Permutation Routing over Sparse Networks Presented by Nithish Kumar - - PowerPoint PPT Presentation

Permutation Routing over Sparse Networks

Nithish Kumar and Shubhang Kulkarni Presented by

Presenter: Nithish Kumar

Routing Algorithm

• Given a network topology, a routing algorithm specifies, for each pair of nodes, a

route - or a sequence of edges - connecting the pair in the network.

• The algorithm may also specify a queuing policy for ordering packets in the

switches' queues. For example. the First In First Out (FIFO) policy orders packets by their order of arrival. The Furthest To Go (FTG) policy orders packets in decreasing order of the number of edges they must stiII cross in the network.

The Model

• We model a communication network by a directed graph on N nodes.

Each node is a routing switch. A directed edge models a communication channel, which connects two adjacent routing switches.

• We consider a synchronous computing model in which

(a) an edge can carry one packet in each time step and (b) a packet can traverse no more than one edge per step.

The Model

• We assume that switches have buffers or queues to store packets

waiting for transmission through each of the switch's outgoing edges.

• The measure of the performance of a routing algorithm on a given

network topology is the maximum time - measured as the number of parallel steps - required to route an arbitrary permutation routing problem.

The Permutation Routing Problem

• In the permutation routing problem
• each node sends exactly one packet
• each node is the address of exactly one packet.
• If the graph is complete, routing the permutation can be done in just
• ne step.
• But most graphs are sparse and packet routing on such networks
• ften leads to congestion and bottlenecks.

Hypercubes

The n-dimensional hypercube (or n-cube) is a network with N = 2n nodes such that node X has a direct connection to node Y if and only if X and Y differ in exactly one bit.

Hypercubes

Permutation Routing in Hypercubes

• We analyse a hypercube with N processors and O(N log N) edges.
• The total number of directed edges in the n-cube is 2nN. since each

node is adjacent to n outgoing and n ingoing edges.

• The diameter of the network is n; that is, there is a directed path of

length up to n connecting any two nodes in the network, and there are pairs of nodes that are not connected by any shorter path(for instance take 00..0 and 11...1).

Bit-Fixing Routing Algorithm

Bit Routing Algorithm – Bad Case

• The above algorithm may lead to lot of congestion in specific
• permutations. Consider the following example.
• Suppose that n is even. Write each source node s as the

concatenation of two binary strings as and bs each of length n/2. Let the destination of s's packet be the concatenation of bs and as. This permutation will take Omega(sqrt(n)).

Two Phase Routing Algorithm

Two Phase Routing Algorithm

In some sense, this is similar in spirit to Quicksort. For a list already sorted in reverse order, Quicksort would take O(n2) comparisons, whereas the expected number of comparisons for a randomly chosen permutation is only O( n log n). Randomizing the data can lead to a better running time for Quicksort. Here, too, randomizing the routes that packets take - by routing them through a random intermediate point - avoids bad initial permutations and leads to good expected performance.

Two Phase Routing Algorithm

The two-phase routing algorithm is executed in parallel by all the packets. The random choices are made independently for each packet. Our analysis holds for any queueing policy that obeys the following natural requirement: if a queue is not empty at the beginning of a time step, some packet is sent along the edge associated with that queue during that time step. We prove that the above routing strategy achieves asymptotically optimal parallel time.

Presenter: Shubhang Kulkarni

v Analyze time for Phase I and argue Phase II by symmetry v Assume no packets start at Phase II until completion of Phase I Theorem

Main Result

In any permutation routing problem, the two-phase routing scheme routes all packets to their destinations in 𝑷(𝒐) time steps with high probability

Technique Overview

Time for Phase I ≤ Time for any packet to reach destination ≤ Number of times its path is touched by any packet ≤ Number of times any path is touched by any packet ≤ (Our Bound) 30𝑜 w.h.p. Pr 𝐵𝑜𝑧 𝑞𝑏𝑢ℎ 𝑢𝑝𝑣𝑑ℎ𝑓𝑒 > 𝑐𝑝𝑣𝑜𝑒 ≤ Pr 𝑓𝑤𝑓𝑜𝑢 + Pr 𝐵𝑜𝑧 𝑞𝑏𝑢ℎ 𝑢𝑝𝑣𝑑ℎ𝑓𝑒 > 𝑐𝑝𝑣𝑜𝑒 𝑓𝑤𝑓𝑜𝑢]. ≤ Very small

