homework2

# homework2 - Dept of Computer Science and Engineering HKUST...

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Dept. of Computer Science and Engineering HKUST CSIT 561 Computer Networks: An Internet Perspective Fall Semester 2009 Homework 2 Name: LU, Lu ID: 09960007 Due Date: Sunday 1 Nov. 2009 11:59 pm Submission Instructions: Please submit your solutions electronically through the LMES website http://lmes2.ust.hk . Please go the CSIT 561 website and look under Assignments. You will see Homework 2 listed there. Please submit your work either in PDF format (preferred), or MS Word format. Problems: 1. Consider the following plot of TCP window size as a function of time. Some special points are (in the form of (time, congestion window size)): A(1, 1), B(7, 64), C(23, 80), D(24, 40), E(30, 46), F(31, 1). Answer the following questions: a) Identify the intervals of time when TCP slow start is operating (before time 30). b) Why the congestion window size increase ratio drops at Point B? c) At point C, is segment loss detected by a triple duplicate ACK or by a timeout? d) What is the value of threshold at point D? e) At point E, is segment loss detected by a triple duplicate ACK or by a timeout? f) What is the value of threshold at point F? CSIT 561 (Fall 2009) Homework 2 1

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Dept. of Computer Science and Engineering HKUST g) Assume that congestion occurs only when the congestion window is equal or larger than 80. The length of a segment is 120 bytes. RTT=0.01s. If there is no timeout occurs, what is the average transmission rate during a long period of time? (Please omit the size of header) Answer a) TCP slow start is operating in the intervals A (1, 1) and E (30, 46) . b) Because point B reaches the threshold . When congestion window size is above threshold, the sender is in congestion-avoidance phase, window grows linearly. c) At point C, segment loss is detected by a triple duplicate ACK . d) The value of the threshold at point D is 40 . e) At point E, segment loss is detected by a timeout . f) The value of the threshold at point F is 23 . g) Let N be the increasing times of congestion window and denote W as the total number of congestion windows, 23 1 16 6 1 ) 64 80 ( 64 log 2 = + + = + - + = N 1287 1160 127 2 16 ) 80 65 ( 2 1 2 1 7 = + = × + + - - = W Then, 56 = = N W w Hence, ms bytes s bytes RTT MSS w Throughput / 672 01 . 0 120 56 = × = × = The average transmission rate is 672 bytes/ms. 2. In Section 3.5.4, we saw that TCP waits until it has received three duplicate ACKs before performing a fast retransmit. Why do you think the TCP designers chose not to perform a fast retransmit after the first duplicate ACK for a segment is received? Answer Suppose packets n, n+1, and n+2 are sent, and that packet n is received and ACKed. If packets n+1 and n+2 are reordered along the end-to-end-path (i.e., are received in the order n+2, n+1) then the receipt of packet n+2 will generate a duplicate ACK for n and would trigger a retransmission under a policy of waiting only for second duplicate ACK for retransmission. By waiting for a triple duplicate ACK, it must be the case that two packets after packet n are correctly received, while n+1 was not received. The designers of the triple duplicate ACK scheme probably felt
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