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Consider transferring an enormous file of L bytes from Host A to Host B. Assume an MSS of 536 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 4 bytes.
b. For the L you obtain 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.

in Computer Networks by Veteran (56.7k points) | 255 views

a) 232 B = 4 GB

b) 248.9 sec

Am I correct?

1 Answer

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(A.) Sequence number field is 32-bit, so 2$^{32}$ sequence number are possible. And 1 sequence number is assigned to 1-byte. Therefore  2$^{32}$ sequence number will be assigned to 2$^{32}$ Bytes. Hence max. value of L  = 2$^{32}$ Bytes.

(B.) Given MSS = 536 Bytes, so number of segment = 2$^{32}/536$.

Given that 66 Byte of headers is added to each segment,

therefore, total amount of headers  =  (2$^{32}/536) * 66 Bytes$

total data to transmit = header + data = (2$^{32}/536) * 66 Bytes$  + 2$^{32}$ Bytes

Bandwidth = $155 Mbps$

Therefore, transmission time = total data * 8 bits / 155* 2$^{20}$ = 449 ms

by (481 points)

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