只寫答案而沒有解釋說明, 扣一半分數. 針對 63.07.72. 這個 IP address,( 以十進位表示, 要寫完整過程 ) (7%) a. 這一個 IP 屬於那個 Class 的網路? 以二進位說明 (%) 其所屬的 IP 網路表示法為何?(2%) 可用 IP 範圍?(2%) 共有幾個 IP 可用?(%) mask 的值為何?(%) b. 將此 IP 網路分成 7 subnets,subnet mask 的值為何?(2%) 請列出第 7 個 subnet 的網路表示法 (2%) 可用 IP 範圍?(2%) 共有幾個 IP 可用?(%) c. 手動設定電腦的網路時, 至少要設定哪三個項目的資訊, 才可以上網?(3%) 2. (a) Draw a figure to show four components of a router (8%) (b) Draw three types of switching fabrics with their names. (% each) (c) What is Head-of-the-Line (HOL) blocking? (3%) (4% total) 3. (a) Consider the two 6-bit words (shown in binary) below. Recall that to compute the Internet checksum of a set of 6-bit words, we compute the one's complement sum of the two words. That is, we add the two numbers together, making sure that any carry into the 7th bit of this initial sum is added back into the 's place of the resulting sum); we then take the one's complement of the result. Compute the Internet checksum value for these two 6-bit words: (8%) 000000 00000 this binary number is 406 decimal (base 0) 00 00 this binary number is 3245 decimal (base 0) (b) With the s complement scheme, how does the receiver detect errors? Is it possible that -bit error will go undetected? (2%) How about a 2-bit error? (2%) (2% total) 4. Draw the flow of the TCP three way handshake to explain its operations. Suppose the initial sequence numbers of the client and the server are and 200, respectively. 必須在圖上分別清楚標示出 TCP 必要的 flag, sequence number, and ACK number. (0%) 5. List and compare two pipelined transport protocols with these two figures. ( 寫出 Window=? 與各標號處的動作 0%) 5 2 6 3 7 4 8
6. (a) Describe how TCP Reno does its congestion control. (8%) (2% total) (b) Answer and justify the following questions. After the 22 th transmission round, is segment loss detected by a triple duplicate ACK or by a timeout? (2%) During what transmission round is the 50 th segment sent? (2%) 7. Consider the TCP procedure for estimating RTT n ( EstimatedRTT = α SampleRTT + ( α) EstimatedRTT ). (a) Why TCP uses this function? (2%) (b) Let SampleRTT n be the most recent sample RTT, let SampleRTT n- be the next most recent sample RTT, and so on. Express EstimatedRTT n in terms of n SampleRTTs if EsitmatedRTT = 0. ( 要有兩次疊代過程 ( 各 2%) 後寫出通式與以 summation 總和符號表示 ( 各 2%) (0% total) 8. (a) Explain how TCP Fast Retransmit works. (6%) (5% total) (b) How TCP does its flow control? (6%) (c) Suppose the TCP sequence number space is of size k. What is the largest allowable sender window w? (3%)
只寫答案而沒有解釋說明, 扣一半分數. 針對 63.07.72. 這個 IP address,( 以十進位表示, 要寫完整過程 ) (7%) a. 這一個 IP 屬於那個 Class 的網路? 以二進位說明 (%) 其所屬的 IP 網路表示法為何?(2%) 可用 IP 範圍?(2%) 共有幾個 IP 可用?(%) mask 的值為何?(%) b. 將此 IP 網路分成 7 subnets,subnet mask 的值為何?(2%) 請列出第 7 個 subnet 的網路表示法 (2%) 可用 IP 範圍?(2%) 共有幾個 IP 可用?(%) c. 手動設定電腦的網路時, 至少要設定哪三個項目的資訊, 才可以上網?(3%) a. 63.07.72. 的二進位表示法為 00. XXXXXXXX.XXXXXXXX.XXXXXXXX, 由前 個 bits 0 可判斷為 Class A 的 IP (%) 此 IP 所屬於的 Class A 的網路表示法為 63.0.0.0 (2%) 所有 host ID 部分的 24 個 bit 的 X 不可以全為 0 或, 因此第一個可用 Host ID 為 00. 00000000.00000000. 0000000 = 63.0.0. (%) 最後一個可用 Host ID 為 00...0 = 63.255.255.254 (%) -> 共有 2 24-2 個可用 Host ID (%) Mask: 255.0.0.0 (%) b. 將此 Class A 網路分成 7 個 subnet, 加上全為 0 與全為 的兩個不能用的 subnet ID, 最少需要 7+2=9 <= 2 4, subnet mask 的值 => 需要 Host ID 的前 4 個 bits 當作 subnet ID 所以新的 subnet mask 是由原本 Class A 的 default subnet mask 255.0.0.0 來改, 改成 255.0000.00000000.00000000=> 255. 