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Consider an IPv4 network. Each host can generate packets at the rate of 500 packets per second. If each packet in the network is identified by unique identification number of 48 bits, then the host wrap around time for generating packets will be ________s
in Computer Networks by Active (3.9k points)
edited by | 455 views
Is it 131.07 sec?
Yes. How did you do it?
Just not able to understand the question, 48 bits to identify host, then why have you taken as the limit for sequence number

2 Answers

+8 votes
Best answer
In IPv4 address size is of 32 bits so there are 2^ 32 hosts in the network.

In 1 sec----->500 packets by each host.

Therefore, in 1 sec,total number of packets present in the network is 500*(2^32).With 48 bits identication number, there are 2^48 packets possible.

So, 1 sec-------------> 500*(2^32) packets

      ? sec<------------- 2^48 packets

i.e. 131.07 sec
by (397 points)
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Had they mentioned 'Each packet has unique identification number then your answer is correct', but they  had mentioned that each host has a unique identification number right?.  @Bikram Sir please reply.

In question they are telling each host is identified by  48 bit number, not each packet. and they are asking about host wrap around time not possible sequence number wrap around.

I think they form question wrongly
0 votes

This can also be done in this way:-

1 second host generate -------> 500 packets (assume 2 9 =500)

Total no. of the packet (using Ipv4 sequence no. 2 32) -----> (2 *9 )* (2*32 )= 2 41 (total packet in the networks )

Time required to eat up 2 48 sequence no. Time required for that = 248/ 241= 2 7 = 128  seconds time

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