hw3s.pdf - EE 284 F Tobagi Autumn 2017-2018 EE284 Homework...

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EE 284 F. Tobagi Autumn 2017-2018 EE284 Homework Assignment No. 3 SOLUTIONS Total Points: 56 Problem 1: Error recovery by means of error detection and/or correction and retransmission (9 points) (a) [2 points] Since A continues to retransmit the frame indefinitely, the number of failed transmissions to transmit a frame can be expressed as a geometric random variable X with probability of success 1 - FER . We can express the probability of X = k for any integer k > = 0 as follows: P [ X = k ] = ( FER ) k (1 - FER ) , k = 0 , 1 , 2 , . . . Finding the expected value for the geometric random variable, we find that the expected value for the number of failed transmissions is: E [ X ] = FER 1 - FER = 1 1 - FER - 1 Additionally, each unsuccessful transmission occupies the link for B W + T 0 sec., while the (last) successful transmission occupies B + C W +2 τ sec. We can thus write the average time for a single frame to be successfully transmitted and acknowledge as: ( 1 1 - FER - 1)( B W + T 0 ) + B + C W + 2 τ Hence, the throughput is 1 divided by this average time, given by: S = 1 ( 1 1 - FER - 1)( B W + T 0 ) + B + C W + 2 τ See the diagram in Figure 1 for an illustration of this transmission process. 1
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Figure 1: Problem 1a (b) [1 point] With the addition of a forward error correction scheme, our error rate is now FER ( r ), but we have also increased the number of bits in each frame. Recall that the coding rate r = number of data bits number of coded bits . Thus, the error correction scheme increases the number of bits in each frame by a factor of 1 /r . Our new expression of the throughput becomes: S = 1 ( 1 1 - FER ( r ) - 1)( B rW + T 0 ) + B + C rW + 2 τ (c) [1 point] Since there is a round trip time of 2 τ between the stations, and the transmis- sion time of the ACK frame is C/rW , the minimum time-out period is given by the following: T min 0 = 2 τ + C rW (d) [1 point] With T 0 = T min 0 , the throughput with forward error correction becomes: S = 1 ( 1 1 - FER ( r ) - 1)( B rW + 2 τ + C rW ) + B + C rW + 2 τ This simplifies to: S = 1 ( 1 1 - FER ( r ) )( B + C rW + 2 τ ) 2
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With τ = 0, we can write throughput as: S = 1 ( 1 1 - FER ( r ) )( B + C rW ) The throughput for when we don’t use forward error correction is the same as the above, except with r = 1 and the frame error rate as FER : S = 1 ( 1 1 - FER )( B + C W ) Forward error correction is desirable when using it can achieve greater throughput. Thus, we can write the condition on FER as: 1 ( 1 1 - FER ( r ) )( B + C rW ) 1 ( 1 1 - FER )( B + C W ) Simplifying this expression, we arrive at the following condition: 1 - FER r (1 - FER ( r )) Or equivalently: FER 1 - r (1 - FER ( r )) (e) [1 point] As τ increases, the propagation time of the frame becomes a larger and larger proportion of the cycle time of a single transmission. This means it is even more fa- vorable to minimize the number of transmissions, even at the cost of a larger frame size. Thus, as τ increases and becomes non-negligible, the condition on FER relaxes, and it becomes favorable to use forward error correction for even lower failure rates than 1 - r (1 - FER ( r )). Equivalently, we can say that as τ increases, the right side of following condition on FER decreases: FER 1 - r (1 - FER ( r )) .
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