Find wrap around time

Question

Consider 100 mbps network with 24 bit sequence number field find the wrap around time for sequence no?

Answer 1

Consider 100 mbps network with 24 bit sequence number field find the wrap around time for sequence no?

1sec = 100 mb

1/ 100m= 1b

as

1byte = 2^24 sequence no.

1bit         = 2^24 * 8 bits

therefore= 2^24 * 8 *1/ 100 m  sec = 1.34 sec

Answer 2

wrap around time= total number of bits/bandwidth

=2^24/100*2^20

=0.16

Comments

How 2^20 came ??

Answer 3

sequence possible=2^24.

wraparound time=(2^24)/100Mbps

Answer 4

Answer will be =0.1455 sec

Explanation::

1000Mbps=10^3*10^6Bits/sec

125x10^6 byte=1 sec

1 Byte=1/125x10^6sec

so:

2^24 byte=2^24/125x10^6= 0.1455sec Approx.