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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.

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Presentation on theme: "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."— Presentation transcript:

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. –t prop =propagation time from S1 to S2. –t frame =time to transmit a frame. –t proc = processing time at each station to react to an event. –t ack =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. T F = total time to send one frame = t frame + t prop + t proc +t ack +t prop +t proc T=total time to send the message (all n frames) = n T F Simplifications: –t proc is negligible, and the ACK frame is short. –Then T= n(2t prop + t frame )

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 t frame –Total Time= T= n(2t prop + t frame ) –U = n t frame / n (2t prop + t frame ) –U = t frame / (2t prop + t frame ) –Let a=t prop /t frame, then U = 1/(1 + 2a).

5 Stop and Wait Flow Control (p.4) Examination of a: –t prop =d / V, where d is the distance and V is the velocity of propagation. –t frame = 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= Mbps, L = 424 bits; optical fiber link with d = 1,000 km. a = 1850; U= –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 (but be careful--ACK2, etc. are not shown.) Figure Utilization as a function of window size and a.

8 Automatic Repeat Request (ARQ) Stop and Wait (with errors): –U=T f / N r T t where N r 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. –N r = 1/(1-P) –U= (1-P)/(1 + 2a) See pg for Go-Back-N and Selective Reject equations. Fig shows relative performance for P=.001.


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