Re: [問題] 如何計算frame size?

看板Network作者freepark (自由‧公園)時間20年前 (2005/03/27 01:12)推噓0(0推 0噓 0→)

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※ 引述《kheng (呆呆)》之銘言： : 1. : Consider a CSMA/CD network running at 1 Gbps over a 1-km calbe with no : repeaters. The signal speed in the cable is 200,000km/sec. What is the : minimum frame size?(note:1Gbps=1,000,000,000 bits/sec) (1/200,000) * 2 * 1,000,000,000 / 8 = 1250 bytes : 2. : A TCP machine is sending windows of 65535 bytes over a 1-Gbps channel that : has a 10-msec one-way delay.What is the maximum throughput achievable? : what is the line efficiency? (65535 * 8) / (0.01 * 2) = 26,214,000 bps 26,214,000 / 1,000,000,000 = 2.6% : 3. : If message, 01110011001, is send and a special bit pattern,01110, is applied : as the leading flag of the message. The generator ploynomial G(x)=x^4+x^2+x+1, : is used for CRC examination. : 1.Is the CRC check used for error detection or error correction? : 2.Find the checksum? : 3.If the frame shall be transmitted as : --------------------------------------------- : |Leading Flag |Message body | CRC checksum | : --------------------------------------------- : Construct the bit pattern the frame to be transmitted(hint:bit stuffing) error detection 0001 01110 0110100110001 0001 not very sure... : 4. : The mzsimum payload of a TCP segment is 65495 bytes. Why was such a : strange number chosen? 65535 - 20 - 20 = 65495 : 請順道寄去我信箱...謝謝 -- 峨嵋弟子們，凡我修練者，切記勿貪婪與急進，此乃一切魔道之起源。 -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 203.70.6.34 ※ 編輯: freepark 來自: 203.67.96.93 (03/27 15:12)