= >?

v 𝑁: Packet v 𝑓: Edge v P: Path (sequence of edges) v 𝑈 𝑁 : Number of time steps for 𝑁 to reach destination v 𝑈 𝑄 : Number of times edges of 𝑄 are used v 𝑂 𝑓 : Number of packets traversing over edge 𝑓 𝑈 𝑁 ≤ E 𝑂(𝑓F)

• FH=

Let 𝑓= 𝑓> ⋯ 𝑓J be the edges traversed by 𝑁 In each time step, 𝑁 halts at a node, or traverses an edge. Let 𝑄 = 𝑓= 𝑓>⋯ 𝑓J be any valid path by the bit fixing algorithm 𝑈 𝑄 = E 𝑂(𝑓F)

• FH=

𝑈 𝑁 ≤ 𝑈(𝑄) Time for a packet is upper bounded by number

• f times its path is touched

Preliminary Notation

Active Packets

Path 𝑄 is a possible packet path if it may be the outcome of the bit fixing algorithm Fix a possible packet path 𝑄 = 𝑤L𝑤= ⋯ 𝑤J. A packet 𝑁 is active at 𝑤F if the following two conditions happen: 𝑤F 𝑤FM= 1 2 ⋯ ⋯ ⋯ 𝑘 Differ in 𝑘QR bit ⋯ ⋯ 𝑜

• 1. 𝑤F and 𝑤FM= differ in 𝑘QR bit
• 2. Packet 𝑁 does not have its 𝑘QR bit fixed by the bit fixing algorithm
• Less than 𝑘 bits of 𝑁 have been fixed
• 𝑁 comes from a distance less than 𝑘 to 𝑤F

A packet is active if it is active at any node 𝑤F

Bounding Active Packets

For every vertex 𝑙 ∈ 𝑂 , define the indicator variable 𝐵V as follows: 𝐵V = 1 … If packet starting at node 𝑙 is active … Otherwise 𝐵V are independent as

• Depend only on intermediate destination
• Choice of intermediate destination is independent for each packet

𝐵 = E 𝐵V

V FH=

Is the total number of active packets 𝐹 𝐵 = ?

Bounding Active Packets

𝑡= 𝑡Z 𝑡

[

𝑡

[\=

𝑡

[M=

𝑢= 𝑡Z 𝑡

[

𝑢[\= 𝑡

[M=

𝑢= 𝑡Z 𝑢[ 𝑢[\= 𝑡

[M=

𝑢= 𝑢Z 𝑢[ 𝑢[\= 𝑢[M= Source of P Destination of P 𝑤F 𝑤FM= 𝑄 v A packet can arrive at 𝑤F only if it starts from ∗ ⋯ ∗ 𝑡

[ ⋯ 𝑡Z

v A packet actually arrives at 𝑤F only if its random destination is (𝑢= ⋯ 𝑢[\= ∗ ⋯ ∗) Pr 𝑏𝑠𝑠𝑗𝑤𝑗𝑜𝑕 𝑏𝑢 𝑤F 𝑏𝑠𝑠𝑗𝑤𝑏𝑚 𝑗𝑡 𝑞𝑝𝑡𝑡𝑗𝑐𝑚𝑓] = 1 2[\= Total number of possible arrivals ≤ 2[\= 𝐹 𝑂𝑣𝑛𝑐𝑓𝑠 𝑝𝑔 𝑏𝑑𝑢𝑗𝑤𝑓 𝑞𝑏𝑑𝑙𝑓𝑢𝑡 𝑞𝑓𝑠 𝑤𝑓𝑠𝑢𝑓𝑦 ≤ 1 𝐹 𝐵 ≤ 1 ⋅ 𝑛 ≤ 𝑜 Pr 𝐵 ≥ 6𝑜 ≤ 2\hZ By Chernoff Bounds We found our event!

