关于分布式鲁棒优化的学习note,用的是沙田苏文藻老师的notes。
Introduction : Optimization Under Uncertainty
Traditional Approach : Deterministic Problem
Given optimization problem
inf x ∈ C f ( x ) f : R n → R C ⊆ R n \begin{array}{ll}
\inf _{x \in C} f(x) & f: \mathbb{R}^n \rightarrow \mathbb{R} \\
& C \subseteq \mathbb{R}^n
\end{array}
inf x ∈ C f ( x ) f : R n → R C ⊆ R n
Typically, data are deterministic . However, in reality, data are often uncertain.
Example:
Consider manufacturing problem:
c j = c_j= c j = cost of j th j^{\text {th }} j th activity
a i j = a_{i j}= a i j = amount of i i i ch commodity per unit of j t h j^{t h} j t h activity
b i = b_i= b i = output requirement of i th i^{\text {th }} i th commodity
min c ⊤ x s.t. A x = b x ≥ 0 \begin{aligned}
\min && \quad & c^{\top} x \\
\text { s.t. } && A x&=b \\
&& x &\geq 0
\end{aligned}
min s.t. A x x c ⊤ x = b ≥ 0
But , c , A c,A c , A can be uncertain !
Q: How to incoporate such data uncertainty into the optimization model?
Idea: Consider the uncertainty as a pertubation vector ξ ∈ R ℓ \xi \in \mathbb{R}^{\ell} ξ ∈ R ℓ
min c ( ξ ) ⊤ x s.t. A ( ξ ) x = b x ≥ 0 \begin{aligned}
\min&& \quad c(\xi)^{\top}& x \\
\text { s.t. }&& A(\xi) x&=b \\
&& x &\geq 0
\end{aligned}
min s.t. c ( ξ ) ⊤ A ( ξ ) x x x = b ≥ 0
Different Points of View
1. Stochastic Optimization
假设pertubation vector ξ \xi ξ 服从一个特定的分布
min E ξ ∼ P [ c ( ξ ) ⊤ x ] s.t. Pr ξ [ A ( ξ ) x = b ( ξ ) ] ⩾ 1 − δ x ⩾ 0 \begin{aligned}
\min&& \mathbb{E}_{\xi \sim \mathbb{P}}&\left[c(\xi)^{\top} x\right]\\
\text { s.t. }&& \operatorname{Pr}_\xi[A(\xi) x=b(\xi)] &\geqslant 1-\delta\\
&& x &\geqslant 0
\end{aligned}
min s.t. E ξ ∼ P P r ξ [ A ( ξ ) x = b ( ξ ) ] x [ c ( ξ ) ⊤ x ] ⩾ 1 − δ ⩾ 0
Contraint 变为chance constraint ,with confidence level 1 − δ 1-\delta 1 − δ
Difficulties:
Hard to evaluate E ξ ∼ P \mathbb{E}_{\xi \sim \mathbb{P}} E ξ ∼ P or Pr ξ \operatorname{Pr}_\xi P r ξ , high dimensional calculation
Usually the distribution is unknown
2. Robust Optimization
与随机优化不同,鲁棒优化不假设一个具体的分布,而是假设一个“可行空间”
ξ \xi ξ belongs to a (bounded) set U ⊆ R l \mathcal{U} \subseteq \mathbb{R}^l U ⊆ R l (uncertainty set)
\begin{align*}
\min \quad & t \\
\text{s.t.} \quad & c^{\top}(\xi) x \leqslant t \quad && \forall \xi \in \mathcal{U} \\
& A(\xi) x = b(\xi) \quad && \forall \xi \in \mathcal{U} \\
& x \geqslant 0
\end{align*}
Thus, a solution ( x ∗ , t ∗ ) \left(x^*, t^*\right) ( x ∗ , t ∗ ) is robust against all possible ξ ∈ U \xi \in \mathcal{U} ξ ∈ U .
Difficulties:
How to chosse U \mathcal{U} U ?
The solution usually too conservative.
3. Distributional Robust Optimization
Somewhere in-between! 即结合了一些随机优化的性质又有robust optimization的鲁棒性
ξ \xi ξ : random vector with distributionP ∗ \mathbb{P}^* P ∗ , P ∗ \mathbb{P}^* P ∗ not exactly known, but some information about it is known.
