1. What are the two services provided by the network layer? It compares its advantages and disadvantages.
The network layer provides a "connection-oriented" virtual circuit service or a "connectionless" datagram service to the transport layer. The former reserves all network resources needed for communication between the two parties. The advantage is that it can provide the promise of service quality. That is, the transmitted packets are free from error, loss, repetition and out of sequence (do not arrive at the end in sequence), and the time limit for packet transmission is also guaranteed. The disadvantage is that the router is complex and the network cost is high; the latter has no network resource barriers, and does its best, with advantages and disadvantages Reciprocity with the former
2. What is the practical significance of network interconnection? What are the common problems that need to be solved when interconnecting the networks?
Network interconnection can expand the scope of user sharing resources and a larger communication area
When network interconnection, the common problems that need to be solved are:
Different maximum packet lengths for different addressing schemes
Different network access mechanisms
Different timeout controls
Different error recovery methods
Different status reporting methods
Different routing techniques
Different user access control
Different services (connection-oriented service and connectionless service)
Different management and control methods
3. As an intermediate device, what is the difference between a repeater, bridge, router and gateway?
The intermediate device is also called an intermediate system or a relay system.
Physical layer relay system: repeater.
Data link layer relay system: network bridge or bridge.
Network layer relay system: router (router).
A mixture of bridges and routers: brouter.
Relay system above the network layer: gateway.
4. Try to briefly explain the functions of the following protocols: IP, ARP, RARP and ICMP.
IP protocol: realize network interconnection. The network with different capabilities participating in the interconnection looks like a unified network from the user. The Internet Protocol IP is one of the two most important protocols in the TCP/IP system, and there are four more protocols used in conjunction with the IP protocol.
ARP protocol: It solves the mapping problem between the IP address and hardware address of a host or router on the same LAN.
RARP: It solves the problem of mapping the hardware address and IP address of a host or router on the same LAN.
ICMP: Provide error reports and query messages to improve the chances of successful IP data delivery
Internet Group Management Protocol IGMP: used to explore and forward the group membership in the local area network.
5. How many types of IP addresses are there? How are each represented? What are the main characteristics of an IP address?
Divided into 5 types of ABCDE; each type of address is composed of two fixed-length fields, one of which is the network number net-id, which indicates the network to which the host (or router) is connected, and the other is the host Host-id, which identifies the host (or router). The net-id field of each type of address is 1, 2, 3, 0, 0 bytes; the host-id field of the host number is 3 bytes, 2 bytes, 1 byte, 4 bytes, and 4 words. Section.
Features:
(1) IP address is a hierarchical address structure. The advantages of the two levels are: First, the IP address management agency only assigns network numbers when assigning IP addresses, and the remaining host numbers are assigned by the unit that gets the network number. This facilitates the management of IP addresses. Second, the router only forwards packets according to the network number connected to the destination host (regardless of the destination host number), which can greatly reduce the number of items in the routing table, thereby reducing the storage space occupied by the routing table . (2) In fact, the IP address is the interface that marks a host (or router) and a link. When a host is connected to two networks at the same time, the host must have two corresponding IP addresses at the same time, and its network number net-id must be different. This kind of host is called a multihomed host. Since a router should be connected to at least two networks (so that it can forward IP datagrams from one network to another), a router should have at least two different IP addresses.
(3) Several LANs connected by repeaters or bridges are still one network, so these LANs all have the same network number net-id.
(4) All networks assigned to the network number net-id, local area networks with a small range, or wide area networks that may cover a large geographic area are equal. 6. Try to calculate the data in Table 4-2 according to the IP address regulations. Solution: 1) In a class A network, the network number occupies seven bits, the number of networks allowed is 2 to the 7th power, which is 128, but the cases of 0 and 127 are excluded, so the maximum number of networks that can be used is 126 , The first network number is 1, and the last network number is 126. The host number occupies 24 bits, then the maximum number of hosts allowed is 2 to the 24th power, which is 16777216, but the case of all 0s and all 1s should be excluded, so the maximum number of hosts that can be used is 16777214. 2) Type B In the network, the network number occupies 14 bits, then the maximum number of networks that can be used is 2 to the 14th power, which is 16384. The first network number is 128.0. Because 127 is to be used for local software loopback testing, starting from 128, The point after the point can
以 容纳2的8次方为256，所以以128为开始的网络号为128.0~~128.255，共256个，以此类 推，第16384个网络号的计算方法是：16384/256=64128+64=192，则可推算出为191.255。主机号占16个 bit, 则允许用的最大主机数为2的16次方，为65536，但是也要除去全0和全 1的情况，所以能用的最大主机数是65534。3）C类网中，网络号占21个bit, 则能用的网络数为2的21次方，为2097152，第一个 网络号是 192.0.0，各个点后的数占一个字节，所以以 192 为开始的网络号为192.0.0~~192.255.255，共256*256=65536，以此类推，第2097152个网络号的计算方法是： 2097152/65536=32192+32=224，be calculated as 223.255.255. The host number occupies 8 bits, and the maximum number of hosts allowed is 2 to the 8th power, which is 256, but the case of all 0s and all 1s should be excluded, so the maximum number of hosts that can be used is 254.7. The difference between IP address and hardware address, why use these two different addresses? The IP address is to assign a 32-bit identifier that is unique throughout the world to each host (or router) connected to the Internet. Therefore, the entire Internet is regarded as a single, abstract network. When data frames are transmitted on the actual network link, hardware addresses must be used in the end. The MAC address is consistent with the hardware to a certain extent, is based on physics, can identify the specific link communication object, and the division of the logical domain by the IP address is not limited by the hardware.
