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hw2sol - EE 284 F Tobagi Autumn 2010-2011 EE284 Homework...

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EE 284 F. Tobagi Autumn 2010-2011 EE284 Homework Assignment No. 2 SOLUTIONS Total Points: 28 Problem 1: Error Correction vs Error Detection (8 points) a) [5 points] Error Correction: [2 points] When we use the error correction mechanism, each frame is transmitted only once. The receiver corrects all errors. For each frame we send D + α bits, where D represents the user’s data and α represents the added error control bits for the error correction mechanism (overhead). In order to calculate the throughput, we have to find the fraction of time to transmit succesful user data in a frame . For the error correction mechanism, each transmission is successful and consists of D bits of data and α bits of overhead. Therefore, the thoughput is: S EC = D/W ( D + α ) /W = D D + α (D+alpha)/W Figure 1: Throughput for the error correction scheme. Error detection: [3 points] When we use the error detection mechanism, the transmitter will re-transmit the frame till it gets an ACK reply. This means that we don’t know exactly 1
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ACK NACK 2tau c/W (D+beta)/W Figure 2: Throughput for the error detection scheme. the time it takes to transmit each frame successfully. It depends on the number of re- transmissions. (See figure 2). To calculate the throughput we have to find the average number of transmissions till the frame is received correctly. Let X be a random variable representing the number of transmissions till the frame is received correctly. Since the frame transmission errors are independent, X is a Geometric random vari- able and the probability that the number of transmissions till the frame is received correctly is given by: P [ X = k ] = p k - 1 (1 p ) The average number of transmissions till the frame is received correctly is: E [ X ] = 1 1 p Using the average number of transmissions, we can calculate the average time it takes to transmit a frame successfully: E ( t suc ) = 1 1 p ( D + β + c W + 2 τ ) Based on figure 2 we can now calculate the throughput: S ED = D/W 1 1 - p 1 W ( D + β + c + 2 τW ) = D (1 p ) D + β + c + 2 τW b) [3 points] Substituting the values D = 1000 bits, α = 200 bits, β = c = 20 bits and W = 100 Kbits/sec: S EC = 1000 1200 = 5 6 2
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S ED = 1000(1 p ) 1040 + 2 τ 10 5
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