Consider an uncertainty set of distributions P \mathscr{P} P :
E.g. :
\begin{align*}
&\min_x \max_{\mathbb{P} \in \mathscr{P}} \mathbb{E}_{\xi \sim \mathbb{P}}[c^{\top}(\xi) x] \\
&\operatorname{inf}_{\mathbb{P} \in \mathscr{P}} \operatorname{Pr}_{\xi \sim \mathbb{P}}[A(\xi) x=b(\xi)] \geqslant 1-\delta
\end{align*}
Difficulties:
How to chosse P \mathscr{P} P ?
Can the resulting problem be solved ?
Example: Robust Linear Constraint
考虑一个简单的robust constrain的例子,体验 “robust” 如何影响并改变constraint
RLC:
a ⊤ x ≤ b a^{\top} x \leq b a ⊤ x ≤ b
∀ ( a , b ) = ( a 0 , b 0 ) ⏟ nominal value + ∑ j = 1 l ξ j ( a j , b j ) ⏟ perturbation ; ξ ∈ U ⊆ R l \forall(a, b)=\underbrace{\left(a^0, b_0\right)}_{\substack{\text { nominal } \\ \text { value }}}+\underbrace{\sum_{j=1}^l \xi_j\left(a^j, b_j\right)}_{\text {perturbation }} ; \qquad \xi \in \mathcal{U}\subseteq \mathbb{R}^l ∀ ( a , b ) = nominal value ( a 0 , b 0 ) + perturbation j = 1 ∑ l ξ j ( a j , b j ) ; ξ ∈ U ⊆ R l
where a j , b j ( j = 0 , 1 , ⋯ , l ) a^j, b_j(j=0,1, \cdots, l) a j , b j ( j = 0 , 1 , ⋯ , l ) are given.
Choice 1 : $\mathcal{U} = $ normal ball
U = { y ∈ R ℓ , ∥ y ∥ ∞ ≤ 1 } U=\left\{y \in \mathbb{R}^{\ell},\|y\|_{\infty} \leq 1\right\} U = { y ∈ R ℓ , ∥ y ∥ ∞ ≤ 1 }
( R L C ) ⇔ ( a 0 ) ⊤ x + ∑ j = 1 ℓ ξ j ( a j ) ⊤ x ⩽ b 0 + ∑ j = 1 ℓ ξ j b j ∀ ∥ ξ ∥ ∞ ≤ 1 ⇔ ∑ j = 1 l ξ j ( ( a j ) ⊤ x − b j ) ⩽ b 0 − ( a 0 ) ⊤ x ∀ ∥ ξ ∥ ∞ ≤ 1 ⇔ max ξ : ∥ ξ ∥ ∞ ≤ 1 ∑ j = 1 l ξ j ( ( a j ) ⊤ x − b j ) ⩽ b 0 − ( a 0 ) ⊤ x ⇔ ∑ j = 1 l ∣ ( a j ) ⊤ x − b j ∣ ≤ b 0 − ( a 0 ) ⊤ x \begin{aligned}
(R L C) &\Leftrightarrow \quad\left(a^0\right)^{\top} x+\sum_{j=1}^{\ell} \xi_j\left(a^j\right)^{\top} x \leqslant b_0+\sum_{j=1}^{\ell} \xi_j b_j \quad \forall\|\xi\|_{\infty} \leq 1\\
& \Leftrightarrow \quad \sum_{j=1}^l \xi_j\left(\left(a^j\right)^{\top} x-b_j\right) \leqslant b_0-\left(a^0\right)^{\top} x \quad \forall\|\xi\|_{\infty} \leq 1 \\
& \Leftrightarrow \quad \max _{\xi:\|\xi\|_{\infty} \leq 1} \sum_{j=1}^l \xi_j\left(\left(a^j\right)^{\top} x-b_j\right) \leqslant b_0-\left(a^0\right)^{\top} x \\
& \Leftrightarrow \quad \sum_{j=1}^l\left|\left(a^j\right)^{\top} x-b_j\right| \leq b_0-\left(a^0\right)^{\top} x
\end{aligned}
( R L C ) ⇔ ( a 0 ) ⊤ x + j = 1 ∑ ℓ ξ j ( a j ) ⊤ x ⩽ b 0 + j = 1 ∑ ℓ ξ j b j ∀ ∥ ξ ∥ ∞ ≤ 1 ⇔ j = 1 ∑ l ξ j ( ( a j ) ⊤ x − b j ) ⩽ b 0 − ( a 0 ) ⊤ x ∀ ∥ ξ ∥ ∞ ≤ 1 ⇔ ξ : ∥ ξ ∥ ∞ ≤ 1 max j = 1 ∑ l ξ j ( ( a j ) ⊤ x − b j ) ⩽ b 0 − ( a 0 ) ⊤ x ⇔ j = 1 ∑ l ∣ ∣ ∣ ∣ ( a j ) ⊤ x − b j ∣ ∣ ∣ ∣ ≤ b 0 − ( a 0 ) ⊤ x
