Answer:
a) There are approximately 2^32 = 4,294,967,296 possible number of the sequence. This number of the sequence does not increase by one with every number of sequences but by byte number of data transferred. Therefore, the MSS size is insignificant. Thus, it can be inferred that the maximum L value is representable by 2^32 ≈ 4.19 Gbytes.
b) ceil(2^32 / 536) = 8,012,999
The segment number is 66 bytes of header joined to every segment to get a cumulative sum of 528,857,934 bytes of header. Therefore, overall number of bytes sent will be 2^32 + 528,857,934 = 4.824 × 10^9 bytes. Thus, we can conclude that the total time taken to send the file will be 249 seconds over a 155~Mbps link.
Explanation:
a) There are approximately 2^32 = 4,294,967,296 possible number of the sequence. This number of the sequence does not increase by one with every number of sequences but by byte number of data transferred. Therefore, the MSS size is insignificant. Thus, it can be inferred that the maximum L value is representable by 2^32 ≈ 4.19 Gbytes.
b) ceil(2^32 / 536) = 8,012,999
The segment number is 66 bytes of header joined to every segment to get a cumulative sum of 528,857,934 bytes of header. Therefore, overall number of bytes sent will be 2^32 + 528,857,934 = 4.824 × 10^9 bytes. Thus, we can conclude that the total time taken to send the file will be 249 seconds over a 155~Mbps link.
