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A Scalable, Commodity Data Center Network Architecture *. Authors: M. Al-Fares, A. Loukissas and A.Vahdat. Presented by: YAN QIAO. *This paper appeared in ACM SIGCOMM 2008, cited by 131 since then. Agenda. Background & Problem Motivation Solution Summary. Background (I).
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A Scalable, Commodity Data Center Network Architecture* Authors: M. Al-Fares, A. Loukissas and A.Vahdat Presented by: YAN QIAO *This paper appeared in ACM SIGCOMM 2008, cited by 131 since then.
Agenda • Background & Problem • Motivation • Solution • Summary
Background (I) • Large data centers of over 10,000 hosts. • Inter-node communication – bottleneck. • Existing tech? Expensive, no TCP/IP. • Problem: To build a scalable, cost-effective data center network.
Background (II) Oversubscription: the ratio of the worst-case achievable aggregate bandwidth among the end hosts to the total bisection bandwidth of a particular communication topology. Typically 2.5:1 (400 Mbps) to 8:1 (125 Mbps).
Background (III) • Fact: The cost for a large cluster network greatly affects design decisions. 48-port GigE switch – $7,000 128-port GigE switch – $700,000
Background (III) • Fact: The cost for a large cluster network greatly affects design decisions. ← Fig: Current cost estimate vs. maximum possible number of hosts for different oversubscription ratios.
Motivation Can we make it cheaper?
Solution (I) • Clos Networks / Fat-Trees
Solution (II) • Addressing: • Pod switches: 10.pod.switch.1 • Core switches: 10.k.j.i • Two level routing tables with central controller.
Solution (IV) • More details... • Flow Classification • Flow Scheduling • Fault Tolerance • Power & Heat • Wire complexity
Experiment Results Fig: Aggregate Bandwidth of the network, as a percentage of ideal bisection bandwidth. The ideal bisection bandwidth for the fat-tree network is 1.536Gbps.
Summary • Fat-tree based architecture, with techs on routing. • Lower cost: switch, power… • Ideas based on this paper: • Doable with new prices? • Combined approach? • Reduce # switches needed.
Thanks! Question? --- YAN QIAO