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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
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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
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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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