occupation 1:   1. Access  shed light on: digital  reader line (DSL)  [pic]  Use existing  reverberate line to commutation office DSLAM  -data  everywhere DSL  head phone line goes to Internet  -voice  all over DSL phone line goes to telephone net  -< 2.5 Mbps upstream   transmission system   roam (typically < 1 Mbps)  -< 24 Mbps downstream transmission rate (typically < 10 Mbps)     2. Access net:  electrify  profits  [pic]  -HFC: hybrid   in writing(p) symbol coax  ? asymmetric: up to 30Mbps downstream transmission rate, 2 Mbps upstream transmission rate  -network of cable, fiber attaches homes to ISP router  ? homes  care  penetration network to cable headend  ? unlike DSL, which has dedicated access to central office   3. Access net: home network  [pic]     line of work 2:  Propagation  armed robbery= 2500000m/250000000  =0.01s   transmittal  storage area= 8000bits/2000000bps (since 1bytes=8 bits)  =0.004s    Because of this the transmission delay is 0.004s, is 2/5 of the  denota   tion delay, the transmission delay cannot be ignored and so that the delay is depends on both propagation and the transmission delay.    Problem 3:   a. packetization time = 48x8bits/64000bps   =0.006s   b. transmission delay = 64000bps x 0.006s / 1000000bps   =0.384ms   c.  sum  sum time = 6ms + 0.384ms + 2ms   =8.384ms    Problem 4:   a. 3Mbps / 150kbps   = 20 users can be supported and 150kbps dedicated.   b. Because each user can only transmits 10% of the time      So, probability = 0.1   c. Probability = Cn120 x 0.1n x 0.9120-n    Problem 5:   a.  [pic]  2 traceroutes with their appropriate averages and standard deviations.     b. The paths may be changed over the hours.   c. The largest delays are likely to occur between  attached ISPs.   d.  [pic]  By comparing the intra-continent and inter-continent results, to connect to other countries outside HK  necessity  more(prenominal) traceroutes.    Problem 6:   a. 8 x 106 / 2 x 106 = 4s      Total time = 4 x 3 = 12s   b.   beatnik    to move the beginning packet from host A to!    the  jump switch = 10000 / 2 x 106      = 0.005s      Time for the   cow dung packet be  estimabley...If you want to get a  practiced essay, order it on our website: BestEssayCheap.com
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