Transport layer

Question

For a network which has maximum packet size is 256 bytes . lifetime of a packet is 20 sec and 8 bit sequence number . Find maximum data rate per connection

Comments

3264 Bps ??

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No. of bits = 256*256*8

bps = 524288/20 = 26214.4 bps ?

http://site.iugaza.edu.ps/djabal/files/Network-Chapter-201.pdf

Answer 1

Maximum packet size  = 256 bytes

Segment lifetime = 30 sec

Bits in sequence number = 8

Maximum number of possible sequence numbers using 8 bits = 256.

So, maximum number of bytes that can be uniquely identified = 256 bytes.

Lifetime of a segment = 20 seconds.

So, maximum amount of data that can be sent in 20 seconds = 256 bytes.

Thus,

Maximum data rate per connection

= 256 bytes / 20 seconds = 12.8 Bytes/ Seconds

Answer 2

8 bit sequence no. means a sender can not send more than 2^8 packets in a lifetime and maximum packet size is 256 bytes

so maximum  number of bits  will be 256*256*8=52488 bits

So maximum data rate will be 52488/20=26214.4 bps

Comments

if seq no is 8 bit then it can send max 2^8 segments in life time is it not?

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Ya it can send only 2^8 segments in a life time....there was some mistake....i have corrected it

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For 8-bit sequence number, we can uniquely number the byte up to 256(0 to 255).
to send 256B, we can send all the bytes in one segment.
So maximum data rate could be 256/ 20 Byte/sec = 12.8 B/Sec

Answer 3

the answer is 12.8 byte per sec.