Unformatted text preview: s about all nodes two hops away and how to reach
them. Eventually, after k advertisements, each node learns about how to reach all nodes k
hops away, assuming of course that no packet losses occur. Hence, it takes d advertisement
intervals before a node discovers routes to all the other nodes, where d is the length of the
longest minimumcost path from the node.
Compared to the distance vector protocol, the path vector protocol consumes more network bandwidth because now each node needs to send not just the cost to the destination,
but also the addresses (or identiﬁers) of the nodes along the best path. In most large realworld networks, the number of links is large compared to the number of nodes, and the
length of the minimumcost paths grows slowly with the number of nodes (typically logarithmically). Thus, for large network, a path vector protocol is a reasonable choice.
We are now in a position to compare the linkstate protocol with the two vector protocols (distancevector and pathvector). 19.9 Summary: Comparing LinkState and Vector Protocols There is nothing either good or bad, but thinking makes it so.
—Hamlet, Act II (scene ii)
Bandwidth consumption. The total number of bytes sent in each linkstate advertisement is quadratic in the number of links, while it is linear in the number of links for the
distancevector protocol.
The advertisement step in the simple distancevector protocol consumes less bandwidth than in the simple linkstate protocol. Suppose that there are n nodes and m links
in the network, and that each [node pathcost] or [neighbor linkcost] tuple in an advertisement takes up k bytes (k might be 6 in practice). Each advertisement also contains a source
address, which (for simplicity) we will ignore.
Then, for distancevector, each node’s advertisement has size kn. Each such advertisement shows up on every link twice, because each node advertises its best path cost to
every destination on each of its link. Hence, the total bandwidth consumed is roughly
2knm/ADVERT INTERVAL bytes/second. SECTION 19.9. SUMMARY: COMPARING LINKSTATE AND VECTOR PROTOCOLS 11 The calculation for linkstate is a bit more involved. The easy part is to observe that
there’s a “origin address” and sequence number of each LSA to improve the efﬁciency
of the ﬂooding process, which isn’t needed in distancevector. If the sequence number
is bytes in size, then because each node broadcasts every other node’s LSA once, the
number of bytes sent is n. However, this is a secondorder effect; most of the bandwidth
is consumed by the rest of the LSA. The rest of the LSA consists of k bytes of information per
neighbor. Across the entire network, this quantity accounts for k (2m) bytes of information,
because the sum of the number of neighbors of each node in the network is 2m. Moreover,
each LSA is rebroadcast once by each node, which means that each LSA shows up twice
on every link. Therefore, the total number of bytes consumed in ﬂooding the LSAs over
the network to all the nodes is k (2m)(2m) = 4km2 . Putting it together with the bandwidth
consumed by the sequence number ﬁeld, we ﬁnd that...
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This document was uploaded on 02/26/2014 for the course CS 6.02 at MIT.
 Fall '13
 HariBalakrishnan

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