utf8''02_Layering

Of data transmiged per unit of kme

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Unformatted text preview: 17 Delay CalculaKon •  PropagaKon Delay: Kme needed for signal to travel the medium, Distance / speed of medium •  Transmission Delay: file size/bandwidth •  Queuing Delay: Kme waiKng in router’s buffer C d1 A d2 R B 18 Example: Problem 1.4 from Book •  Transfer 1.5 MB file, assuming RTT of 80 ms, a packet size of 1- KB and an iniKal “handshake” of 2xRTT •  Bandwidth is 10 Mbps and data packets can be sent conKnuously A B request RTT reply confirm Ack Tt Tp RTT = 80 ms Tt = 1024x8 bits/107 bits/s = 0.8192 ms Tp = 40 ms # of packets = 1000000/1024 = 976.56 D = 2xRTT + 976.56xTt + Tp = 160 + 800 + 40 ms = 1000 ms = 1s . . . t 19 Example: Problem 1.4 from Book •  Transfer 1,5 MB file, assuming RTT of 80 ms, a packet size of 1- KB and an iniKal “handshake” of 2xRTT •  ANer sending each packet must wait one RTT A B request RTT reply confirm Ack Tt RTT RTT = 80 ms Tt = 1024x8 bits/107 bits/s = 0.8192 ms Tp = 40 ms # of packets = 1536 (1.5 x 1024) D = 2xRTT + 1535x(Tt +RTT)+ Tt+Tp = 160 + 124,057 + 0.8192 + 40 ms = 124.258 s . . . t 20 Example: Problem 1.4 from Book •  Transfer 1,5 MB file, assuming RTT of 80 ms, a packet size of 1- KB and an iniKal “handshake” of 2xRTT •  Can transmit 20 packets per one RTT A B request RTT reply confirm Ack RTT RTT = 80 ms Tt = 0 ms Tp = 40 ms # of packets = 1536 (1.5 x 1024) D = 2xRTT + 76xRTT + Tp = 160 + 6080 + 40 ms = 6.28 s . . . t 21 Example: Problem 1.4 from Book •  Transfer 1,5 MB file, assuming RTT of 80 ms, a packet size of 1- KB and an iniKal “handshake” of 2xRTT •  Send one packet aNer first RTT, two packets aNer second RTT, and third packets aNer third RTT and so on A B request RTT reply confirm Ack RTT . . . t RTT = 80 ms Tt = 0 ms Tp = 40 ms # of packets = 1536 (1.5 x 1024) # of waits (1+2+…2n = 2n+1 - 1) 211 - 1 =2047 packets, n = 10 D = 2xRTT + 10xRTT + Tp = 160 + 800 + 40 ms = 1 s 22 Latency vs. Bandwidth •  Importance depends on applicaKon –  1 byte file, 1ms/1Mbps vs. 100ms/100Mbps 1 ms + 8μs = 1....
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