Register

Find wrap around time

• edited by
8,893 views

4 Answers

Best answer
16 16 votes
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
• selected by
2 2 votes
sequence possible=2^24.

wraparound time=(2^24)/100Mbps
1 1 vote
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.
Voting & Accepting an answer helps improve the community
← Previous · denormalized rangeNext → · Q6 ch-5 me workbook
← Previous in category · stop and waitNext in category → · TIFR CSE 2011 | Part...
Position:
Show:

Related questions

1 1 vote
1 1 answer
1.9k
1.9k views
arch asked Dec 11, 2017
1,896 views
wrap around time
0 0 votes
2 2 answers
1.5k
1.5k views
Neal Caffery asked Jan 12, 2017
1,459 views
wrap around time
2 persons are using a channel like TCP. In their link they have used 64-bit sequence numbers with a link speed of 75 Tbps. How long can a packet stay in link without worr...
3 3 votes
2 answers 2 answers
5.3k
5.3k views
Akriti sood asked Oct 23, 2016
5,285 views
computer networks (wrap around time)
A TCP segments of size(excluding headers) 1500 bytes with life time of a packet as 120 sec is sent over a TCP connection. There is no wrap around time for sequence number...
3 3 votes
1 answers 1 answer
6.7k
6.7k views
shane.126 asked Oct 13, 2017
6,702 views
Computer Networks - Transport Layer - Sequence number wrap around
TCP operates over a 40 Gbps link. If TCP uses the full bandwidth continuously, how long (in msec) would it take the sequence number to wrap around completely?