Compared to the distance vector protocol the path

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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 minimum-cost 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 minimum-cost 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 link-state protocol with the two vector protocols (distance-vector and path-vector). ￿ 19.9 Summary: Comparing Link-State 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 link-state advertisement is quadratic in the number of links, while it is linear in the number of links for the distance-vector protocol. The advertisement step in the simple distance-vector protocol consumes less bandwidth than in the simple link-state 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 distance-vector, 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 LINK-STATE AND VECTOR PROTOCOLS 11 The calculation for link-state 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 distance-vector. 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 second-order 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 re-broadcast 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.

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