Stochastic makespan minimization in structured set systems

Abstract

We study stochastic combinatorial optimization problems where the objective is to minimize the expected maximum load (a.k.a. the makespan). In this framework, we have a set of n tasks and m resources, where each task j uses some subset of the resources. Tasks have random sizes $X_j$, and our goal is to non-adaptively select t tasks to minimize the expected maximum load over all resources, where the load on any resource i is the total size of all selected tasks that use i. For example, when resources are points and tasks are intervals in a line, we obtain an $O(\log \log m)$-approximation algorithm. Our technique is also applicable to other problems with some geometric structure in the relation between tasks and resources; e.g., packing paths, rectangles, and “fat” objects. Our approach uses a strong LP relaxation using the cumulant generating functions of the random variables. We also show that this LP has an $\varOmega (\log ^* m)$ integrality gap, even for the problem of selecting intervals on a line; here $\log ^* m$ is the iterated logarithm function.

Appendices

A Scaling the optimal value

Suppose that $\mathcal{A}$ is a polynomial algorithm that given any GenMakespan instance, returns one of the following:

• a solution of objective at most $\rho$, or

• a certificate that the optimal value is more than 1.

Using this, we provide a polynomial time $O(\rho )$-approximation algorithm for GenMakespan. Observe that the optimal value $\mathsf {OPT}$ of GenMakespan lies between $L:=\min _{j\in [n]} \mathbb {E}[X_j]$ and $U:=n\cdot \max _{j\in [n]} \mathbb {E}[X_j]$. It follows that $\frac{B^*}{2}\le \mathsf {OPT} \le B^*$ for some value $B^*$ in the set:

$$\begin{aligned} \mathcal{G} := \left\{ 2^\ell \cdot L: 0 \le \ell \le \log _2 (U/L) +1, \, \ell \in \mathbb {Z}\right\} \end{aligned}$$

For each $B\in \mathcal{G}$, consider the modified GenMakespan instance with r.v. $X_j/B$ for each task j; and run algorithm $\mathcal{A}$ on this instance. Finally, return the solution with the smallest objective obtained over all $B\in \mathcal{G}$. Note that when $B=B^*$, the optimal value of the modified GenMakespan instance is at most 1: so algorithm $\mathcal{A}$ must find a solution with (modified) objective at most $\rho$, i.e., the expected makespan under the original r.v.s $\{X_j\}$ is at most $\rho \cdot B^*\le 2\rho \cdot \mathsf {OPT}$. The number of times we call algorithm $\mathcal{A}$ is $O(\log (U/L))$. Note that $L\ge s_{min}$ and $U\le n\cdot s_{max}$ where $s_{min}$ and $s_{max}$ are the minimum and maximum values that the r.v.s take. So, $\log (U/L) = \log (n\frac{s_{max}}{s_{min}})$, which is polynomial in the instance size. Hence, we obtain a polynomial time $2\rho$-approximation algorithm for GenMakespan.

B The $\alpha$-packable property for rectangles and fat objects

In this section, we relate the $\alpha$-packable property of a set system to the the intergrality gap of the natural LP relaxation for maximum (weighted) independent set for the set system. Using known integrality gap results for maximum independent set for axis-parallel rectangles and fat objects, we can show $\alpha$-packability of the corresponding set systems for suitable values of $\alpha$.

Recall the setting in the $\alpha$-packable property. There is a set system $([n], \mathcal{L})$ with size $s_j \ge 0$ and reward $r_j$ for each element $j \in [n]$, and a bound $\theta \ge \max _j s_j$. We are interested in the integrality gap (and a polynomial-time rounding algorithm) for LP (2), restated below.

$$\begin{aligned} \max \bigg \{ \sum _{j \in [n]} r_j\cdot y_j \,:\, \sum _{j\in L} s_j\cdot y_j \le \theta , \, \, \forall L \in \mathcal{L}; \ 0\le y_j\le 1, \, \, \forall j \in [n] \bigg \}. \end{aligned}$$

When all sizes $s_j=1$ and the bound $\theta =1$, we obtain the independent set LP:

$$\begin{aligned} \max \bigg \{ \sum _{j \in [n]} r_j\cdot y_j \,:\, \sum _{j\in L} y_j \le 1, \, \, \forall L \in \mathcal{L}; \ 0\le y_j\le 1, \, \, \forall j \in [n] \bigg \}. \end{aligned}$$

(20)

Note that the corresponding integral problem involves selecting a max-reward subset of disjoint elements. (Elements e and f are disjoint if there is no set $L\in \mathcal{L}$ with $e,f\in L$.)

Theorem 9

Suppose that the independent set LP (20) has integrality gap $\rho$ and an associated polynomial time rounding algorithm. Then, the set-system is $O(\rho \cdot \log \log m)$-packable.

