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Appendix 7A Performance Issues

Appendix 7A Performance Issues. Stop and Wait Flow Control. Terms: S1, S2--two stations that are communicating. F1,F2,...Fn--the frames that make up a long message. t prop =propagation time from S1 to S2. t frame =time to transmit a frame.

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Appendix 7A Performance Issues

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  1. Appendix 7A Performance Issues

  2. Stop and Wait Flow Control • Terms: • S1, S2--two stations that are communicating. • F1,F2,...Fn--the frames that make up a long message. • tprop=propagation time from S1 to S2. • tframe=time to transmit a frame. • tproc= processing time at each station to react to an event. • tack=time to transmit a positive ACK.

  3. Stop and Wait Flow Control (p.2) • Send message as short frames: F1,F2,...,Fn. • Each frame is sent and acknowledged. • TF= total time to send one frame = tframe+ tprop+ tproc +tack +tprop+tproc • T=total time to send the message (all n frames) = n TF • Simplifications: • tproc is negligible, and the ACK frame is short . • Then T= n(2tprop + tframe)

  4. Stop and Wait Flow Control (p.3) • Define utilization (U) of the line as the time used for transmitting the message frames divided by the total time. • Time to transmit all frames = n tframe • Total Time= T= n(2tprop + tframe) • U = n tframe/ n (2tprop + tframe) • U = tframe/ (2tprop + tframe) • Let a=tprop/tframe, then U = 1/(1 + 2a).

  5. Stop and Wait Flow Control (p.4) • Examination of a: • tprop=d / V, where d is the distance and V is the velocity of propagation. • tframe = L/R where L is the length of the frame in bits and R is the data rate. • a = Rd/VL = (R{d/V}) / L

  6. Stop and Wait Flow Control (p.5) • Example 7.6 Stop and Wait Utilization • a. ATM (WAN) • R=155.52 Mbps, L = 424 bits; optical fiber link with d = 1,000 km. • a = 1850; U= .00027. • b. LAN • R=10 to 100Mbps, d=.1 to 10km. • a ranges from .005 to .5 and U= .5 to .99 • c. Voice Grade Line • R = 56,000 bps; L = 1,000 bits. • a ranges from .26 to 1, depending on distance.

  7. Sliding Window Control • Let the window size be W; this provides for greater efficiency (Eq. 7.6) • For W  2a+1 : U =1 (Window size is never exhausted.) • For W< 2a + 1: U = W/ 2a + 1 (Think about one ACK for all W frames.) • See Fig. 7.11 (but be careful--ACK2, etc. are not shown.) • Figure 7.12-- Utilization as a function of window size and a.

  8. Automatic Repeat Request (ARQ) • Stop and Wait (with errors): • U=Tf/ NrTt where Nr is the average number of times that a frame is retransmitted. • Let P be the probability that a frame is in error and assume that ACKs and NAKs are never in error. • Nr = 1/(1-P) • U= (1-P)/(1 + 2a) • See pg. 238-239 for Go-Back-N and Selective Reject equations. • Fig. 7.13 shows relative performance for P=.001.

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