Machine Covering in the Random-Order Model

Abstract

In the Online Machine Covering problem, jobs, defined by their sizes, arrive one by one and have to be assigned to m parallel and identical machines, with the goal of maximizing the load of the least-loaded machine. Unfortunately, the classical model allows only fairly pessimistic performance guarantees: The best possible deterministic ratio of m is achieved by the Greedy-strategy, and the best known randomized algorithm has competitive ratio ${\tilde{O}}(\sqrt{m})$, which cannot be improved by more than a logarithmic factor. Modern results try to mitigate this by studying semi-online models, where additional information about the job sequence is revealed in advance or extra resources are provided to the online algorithm. In this work, we study the Machine Covering problem in the recently popular random-order model. Here, no extra resources are present but, instead, the adversary is weakened in that it can only decide upon the input set while jobs are revealed uniformly at random. It is particularly relevant to Machine Covering where lower bounds are usually associated to highly structured input sequences. We first analyze Graham’s Greedy-strategy in this context and establish that its competitive ratio decreases slightly to $\Theta \left( \frac{m}{\log (m)}\right)$, which is asymptotically tight. Then, as our main result, we present an improved ${\tilde{O}}(\root 4 \of {m})$-competitive algorithm for the problem. This result is achieved by exploiting the extra information coming from the random order of the jobs, using sampling techniques to devise an improved mechanism to distinguish jobs that are relatively large from small ones. We complement this result with a first lower bound, showing that no algorithm can have a competitive ratio of $O\left( \frac{\log (m)}{\log \log (m)}\right)$ in the random-order model. This lower bound is achieved by studying a novel variant of the Secretary problem, which could be of independent interest.

Knowing n Does not Help in the Adversarial Model.

As common in the area of online algorithms, we ignore computability issues. Formally, a deterministic online algorithm that does not know n is a function that receives as input a sequence of jobs $J_1,\ldots ,J_{n'}$, which corresponds to the information the algorithm has at the moment of scheduling $J_{n'}$, and returns a machine M onto which the job $J_{n'}$ is scheduled. Similarly, a deterministic online algorithm that knows n receives as input a tuple $n, J_1,\ldots ,J_{n'}$ and returns the machine M onto which job $J_{n'}$ is scheduled. Here, $n\ge n'$ is the total length of the input. Randomized algorithms are simply probability distributions over deterministic algorithms.

Proposition 19

Let A be any (possibly randomized) online algorithm that is c-competitive in the adversarial model knowing n beforehand. Then there is a strategy B that is c-competitive not knowing n. If A is deterministic, the strategy B can be chosen to be deterministic too.

Proof

Let us denote the algorithm A that knows beforehand that the input will have length n by A[n]. This algorithm outputs schedules for any sequence of jobs having length at most n. Given some number $n'\in {\mathbb {N}}$, let ${\mathcal {S}} [n']$ be the set of all probability distributions over possible schedules of $n'$ jobs. Schedules of $n'$ jobs are nothing but assignments $\{1,\ldots n'\}\rightarrow \{1,\ldots m\}$. Let F be the set of all such maps. We can then identify ${\mathcal {S}} [n']$ with a compact subspace of $[0,1]^{F}\subset {\mathbb {R}} ^F$. In fact it is the standard simplex spanned by the set ${\mathcal {E}} [n']\subset {\mathcal {S}} [n']$ consisting of probability distributions that choose one schedule with probability 1, or, geometrically speaking, the set of standard base vectors in ${\mathbb {R}} ^{M}$. For $n'>0$, forgetting how we scheduled the last job yields a canonical continuous map $r_{n'}:{\mathcal {S}} [n']\rightarrow {\mathcal {S}} [n'-1]$.

Given jobs $J_1,\ldots J_{n'}$ and $n\ge n'$, let $s_{n'}^n=s_{n'}^n(J_1,\ldots J_{n'})\in {\mathcal {S}} [n']$ be the probability distribution of schedules obtained by algorithm A[n] on the job sequence $J_1,\ldots , J_{n'}$. Consider the set of points $s_{n'}^n$ with $n\ge n'$. We pick inductively a limit point $s_{n'}$ of this point set such that $r_{n'}(s_{n'})=s_{n'-1}$ (this is possible because $r_{n'}$ is continuous). The choice of $s_{n'}$ only depends on the jobs $J_1,\ldots J_{n'}$, hence can be picked ’online’. Now all our algorithm B has to do is to maintain its schedule at time $n'$ according to the probability distribution $s_{n'}$. The condition $r_{n'}(s_{n'})=s_{n'-1}$ ensures that this is possible.

We will prove that this algorithm is c-competitive. Pick $\varepsilon >0$ and any input sequence $J_1,\ldots J_{n'}$. As $s_{n'}$ is a limit point of the set of points $s_{n'}^n$ there exists $n\ge n'$ such that $(1-\varepsilon )s_{n'}^n\le s_{n'}$. In other words, if algorithm A[n] obtains schedule F after treating $J_1,\ldots J_{n'}$ with probability p, then B obtains the same schedule with probability at least $(1-\varepsilon )p$. In particular, the expected minimum load of B on $J_1,\ldots J_{n'}$ is at least $(1-\varepsilon )$ times the minimum load of A[n] on that sequence. The latter minimum load is at least $c\cdot {\text {OPT}}(J_1,\ldots J_{n'})$ since A[n] is c-competitive. We have thus shown that the expected minimum load of B is at least $(1-\varepsilon )c\cdot {\text {OPT}}$. The proposition follows by taking the limit $\varepsilon \rightarrow 0$.

If A is deterministic, all the points $s_{n'}^n$ lie in the closed, in fact discrete, subset ${\mathcal {E}} [n']\subset {\mathcal {S}} [n']$ of probability distributions simply picking one schedule with probability 1. All limit points, in particular the points $s_{n'}$, lie in ${\mathcal {E}} [n']$ too. Thus, B will also be deterministic. $\square$