(Almost) Tight bounds and existence theorems for single-commodity confluent flows
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Citations
On the effect of forwarding table size on SDN network utilization
The Fluid Mechanics of Liquid Democracy
Maximum Edge-Disjoint Paths in Planar Graphs with Congestion 2
Single sign-on for unmanaged mobile devices
References
Flows in networks
A threshold of ln n for approximating set cover
Maximal Flow Through a Network
Related Papers (5)
Frequently Asked Questions (17)
Q2. How can a confluent flow be found in polynomial time?
A splittable flow with minimum congestion can, in fact, be found in strongly polynomial time by reducing to the perfect sharing problem, which can be solved using standard parametric maximum flow techniques [9].
Q3. What is the naive analysis of the graph?
In each iteration, the size of the graph decreases, thus yielding a polynomial bound on the running time of Conflt (a naive analysis yields an O(m2 + mn log(n2/m)) bound).
Q4. How do the authors complete the proof of Theorem 29?
To complete the proof of Theorem 29, assume by way of contradiction that φ misses an interior point p ∈ ∆r, and let π : ∆r −{p} → ∂∆r denote the continuous map which radially projects each point q of ∆r − {p} to the boundary by drawing a ray from p through q and continuing until the ray hits ∂∆r.
Q5. what is the vanishing criterion of the theorem 29?
Proof of Theorem 29: Suppose the authors are given a map φ : X → ∆r satisfying the homological vanishing criterion of Theorem 29, i.e. for any face σs ⊆ ∆r, the inverse image Y = φ−1(σs) satisfies H̃0(Y ) = . . . = H̃s−1(Y ).
Q6. How does the algorithm resolve the gap between confluence and splittability?
Hk = ln k+γ−o(1), where γ is Euler’s constant, the authors have resolved the gap between confluence and splittability to within an additive constant less than 1.
Q7. What is the main result of the multi-sink confluent flow problem?
In addition to being a useful reformulation of the original single-commodity confluent flow problem, the multi-sink version generalizes a classic graph partitioning question studied more than three decades ago, as the authors discuss below, and is thus of independent interest.
Q8. Why is it not possible to minimize congestion in Internet routing?
because of the confluence constraint on routing paths, it is not possible to minimize congestion in Internet routing simply by solving traditional multi-commodity flow problems.
Q9. How do the authors ensure that the optimal splittable flow is 1?
The authors will assume, without loss of generality, that the congestion of this optimal splittable flow is 1, since this property can always be ensured by scaling all demands and flows by the congestion value.
Q10. What is the way to solve the gap between confluent and splittable flows?
While the bound of 1+ln(k) on the ratio between the congestion of confluent and splittable flows is existentially tight up to an additive constant, it is natural to wonder if there are interesting classes of graphs for which the gap is smaller.•
Q11. How can the authors estimate the flow of unsplittable flow to within a constant factor?
Both the congestion minimization and demand maximization versions of unsplittable flow may be approximated to within a constant factor using the algorithms of [3, 15].
Q12. What is the simplest way to denote the congestion induced by a flow?
S can be joined to S by k paths which are disjoint except for the common vertex v. Throughout this section, the authors will use the notation Cf (v) to denote the congestion induced by a flow f at a vertex v; the subscript f will be omitted when it is clear from context.
Q13. What is the meaning of the term near-arborescence?
In other words, a near-arborescence is a convex combination of arborescences, any of which can be transformed into any other by disconnecting and reattaching some leaves.
Q14. How is the congestion of the optimal confluent flow calculated?
The authors first present an instance where the congestion of the optimal confluent flow is at least Hk times that of the optimal splittable flow.
Q15. what is the vanishing of the required homology groups?
In this graph, no v 6= â can be separated from â by removing fewer than s+1 vertices, so Theorem 28 ensures the vanishing of the required homology groups.
Q16. how many arborescences are obtained by taking a vertex v with three outgoing?
K contains a triangle for every three arborescences obtained by taking a vertex v with three outgoing edges e, e′, e′′, and an arborescence A in G − {v}, and joining v to A using each of the three edges; and K contains a square for every four arborescences obtained by taking a pair of vertices v1, v2, each with two outgoing edges, and an arborescence A in G − {v1, v2}, and joining v1, v2 to A in each of the four possible combinations.
Q17. What is the common example of ad hoc networks of Wi-Fi access points?
The wireless domain provides another example where ad hoc networks of Wi-Fi access points act as forwarding agents back to the wired access point that connects to the Internet.