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
 |
6.18 Let SeqNumToAck denote the largest sequence number not yet acknowledged, such that all frames with sequence number less than SeqNumToAck have been received. Which of the following statements are correct for the “cumulative acknowledgement” ? 令 SeqNumToAck 表示尚未回覆的訊框最大序號, 也就是所有序號小於SeqNumToAck 的訊框都已被接收.下列哪些有關“累積回覆”的敘述是正確的 ? |
| (a) The receiver acknowledges SeqNumToAck to sender if a frame with sequence number of SeqNumToAck + k, k > 0, is received. | |
| (b) The receiver acknowledges SeqNumToAck+k to sender if a frame with sequence number of SeqNumToAck + k, k > 0, is received. | |
| (c) The receiver then sets LFR = SeqNumToAck - 1 and LAF = LFR + RWS. 接收端會設定 LFR = SeqNumToAck - 1 與 LAF = LFR + RWS. | |
| (d) The receiver then sets LFR = SeqNumToAck and LAF = LFR + RWS.接收端會設定 LFR = SeqNumToAck 與 LAF = LFR + RWS. | |
| (e) The receiver then sets LFR = SeqNumToAck + 1 and LAF = LFR + RWS-1. 接收端會設定 LFR = SeqNumToAck + 1 與 LAF = LFR + RWS-1. |
|
6.19 Consider the following communication sequence between sender and receiver. Assume the “cumulative acknowledgement” scheme is used and the frame with sequence number of 2 is lost. The lost frame is retransmitted again when the timer timeouts.
|
| a = | |
| b = | |
| c = | |
| d = | |
| e = |
檔案上傳中...