 + (f (x, y i ) − uJ )2 + f (x, y i ) − uJ 2(1 − αJ ) i=1

s. t.

(x, u1 , . . . , uJ ) ∈ X × IR × · · · × IR, (3.42)

where ε > 0 is the smoothing parameter and driven to zero. Note that the size of the smoothed problem (3.42) keeps the same as (3.41) and the solution of (3.42) tends to that of problem (3.41) as ε tends to zero. Regarding the computational aspect, we can employ some well-developed smoothing Newton methods established in the past few years for solving (3.42), see [5, 15] for instance.

4

Computational Examples in Logistics Management

In this section, we first consider a problem concerning the supply chain of a wine company. where the random variable follows a discrete distribution. Then, we discuss a distribution mix problem under uncertain demand with a continuous distribution.

4.1

Supply Chain Management of a Wine Company

This example is modified from Yu and Li [28]. In the company’s supply chain, there are four raw material providers (uniform-quality wine in bulk) located in A, B, C, and D, and three bottling plants located in E, F, and G, and three distribution warehouses located in three different cities L, M, and N, respectively. We suppose that each market demand merely depends on the local economic condition, which is treated as an uncertain factor following a discrete distribution with four situations: boom, good, fair, or poor and associated probabilities of 0.45, 0.25, 0.17, or 0.13, respectively. 15

The unit production costs of three bottling plants and market demands from customers under each situation are listed in Table 1.

Economy situation

Demands L M N

Unit product costs E F G

Likelihood of economic situation

Boom

400

188

200

755

650

700

0.45

Good

350

161

185

700

600

650

0.25

Fair

280

150

160

675

580

620

0.17

Poor

240

143

130

650

570

600

0.13

Table 1: Basic Problem Profile Notations • i – index of four wineries, i = 1, 2, 3, 4 • j – index of the three distribution centers, j = 1, 2, 3 • k – index of three bottling plants, k = 1, 2, 3 • s – uncertain economic situation with four scenarios {Boom, Good, F air, P oor} denoted by {1, 2, 3, 4} each associated with the probability ps , s ∈ {1, 2, 3, 4} • xik – amount of bulk wine to be shipped from winery i to bottling plant k 1 – delivery cost for shipment from winery i to bottling plant k • qij

• ρi – capacity of bulk wine of winery i • yk – amount of bulk wine to be bottled in bottling plant k • zkj – amount of bottled wine shipped from bottling plant k to distribution center j 2 – delivery cost from bottling plant k to distribution center j • qkj

• νk – production capacity of bottling plant k • h1k – unit inventory cost for bulk wine in bottling plant k • πk – storage capacity of bulk wine in bottling plant k • %k – storage capacity of bottled wine in bottling plant k • h2k – unit inventory cost for bottled wine in bottling plant k • uk – stock of bulk wine in bottling plant k carried from previous weeks • vk – stock of bottled wine in bottling plant k carried from previous weeks

16

• cj (s) – uncertain unit cost for bottling wine in bottling plant j • Dj (s) – the uncertain market demand from the distribution center j • ξ(s): underlying random variables of the problem, i.e., ξ(s) = (D(s), c(s)) where D(s) = (D1 (s), D2 (s), D3 (s)) and c(s) = (c1 (s), c2 (s), c3 (s)) Problem Formulation We suppose that the company would like to determine the optimal amounts of raw materials ordered from the providers, bottled wine produced in the plants, amounts of bottled wine shipped to the suppliers together with the optimal stock levels of the raw materials and bottled wine, so as to minimize the total cost using the measure of conditional value at risk, subject to the satisfaction of necessary requirements on the resource capacity of the company and the demand of the suppliers of the supply chain. In general, the total cost f concerning the problem includes the following three parts: i.e., transportation cost fT , production cost fP , and inventory cost fI , which are stated as follows. fT (x, y, z, u, v, ξ(s)) :=

4 X 3 X

1 qij xij +

i=1 j=1

fP (x, y, z, u, v, ξ(s)) :=

3 X

3 X 3 X

2 zjk , qjk

j=1 k=1

cj (s)yj ,

j=1

fI (x, y, z, u, v, ξ(s)) :=

3 X

h1k

k=1

µX 4

¶ xik − yk + uk

+

i=1

3 X

µ h2k

yk + vk −

3 X

¶ zkj .

j=1

k=1

Then, the total cost function gives f (x, y, z, u, v, , ξ(s)) = fT (x, y, z, u, v, ξ(s)) + fP (x, y, z, u, v, ξ(s)) + fI (x, y, z, u, v, ξ(s)) 4 X 3 3 X 3 3 X X X 1 2 = qij xij + qjk zjk + cj (s)yj i=1 j=1

