can transmit 20 packets per one

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Unformatted text preview: 008ms, 100ms + 0.08μs =100 ms. –  1GB file, 1ms/1Mbps vs. 100ms/100Mbps 1ms + 10243 x 8 /106 = 2.38h + 1ms, 100ms + 85 s 23 Bandwidth x Delay Product •  The amount of data (bits or bytes) “in the pipe” •  Example: 100Mbps x 10ms = 1 Mbit •  The amount of data sent before first bit arrives •  Usually use RTT as delay: amount of data before a possible receiver reply arrives 24 High- Speed Networks Link Type Bandwidth Distance RTT Delay x BW Dial- up 56 kbps 10 km 87 μs 5 bits Wireless LAN 54 Mbps 50 m 0.33 μs 18 bits Satellite link 45 Mbps 35,000 km 230 ms 10 Mb Cross- country fiber 10 Gbps 4,000 km 40 ms 400 Mb •  Infinite bandwidth –  PropagaKon delay dominates –  Throughput = Transfer size/Transfer Kme –  Transfer Kme = RTT + Transfer size/Bandwidth –  1MB file across 1Gbps line with 100ms RTT =74.1 25 Mbps CompuKng ApplicaKon Bandwidth •  FTP can uKlize enKre BW available •  Video- on- demand may specify upper limit (only what’s needed) •  Example: res: 352x240 pixels, 24- bit color, 30 fps –  Each frame is (352 x 240 x 24)/8 =247.5 KB –  Total required BW = 352 x 240 x 24 x 30 = 60.8 Mbps 26 Network Jiger •  Variability in the delay between packets •  Video- on- demand applicaKon: If jiger is known, applicaKon can decide how much buffering is needed •  Example: jiger is 50 ms per frame and I have 10 s of video at 30fps. •  If Y frames buffered, video can play uninterrupted for Y x 1/30 s. •  The last frame will arrive 50 x (10 x 30 – Y) ms aNer video start, worst case Y/30 = 50 x (300 – Y) à༎ Y = 180 frames 27 Example: Problem 1.19 from Book •  1 – Gbps Ethernet with a s- a- f switch in the path and a packet size of 5,000 bits. Tp = 10 μs, switch transmits immediately aNer recepKon A S B 1st bit: Kme 0 Last bit: 5μs Tp Last bit rec: 15μs Last bit sent: 20μs Last bit rec: 30μs t 28 Example: Problem 1.19 from Book •  1 – Gbps Ethernet with a s- a- f switch in the path and a packet size of 5,000 bits. Tp = 10 μs, 3 switches in between A and B •  4 links equal to 4 Tp delay •  4 transmissions equal to 4 Tt delay •  Total = 4Tp + 4Tt = 60 μs •  Three switches, each transmits aNer 128 bits are received •  Total = 4Tp + Tt + 3x128/109 = 40μs + 5μs + 0.384μs = 45.384μs 29...
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This note was uploaded on 09/27/2013 for the course ECE 578 taught by Professor Srini during the Fall '07 term at University of Arizona- Tucson.

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