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1、Chapter 44.1,4.2,4.3节后作业1. (R3)What are the two most important network-layer functions in a datagram network? What are the three most important network-layer functions in a virtual-circuit network?答:forwarding and routing.forwarding, routing and connection setup.2. (R2)What is the difference between
2、 routing and forwarding?答:Forwarding is about moving a packet from a routers input link to the appropriate output link. Routing is about determining the end-to-routes between sources and destinations.3. (R9)Describe how packet loss can occur at input ports. Describe how packet loss at input ports ca
3、n be eliminated (without using infinite buffers).答:The queue size at the input port grows large because of slow switching fabric speed and exhausting routers buffer space. It can be eliminated if the switching fabric speed is at least n times as fast as the input line speed, where n is the number of
4、 input ports.4.4,4.5,4.6,4.7节后作业4. (R12)What is the 32-bit binary equivalent of the IP address 223.1.3.27?答: 5. Do routers have IP addresses? If so, how many?答:Yes. They have one address for each interface.6. Consider Figure 4.34. Starting with the original table in D, suppose that D receives from A
5、 the following advertisement. Will the table in D change? If so how?答:No. If the advertisement said that A were only 4 hops away from z by way of C, then D would indeed change its forwarding table.7. (P8)Consider a datagram network using 32-bit host addresses. Suppose a router has five links, number
6、ed 0 through 4, and packets are to be forwarded to the link interfaces as follows:Destination Address Range Link Interface Through 0 Through 1 Through 2 Through 3 Otherwise 4a. Provide a forwarding table that has five entries, uses longest prefix matching, and forwards packets to the correct link in
7、terfaces.b. Describe how your forwarding table determines the appropriate link interface for datagrams with destination addresses: 答:a.IPLink interface 0 123Otherwise4b. Prefix match for first address is 4th entry: link interface 3 Prefix match for second address is 2nd entry: link interface 1 Prefi
8、x match for first address is 3rd entry: link interface 2 8. (P9)Consider a datagram network using 4 bit host addresses. Suppose a router uses longest prefix matching and has the following forwarding table:Prefix MatchInterface000011102113For each of the four interfaces, give the associated range of
9、destination host addresses and the number of addresses in the range.答:Destination Address RangesNumber of addressLink Interface6406416426439. (P11)Consider a router that interconnects three subnets: Subnet 1, Subnet 2, and Subnet 3. Suppose all of the interfaces in each of these three subnets are re
10、quired to have the prefix 220.2.240/20. Also suppose that Subnet 1 is required to support up to 2000 interfaces, and Subnets 2 and 3 are each required to support up to 1000 interfaces. Provide three network addresses (of the form a.b.c.d/x) that satisfy these constraints.答:223.1.17.0/25223.1.17.128/
11、26223.1.17.192/2610. (P14)Consider a subnet with prefix 101.101.101.64/26. Give an example of one IP address (of form xxx.xxx.xxx.xxx) that can be assigned to this network. Suppose an ISP owns the block of addresses of the form 101.101.128/17. Suppose it wants to create four subnets from this block,
12、 with each block having the same number of IP addresses. What are the prefixes (of form a.b.c.d/x) for the four subnets?答:The IP address in range is from 101.101.101.64 to 101.101.101.128The four equal size subnets (extends 18 and 19)101.101.128/19101.101.160/19101.101.192/19101.101.224/1911. (P15)C
13、onsider the topology shown in Figure 4.17. Denote the three subnets with hosts (starting clockwise at l2:00) as Networks A, B, and C. Denote the subnets without hosts as Networks D, E, and F.FEDCBAa. Assign network addresses to each of these six subnets, with the following constraints: All addresses
14、 must be allocated from214.97.254/23; Subnet A should have enough addresses to support 250 interfaces; Subnet B should have enough addresses to support 120 interfaces; and Subnet C should have enough addresses to support 120 interfaces. Of course, subnets D, E and F should each be able to support tw
15、o interfaces. For each subnet, the assignment should take the form a.b.c.d/x or a.b.c.d/x - e.f.g.h/y.b. Using your answer to part (a), provide the forwarding tables (using longest prefix matching)for each of the three routers. 答:a. From 214.97.254/23, the assignments areSubnet A: 214.97.255/24(28=2
16、56 addresses)Subnet B: 214.97.254.0/25-214.97.254.0/29(27-8=120 addresses)Subnet C: 214.97.254.128/25(27=128 addresses)Subnet D: 214.97.254.0/31(21=2 addresses)Subnet E: 214.97.254.2/31(21=2 addresses)Subnet F: 214.97.254.4/30(22=4 addresses)b. Assume that no datagrams have router interface as ultim
17、ate destinations. Also, label D,E,F for the upper-right, bottom, and upper-left interior subnets, respectively.Router1:Outgoing Interface Longest Prefix MatchA F D Router2:Outgoing InterfaceLongest Prefix MatchF E C Router3:Outgoing InterfaceLongest Prefix MatchB D E 12. (P16)Consider sending a 4,00
18、0-byte datagram into a link that has an MTU of 400 bytes. Suppose the original datagram is stamped with the identification number 422-How many fragments are generated? What are their characteristics?答:The maximum size of data field in each fragment = 380(20 bytes IP header). Thus the number of requi
19、red fragments = (4000-20)380=11Each fragment will have identification number 422. Each fragment except the last one will be of size 500 bytes (including IP header). The last datagram will be of size 120 bytes (including IP header). The offsets of 7fragments will be 0,60,120,180,240,300,360. Each of
20、the first 6 fragments will have flag = 1; the last fragment will have flag = 0 .13. (P25)Consider the network fragment shown below. -r has only two attached neighbors, w and y. w has a minimum-cost path to destination u (not shown) of 5,and y has a minimum-cost path to u of 6. The complete paths from w and y to z (and between w and y) are not shown. All link costs in the
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