以下答案由我提供，題目由徐子祥打的
因為今天趙老師不給問問題，所以下面解答僅供參考
題目有錯請找打題目的人(不要找我)
希望這次考題不會差很多 大家都能有不錯的成績
以下答案有錯或不懂 歡迎與我討論
以下答案只有第二題丁不確定其他都為正解(應該是...)
1.Find the class of the following IP addresses (please showing the reason)?答:
甲、208.34.54.12
乙、238.34.2.1
丙、114.34.2.8
丁、129.14.6.8
戊、241.34.2.8
由10進位的分法可知
0~127 | Class A |
128~191 | Class B |
192~223 | Class C |
224~239 | Class D |
240~255 | Class E |
所以答案為
甲、Class C
乙、Class D
丙、Class A
丁、Class B
戊、Class E
2.In a subnet, we know that the IP address of one of the hosts and its mask as given below:答:
IP address:125.134.112.66
Mask:255.255.224.0
甲、What is the network address of this subnet?
乙、What is the broadcast address of this subnet?
丙、How many IP addresses are allocated in this subnet?
丁、How many IP address are available actually for use in this subnet?
(please show your calculation details, not simply answer.)
甲.
將IP address與Mask兩者作邏輯AND運算即可求得答案
所得為125.134.96.0
乙.
125.134.127.255
丙.
213=8192 (請自己把224換成二進位，然後數出有幾個0，再經由二的冪次方就是IP數)
丁.
8192-3=8189 (3為路由器、廣播、網路位址)[因不確定是減2還是減3，考試時請註明說如果有+路由器時就減3]
3.An Ethernet MAC sublayer receivers 1508 bytes of data from the LLC sublayer. Can the data be encapsulated in one frame? If not, how many frames need to be sent? What is the size of the data in each frame?答:
最大的封包資料長度只能有1500bytes
最小的封包資料長度只能有46bytes
所以答案是需要兩個frames(訊框) [注:因為1508大於1500，所以要拆成兩個]
故 第一個封包含有 1500 bytes LLC資料 亦即1518 bytes MAC
第二個封包含有 46 bytes LLC資料 亦即 64 bytes MAC
答:4.How does the CSMA/CD solve the “hidden terminals” problem? “Exposed terminals” problem?
因為之前有發過文章解釋過了，所以請看以前的文章，謝謝~!
，有錯請跟我聯絡討論~謝謝!
