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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 numbers and sender sends the packets continuously. Consider TCP, IP and Ethernet overhead into consideration, what is the required data rate?

1. 299 Mbps

2. 319Mbps

3. 288 Mbps

4. 240 Mbps
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As Sequence no field in TCP is 32 bit so Max TCP packet could have 232 Byte data(As TCP Byte Oriented)

Now Segment size is 1500B

So no of segment=(232/1500).

Now in each segment we will have TCP Header(20B),IP Header(20B) and Frame Header(18 B) added as overhead.

So extra Overhead=(232/1500)*58 B

So total Size of Packet=232+(232/1500)*58B

As no wraparound of Sequence Number So this would be transfered within life time of packet i.e 120 sec

So Data Rate=[{232+(232/1500)*58}8]/120 =297Mbps

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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 numbers and sender sends the packets continuously. Consider TCP, IP and Ethernet overhead into consideration, what is the required data rate?

ANS:. 299 Mbps

TCP segment size = 1500 bytes(excluding header)

TCP segment with header= 1500 + 20 +20 +18=1558 bytes

TCP segemnt is havinfg maximum 2^32 size

TCP segment withoutt header = 2^32 bytes*8-------------------1

TCP with header = 2^32*58/1558 bytes *8---------------------------2

add 1 and 2 and divide by 120= you got 296990464.3 = 297Mbps

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