+

3 X k=1

h1k

µX 4

j=1 k=1

xik − yk + uk

i=1

j=1

¶ +

3 X k=1

µ h2k

yk + vk −

3 X

¶ zkj .

j=1

Here, we treat f as the loss function discussed earlier in which a random factor ξ actually follows a discrete distribution. Then, the CVaR minimization can be formulated as follows: min CVaRα (x, y, z, u, v) s.t. xi1 + xi2 + xi3 ≤ ρi , i = 1, 2, 3, 4, yk ≤ νk , k = 1, 2, 3, 0 ≤ x1k + x2k + x3k + x4k + uk − yk ≤ πk , k = 1, 2, 3, 0 ≤ yk + vk − zk1 − zk2 − zk3 ≤ %k , k = 1, 2, 3, 0 ≤ z1j + z2j + z3j − Dj (s), ∀ s ∈ {1, 2, 3, 4}, j = 1, 2, 3, xik , yk , zkj , uk , vk ≥ 0, i = 1, 2, 3, 4, k = 1, 2, 3, j = 1, 2, 3. 17

(4.43)

As discussed earlier, (4.43) is equivalent to 1 E[f (x, y, z, u, v, ξ) − γ]+ 1−α s.t. xi1 + xi2 + xi3 ≤ ρi , i = 1, 2, 3, 4, yk ≤ νk , k = 1, 2, 3, 0 ≤ x1k + x2k + x3k + x4k + uk − yk ≤ πk , k = 1, 2, 3, 0 ≤ yk + vk − zk1 − zk2 − zk3 ≤ %k , k = 1, 2, 3, 0 ≤ z1j + z2j + z3j − Dj (s), ∀ s ∈ {1, 2, 3, 4}, j = 1, 2, 3, γ ∈ IR, xik , yk , zkj , uk , vk ≥ 0, i = 1, 2, 3, 4, k = 1, 2, 3, j = 1, 2, 3.

min γ +

(4.44)

Noticing that ξ has a discrete distribution with four scenarios, we then derive the following problem using the CHKS smoothing function: hp i P4 1 2 + (f (x, y, z, u, v, ξ(s)) − γ)2 + f (x, y, z, u, v, ξ(s)) − γ p 4ε s 2(1 − α) s=1 s.t. xi1 + xi2 + xi3 ≤ ρi , i = 1, 2, 3, 4, yk ≤ νk , k = 1, 2, 3, (4.45) 0 ≤ x1k + x2k + x3k + x4k + uk − yk ≤ πk , k = 1, 2, 3, 0 ≤ yk + vk − zk1 − zk2 − zk3 ≤ %k , k = 1, 2, 3, 0 ≤ z1j + z2j + z3j − Dj (s), ∀ s ∈ {1, 2, 3, 4}, j = 1, 2, 3, γ ∈ IR, xik , yk , zkj , uk , vk ≥ 0, i = 1, 2, 3, 4, k = 1, 2, 3, j = 1, 2, 3.

min γ +

Numerical Results The numerical test is carried out by implementing mathematical programming codes in MATLAB 6.5 installed in a PC with the Windows XP operating system. We use the MATLAB built-in solver fmincon for solving the associated problems. We set the smoothing parameter ε = 10−5 and the confidence level α = 0.98. In Table 2, we report the amounts of bulk wine to be shipped to the plants and bottled wine in the plants x, y, the amount of bottled wine shipped to the distributors z, and the corresponding delivery cost, the production cost, the inventory holding cost, the total cost, the minimum value of CVaRα . Note that, in this example we set ε = 10−5 , which is a very small value. However, for general smoothing optimization problems, it might be difficult to obtain the desired solution with such a small parameter. In this case, people are suggested to explore other techniques and methods in numerical optimization to tackle this

18

issue. The values of the parameters are set below: h1 = (75, 60, 69) h2 = (125, 100, 108) ρ = (300, 150, 200, 150) ν = (340, 260, 300) π = (150, 100, 120) % = (50, 50, 50) q1 = (66, 156, 65, 62, 151, 60, 84, 175, 88, 111, 101, 109) q2 = (201, 301, 700, 693, 533, 362, 165, 306, 598) Item

Results

x

(0, 0, 25.36, 29.93, 32.53, 25.35, 29.93, 32.52, 25.35, 29.9332.52, 25.35)

y

(119.74, 117.67, 123.36)

z

(136.51, 64.06, 68.16, 135.98, 63.53, 67.64, 137.46, 65.02, 69.12)

