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UCB Packet Dynamics Jean Walrand U.C. Berkeley www.eecs.berkeley.edu/~wlr.

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

1 UCB Packet Dynamics Jean Walrand U.C. Berkeley www.eecs.berkeley.edu/~wlr

2 UCB Outline Packet Delay Queuing Delay

3 UCB Packet Delay Packet Delay = TRANS + PROP + QD TRANS = Transmission Time = Number of bits/Rate = P/R PROP = Propagation Time = Distance/Speed = D/S Speed = 5  s/km in fiber 4  s/km in copper 3.3  s/km in space QD = Queuing Delay This delay depends on the load of the router and burstiness of the traffic.

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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