CS/ECE 715 Spring 2004 Homework 6 Solution(Due date: April 1)

Transcription

1 CS/ECE 715 Sprig 004 Homework 6 Solutio(Due date: April 1) Problem 1. A commuicatio lik provides 1 Mbps for commuicatios betwee the earth ad the moo. The lik seds color images from the moo. Each image cosists of 10,000x10,000 pixels, ad 16 bits are used for each of the three color compoets of each pixel. (a) How may images/secod ca be trasmitted over the lik? (b) If each image is trasmitted as a sigle block, how log does it take to get a ackowledgemet back from earth? The distace betwee earth ad the moo is approximately 375,000 km. (c) Suppose that the bit error rate is, compare Go-Back-N ad Selective Repeat ARQ i terms of their ability to provide reliable trasfer of these images from the moo to earth. Optimize the frame size for each case usig trial ad error. Assume a header overhead of 64 bits i a frame of size greater tha 64 bits. [Solutio] 10 5 (a) The umber of images that ca be trasmitted per secod is images/secod (1) = (b) The total time to get a ackowledgmet from earth, assumig that is t o = t f + t prop = = sec/img () Note if each image is trasmitted i a sigle block, becomes isigificat compared to. (c) The difficulty here is that if we trasmit a etire image as a sigle block, the the average umber of errors i a block is 48 x 10 8 x 10-5 = Thus every trasmissio fails. Clearly we eed to trasmit usig a smaller frame size, say. Qualitatively, the followig happes as we vary. If is very small, the the probability of frame error is small, but the overhead due to the header o will become sigificat. If becomes too large, the the efficiecy drops because of frequet retrasmissios. Thus there must be a itermediate value of that optimizes efficiecy. We will fid this value by trial ad error. We will assume a header overhead of 64 bits i a frame of size greater tha 64 bits. Each frame carries bits of iformatio, so the required widow size for a propagatio delay of.5 secods is the W s =.5x10 6 /+1 The probability of frame error is the t prop t ACK t f t f 1

2 P f 1 ( ) 1 e 10 5 = The efficiecy for Go-Back-N is give by (3) ( 1 P f ) η GBN = = 1 + ( W s 1)P f e e ( ) (4) Usig similar assumptios, the efficiecy of Selective Repeat ARQ is give by η SR ( 1 P f ) = = e (5) 1 Trasmissio efficiecy vs. frame size Trasmissio Efficiecy Go-Back-N Selective Repeat Frame Size (bytes) The figure above shows the efficiecy of Go-Back-N ad Selective Repeat ARQ as a fuctio of frame size. Go-Back-N has very low efficiecy (always below 10%) for all values of. Selective Repeat achieves a maximum of about 95% efficiecy at aroud =500 bits. Note that the optimum value of is depedet o the value of o. Problem. Fid the optimum frame legth f that maximized trasmissio efficiecy for a chael with radom bit errors by takig the derivative ad settig it to zero for the followig protocols: (a) Stop-ad-Wait ARQ (b) Go-Back-N ARQ (c) Selective Repeat ARQ

3 (d) Fid the optimum frame legth for a 1 Mbps chael with 10 ms reactio time, 5-byte overhead, 5-byte ACK frame, ad p = 10 4, 10 5 ad [Solutio] (a) Stop-ad-wait ARQ η ( 1 P b ) f f o = f + a + ( t proc + t prop )R Let a = 1 P b, b = a + ( t proc + t prop )R d f f o a f [ la ( a f o ) + 1] ( f + b) a f ( f o ) = d f f + b ( f + b) = 0 b+ f + ( b o ) f o la o b = 0 o b ( b o ) 4 b + o ± la o b f = (b) Go-back-N ARQ (6) (7) (8) (9) η ( 1 P b ) f f o = (10) f f 1 ( 1 P b ) f [ ] Rt ( proc + t prop ) f Let a = 1 P b, b = R( t proc + t prop ) ad use the approximatio 1 ( 1 P b ) f f P b (which is valid whe f P b «1 ), the Take the derivative ad equatig the umerator to zero, we fid that which leads to the quadratic equatio a f ( η f o ) = f ( 1 + P b b) a f ( f o ) la a f ( + ) f a f ( f o ) = 0 f (11) (1) which give the solutio f o f + o la = 0 (13) (c) Selective repeat o + o o l a f = (14) 3

4 dη d f = 0 leads to η ( 1 P b ) f 1 o = ---- (15) f f f o + o l( 1 P b ) = 0, the (16) (d) R = o + o 4 o l( 1 P b ) o + o o la f = = , where a = 1 P. (17) b o = 00 a = 00 P b = 10 4, 10 5, 10 6 a = 1 P b t proc + t prop = 10 ms b = a + ( t proc + t prop )R = 000 Table 1: Optimum packet size P b Stop-ad-wait Go-back-N Selective Repeat Problem 3. A data-lik layer protocol that uses ON/OFF cotrol executes o a 1.5Mbps lik. The seder emits data at this full data rate ad uses frames of size 1000 bytes. Oe-way propagatio delay is 0ms. The rate at which the receiver empties its buffer, R rcv for differet time itervals. Assume the receiver buffer size is bytes., is show i the figure below R rcv = 1.5Mbps R rcv = 0.7Mbps R rcv = 1.5Mbps t=0 t=1sec t=sec 1. Does the receiver sed a OFF sigal to avoid packet loss at ay of these time istats? a sec 4

5 b sec c..06 sec d sec. How full is the receiver buffer at t=1sec? a bytes b bytes c bytes d. frames [Solutio] Whe the buffer at the receiver has oly t prop ( R sd R rcv ) bytes left, it will sed a OFF sigal. This umber is = 4000 bytes The seder seds at 1.5Mbps ad the receiver empties its buffer at the same rate up to t=1 sec. From 1 sec, the latter rate drops to 0.7Mbps. Therefore it will take t = ( ) 8 = µs for the receiver buffer to reach a poit whe there is oly ( ) space for 4000 bytes left i the buffer. Therefore at t=1sec+60ms, the receiver will geerate a OFF sigal. Aswer to part is 0 bytes sice the receiver buffer is beig depleted at the same rate as the seder. 5