Online covering with $\ell _q$-norm objectives and applications to network design

Abstract

We consider fractional online covering problems with $\ell _q$-norm objectives as well as its dual packing problems. The problem of interest is of the form $\min \{ f(x) \,:\, Ax\ge 1, x\ge 0\}$ where $f(x)=\sum _{e} c_e \Vert x(S_e)\Vert _{q_e}$ is the weighted sum of $\ell _q$-norms and A is a non-negative matrix. The rows of A (i.e. covering constraints) arrive online over time. We provide an online $O(\log d+\log \rho )$-competitive algorithm where $\rho = \frac{ a_{\max }}{ a_{\min }}$ and d is the maximum of the row sparsity of A and $\max |S_e|$. This is based on the online primal-dual framework where we use the dual of the above convex program. Our result is nearly tight (even in the linear special case), and it expands the class of convex programs that admit online algorithms. We also provide two applications where such convex programs arise as relaxations of discrete optimization problems, for which our result leads to good online algorithms. In particular, we obtain an improved online algorithm (by two logarithmic factors) for non-uniform buy-at-bulk network design and a poly-logarithmic competitive ratio for throughput maximization under $\ell _p$-norm capacities.

Appendices

Deriving $f^*_e(\cdot )$

Recall that $f_e(x)=c_e\Vert x(S_e)\Vert _{q_e}$ where $x\in \mathbb {R}^n_+$ and $S_e\subseteq [n]$. For any $\mu \in \mathbb {R}^n_+$ we have

$$\begin{aligned} f^*_e(\mu )= & {} \sup _{x \in \mathbb {R}^n_+} \,\mu ^T x - c_e\Vert x(S_e)\Vert _{q_e} \\= & {} \sup _{x \in \mathbb {R}^n_+} \left\{ \mu ({\overline{S}}_e)^T x({\overline{S}}_e) \, + \,\mu _e(S_e)^Tx(S_e)-c_e\Vert x(S_e)\Vert _{q_e} \right\} . \end{aligned}$$

If $\mu ({\overline{S}}_e) \ne {\mathbf {0}}$ then it is clear that $f_e^*(\mu )=\infty$. So we assume $\mu ({\overline{S}}_e) = {\mathbf {0}}$, in which case

$$\begin{aligned} f_e^*(\mu ) = \sup _{y \in \mathbb {R}^{|S_e|}_+} \left\{ \mu (S_e)^T y -c_e\Vert y\Vert _{q_e}\right\} . \end{aligned}$$

Let $\Vert \cdot \Vert _{p_e}$ be the dual norm of $\Vert \cdot \Vert _{q_e}$. By the definition of the dual norm, if $\Vert \mu (S_e)\Vert _{p_e}>c_e$, there exists $z\in \mathbb {R}^{|S_e|}$ with $\Vert z\Vert _{q_e}\le 1$ such that $\mu (S_e)^Tz>c_e$. As $\mu \ge 0$, we can ensure $z\ge 0$. Then taking $y=tz$ as $t\rightarrow \infty$, we have

$$\begin{aligned} f_e^*(\mu ) \ge \mu (S_e)^T y -c_e\Vert y\Vert _{q_e} \,=\, t(\mu (S_e)^Tz-c_e\Vert z\Vert _{q_e})\rightarrow \infty . \end{aligned}$$

On the other hand, if $\Vert \mu (S_e) \Vert _{p_e}\le c_e$, then by Hölder’s inequality, for any $y\in \mathbb {R}^{|S_e|}_+$,

$$\begin{aligned} \mu (S_e)^T y \le \Vert \mu (S_e) \Vert _{p_e}\Vert y\Vert _{q_e}\le c_e \Vert y\Vert _{p_e}, \end{aligned}$$

which implies that $f_e^*(\mu )=0$.

Summarizing the above cases, we have for any $\mu \in \mathbb {R}^n_+$:

$$\begin{aligned} f^*_e(\mu )={\left\{ \begin{array}{ll} 0,\quad \text{ if } \Vert \mu (S_e)\Vert _{p_e}\le c_e \text{ and } \mu ({\overline{S}}_e)={\mathbf {0}},\\ \infty ,\quad \text{ otherwise }. \end{array}\right. } \end{aligned}$$

Limitations of previous approaches in handling $\ell _q$-norm objectives.

