spanning tree and RBF flooding.pdf

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Winter 2009 CMPE 150 Due February 24, 2009

PEM Assignment #5 Solutions

1. For the network of Figure 5.7a (and ignore the line weights), suppose flooding is used for the

routing algorithm. If a packet is sent from E to D, with a maximum hop count of 3, list all the

routes it will take. How many packets are sent in this flooding?

Hop 1, Hop 2, Hop 3. 11 packets are sent in this flooding

2. For the subnet of Figure 5.13(a), using distance vector routing. As router B, build the router

table for that node having received the following vectors: from A (0,4,6,8,6,10), from C

(7,2,0,5,2,8) and from F (11,4,8,5,7,0). B has measured its routes to its neighbors A, C and F

and gets 2, 3 and 2 respectively. Compute B’s new routing table, showing both the outgoing

line and the expected delay.

2



3. For the subnet of Figure 5.6(a), draw a sink tree for router B. Using this subtree, how many packets are generated by a broadcast from B using:

a. reverse path forwarding

b. the sink tree?

Packets generated by broadcast from B:

a) Reverse Path Forwarding : 28 packets

b) Sink Tree: 14 packets(obtained by counting the links on the tree)

Sink Tree at Node BSink Tree at Node B (Re-ordered)

Tree built by reverse path forwarding. Circles in blue indicates paths that are not found in the sink treefor node B. Packets received on these paths are duplicates and are not forwarded



4. Referring to Figure 5-20, suppose that node C has just rebooted and has no routing informationin its tables. It immediately needs a route to node H. It sends out broadcasts with TTL of 1, 2, 3,

and so on. How many rounds does it take to find the route?

Node C is 3 hops away from H, so it takes 3 rounds for it to find the route.

5. Suppose an ATM network uses a token bucket scheme for traffic shaping. If a new token is put

into the bucket every 10 µsec, and is good for one 64K Byte packet, what is the maximumsustainable data rate?

10 µsec = 64 x 8 x 103 bits 1sec = (64 x 8 x 10

3) / 10 x 10

-6

1 sec = 51.2x109

The maximum sustainable data rate is 51.2Gbps

6. An IP datagram reassembly algorithm uses a timer to avoid having a lost fragment tie up buffersindefinitely. Suppose that a datagram was fragmented into three fragments, and the first two

arrive, but the timer goes off before the last fragment arrives. The algorithm has discarded thefirst two fragments, now it has – finally – the missing third fragment. What does it do with it?

As far as the receiver is concerned, this is a part of a new datagram, since no other parts ofit are known. It will therefore be queued up until the rest show up. If they do not, this one

will time out too.

8. Convert the IP address whose hexadecimal representation is C0A8000D.

11000000.10101000.00000000.00001101 = 192.168.0.13

9. What does this address tell you about its location?

This is a private IP address and the host is located in a private network. ie, it is not directly

reachable from the public internet.



10. A router has the following (CIDR) entries in its routing table:

Address/mask Next hop

128.114.56.0/22 Interface 0

128.114.60.0/22 Interface 1

192.168.30/23 Router 1

Default Router 2 For packets with the following IP addresses, show where therouter will send the packet

(a) 128.114.63.09

(b) 128.114.57.11(c) 128.114.52.02

(d) 192.168.33.05(e) 192.168.31.06

Solution – Convert the IP address to bits and then make an AND with the subnet mask of the

interface whose address is closest to that of the IP address. The results of the AND will give you

the network address and the interface to send the packet to.

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spanning tree and RBF flooding.pdf