{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

HW3F11-key

# Client b initiates a http session with server s

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: iii) Host A and B are communicating over a TCP connection, and Host B has already received from A all bytes up through byte 6500. A now sends two segments to B back-to-back. The first segment has sequence number 6501 and contains 200 bytes of data. The second contains 400 bytes of data. What is sequence number of the second segment? _6701_ If the segments arrive in order, and B sends ACKs to both packets, what are acknowledgement numbers in the first_6701_ and second_7101_ACKs? If the second segment arrives before the first segment, what are acknowledgement numbers in the first_6501_ and second_7101_ ACKs? If the first segment is lost, what is the acknowledgement number is the ACK packet sent in response to the arrival of the second segment? 6501__ iv) If the average of the RTT is estimated to be 100msec and the average of the deviation of the RTT is estimated to be 20msec, what is the resulting RTO?__ 180_ 3 2. TCP Congestion Control [15 points] Use the approximate equation (derived in class) for throughput as a function of drop rate: 1.5 = Assume an RTT of 40msec and an MSS of 1000bytes. In the questions to follow, give answers to three significant figures and ignore IP and TCP headers in your calculations. i). What drop rate p would lead to a throughput of 1Gbps?_ 6 x 10-8_ ii) What drop rate p would lead to a throughput of 10 Gbps?_ 6 x 10-10_ iii) If the connection is sending data at a rate of 10Gbps, how long on average is the time interval between drops?__ 1333sec___ iv) What window size W (measured in terms of MSSes) would be required to maintain a sending rate of 10Gbps?_ _(rounded down to the nearest integer) There are two acceptable answers. The first, is that the minimum fixed window needed to sustain a certain throughput T must satisfy the equation T = W*MSS/RTT. That gives the answer of 50,000. The second acceptable answer is that in AIMD, the peak window size that yields a certain throughput satisfies the equation T=(3/4) W*MSS/RTT (i.e. AIMD would sawtooth between W/2 and W). This yields 66,666. v) If a connection suff...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online