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UCB Sharing a Channel Jean Walrand U.C. Berkeley www.eecs.berkeley.edu/~wlr.

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Presentation on theme: "UCB Sharing a Channel Jean Walrand U.C. Berkeley www.eecs.berkeley.edu/~wlr."— Presentation transcript:

1 UCB Sharing a Channel Jean Walrand U.C. Berkeley www.eecs.berkeley.edu/~wlr

2 UCB Outline Channel TDM FDM CDMA Statistical Multiplexing Aloha: Slotted, Unslotted, Reservation CSMA/CA Collision Avoidance

3 UCB Channel Examples: Wireless, copper, fiber Can transmit range of frequencies f1f1 f2f2 Gain(f) f

4 UCB Packet Delay - continued Illustration

5 UCB Packet Delay - continued Examples: Link 1: Long Fiber Link 2: CopperLink 3: Wireless Length L (km)80410 Speed (  s/km) 543.3 Transmission Rate 1Gbps1Mbps10kbps TRANS 1s1s 1ms100ms PROP 400  s16  s33  s

6 UCB Queuing Delay Isolated Packets: Time t X(t) = number of bits at time t X(t) P P/R R bps QD = 0

7 UCB Queuing Delay (continued) Packet Bursts: Time t X(t) = number of bits at time t X(t) P P/R R bps Average value of QD = (0 + 1 + 2)TRANS/3 = TRANS For an isolated burst of size N: = (0 + 1 + … + N - 1 )TRANS/N = (N – 1)TRANS/2 P P P/R QD2 QD3 QD = (N – 1)TRANS/2

8 UCB Queuing Delay (continued) Rule of thumb …. 80% utilization => QD  4TRANS Example: P = 1,000 bits; 80% utilization on each link QD  4[10ms + (n – 1)0.01ms + 10ms]  80ms PROP = 5  1500  s = 7.5ms TRANS = 10ms + (n – 1)0.01ms + 10ms  20ms => Delay  108ms


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