Bounding Path Time

v We now know that Pr 𝑏𝑑𝑢𝑗𝑤𝑓 𝑞𝑏𝑑𝑙𝑓𝑢𝑡 ≥ 6𝑜 ≤ 𝑡𝑝𝑛𝑓𝑢ℎ𝑗𝑜𝑕 𝑢𝑗𝑜𝑧 v We can now bound the time for Phase I conditioning on high probability complement event v Use following inequality to get final bound: Pr 𝑄 = Pr 𝑄 𝑅] Pr 𝑅 + Pr 𝑄 𝑅 j] Pr 𝑅 j ≤ Pr 𝑅 + Pr 𝑄 𝑅 j] P ∶ 𝐹vent that T P ≥ 𝑑𝑜 Q ∶ 𝐹𝑤𝑓𝑜𝑢 𝑢ℎ𝑏𝑢 𝐵 ≥ 6𝑜

Bounding Path Time

Observation 1: Observation 2: Due to the ordering of bit fixing, a packet that leaves 𝑄 cannot return to 𝑄 in this phase An active packet at 𝑤F, crosses the edge (𝑤F, 𝑤FM=) with probability at most

= >

Probability that the active packets cross edges of 𝑄 more than 30𝑜 times is less than probability that a fair coin flipped 36𝑜 times comes up heads fewer than 6𝑜 times Lemma: Can be thought of as flipping fair independent coins to generate next co-ordinate

Bounding Path Time

Trial: each point in the algorithm where an active packet on path P, might cross an edge of P trial successful if packet leaves path failure if packet stays on path Since packet leaves 𝑄 on a successful trial, there can be at most 6𝑜 successful trials (due to our condition event!) Let 𝑌F be the number of transitions that the 𝑗QR packet makes on 𝑄. Then 𝑌F is a geometric random random variable with probability of success

= >

Let 𝑇hZ be an upper bound on the total path time 𝑇hZ = E 𝑌F

hZ FH=

Bounding Path Time

Let 𝑇hZ be an upper bound on the total path time 𝑇hZ = E 𝑌F

hZ FH=

Thus, Pr 𝑈 𝑄 ≥ 30𝑜 𝐵 ≤ 6𝑜] ≤ Pr 𝑇hZ ≥ 30𝑜 ≤ Pr 𝑎 ≤ 6 ≤ 𝑓\

wx? y z y y

… by Chernoff bounds Let 𝑎 be the number of heads in 36n fair coin flips On simplifying we get that: 𝑓\

wx? y z y y

= 𝑓\{Z ≤ 2\|Z\= Time for our favorite inequality to get the desired bound!

Bounding Path Time

Pr 𝑄 = Pr 𝑄 𝑅] Pr 𝑅 + Pr 𝑄 𝑅 j] Pr 𝑅 j ≤ Pr 𝑅 + Pr 𝑄 𝑅 j] Pr 𝑈 𝑄 ≥ 30𝑜 ≤ Pr 𝐵 > 6𝑜 + Pr 𝑈 𝑄 ≥ 30𝑜 𝐵 ≤ 6𝑜] ≤ 2\hZ + 2\|Z \= ≤ 2\|Z Due to bit fixing, a source and destination uniquely determines a possible path. Thus 2Z ⋅ 2Z = 2>Z possible paths Union bounding over all these paths, we get probability that any possible packet path needing time more than 30𝑜 is at most 2>Z ⋅ 2\|Z = 2\Z = 𝑃(𝑂\=)

Phase II

v Can be viewed as running Phase I backwards v Instead of packets starting at given source and routing to a random sink, they start at a random origin and end at a given destination. v Hence no packet spends more than 30n steps in Phase II with probability 1 − 𝑃(𝑂\=) v We may remove the assumption that packets begin Phase II only after Phase I has completed v Total number of packet traversals across the edges of any packet path during Phase I and Phase II together is bounded by 60𝑜 = 𝑃(𝑜) with probability 1 − 𝑃(𝑂\=) v The bound holds regardless of how the phases interact v This concludes the proof!