Homework Chapter

1、Homework Chapter 41. In the text we have used the term connection-oriented service to describe a transport-layer service and connection service for a network-layer service. Why the subtle shades in terminology?在文中，我們所用的術(shù)語(yǔ)面向連接的服務(wù)來(lái)描述傳輸層服務(wù)和連接服務(wù)為網(wǎng)絡(luò)層的服務(wù)。為什么術(shù)語(yǔ)中細(xì)微的色調(diào)？In a virtual circuit network, there is

2、an end-to-end connection in the sense that each router along the path must maintain state for the connection; hence the terminology connection service. In a connection-oriented transport service over a connectionless network layer, such as TCP over IP, the end systems maintain connection state; howe

3、ver the routers have no notion of any connections; hence the terminology connection-oriented service.在一個(gè)虛電路網(wǎng)絡(luò)，有在這個(gè)意義上，沿路徑的每個(gè)路由器必須維護(hù)狀態(tài)為連接端至端連接;即術(shù)語(yǔ)連接服務(wù)。在通過(guò)無(wú)連接網(wǎng)絡(luò)層，例如TCP到 IP上一個(gè)面向連接的傳輸服務(wù)，所述終端系統(tǒng)保持連接狀態(tài);但是，路由器沒(méi)有任何連接的概念;因此術(shù)語(yǔ)面向連接的服務(wù)。2. Consider a datagram network using 32-bit host addresses. Suppose a router

4、has four links, numbered 0 through 3, and packets are to be forwarded to the link interfaces as follows:考慮使用32位主機(jī)地址的數(shù)據(jù)包網(wǎng)絡(luò)。假設(shè)路由器具有四個(gè)鏈接，編號(hào)為0到3，并且分組將被轉(zhuǎn)發(fā)到鏈路接口如下：Destination Address RangeLink Interface11100000 00000000 00000000 00000000through011100000 00111111 11111111 1111111111100000 01000000 00000000

5、 00000000through111100000 01000000 11111111 1111111111100000 01000001 00000000 00000000through211100001 01111111 11111111 11111111otherwise3a. Provide a forwarding table that has four entries, uses longest prefix matching, and forwards packets to the correct link interfaces.b. Describe how your forw

6、arding table determines the appropriate link interface for datagrams with destination addresses:11001000 10010001 01010001 0101010111100001 01000000 11000011 0011110011100001 10000000 00010001 011101111.提供一個(gè)轉(zhuǎn)發(fā)表，有四個(gè)條目，使用最長(zhǎng)前綴匹配，并且報(bào)文轉(zhuǎn)發(fā)到正確的鏈路接口。2.描述你如何根據(jù)轉(zhuǎn)發(fā)表，決定了數(shù)據(jù)報(bào)的目的地址相應(yīng)的鏈路接口：3. Consider a datagram netw

7、ork using 8-bit host addresses. Suppose a router uses longest prefix matching and has the following forwarding table:考慮使用8位主機(jī)地址的數(shù)據(jù)包網(wǎng)絡(luò)。假設(shè)一臺(tái)路由器使用最長(zhǎng)前綴匹配，并具有下列轉(zhuǎn)發(fā)表：For each of the four interfaces, give the associated range of destination host addresses and the number of addresses in the range.對(duì)于每4個(gè)接口，給目的

8、地的主機(jī)地址相關(guān)聯(lián)的范圍，并在范圍內(nèi)的地址的數(shù)目。4. Consider sending a 2400-byte datagram into a link that has an MTU of 700 bytes. Suppose the original datagram is stamped with the identification number 422. How many fragments are generated? What are the values in the various fields in the IP datagram(s) generated related

9、 to fragmentation?考慮派遣一個(gè)2400字節(jié)的數(shù)據(jù)包到具有700字節(jié)的MTU的鏈接。假設(shè)原始數(shù)據(jù)包上印有識(shí)別號(hào)碼422。產(chǎn)生多少碎片？什么是在IP數(shù)據(jù)報(bào)（多個(gè)）的各個(gè)領(lǐng)域中的值產(chǎn)生相關(guān)碎片？5. Suppose datagrams are limited to 1500 bytes (including header) between source Host A and destination Host B. Assuming a 20-byte IP header, how many datagrams would be required to send an MP3 con

10、sisting of 5 million bytes? Explain how you computed your answer.假設(shè)數(shù)據(jù)報(bào)被限制在1500字節(jié)（包括字頭）在源主機(jī)A和目標(biāo)主機(jī)B之間.假設(shè)一個(gè)20字節(jié)的IP報(bào)頭，有多少數(shù)據(jù)包將被要求送往一個(gè)有5億字節(jié)的MP3？解釋你如何計(jì)算你的答案。6. Consider the network setup in following 考慮下圖中的網(wǎng)絡(luò)設(shè)置。假設(shè)在ISP代替分配路由器的地址24.34.112.235和所述家庭網(wǎng)絡(luò)的網(wǎng)絡(luò)地址為192.168.1/ 24。a. Assign addresses to all interfaces in

11、 the home network.b. Suppose each host has two ongoing TCP connections, all to port 80 at host 128.119.40.86. Provide the six corresponding entries in the NAT translation table.1.指定地址在家庭網(wǎng)絡(luò)中的所有接口。2.假定每個(gè)主機(jī)有兩個(gè)正在進(jìn)行的TCP連接，所有80端口的主機(jī)128.119.40.86。提供NAT轉(zhuǎn)換表的六個(gè)相應(yīng)的條目。7. Consider the following network. With the

12、 indicated link costs, use Dijkstras shortest-path algorithm to compute the shortest path from x to all network nodes. And compute the shortest path from t to all network nodes. Show how the algorithm works by computing a table.考慮下面的網(wǎng)絡(luò)。用所示鏈路成本，使用Dijkstra的最短路徑算法來(lái)從x到所有的網(wǎng)絡(luò)節(jié)點(diǎn)計(jì)算最短路徑。并從t到所有的網(wǎng)絡(luò)節(jié)點(diǎn)計(jì)算最短路徑。顯示如何算法通過(guò)計(jì)算表。Answer:StepND(t),p(t)D(u),p(u)D(v),p(v)D(w),p(w)D(y),p(y)D(z),p(z)0x123456StepND(x), p(x)D(u),p(u)D(v),p(v)D(w),p(w)D(y),p(y)D(z),p(z)0t1234568. Consider the network shown below, and assume that eac