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hw6key - EEE459 Homework 6 answer key 10 points/problem: Ch...

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Unformatted text preview: EEE459 Homework 6 answer key 10 points/problem: Ch 3: 34, 35, 36, and 37 5 points/review question: Ch 4: 13 Problems 34. Object size O = 100 kbytes , MSS S = 536 bytes , RTT = 0 . 1 s , W is static. The minimum possible latency is = 2 RTT + O/R . The minimum window size W that achieves this latency is the window size where we receive the ACK for the first segment before the server transmits the last segment, i.e. S/R + RTT < W S/R W > 1 + RTT S/R R min. latency W a) 28kbps 28.77s 2 b) 100kbps 8.2s 4 c) 1Mbps 1s 25 d) 10Mbps 0.28s 235 (The answer to part (d) obtained by applying the above formula is somewhat misleading because the object is broken into 187 segments. Thus, W = 187 would be sufficient to send the entire object without stalling.) 35. a) K = min { k : 3 + 3 1 + + 3 k- 1 O/S } = min { k : 3 k- 1 2 O/S } = min { k : k log 3 (2 O/S + 1) } = d log 3 (2 O/S + 1) e b) Q = max { k : RTT + S R S R 3 k- 1 } = max { k : RTT S/R S/R 3 k- 1 } = max { k :...
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This note was uploaded on 11/01/2009 for the course EEE 459 taught by Professor Martinreisslein during the Spring '09 term at ASU.

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hw6key - EEE459 Homework 6 answer key 10 points/problem: Ch...

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