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)
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
※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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6.21. In order to distinguish between different frames of the same sequence number, the number of possible sequence numbers must be larger than the number of allowed outstanding frames. Let MaxSeqNum be the number of total available sequence numbers. In case of SWS + 1 ≤ MaxSeqNum, which of the following statements are correct ? |
| (a) It is sufficient to distinguish between different frames of the same sequence number. 這樣就足以分辨出有相同序號的不同訊框. | |
| (b) It is NOT sufficient to distinguish between different frames of the same sequence number. 這樣不足以分辨出有相同序號的不同訊框. | |
| (c) If RWS = 1 then it is sufficient to distinguish between different frames of the same sequence number. 若 RWS = 1, 則足以分辨出有相同序號的不同訊框. | |
| (d) If RWS = SWS then it is NOT sufficient to distinguish between different frames of the same sequence number. 若 RWS = SWS, 則不足以分辨出有相同序號的不同訊框. | |
| (e) If RWS = SWS then it is sufficient to distinguish between different frames of the same sequence number. 若 RWS = SWS, 則足以分辨出有相同序號的不同訊框. |
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