Proof

The proof proceeds in several steps: (i) we first consider the special case when all $s_j$ values are 1, but the parameter $\theta$ can be arbitrary, (ii) secondly, when $\theta \gg s_j$ for all j (by more than a $\log m$ factor), we use randomized rounding, and (iii) finally, for the general case, we use a standard bucketing trick to create $O(\log \log m)$ groups, and show that one of the above two steps will work for each of these groups.

We give details of the first step. We show a rounding algorithm for the following LP, and show that its integrality gap is at most $2 \rho$:

$$\begin{aligned} \max \bigg \{ \sum _{j \in [n]} r_j\cdot y_j \,:\, \sum _{j\in L} y_j \le b, \, \, \forall L \in \mathcal{L}; \ 0\le y_j\le 1, \, \, \forall j \in [n] \bigg \}. \end{aligned}$$

(21)

Here, we assume that $b\ge 1$ is integer. Note that this is a special case of the LP (2) used in the $\alpha$-packable condition.

(i) Rounding for the LP (21):    We combine the rounding algorithm for the independent set LP relaxation (20) with a greedy strategy to round a feasible solution to the LP (21). Let y be a feasible (fractional) solution to the latter LP. We define $\bar{y}=y/b$, which is a feasible solution to the independent set LP relaxation (20).

We build the solution $T\subseteq [n]$ iteratively; initially $T=\emptyset$. For each iteration $k=1,\cdots b$, we perform the following steps:

• 1. Consider the solution $\bar{y}$ restricted to $[n]{\setminus } T$. Since this is a feasible solution to the independent set LP (20), we use the independent set rounding algorithm to obtain an integral solution $S_k\subseteq [n]{\setminus } T$.

• 2. Update $T\leftarrow T\cup S_k$.

As $\{S_k\}$ are disjoint subsets, $T=\cup _{k=1}^b S_k$ is a feasible integral solution to (21).

We now analyze the reward of the solution T. For any subset $U\subseteq [n]$ let $Y(U):= \sum _{j \in U} r_j\cdot y_j$ be the LP-value restricted to U. Consider the two cases:

• Suppose $Y([n]{\setminus } T)\ge \frac{1}{2} \cdot Y([n])$ at the end of the algorithm. It follows that $Y([n]{\setminus } T)\ge \frac{1}{2} \cdot Y([n])$ in each iteration k. Consider the LP solution $\bar{y}$ restricted to $[n]{\setminus } T$ (in iteration k). Since the rounding algorithm for the independent set LP relaxation has integrality gap $\rho$,

  $$\begin{aligned} r(S_k)\ge \frac{1}{\rho }\sum _{j\in [n]{\setminus } T} r_j\cdot \bar{y}_j = \frac{1}{\rho b} Y([n]{\setminus } T) \ge \frac{Y([n])}{2\rho b}. \end{aligned}$$

  Adding over all b iterations, $r(T)=\sum _{k=1}^b r(S_k)\ge \frac{Y([n])}{2\rho }$.

• Suppose $Y([n]{\setminus } T)<\frac{1}{2} \cdot Y([n])$ at the end of the algorithm. Then,

  $$\begin{aligned} r(T)\ge Y(T) = Y([n]) - Y([n]{\setminus } T)>\frac{1}{2} \cdot Y([n]). \end{aligned}$$

In either case, we obtain that the algorithm’s reward $r(T)\ge \frac{1}{2\rho }\cdot Y([n])$. This proves that the integrality gap of (21) is at most $2\rho$.

(ii) Randomized Rounding for large $\theta$.    Let $\tau$ denote $\frac{\theta }{2\log m}$. Consider the special case when $s_j \le \tau$ for all $j \in [n]$. In this case, the LP relaxation (2) is a special case of packing integer programs (PIPs), studied in [19]. Theorem 3.7 in [19] implies an $O(m^{1/P})$ integrality gap for the LP (2), where

$$\begin{aligned} P=\frac{\theta }{\max _{j\in [n]} s_j} \ge \frac{\theta }{\tau }=2\log m. \end{aligned}$$

Therefore, the LP (2) has constant integrality gap in this special case.

(iii) Geometric grouping for the general case.    We now consider the LP (2) in the general setting. Let y be a fractional solution to this LP. As define above, $\tau := \frac{\theta }{2\log m}$. We first partition the elements into groups based on their sizes as follows:

$$\begin{aligned} G_k := \left\{ \begin{array}{ll} \{j\in [n] : s_j< \tau \} &{} \text{ if } k=0 \\ \{j\in [n] : 2^{k-1} \tau \le s_j < 2^k \tau \} &{} \text{ if } k\ge 1 \end{array}\right. . \end{aligned}$$

Note that the number of groups is $K=O(\log \log m)$ as $\max _j s_j\le \theta$. We handle each group separately, and pick the maximum reward solution across the K groups.