This can be converted into a finite system linear inequalities
Choice 2 : U = { y < R l : ∥ y ∥ 2 ⩽ 1 } \mathcal{U} = \left\{y<\mathbb{R}^l:\|y\|_2 \leqslant 1\right\} U = { y < R l : ∥ y ∥ 2 ⩽ 1 }
( R L C ) ⇔ max ξ : ∥ ξ ∥ 2 ≤ 1 ξ j ( ( a j ) ⊤ x − b j ) ⏟ u ⊤ ξ u j = ( a j ) ⊤ x − b j ⩽ b 0 − ( a 0 ) ⊤ x ⩽ ∥ u ∥ 2 ⋅ ∥ ξ ∥ 2 (Cauchy-Schwarz) ⩽ ∥ u ∥ 2 ( ∥ ξ ∥ 2 ⩽ 1 ) ⇔ [ ∑ j = 1 l ( ( a j ) ⊤ x − b j ) 2 ] 1 / 2 ⩽ b 0 − ( a 0 ) ⊤ x \begin{aligned}
& (R L C) \Leftrightarrow \quad \max _{\xi:\|\xi\|_2 \leq 1} \underbrace{\xi_j\left(\left(a^j\right)^{\top} x-b_j\right)}_{u^{\top} \xi \quad u_j=\left(a^j\right)^{\top} x-b_j} \leqslant b_0-\left(a^0\right)^{\top} x \\
& \leqslant\|u\|_2 \cdot\|\xi\|_2 \quad \text { (Cauchy-Schwarz) } \\
& \leqslant\|u\|_2 \quad\left(\|\xi\|_2 \leqslant 1\right) \\
& \Leftrightarrow \quad\left[\sum_{j=1}^l\left(\left(a^j\right)^{\top} x-b_j\right)^2\right]^{1 / 2} \leqslant b_0-\left(a^0\right)^{\top} x \\
&
\end{aligned}
( R L C ) ⇔ ξ : ∥ ξ ∥ 2 ≤ 1 max u ⊤ ξ u j = ( a j ) ⊤ x − b j ξ j ( ( a j ) ⊤ x − b j ) ⩽ b 0 − ( a 0 ) ⊤ x ⩽ ∥ u ∥ 2 ⋅ ∥ ξ ∥ 2 (Cauchy-Schwarz) ⩽ ∥ u ∥ 2 ( ∥ ξ ∥ 2 ⩽ 1 ) ⇔ [ j = 1 ∑ l ( ( a j ) ⊤ x − b j ) 2 ] 1 / 2 ⩽ b 0 − ( a 0 ) ⊤ x
So far, we seem to be lucky in that max ξ ∈ u [ ⋯ ] \max _{\xi \in u}[\cdots] max ξ ∈ u [ ⋯ ] has a closed form. In general, we do not expect this
Choice 3 U = { y ∈ R l : P y ⩽ q } U=\left\{y \in \mathbb{R}^l: P y \leqslant q\right\} U = { y ∈ R l : P y ⩽ q } (a polyhedron)
( R L C ) ⇔ max ξ : P ξ ≤ q ∑ j = 1 l ξ j ( ( a j ) ⊤ x − b j ) ≤ b 0 − ( a 0 ) ⊤ x ⇔ ? ? ? \begin{aligned}
(R L C) & \Leftrightarrow \max _{\xi: P \xi \leq q} \sum_{j=1}^l \xi_j\left(\left(a^j\right)^{\top} x-b_j\right) \leq b_0-\left( a^0\right)^{\top} x \\
& \Leftrightarrow \quad ? ? ?
\end{aligned}
( R L C ) ⇔ ξ : P ξ ≤ q max j = 1 ∑ l ξ j ( ( a j ) ⊤ x − b j ) ≤ b 0 − ( a 0 ) ⊤ x ⇔ ? ? ?
Perhaps LP duality can be applied here.
Summary
Lessons learned
Complexity of the robust constraint depends on U \mathcal U U
duality theory is helpful in reformulating robust constraints into finite system of constraints
even for robust linear constraints, their reformulations may not be linear → \rightarrow → need nonlinear optimization techniques