Consider transferring an enormous file of L bytes from Host A to Host B. Assume
ID: 3604033 • Letter: C
Question
Consider transferring an enormous file of L bytes from Host A to Host B. Assume an MSS of 1024 bytes. a. What is the maximum value of L such that TCP sequence numbers are not exhausted? Recall that the TCP sequence number field has four bytes. b. For the L you obtained in (a), find how long it takes to transmit the file. Assume that a total of 66 bytes of transport, network and data link header are added to each segment before the resulting packet is sent out over a 155 Mbps link. Ignore flow control and congestion control so A can pump out the segments back to back and continuously.
Explanation / Answer
Solution:
a)
As we know that
=> 1byte is equal to 8 bits
Now since we have 4 bytes here, using conversion we have 32 bits.
Then maximum value of L will be= 2^32= 4.294.967.296 bits
b)
Number of segments= ceil((2^32)/1024)= 4194304
total number of bytes sent over 155 Mbps link= 4194304*66= 276824064 bytes
Total number of bits transmitted= ((2^32) + 276824064)= 4571791360 bytes
Time to transfer this data through 155 Mbps link= (4571791360 *8)/(155 * 10^6) bps
= 235.963425032= 236 seconds
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