CVaRα

1.04

VaRα

0.55

delivery cost

360 510

production cost

239 410

holding cost

3 570

total cost

603 490 Table 2: Numerical results using the smoothing method

4.2

A Distribution Mix Problem with Uncertain Demand

In this subsection, we investigate another application of CVaR to the logistics industry. The problem under consideration is the distribution mix problem with uncertain demand. We attempt to investigate the service levels of firms in supply chains with uncertain demand, using the CVaR risk measure. In a typical supply chain, involving parts delivery and open market sales, for example, it is normal for companies to keep an inordinate amount of stock, anticipating customer demand and responding to that demand in the shortest time possible. At the same time, the firm has an obligation to keep a certain amount of stock either as policy or stipulation. In what follows, we discuss briefly the demand of the company and frame the problem. Problem Formulation In this particular problem, the monthly demand for parts is assumed to be unusually spiky 19

and the company has to make a decision on the optimal amount of stock to hold without compromising service level. So, the decision variables in our case would be, x1 , x2 , . . . , xn , which would represent the amount of stock to keep for item 1, . . . , item n, to at least meet demand for that period. It is assumed that when the customer demands are unmet, they are lost forever and the company bears that loss. Hence, it is desirable for companies to minimize the total loss subject to some necessary constraints. Let x = (x1 , x2 , · · · , xn )T , y = (y1 , y2 , . . . , yn )T with yi representing the random demand of item i from the customers, and set p = (p1 , p2 , . . . , pn )T where each pi denotes the unit sale price of item i, i = 1, 2, . . . , n. Then, the failure to provide for xi amount of stock of item i is max{0, yi − xi } when it is needed by the customers, i = 1, 2, . . . , n. Thereby, the loss function f associated with the total n items is given as f (x, y) :=

n X

pi max{0, yi − xi }.

i=1

Given a confidence level α ∈ (0, 1), the objective of the problem is to minimize the conditional value of risk associated with the loss f . For simplicity, here the feasible set is denoted by a compact set X ∈ IRn , which is defined by the underlying constraint functions. Hence, this problem can be formulated as follows: min{CVaRα (x) | x ∈ X}.

(4.46)

We further assume that the random demand y from the customers follows a certain continuous distribution. Note that the above problem is equivalent to ½ ¾ 1 + min u + E[f (x, y) − u] | (x, u) ∈ X × IR . (4.47) 1−α Note that model (4.46) is a very general formulation for the distribution mix problem in that here set X can be broad enough to contain the necessary constraints depending on some specific requirements of the entire problem in practice. For the sake of simplicity in analysis, here we use the abstract set X in the model. Further, in many practical cases, most service units of a firm would normally assign a 95% confidence level of meeting a demand. This means that we can impose α = 0.95 in the model. Numerical Results We have carried out numerical tests on the distribution mix problem with uncertain demand and report some numerical results. The test code is written in MATLAB 6.5 installed in a PC with Windows XP operating system. We use the MATLAB built-in solver fmincon for solving the associated problems. Note that the smoothing SAA program of (4.47) can be written as:   ¶ N µq  X 1 4ε2 + (f (x, y j , ε) − u)2 + f (x, y j , ε) − u min , (4.48) u+  2N (1 − α) (x,u)∈X×IR  j=1

20

where {y 1 , . . . , y N } is an i.i.d sample of y, ε ³> 0 is the smoothing parameter driven to zero in ´ p 1 Pn the computation, and f (x, y, ε) := 2 i=1 pi 4ε2 + (yi − xi )2 + yi − xi . Set ψ(x, u, y, ε) := u +

³p ´ 1 4ε2 + (f (x, y, ε) − u)2 + f (x, y, ε) − u . 2(1 − α)

Then, the smoothing problem of (4.47) can be rewritten as min {E[ψ(x, u, y, ε)] | (x, u) ∈ X × IR} .

(4.49)

The following methodology of constructing statistical lower and upper bounds can be found in [13, 27]. Given ε > 0, let v(ε) denote the optimal value of problem (4.49) and vN (ε) the optimal value of (4.48). It is known that E[vN (ε)] ≤ v(ε). To estimate the expected value E[vN (ε)], we generate M independent samples of y, j = 1, 2, . . . , M , each of size N . For each sample j, solve the corresponding SAA problem ( ) N 1 X min ψ(x, u, y ij , ε) | (x, u) ∈ X × IR , N i=1

j (ε), j = 1, 2, . . . , M , denote the where {y 1j , y 2j , . . . , y Nj } denotes the sample j of y. Let vN 1 PM j corresponding optimal value of the above problem. Compute LN,M (ε) := v (ε), which M j=1 N is an unbiased estimate of E[vN (ε)]. Then, LN,M provides a statistical lower bound of v(ε). An estimate of variance of the estimator LN,M (ε) can be computed as M

s2L (M, ε)