Consider a group $G_k, k\ge 1$. Consider the fractional solution z defined as:

$$\begin{aligned} z_j = \left\{ \begin{array}{ll} y_j/4 &{} \text{ if } j\in G_k \\ 0 &{} \text{ otherwise } \end{array}\right. \end{aligned}$$

We claim that z is a feasible solution to the LP (21) restricted to $G_k$, and a suitable value of b. Indeed, consider any $L\in \mathcal{L}$. Then,

$$\begin{aligned} \sum _{j\in L}z_j \le \frac{1}{2^{k-1}\tau } \sum _{j\in G_k\cap L} s_j\cdot z_j = \frac{1}{2^{k+1}\tau } \sum _{j\in G_k\cap L} s_j\cdot y_j\le \frac{\theta }{2^{k+1}\tau }\le \frac{c}{2} \le \lfloor c\rfloor , \end{aligned}$$

where we have used the fact that y is a feasible solution to (2), and $c := \theta / \max _{j\in G_k} s_j \ge \max \{1 , \frac{\theta }{2^k\tau }\}.$ It follows that z is a feasible solution to the LP (21) where $b=\lfloor c\rfloor$. Hence, using the rounding algorithm for (21) mentioned in the first step above, we obtain a solution $V_k\subseteq G_k$ with reward

$$\begin{aligned} r(V_k)\ge \frac{1}{2\rho } \sum _{j \in G_k} r_j\cdot z_j \ge \frac{1}{8\rho } \sum _{j\in G_k} r_j y_j. \end{aligned}$$

Moreover, for each $L\in \mathcal{L}$, we have $|V_k\cap L|\le b$. Hence, for any $L\in \mathcal{L}$,

$$\begin{aligned} \sum _{j\in V_k\cap L} s_j\le \left( \max _{j\in G_k} s_j \right) \cdot |V_k\cap L| =\frac{\theta }{ c} \cdot |V_k\cap L| \le \frac{\theta b}{ c}\le \theta . \end{aligned}$$

Thus, $V_k$ is a feasible integral solution to (2).

Finally we consider the case $k=0$, i.e., the group $G_0$. As argued in the second step above, we obtain a solution $V_0 \subseteq G_0$ with reward $r(V_0)\ge \frac{1}{\sigma }\cdot \sum _{j\in G_0} r_j y_j$ where $\sigma \ge 1$ is constant. It follows that $V_0$ is an integral solution to (2) as well.

Finally, choosing the best solution from $\{V_k\}$ over all groups, we obtain reward at least

$$\begin{aligned} \max _k r(V_k) \ge \frac{1}{K} \sum _{k} r(V_k) \ge \frac{1}{K\cdot \max (8\rho , \sigma )}\sum _k \sum _{j\in G_k} r_j y_j = \frac{1}{K\cdot \max (8\rho , \sigma )} \sum _{j\in [n]} r_j y_j. \end{aligned}$$

This proves that the integrality gap of (2) is $O(\rho \log \log m)$. $\square$

We now combine Theorem 9 with known results on maximum weight independent sets for rectangles and fat objects, to prove their $\alpha$-packable property.

Corollary 5

The set-system where tasks are n axis-aligned rectangles in the plane and resources are all points in the plane, is $O((\log \log n)^2)$-packable.

Proof

The weighted independent set problem for rectangles in the plane has an LP-based $O(\log \log n)$ approximation [7]. Combined with Theorem 9 and the fact that the number of points m can be ensured to be poly(n) (see Sect. 4.3), we obtain that the set-system is $O((\log \log n)^2)$-packable. $\square$

Corollary 6

The set-system where tasks are n disks (of arbitrary radii) in the plane and resources are all points in the plane, is $O(\log \log n)$-packable.

Proof

There is an LP-based O(u(n)/n)-approximation algorithm for weighted independent set on set-systems where the “union complexity” of n objects is at most u(n) [9]. See the survey [2] for more details on union complexity. The union complexity of disks (of arbitrary radii) is O(n). So there is an LP-based O(1)-approximation algorithm for weighted independent set. Combined with Theorem 9 and that $m=poly(n)$, the result follows. $\square$

Corollary 7

The set-system where tasks are n fat triangles in the plane and resources are all points in the plane, is $O(\log ^* n \cdot \log \log n)$-packable.

Proof

The union complexity of fat triangles is $u(n)=O(n\, \log ^* n)$ [3]. Using the result from [9], we obtain an LP-based $O(\log ^* n)$-approximation algorithm for the weighted independent set problem. Using Theorem 9 and that $m=poly(n)$, the result follows. $\square$