240.0.0 (2%) subnet 的 ID 要從此 Class A Network ID 00.XXXXXXXX.XXXXXXXX.XXXXXXXX 來改, 需要 Host ID 的前 4 個 bits 當作 subnet ID, 不可全為 0 或 因此第 7 個 subnet ID 為 00. 00000.00000000.00000000 => 63.2.0.0(2%) 因此第一個可用 Host ID 為 00.00000.00000000. 0000000 = 63.2.0. (%) 最後一個可用 Host ID 為 00.0.. 0 = 63.27.255.254 (%) -> 共有 2 20-2 個可用 Host ID (%) c. IP address, subnet mask, default gateway (3%) 2. (a) Draw a figure to show four components of a router (8%) (b) Draw three types of switching fabrics with their names. (% each) (c) What is Head-of-the-Line (HOL) blocking? (3%) (4% total) (a) (2% each, 8% total) (b) (3%) switching via memory; (%) switching via a bus; (%) switching via an interconnection network (%)
(c) queued datagram at front of queue prevents others in queue from moving forward (3%) 3. (a) Consider the two 6-bit words (shown in binary) below. Recall that to compute the Internet checksum of a set of 6-bit words, we compute the one's complement sum of the two words. That is, we add the two numbers together, making sure that any carry into the 7th bit of this initial sum is added back into the 's place of the resulting sum); we then take the one's complement of the result. Compute the Internet checksum value for these two 6-bit words: (8%) 000000 00000 this binary number is 406 decimal (base 0) 00 00 this binary number is 3245 decimal (base 0) (c) With the s complement scheme, how does the receiver detect errors? Is it possible that -bit error will go undetected? (2%) How about a 2-bit error? (2%) (2% total) (a) When we add these first two numbers together, we get: 000000 00000 00 00 -------- -------- 000 00000 note the carry into the 7th bit (4%) Since there is a carry, we need to add a in the ones place (rightmost bit) of the rightmost 6-bit quantity above, giving: 000 00000 (2%) and then we need to take the ones complement of this value, to get the Internet checksum: 00000 000 (2%)
(b) All one-bit errors will be detected (2%), but two-bit errors can be undetected (2%) (e.g., if the last digit of the first word is converted to a 0 and the last digit of the second word is converted to a ). 4. Draw the flow of the TCP three way handshake to explain its operations. Suppose the initial sequence numbers of the client and the server are and 200, respectively. 必須在圖上分別清楚標示出 TCP 必要的 flag, sequence number, and ACK number. (0%) Three way handshake: Step : client host sends TCP SYN segment to server ( 搭配圖要正確 2%) Step 2: server host receives SYN, replies with SYNACK segment (4%) Step 3: client receives SYNACK, replies with ACK segment, which may contain data (4%) client server Connection request SYN bit=, seq= SYN bit=ack bit=, seq=200, ack=2 Connection granted ACK SYN bit=0, ACK bit= seq=2 Ack=20 上圖每個符號含內容 分, 標示不全者, 視狀況扣分, 共 0 分 5. List and compare two pipelined transport protocols with these two figures. ( 寫出 Window=? 與各標號處的動作 0%) Go-back-N (5%) window of up to N, consecutive unack ed pkts allowed (window = 4) (%) () ACK-only: always send ACK for correctly-received pkt with highest in-order seq # (%) (2) out-of-order pkt: discard (don t buffer) -> no receiver buffering! (%) Re-ACK pkt with highest in-order seq # (%) (3) timeout(n): retransmit pkt n and all higher seq # pkts in window (%) (4) deliver in-order segments to upper layer. (%)