X j 1 := (vN (ε) − LN,M (ε))2 . M (M − 1) j=1

Let v˜(x, u, ε) = E[ψ(x, u, y, ε)]. An upper bound for the optimal value v(ε) can be obtained by the fact that v˜(¯ x, u ¯, ε) ≥ v(ε) for any feasible point (¯ x, u ¯). Hence, by choosing (¯ x, u ¯) to be a near-optimal solution, for example, by solving one SAA problem and using an unbiased estimator of v˜(¯ x, u ¯, ε), we can obtain an estimate of an upper bound for v(ε). To do so, we 1 P 0 0 generate an i.i.d sample {y 1 , . . . , y N } of y, let v˜N 0 (¯ x, u ¯, ε) := 0 N x, u ¯, y i , ε). Then, we i=1 ψ(¯ N have E[˜ vN 0 (¯ x, u ¯, ε)] = v˜(¯ x, u ¯, ε), thereby, v˜N 0 (¯ x, u ¯, ε) is an estimate of an upper bound on v(ε). An estimator of variance of the estimator v˜N 0 (¯ x, u ¯, ε) can be computed as 0

s2U (¯ x, u ¯, N 0 , ε)

M X ¡ ¢2 1 := 0 0 ψ(¯ x, u ¯, y i , ε) − v˜N 0 (¯ x, u ¯, ε) . N (N − 1) j=1

Using the lower bound estimate and the objective function value estimate as discussed above, we compute an estimate of the optimality gap of the solution (¯ x, u ¯) and the corresponding estimated variance as follows: GapN,M,N 0 (¯ x, u ¯) := v˜N 0 (¯ x, u ¯, ε) − LN,M (ε), 21

2 SGap := s2L (M, ε) + s2U (¯ x, u ¯, N 0 , ε).

In the numerical test, we consider the distribution mix problem with two different items. The associated random demands y1 and y2 are independent, both having uniform distribution on the closed interval [3.5, 7]. The prices are set as p1 = 0.2 and p2 = 0.3 and X = IR2+ . We conduct our test with different values of the smoothing parameter ε and sample sizes N , M and N 0 . We report the lower and upper bounds, LN,M and v˜N 0 , of v(ε), the sample variances, sL , sU , and the estimate of the optimality gap, Gap, of the solution (¯ xjN , u ¯jN ), the variance of the gap estimator SGap . The test results are displayed in Table 1. The sample points are taken for calculating the lower and upper bounds are N × M and N 0 , respectively. For instance, in the second row of Table 1, the number of sample points for computing LN,M equals to 1500 × 100 = 150000 and the number of sample points for deriving sU is 4000. α .95 .95 .95 .98 .98 .98

ε 10−4 10−5 10−6 10−4 10−5 10−6

N 1500 2000 2000 1500 2000 2000

M 100 100 100 100 100 100

N0 4000 5000 6000 4000 5000 6000

LN,M 1.09e-004 2.20e-005 2.29e-005 3.78e-005 1.77e-005 1.72e-005

sL 1.54e-006 2.83e-006 3.80e-006 2.37e-006 3.45e-005 4.06e-006

(¯ xjN , u ¯jN ) (4.17, 6.01, 0.01) (4.17, 6.00, 0.00) (4.17, 6.00, 0.00) (5.07, 7.35, 0.00) (5.06, 7.34, 0.00) (5.03, 7.30, 0.00)

v˜N 0 1.23e-004 2.34e-005 2.42e-005 3.92e-005 1.91e-005 1.83e-005

sU 1.32e-005 6.23e-007 5.75e-007 1.23e-006 1.09e-008 1.05e-008

Gap 1.78e-006 1.40e-006 1.30e-006 1.41e-006 1.40e-006 1.14e-006

SGap 1.32e-005 2.89e-006 3.84e-006 2.67e-006 3.45e-005 4.06e-006

Table 3: Numerical results using the smoothing SAA method

In conclusion, the above two numerical results show that the smoothing SAA method is easy to implement for solving problems involving the CVaR as a risk measure. Acknowledgement. The authors would like to thank the two anonymous referees for their constructive comments, which help to improve the presentation of the paper.

