## 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, given a set of intervals in time, with each interval *j* having random load \(X_j\), how do we choose *t* intervals to minimize the expected maximum load at any time? 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. Specifically, we give an \(O(\log \log m)\)-approximation algorithm for all these problems.

Our approach uses a strong LP relaxation using the cumulant generating functions of the random variables. We also show an LP integrality gap of \(\varOmega (\log ^* m)\) even for the problem of selecting intervals on a line.

All missing proofs and details can be found in the full version [11].

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## Notes

- 1.
The support of vector \(z\in \mathbb {R}^n_+\) is \(\{j\in [n] : z_j>0\}\) which corresponds to its positive entries.

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## A Analysis Outline

### A Analysis Outline

We now show that the expected makespan for the solution produced by the rounding algorithm above is \(O(\alpha \lambda \rho )\), where \(\rho =\log \log m\) is the number of classes. In particular, we show that the expected makespan (taken over all resources) due to tasks of each class \(\ell \) is \(O(\alpha \lambda )\).

Using the terminating condition in line 5, we can show:

### Lemma 6

For any class \(\ell \), \(0\le \ell \le \rho \), and resource \(i\in [m]\),

Next, the sets \(D_\ell \) cannot become too large (as a function of \(\ell \)).

### Lemma 7

For any \(\ell \), \(0 \le \ell \le \rho ,\) \(|D_\ell |\le k^2\), where \(k=2^{2^{\ell }}\). So \(|M_\ell |\le k^p\) for some constant *p*.

### Proof

The claim is trivial for the last class \(\ell =\rho \) as \(k\ge m\) in this case. Now consider any class \(\ell <\rho \). For each \(i \in D_\ell \), we know \(\sum _{j \in \widetilde{L_i}}\beta _{k^2}(X_j')\cdot y_j> 2b\), where \(\widetilde{L_i}\) is as defined in line 7. Moreover, the subsets \(\{\widetilde{L_i} : i\in D_\ell \}\) are disjoint as the set *J* gets updated (in line 7) after adding each \(i\in D_\ell \). Suppose, for the sake of contradiction, that \(|D_\ell |>k^2\). Then let \(K\subseteq D_\ell \) be any set of size \(k^2\). By the LP constraint (8) on this subset *K*,

which is a contradiction. This proves the first part of the claim. Finally, the \(\lambda \)-safe property implies that \(|M_\ell |\) is polynomially bounded by \(|D_\ell |\). \(\blacksquare \)

Using the definition of the DetCost instance and Lemma 6, we can show:

### Lemma 8

The fractional solution \(\bar{y}\) is feasible for the LP relaxation (10) corresponding to the DetCost instance \({\mathcal {C}}\). Moreover, \(\theta \ge \max _j s_j\) and \(\psi \ge \max _j c_j\).

The above lemmas show that the algorithm is well-defined, so we can indeed use Theorem 4 to round \(\bar{y}\) into an integer solution. Recall that our final solution is \(N = N_H\cup N_L\). The next two lemmas follow from this rounding step.

### Lemma 9

\(|N_H|+|N_L|\ge t\).

### Lemma 10

For any class \(\ell \le \rho \) and resource \(i\in M_\ell \),

We now focus on a particular class \(\ell \le \rho \) and show that the expected makespan due to tasks in \(N\cap J_\ell \) is small. Recall that \(k=2^{2^\ell }\). Let \(N_\ell := N\cap J_\ell \) and let \(\mathsf {Load}^{(\ell )}_i := \sum _{j\in N_\ell \cap L_i} X'_{j}\) be the load on any resource \(i\in [m]\) due to class-\(\ell \) tasks. We can now bound the makespan due to the truncated random variables.

### Lemma 11

For any class \(\ell \le \rho \), \(\mathbb {E}\left[ \max _{i\in M_\ell } \mathsf {Load}^{(\ell )}_i\right] \le 4\bar{\alpha } b +O(1)\), and

### Proof

Consider a resource \(i \in M_\ell \). Lemmas 10 and 3 imply:

By a union bound, we get

where *p* is the constant from Lemma 7. So the expectation

which completes the proof of the first statement.

We now prove the second statement. Consider any class \(\ell <\rho \): by definition of \(J_\ell \), we know that \(J_\ell \subseteq L(D_\ell )\). So the \(\lambda \)-safe property implies that for every resource *i* there is a subset \(R_i\subseteq M_\ell \) of size at most \(\lambda \) such that \(L_i\cap J_\ell \subseteq L(R_i)\cap J_\ell \). Because \(N_\ell \subseteq J_\ell \), we also have \(L_i\cap N_\ell \subseteq L(R_i)\cap N_\ell \). Therefore,

Taking expectation on both sides, we obtain the desired result.

Finally, for the last class \(\ell =\rho \), note that any task in \(J_\rho \) loads the resources in \(D_\rho = M_\rho \) only. Therefore, \(\max _{i=1}^m \mathsf {Load}^{(\ell )}_i = \max _{z\in M_\ell } \mathsf {Load}^{(\ell )}_z\). The desired result now follows by taking expectation on both sides. \(\blacksquare \)

Using Lemma 11 for all \(\rho \) classes, it follows that the expected makespan due to all truncated r.v.s is \(O(\alpha \lambda \rho )\). For the exceptional random variables, we use:

### Lemma 12

\(\mathbb {E}\left[ \sum _{j\in N} X''_j \right] = \sum _{j\in N} c_j \le 4 \bar{\alpha }\).

Adding the contributions from truncated and exceptional r.v.s, the overall expected makespan is \(O(\alpha \lambda \rho )\), which completes the proof of Theorem 2.

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Gupta, A., Kumar, A., Nagarajan, V., Shen, X. (2020). Stochastic Makespan Minimization in Structured Set Systems (Extended Abstract). In: Bienstock, D., Zambelli, G. (eds) Integer Programming and Combinatorial Optimization. IPCO 2020. Lecture Notes in Computer Science(), vol 12125. Springer, Cham. https://doi.org/10.1007/978-3-030-45771-6_13

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