The general convex covering problem is

$$\begin{aligned} \min \, \left\{ f(x) \,:\, Ax\ge {\mathbf {1}},\, {x\ge {\mathbf {0}}}\right\} , \end{aligned}$$

where $f:\mathbb {R}^n_+\rightarrow \mathbb {R}_+$ is a convex function and $A\in \mathbb {R}^{m\times n}_+$. Its dual is:

$$\begin{aligned} \max \, \left\{ \sum _{k=1}^m y_k - f^*(\mu ) \, :\, A^Ty=\mu ,\, y\ge {\mathbf {0}} \right\} , \end{aligned}$$

where $f^*(\mu ) = \max _{x\in \mathbb {R}^n_+} \{ \mu ^Tx - f(x)\}$ is the Fenchel conjugate of f. When f is the sum of $\ell _q$-norms, these primal-dual convex programs reduce to (P) and (D).

We restrict the discussion of prior techniques to functions f with $\max _{x\in \mathbb {R}^n_+} \frac{x^T \nabla f(x)}{f(x)} \le 1$ because this condition is satisfied by sums of $\ell _q$ norms.¹ At a high level, the analysis in [6] uses the gradient monotonicity to prove a pointwise upper bound $A^Ty\le \nabla f(\bar{x})$ where $\bar{x}$ is the final primal solution. This allows them to lower bound the dual objective by $\sum _{k=1}^m y_k$ because $f^*(\nabla f(\bar{x}))\le 0$ for any $\bar{x}$ (see Lemma 4(d) in [6]). Moreover, proving the pointwise upper bound $A^Ty\le \nabla f(\bar{x})$ is similar to the task of showing dual feasibility in the linear case [15, 26] where $\nabla f(\bar{x})$ corresponds to the (fixed) primal cost coefficients.

Below we give a simple example with an $\ell _q$-norm objective where the pointwise upper bound $A^Ty\le \nabla f(\bar{x})$ is not satisfied by the online primal-dual algorithm unless the dual solution y is scaled down by a large (i.e. polynomial) factor. This means that one cannot obtain a sub-polynomial competitive ratio for (P) using this approach directly.

Consider an instance with objective function $f(x)=\Vert x\Vert _2=\sqrt{\sum _{i=1}^n x_i^2}$. So the gradient $\nabla f(x) = x/\Vert x\Vert _2$ which is not monotone. There are $m=\sqrt{n}$ covering constraints, where the $k^{th}$ constraint is $\sum _{i=m(k-1)+1}^{km} x_i \ge 1$. Note that each variable appears in only one constraint. Let P be the value of the primal objective and D be the value of the dual objective at any time. Suppose that the rate of increase of the primal objective is at most $\alpha$ times that of the dual; $\alpha$ corresponds to the competitive ratio in the online primal-dual algorithm. Upon arrival of any constraint k, it follows from the primal updates that all the variables $\{x_i\}_{i=m(k-1)+1}^{km}$ increase from 0 to $\frac{1}{m}$. So the increase in P due to constraint k is $(\sqrt{k}-\sqrt{k-1})\frac{1}{\sqrt{m}}$ for iteration k. This means that the increase in D is at least $\frac{1}{\alpha }(\sqrt{k}-\sqrt{k-1})\frac{1}{\sqrt{m}}$, and so $y_k\ge \frac{1}{\alpha }(\sqrt{k}-\sqrt{k-1})\frac{1}{\sqrt{m}}$. Finally, since $\bar{x}=\frac{1}{m}{\mathbf {1}}$, we know that $\nabla f(\bar{x})=\frac{1}{m}{\mathbf {1}}$ (recall $n=m^2$). On the other hand, $(A^Ty)_1 = y_1\ge \frac{1}{\alpha \sqrt{m}}$. Therefore, in order to guarantee $A^Ty\le \nabla f(\bar{x})$ we must have $\alpha \ge \sqrt{m} = n^{1/4}$.