References [1] Z. Artstein and R. A. Vitale, A strong law of large numbers for random compact sets, Annals of Probability, 3 (1975), pp. 879–882. [2] F. Anderson, H. Mausser, D. Rosen, and S. Uryasev, Credit risk optimization with conditional Value-at-Risk criterion, Mathematical Programming, 89(2001), pp. 273–291. [3] P. Artzner, F. Delbaen, J.M. Eber, and D. Heath, Coherent measures of risk, Mathematical Finance, 9(1999), pp. 203–228. [4] E. Bogentoft, H. E. Romeijn, and S. Uryasev, Asset/liability management for pension funds using CVaR constraints, The Journal of Risk Finance, 3(2001), pp. 57–71. [5] X. Chen, L. Qi, and D. Sun, Global and superlinear convergence of the smoothing Newton’s method and its application to general box constrained variational inequalities, Mathematics of Computation, 67(1998), pp. 519–540. 22

[6] F. H. Clarke, Optimization and Nonsmooth Analysis, John Wiley and Sons, New York, 1983. [7] W. Hurlimann, Conditional Value-at-Risk bounds for compound Poisson risks and a normal approximation, Journal of Applied Mathematics, 3(2003), pp. 141–153. [8] P. Jorion, Value At Risk: the New Benchmark for Controlling Market Risk, McGraw-Hill, New York, 1997; McGraw-Hill International Edition, 2001. [9] P. Krokhmal, J. Palmquist, and S. Uryasev, Portfolio optimization with conditional Value-At-Risk objective and constraints, The Journal of Risk, 4(2002), pp. 43–68. ¨zi-Bay and J. Mayer, Computational aspects of minimizing conditional value-at[10] A. Ku risk, Computational Management Science, 3(2006), pp. 3–27. [11] G. Lin, X. Chen, and M. Fukushima, Solving stochastic mathematical programs with equilibrium constraints via approximation and smoothing implicit programming with penalization, Mathematical Programming, 116 (2009), pp. 343–368. [12] F. Meng, J. Sun, and M. Goh, Stochastic optimization problems with CVaR risk measure and their sample average approximation, working paper, National University of Singapore, 2009. [13] F. Meng and H. Xu, A regularized sample average approximation method for stochastic mathematical programs with nonsmooth equality constraints, SIAM Journal on Optimization, 17(2006), pp. 891–919. [14] K. Natarajan, D. Pachamanova, and M. Sim, Incorporating asymmetric distributional information in robust Value-at-Risk optimization, Management Science, 54(2008), pp–585. [15] L. Qi, D. Sun, and G. Zhou, A new look at smoothing Newton methods for nonlinear complementarity problems and box constrained variational inequalities, Mathematical Programming, 87(2000), pp. 1–35. [16] L. Qi and J. Sun, A nonsmooth version of Newton’s method, Mathematical Programming, 58(1993), pp. 353–367. [17] R. T. Rockafellar, Convex Analysis, Princeton University, 1970. [18] R. T. Rockafellar, Coherent approaches to risk in optimization under uncertainty, Tutorials in Operations Research, INFORMS, 38–61, 2007. [19] R. T. Rockafellar and S. Uryasev, Conditional Value-at-Risk for general loss distributions, Journal of Banking & Finance, 26(2002), pp. 1443–1471. [20] R. T. Rockafellar and S. Uryasev, Optimization of conditional Value-at-Risk, Journal of Risk, 2(2000), pp. 21–41. 23

[21] R. Y. Rubinstein and A. Shapiro, Discrete Event Systems: Sensitivity Analysis and Stochastic Optimization by the Score Function Method, John Wiley and Sons, New York, 1993. ´ski and A. Shapiro, Optimization of convex risk functions, Mathematics of [22] A. Ruszczyn Operations Research, 31(2006), pp. 433-452. [23] A. Shapiro, D. Dentcheva, and A. Ruszczy´ nski, Lectures on Stochastic Programming: Modelling and Theory. SIAM Philadelphia, 2009. [24] A. Shapiro, Stochastic Programming by Monte Carlo Simulation Methods, Published electronically in: Stochastic Programming E-Print Series, 2000. [25] A. Shapiro, Stochastic mathematical programs with equilibrium constraints, Journal of Optimization Theory and Application, 128 (2006), pp. 223–243. [26] D. Tasche, Expected shortfall and beyond, Journal of Banking and Finance, 6(2002), pp. 1519–1533. [27] H. Xu and F. Meng, Convergence analysis of sample average approximation methods for a class of stochastic mathematical programs with equality constraints, Mathematics of Operations Research, 32(2007), pp. 648–668. [28] C.-S. Yu and H. Li, A robust optimization model for stochastic logistic problems, International Journal of Production Economics, 64(2000), pp. 385–397.

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