EE450_Discussion8_Fall08

EE450_Discussion8_Fall08 - Discussion #8 EE450, 10/15/2008...

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1 Discussion #8 EE450, 10/15/2008 Sample Problems: -ARQ protocols (Sliding window) Shirin Ebrahimi
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2 Problem#1: Description ± Draw a timeline diagram for the sliding window algorithm with SWS=RWS=4 frames for two given scenarios using the following assumptions: 1. The receiver sends a duplicate acknowledgement if it does not receive the expected frame. For example it sends dupack[2] when it expects to see frame[2] but receives frame[3] instead. 2. The receiver sends a cumulative acknowledgement after it receives all the outstanding frames. For example it sends ACK[5] when it receives the lost frame[2] after it already received frame[3], frame[4] and frame[5]. 3. The timeout interval is about 2RTT.
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3 Problem#1: Two Scenarios a. Frame[2] is lost. Retransmission takes place upon timeout (as usual). b. Frame[2] is lost. Retransmission takes place either upon receipt of the first dupack or upon timeout. Does this scheme reduce the transaction time? Note that some end-to-end protocols (e.g. variants of TCP) use a similar scheme for fast retransmission.
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4 Problem#1 Solution: Scenario (a) Transmission time Frame[2] Ack[1] time time Sender Receiver Frame[3] Frame[4] Frame[5] Frame[1] Dupack[2] Dupack[2] Dupack[2] Timeout=2RTT Frame[2] Ack[5] Frame[6] Ack[6] T trans + RTT T trans + RTT
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5 Frame[2] Ack[1] time time Problem#1 Solution: Scenario (b) Sender Receiver Frame[3] Frame[4] Frame[5] Frame[1] Dupack[2] Dupack[2] Dupack[2] Frame[2] Ack[5] Frame[6] Ack[6] T trans + RTT T trans + RTT Transmission time
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6 Comparing the two scenarios (a) Delay1= 2Ttrans + 3RTT (b) Delay2= 4T trans + 2RTT Frame[2] Ack[1] Frame[3] Frame[4] Frame[5] Frame[1] Dupack[2] Dupack[2] Dupack[2] Timeout=2RTT Frame[2] Ack[5] Frame[6] Ack[6] T trans + RTT Frame[2] Ack[1] Frame[3] Frame[4] Frame[5] Frame[1] Dupack[2] Dupack[2] Dupack[2] Frame[2] Ack[5] Frame[6] Ack[6] 2T trans 2T trans + 2RTT Transmission time Transmission time T trans
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7 Comparing the transaction time (transfer delay) in the two scenarios RTT=2T prop Delay1= 2T trans + 3RTT= 2T trans + 6T prop Delay2= 4T trans + 2RTT=4T trans + 4T prop ? 2T trans + 6T prop > 4T trans + 4T prop ? 2T prop > 2T trans If Tprop > Ttrans , in Scenario (b) the transaction time is reduced!
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Problem#2: Description ± A is connected to B via an intermediate router R. ± The A-R and R-B links each accept and transmit only one packet per second in each direction (so two packets take 2 seconds). ± The two directions transmit independently.
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EE450_Discussion8_Fall08 - Discussion #8 EE450, 10/15/2008...

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