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

6.22. Consider an example that we have eight available sequence numbers: 0, 1, 2, 3, 4, 5, 6, and 7. Assume that RWS = SWS = 7. Now, we have the following steps:

  • Step 1. Sender sends frames 0, 1, …, 6
  • Step 2. Receiver receives frames 0, 1, … ,6, and acknowledges 0, 1, …, 6.
  • Step 3. All ACKs (0, 1, …, 6) are lost
  • Step 4. Sender finally timeouts and retransmits frames 0, 1, …, 6

假設有 8 個可使用的序號: 0, 1, 2, 3, 4, 5, 6, 7 且 RWS = SWS = 7. 經過以下步驟:

  • 步驟 1. 傳送端送出訊框 0, 1, …, 6
  • 步驟 2. 接收端收到訊框 0, 1, … ,6, 且肯定回覆 0, 1, …, 6.
  • 步驟 3 所有肯定回覆 (0, 1, …, 6) 皆遺失
  • 步驟 4. 傳送端等待逾時且重送訊框 0, 1, …, 6

Then, how many frames retransmitted in Step 4 will be accepted by receiver ?  則有多少在步驟 4 重送的訊框會被接收端接受 ?

Ans :___a____

How many frames retransmitted in Step 4 will be discarded by receiver ?有多少在步驟 4 重送的訊框會被接收端丟棄 ?

Ans:____b____

 

a =
b =

6.23  Consider an example that we have eight available sequence numbers: 0, 1, 2, 3, 4, 5, 6, and 7. Assume that RWS = SWS = 5. Now, we have the following steps:

  • Step 1. Sender sends frames 0, 1, …, 4
  • Step 2. Receiver receives frames 0, 1, … ,4, and acknowledges 0, 1, …, 4.
  • Step 3. All ACKs (0, 1, …, 4) are lost
  • Step 4. Sender finally timeouts and retransmits frames 0, 1, …,4

假設有 8 個可使用的序號: 0, 1, 2, 3, 4, 5, 6, 7. 假設 RWS = SWS = 5. 經過下列步驟:

  • 步驟 1. 傳送端送出訊框 0, 1, …, 4
  • 步驟 2. 接收端收到訊框 0, 1, … ,4, 且肯定回覆 0, 1, …, 4.
  • 步驟 3 所有肯定回覆 (0, 1, …, 4) 皆遺失
  • 步驟 4. 傳送端等待逾時且重新傳送訊框 0, 1, …,4

Then, how many frames retransmitted in Step 4 will be accepted by receiver ? 則有多少在步驟 4 重送的訊框會被皆端接受?

Ans: ____a______

How many frames retransmitted in Step 4 will be discarded by receiver ? 有多少在步驟 4 重送的訊框會被接收端丟棄?

Ans: ____b____

a =
b =

6.24. Consider an example that we have eight available sequence numbers: 0, 1, 2, 3, 4, 5, 6, and 7. Assume that RWS = SWS = 4. Now, we have the following steps:

  • Step 1. Sender sends frames 0, 1, 2,3
  • Step 2. Receiver receives frames 0, 1, 2, 3 and acknowledges 0, 1, 2,3.
  • Step 3. All ACKs (0, 1, 2, 3) are lost
  • Step 4. Sender finally timeouts and retransmits frames 0, 1, 2, 3

假設有 8 個可使用的序號: 0, 1, 2, 3, 4, 5, 6, 7. 且 RWS = SWS = 4. 經過以下步驟:

  • 步驟 1. 傳送端送出訊框 0, 1, 2, 3
  • 步驟 2. 接收端收到訊框 0, 1, 2, 3 且肯定回覆 0, 1, 2, 3.
  • 步驟 3. 所有肯定回覆 (0, 1, 2, 3) 皆遺失
  • 步驟 4. 傳送端等待逾時且重送訊框 0, 1, 2, 3

Then, how many frames retransmitted in Step 4 will be accepted by receiver ? 則有多少在步驟 4 重送的封包會被接收端接受 ?

Ans:  ____(a)______

How many frames retransmitted in Step 4 will be discarded by receiver ?有多少在步驟 4 重送的封包會被接收端丟棄 ?

Ans: ____(b)____

a =
b =

6.25 Let MaxSeqNum be the total number of available sequence numbers. Assume RWS = SWS, then which of the following relationship is sufficient to distinguish between different frames of the same sequence number ?
令 MaxSeqNum 表示可使用的序號的總數. 假設 RWS = SWS, 則下列哪組關係可以分辨出擁有相同序號的不同訊框 ?

(a) SWS + 1 ≤ MaxSeqNum
(b) SWS ≤ (MaxSeqNum+1)/2
(c) SWS < (MaxSeqNum+1)/2
(d) SWS < MaxSeqNum x 2
(e) SWS = (MaxSeqNum+1)/2

6.26  Consider an example that we have seven available sequence numbers: 0, 1, 2, 3, 4, 5, and 6. Assume that RWS = SWS = 4. Now, we have the following steps:

  • Step 1. Sender sends frames 0, 1, 2, 3
  • Step 2. Receiver receives frames 0, 1, 2, 3 and acknowledges 0, 1, 2, 3.
  • Step 3. All ACKs (0, 1, 2, 3) are lost
  • Step 4. Sender finally timeouts and retransmits frames 0, 1, 2, 3

考慮有 7 個可使用的序號: 0, 1, 2, 3, 4, 5, 6. 假設 RWS = SWS = 4. 我們進行以下步驟:

  • 步驟 1. 傳送端送出訊框 0, 1, 2, 3
  • 步驟 2. 接收端收到訊框 0, 1, 2, 3 和肯定回覆 0, 1, 2, 3.
  • 步驟 3. 所有肯定回覆 (0, 1, 2, 3) 都遺失
  • 步驟 4. 傳送端等待逾時且重傳訊框 0, 1, 2, 3

Then, how many frames retransmitted in Step 4 will be accepted by receiver ? 則有多少在步驟 4 重送的訊框會被接收端接受 ?

Ans: ____(a)______

How many frames retransmitted in Step 4 will be discarded by receiver ? 有多少在步驟 4 重送的訊框會被接收端丟棄 ?

Ans: ____(b)______

a =
b =

6.27  Which of the following are the major features of the sliding window protocol ? 下列哪些是滑動視窗協議的主要特性?

(a) unreliable transmission 不可靠的傳輸
(b) reliable transmission 可靠的傳輸
(c) The frame order can be preserved. 訊框順序可以保持
(d) The frame order can not be preserved 訊框順序無法保持
(e) The receiver can do flow control by setting the value of RWS 接收端可以利用設定 RWS 的值來達到流量控制

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