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A New Method for PMTU Discovery Using Split Path MTUs

IP.com Disclosure Number: IPCOM000033372D
Original Publication Date: 2004-Dec-08
Included in the Prior Art Database: 2004-Dec-08
Document File: 2 page(s) / 51K

Publishing Venue

IBM

Abstract

Path Maximum Transmission Unit (PMTU) Discovery is the mechanism by which hosts discover the maximum message unit size that can be supported on a path without requiring fragmentation. However, the PMTU represents the LEAST Maximum Transmission Unit (MTU) size along that path. While this eliminates the need for fragmentation, it still generates a considerable number of packets and does not make optimal use of the cumulative MTUs along the path. Proposed is a split-path MTU discovery mechanism that makes optimal use of cumulative path-base MTU to reduce the number of packets in the network based on the hop count for the path. A new metric is introduced - the packet-hop count for a path, which represents the number of hops made by each packet along the path, from the source to the destination. The lower this value, the lesser number of packets going in the internet, and greater the performance of the overall data transfer. The goal is to provide a method to reduce this packet-hop count variable for a given data transfer.

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A New Method for PMTU Discovery Using Split Path MTUs

The Split-Path MTU discovery mechanism works as described below:

S

4K

PMTU = 1.5 K

Path1, PMTU1 = 1.5K, hops = 4

Path2, PMTU2 = 9K, hops = 6

Case1: To send 18K data, 12 pkts over 10 hops (120)

Cas e2: 12 pkts over 4 hops (path1) & 2 over 6 hops (path2) for total of 60 pkt-hop

R 1

1.5K

R2

9K

R3

2K

R4

9K

9K

R5

R6

10K

R 7

9K

R8

16K

R9

9K

D

Considering the path in the figure above, we observe that current PMTU discovery will find the PMTU as 1.5 Kbytes. Since the path has 10 hops, the total number of packet-hops (the number of hops each packet makes, for every packet) for a message of size 18 Kbytes is ((18K /1.5K) * 10 ) = 120.

If we however, split this path into two, as in the figure, we observe that path#1 has a PMTU of 1.5 Kbytes and path#2 has PMTU of 9 Kbytes. Assuming that messages of size 1.5 Kbytes are sent over path#1 with 4 hops and of size 9 Kbytes over path#2 with 6 hops, the total packet-hop count is ((18K /1.5K) * 4 ) + ((18K/9K)* 6) = 48+ 12 = 60 for the above message of size 18 Kbytes. As can be seen, the packet-hop count decreased from 120 to 60.

The above calculation can be summarized as 18K( (4/1.5K) + (6/9K)). So for a given message size(18k), our goal is to reduce the

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((h1/p1) + (h2/p2)), where h1, h2 are the hop counts and p1, p2 are the PMTU values for each section of the path.

The following algorithm is proposed:

For each path between hosts A & B

For each...