課程影片與練習
課程簡介
第1週：網路架構與七層參考模式簡介
第2週：IEEE 802.3 乙太網路 (Ethernet)
※觀看影片請點擊影片名稱後方的"YOUTUBE"或"YOUKU"，
觀看YOUKU影片，請輸入密碼：mooc01004
第3週：IEEE 802.11 無線區域網路 (Wireless LAN)
第4週：IEEE 802.1D 交換機的擴張樹演算法 (Spanning Tree Algorithm)
第5週：IEEE 802.1Q 虛擬區域網路 (Virtual LAN)
第6週：區域網路可靠傳輸技術 －－ 滑動視窗技術
第7週：路由器運作原理以及網路互連技術
第8週：TCP 與網路阻塞偵測與控制技術
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2.21. Which of the following are correct for CSMA/CD (Carrier Sense Multiple Access with Collision Detection) protocol ? 下列哪些對於CSMA/CD協議的敘述是正確的 ?
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| (a)Listen (carrier sense) before transmit. 傳輸前先聽 | |
| (b)If channel is sensed idle, then transmit a frame and stop the transmission when a collision is detected. 假如偵測頻道為空閒時，就開始傳送訊框，一旦發現碰撞發生，立即停止傳送 | |
| (c)If channel is sensed idle, then transmit entire frame without considering if frame collision is detected. 假如偵測頻道為空閒時，就開始傳送整個訊框，無論是否偵測到訊框碰撞 | |
| (d)If channel is sensed busy, then defer transmission for a random time. 假如偵測頻道為忙碌時，則延遲一段亂數時間後再傳送 | |
| (e)If channel is sensed busy, then defer transmission until the channel is idle. 假如偵測頻道為忙碌時，則延遲到頻道變成空閒時再傳送 |
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2.22. Which of the following are correct when a “collision” happens for CSMA/CD protocol ?針對CSMA/CD 協議，當碰撞發生時，下列敘述哪些是正確的？
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| (a)A jamming sequence (frame) will be sent to enforce the collision. 擾亂訊號會被送出，以強迫發生碰撞 | |
| (b)The entire frame will still be transmitted. 整個訊框仍會被傳送 | |
| (c)The transmission is stopped and after a random delay, the frame is transmitted again immediately. 傳輸會先停止，經過延遲一段亂數的時間後，會立即傳送該訊框 | |
| (d)The transmission is stopped and after a random delay, the frame is transmitted again from the carrier sense step. 傳輸會先停止，經過延遲一段亂數的時間後，會由載波偵測的步驟開始傳送該訊框 | |
| (e)The transmission is stopped. But the frame is transmitted again as long as the channel becomes idle. 傳輸會先停止，但只要頻道變成空閒，訊框就會立即再傳送出去 |
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2.23. The “collision window” of the CSMA/CD protocol is time required to detect a collision. Which of the following are correct for collision window ? Assume the one way propagation time of the network is “a”.CSMA/CD 協定中的”衝撞視窗”是指偵測碰撞所需要的時間。下列哪些有關衝撞視窗的敘述是正確的？假設網路最遠兩端單向的訊號傳遞時間為‘a’
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| (a)The collision window is equal to a 衝撞視窗等於a | |
| (b)The collision window is equal to 2a 衝撞視窗等於2a | |
| (c)The worst case scenario to detect a collision happens when the two hosts are at opposite ends of the Ethernet 偵測碰撞之最差情況(需要最長時間)是當兩個主機位於乙太網路的兩端 | |
| (d)The worst case scenario to detect a collision happens when the two hosts are close to each other. 偵測碰撞之最差情況(需要最長時間)是當兩個主機非常靠近 | |
| (e)The best case scenario to detect a collision happens when the two hosts are close to each other. 偵測碰撞之最佳情況(需要最短時間)是當兩個主機非常靠近 |
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2.24. Consider that a maximally configured Ethernet is 2500 m long, and there may be up to four repeaters between any two hosts, the round trip delay has been determined to be 51.2 us. For 10Mbps Ethernet (such as 10base5), what is the minimum length of a frame in order to detect all possible collisions ? ___________ bits
假設最長的乙太網路最長為2500公尺，在任意兩主機之間可能最多可以有4個中繼器，而訊號來回的延遲時間為51.2 us。針對 10Mbps 乙太網路(如10base5)，為了能偵測到所有可能的碰撞，一個訊框長度至少要有多少位元 ? ＿＿＿＿
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