EE450-Discussion14-Fall10

# EE450-Discussion14-Fall10 - 1 Discussion#14 EE450 Sample...

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Unformatted text preview: 1 Discussion #14 EE450, 11/24/2010 Sample Problems -TCP Congestion Control 2 Review: TCP Congestion Control MaxWindow = MIN (CongestionWindow, AdvertisedWindow) EffectiveWindow = MaxWindow – (LastByteSent- LastByteAcked) Increment = MSS x (MSS / CongestionWindow) CongestionWindow + = Increment 3 Problem #4 Consider a simple congestion-control algorithm that uses linear increase and multiplicative decrease but not slowstart, that works in units of packets rather than bytes, and that starts each connection with a congestion windows equal to one packet. Give a detailed sketch of this algorithm. Assume the delay is latency only, and that when a group of packets is sent, only a single ACK is returned. Plot the congestion window as a function of round-trip times for the situation in which the following packets are lost: 9, 25, 30, 38 and 50. For simplicity, assume a perfect timeout mechanism that detects a lost packet exactly 1 RTT after it is transmitted. 4 Solution The window size is initially 1; when we get the first ACK it increases to 2. At the beginning Of the second RTT we send packets 2 and 3. When we get their ACKs we increase the window size to 3 and send packets 4, 5 and 6. When these ACKs arrive the window size becomes 4. Now, at the beginning of the fourth RTT, we send packets 7, 8, 9, and 10; by hypothesis packet 9 is lost. So, at the end of the fourth RTT we have a timeout and the window size is reduced to 4/2 = 2. Continuing we have: RTT 5 6 7 8 9 Sent 9-10 11-13 14-17 18-22 23-28 Again the congestion window increases up until packet 25 is lost, when it is halved, to 3, at the end of the ninth RTT. 5 Solution The plot below shows the window size vs. RTT....
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EE450-Discussion14-Fall10 - 1 Discussion#14 EE450 Sample...

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