Fundamenals of Compuer Neworks ECE 478/578 Lecure #3 Insrucor: Loukas Lazos Dep of Elecrical and Compuer Engineering Universiy of rizona Nework Performance Merics andwidh moun of daa ransmied per uni of ime; per link, or end-o-end Unis 1K = 2 10 byes, 1Mbps = 10 6 bis per sec How many K/sec is a 1Mbps line? How many M/sec? Throughpu Daa rae delivered by he a link, connecion or nework Per link or end-o-end, same unis as andwidh 2 Laency or Delay Time for sending daa from hos o (in sec, msec, or μsec) Per link or end-o-end Usually consiss of T : Transmission delay T p : Propagaion delay T q : Queuing delay Round Trip Time () : ime o send a message from o and back Imporan for flow conrol mechanisms 3 1
Delay Calculaion T : Transmission Delay: file size/bandwidh T p : Propagaion Delay: ime needed for signal o ravel he medium, Disance / speed of medium T q : Queuing Delay: ime waiing in rouer s buffer C d 1 d 2 R 4 Example: Problem 1.6 from ook Transfer 1,5 M file, assuming of 80 ms, a packe size of 1-K and an iniial handshake of 2x andwidh is 10 Mbps and daa packes can be sen coninuously reques ck = 80 ms T = 1024x8 bis/10 7 bis/s = 0.8192 ms # of packes = 1536 (1.5 x 1024) T T p D = 2x + 1536xT + T p = 160 + 1258.29 + 40 ms = 1.458 s 5 Example: Problem 1.6 from ook Transfer 1,5 M file, assuming of 80 ms, a packe size of 1-K and an iniial handshake of 2x fer sending each packe mus wai one reques ck = 80 ms T = 1024x8 bis/10 7 bis/s = 0.8192 ms # of packes = 1536 (1.5 x 1024) T D = 2x + 1535x(T +)+ T +T p = 160 + 124,057 + 0.8192 + 40 ms = 124.258 s 6 2
Example: Problem 1.6 from ook Transfer 1,5 M file, assuming of 80 ms, a packe size of 1-K and an iniial handshake of 2x Only 20 packes can be send per, bu infiniely fas reques ck = 80 ms T = 0 ms # of packes = 1536 (1.5 x 1024) D = 2x + 76x + T p = 160 + 6080 + 40 ms = 6.28 s 7 Example: Problem 1.6 from ook Transfer 1,5 M file, assuming of 80 ms, a packe size of 1-K and an iniial handshake of 2x 1 s one packe, 2 wo packes Infinie ransmission rae reques ck = 80 ms T = 0 ms # of packes = 1536 (1.5 x 1024) # of wais (1+2+ 2 n = 2 n+1-1) 2 11-1 =2047 packes, n = 10 D = 2x + 10x + T p = 160 + 800 + 40 ms = 1 s 8 Laency vs. andwidh Imporance depends on applicaion 1 bye file, 1ms/1Mbps vs. 100ms/100Mbps 1 ms + 8μs = 1.008ms, 100ms + 0.08μs =100 ms. 1G file, 1ms/1Mbps vs. 100ms/100Mbps 1ms + 1024 3 x 8 /106 = 2.38h + 1ms, 100ms + 85 s 9 3
andwidh x Delay Produc The amoun of daa (bis or byes) in he pipe Example: 100Mbps x 10ms = 1 Mbi The amoun of daa sen before firs bi arrives Usually use as delay: amoun of daa before a from a receiver arrives o he sender 10 High-Speed Neworks Link Type andwidh Disance Delay x W Dial-up 56 kbps 10 km 87 μs 5 bis Wireless LN 54 Mbps 50 m 0.33 μs 18 bis Saellie link 45 Mbps 35,000 km 230 ms 10 Mb Cross-counry fiber 10 Gbps 4,000 km 40 ms 400 Mb Infinie bandwidh Propagaion delay dominaes Throughpu = Transfer size/transfer ime Transfer ime = + Transfer size/andwidh 1M file across 1Gbps line wih 100ms, Throughpu is 74.1 Mbps 11 Compuing pplicaion andwidh FTP can uilize enire W available Video-on-demand may specify upper limi (only wha s needed) Example: res: 352x240 pixels, 24-bi color, 30 fps Each frame is (352 x 240 x 24)/8 =247.5 K Toal required W = 352 x 240 x 24 x 30 = 60.8 Mbps 12 4
Nework Jier Variabiliy in he delay beween packes Video-on-demand applicaion: If jier is known, applicaion can decide how much buffering is needed Example: jier is 50ms per frame and 10s video a 30fps mus be ransmied. If Y frames buffered, video can play uninerruped for Y x 1/30s. The las frame will arrive 50 x (10 x 30 Y) ms afer video sar, wors case Y/30 = 50 x (300 Y) Y = 180 frames 13 Example: Problem 1.19 from ook 1 Gbps Eherne wih a s-a-f swich in he pah and a packe size of 5,000 bis. T p = 10 μs, swich ransmis immediaely afer recepion 1 s bi: ime 0 Las bi: 5μs T p Las bi rec: 15μs Las bi sen: 20μs S Las bi rec: 30μs 14 Example: Problem 1.19 from ook 1 Gbps Eherne wih a s-a-f swich in he pah and a packe size of 5,000 bis. T p = 10 μs, 3 swiches in beween and 4 links equal o 4 T p delay 4 ransmissions equal o 4 T delay Toal: 4T p + 4T = 60 μs Three swiches, each ransmis afer 128 bis are received Toal: 4T p + T + 3x128/10 9 = 40μs + 5μs + 0.384μs = 45.384μs 15 5