414Lecture10solutions

414Lecture10solutions - Throughput Concrete Example • ...

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Unformatted text preview: Throughput Concrete Example •  •  •  •  One user, link with bandwidth R = 1 Mbps RTT = 100ms MSS = 1000 bits Assume packets dropped as soon as throughput exceeds R (no queuing) •  Always see triple duplicate ACK (no timeouts) •  How long does it take to reach steady state? (seconds) •  What is the average throughput in steady state? (Calculate explicitly. Not using the macroscopic eqn.) 2010 Michael Rabbat ECSE 414, Lecture 10 5 Solutions 1 •  How long to reach steady state? •  Sending 1 segment in 1 transmission round gives a transmission rate of MSS/RTT bps = 104 bps •  Sending k segments gives rate k x 104 bps •  This reaches R=106 bps when k ≥ 100 •  In slow start, we send 1 segment in the first round, 2 in the second, 4 in the third, 2m in the m-th •  When m=8, we send 128 segments and exceed R •  Each transmission round lasts RTT=100ms, so this occurs after 800ms 2010 Michael Rabbat ECSE 414, Lecture 10 6 Solutions 2 •  •  •  •  •  In steady state, assume we experience loss when sending exactly 100 segments Then we drop down to 50 and climb back up, increasing congestion window by 1 every round This takes a total of 51 rounds (from 50 up to 100) To calculate average throughput, we first calculate the total number of bits transmitted over these 51 rounds, and then divide by 51 RTTs 100 50 ￿ ￿ The total number of bits is given by k MSS = (50 + k ) k=50 k=0 = = = 51(50) + k=1 51(50) 51(50) + 2 ￿￿ 3 51(50) MSS 2 50 ￿ k •  The total time is 51 RSS, and so the average throughput is (3/2)(51)(50) MSS = 7.5 × 105 bps 51 RTT 2010 Michael Rabbat ECSE 414, Lecture 10 7 ...
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