5 2 6 3 7 4 8 Selective Repeat (4%) (5) receiver individually acknowledges all correctly received pkts (%) (6) buffers out-of order pkts (%) (7) sender only resends pkts for which ACK not received when timeout (%) (8) deliver total in-order pkts to upper layer (%) 6. (a) Describe how TCP Reno does its congestion control. (8%) (2% total) (b) Answer and justify the following questions. After the 22 th transmission round, is segment loss detected by a triple duplicate ACK or by a timeout? (2%) During what transmission round is the 50 th segment sent? (2%) (8%) When CongWin is below Threshold (%), sender in slow-start phase, window grows exponentially (%). When CongWin is above Threshold (%), sender is in congestion-avoidance phase, window grows linearly (%). When a triple duplicate ACK occurs (%), Threshold set to CongWin/2 and CongWin set to Threshold (%). When timeout occurs (%), Threshold set to CongWin/2 and CongWin is set to MSS (%). (b) (8%) a. After the 22 th transmission round, packet loss is recognized by a timeout. (2%)
b. During the st transmission round, packet is sent; packet 2-3 are sent in the 2 nd transmission round; packets 4-7 are sent in the 3 rd transmission round; packets 8-5 are sent in the 4 th transmission round; packets 6-3 are sent in the 5 th transmission round; packets 32-63 are sent in the 6 th transmission round; packets 64 96 are sent in the 7 th transmission round. Thus packet 50 is sent in the 6 th transmission round. ( 說明 %, 答案 %, 共 2%) 7. Consider the TCP procedure for estimating RTT n ( EstimatedRTT = α SampleRTT + ( α) EstimatedRTT ). (a) Why TCP uses this function? (2%) (b) Let SampleRTT n be the most recent sample RTT, let SampleRTT n- be the next most recent sample RTT, and so on. Express EstimatedRTT n in terms of n SampleRTTs if EsitmatedRTT = 0. ( 要有兩次疊代過程 ( 各 2%) 後寫出通式與以 summation 總和符號表示 ( 各 2%) (0% total) (a) Exponential weighted moving average => influence of past sample decreases exponentially fast. 據測量出來的 SampleRTT, 估計下一次的 EstimatedRTT, 用來設定下一次的 Timeout 時間 (2%) (b) n EstimatedRTT = α SampleRTT + ( α) EstimatedRTT n 2 n 2 = α SampleRTT + ( α) [ α SampleRTT + ( α) EsitmatedRTT ] n 2 2 n 2 = α SampleRTT + α( α) SampleRTT + ( α) EsitmatedRTT n 2 2 = α SampleRTT + α( α) SampleRTT + ( α) n 3 n 3 [ α SampleRTT + ( α) EsitmatedRTT ] n 2 2 = α SampleRTT + α( α) SampleRTT + α( α) n 3 4 n 3 SampleRTT + ( α) EsitmatedRTT =... n 2 = α SampleRTT + α( α) SampleRTT + 2 n 3 n 2 n ( ) α( α) SampleRTT +... + α( α) SampleRTT n ( ) + ( α) EsitmatedRTT == α ( α) SampleRTT + ( α) EsitmatedRTT j n j j= ( j n j α) SampleRTT ( EsitmatedRTT 0) j= = α = 8. (a) Explain how TCP Fast Retransmit works. (6%) (5% total) (b) How TCP does its flow control? (6%) (c) Suppose the TCP sequence number space is of size k. What is the largest allowable sender window w? (3%) (a) Explain how TCP Fast Retransmit works. (6%) If sender receives 3 ACKs for the same data, (2%) it supposes that segment after ACKed data was lost (2%): resend segment before timer expires (2%) (6% total) (b) How TCP does its flow control? (6%) Rcvr advertises spare room by including value of RcvWindow in segments (2%) Sender limits unacked data to RcvWindow (2%) for guaranteeing receive buffer doesn t overflow (2%) (c) The sequence number space must be at least twice as large as the window size, k 2w. (3%)