8. What are the main differences between the IP address scheme and my country's telephone number system?
Independent of the geographic distribution of the network
9. (1) What does the subnet mask 255.255.255.0 mean?
There are three meanings.
One is the subnet mask of a class A network. For the IP address of a class A network, the first 8 bits represent the network number, the last 24 bits represent the host number, and the subnet mask 255.255.255.0 represents the first 8 bits. It is the network number, the middle 16 bits are used to divide the subnet segment, and the last 8 bits are the host number.
The second case is a class B network. For the IP address of a class B network, the first 16 bits represent the network number, the last 16 bits represent the host number, and the subnet mask 255.255.255.0 means the first 16 bits are the network number, and the middle 8 Bits are used to divide the subnet segment, and the last 8 bits are the host number.
The third case is a class C network, and this subnet mask is the default subnet mask of the class C network.
(2) The current mask of a network is 255.255.255.248. How many hosts can be connected to this network? 255.255.255.248 is 11111111.11111111111.11111111.11111000. The host on each subnet is (2^3)=6 and the number of mask bits is 29. The network can connect to 8 hosts, after deducting all 1s and all 0s, 6 hosts.
(3) The subnet-id numbers of a type A network and a type B network are 16 1 and 8 1 respectively. What is the difference between the two subnet masks?
Class A network: 11111111 11111111 11111111 00000000
Given the subnet number (16-bit "1"), the subnet mask is 255.255.255.0
Class B network 11111111 11111111 11111111 00000000
Given the subnet number (8-digit "1"), the subnet mask is 255.255.255.0 but the number of subnets is different
(4) The subnet mask of a Class B address is 255.255.240.0. What is the maximum number of hosts on each subnet?
(240)10=(128+64+32+16)10=(11110000)2 The number of digits of Host-id is 4+8=12, therefore, the maximum number of hosts is: 2 12 -2=4096-2=4094
11111111.11111111.11110000.00000000 Number of hosts 2^12-2
(5) The subnet mask of a Class A network is 255.255.0.255; is it a valid subnet mask?
Is 10111111 11111111 00000000 11111111
(6) The hexadecimal representation of an IP address is C2.2F.14.81, try to convert it into dotted decimal form. What kind of IP address is this address?
C2 2F 14 81--à(12*16+2).(2*16+15).(16+4).(8*16+1)---à194.47.20.129 C2 2F 14 81 ---à11000010.00101111.00010100.10000001Class C address
(7) Is there any practical meaning to use subnet mask for class C network? why?
It has practical meaning. Of the 32 bits of the IP address of the C subnet, the first 24 bits are used to determine the network number, and the last 8 bits are used to determine the host number. If you divide the subnet, you can choose the high bit of the last 8 bits, which can be The network is further divided and the content of the routing table is not increased, but the cost is that the number of hosts is believed to decrease.
10. Try to identify the network type of the following IP address.
（1）128.36.199.3 （2）21.12.240.17 （3）183.194.76.253 （4）
192.12.69.248 （5）89.3.0.1 （6）200.3.6.2
(2) and (5) are category A, (1) and (3) are category B, and (4) and (6) are category C.
11. The header checksum in the IP datagram does not check the data in the datagram. What is the biggest benefit of doing this? What is the disadvantage?
Errors in the header are more serious than errors in the data. For example, a bad address may cause the packet to be posted to the wrong host. Many hosts do not check whether the packets delivered to them are actually intended to be delivered to them. They assume that the network will never deliver to them packets that were originally intended for another host. The data does not participate in the calculation of the checksum, because it is expensive to do so, and the upper layer protocol usually does this kind of check work in the past, which causes duplication and redundancy. Therefore, this can speed up the forwarding of packets, but errors in the data part cannot be detected early.
12. When a router finds an error in the checksum of an IP datagram, why does it adopt the method of discarding it instead of requiring the source station to retransmit the datagram? Why not use the CRC check code to calculate the first checksum?
Answer: The error correction control is executed by the upper layer (transport layer). The source station address in the IP header may also be wrong. Please retransmit the datagram with the wrong source address. It is meaningless. It does not use CRC to simplify the decoding calculation and improve the throughput of the router.
13. Suppose that the IP datagram uses a fixed header, and the specific values of each field are shown in the figure (except for the IP address, all are expressed in decimal). Try the binary calculation method to calculate the value (in binary representation) that should be written into the first checksum field.
4 5 0 28 1 0 0 4 17 10.12.14.5 12.6.7.9
1000101 00000000 00000000-00011100
00000000 00000001 00000000-00000000
00000100 00010001 xxxxxxxx xxxxxxxx
00001010 00001100 00001110 00000101
00001100 00000110 00000111 00001001 for binary checksum (XOR) 01110100 01001110 inverted code 10001011 10110001
14. Recalculate the above question, but use the hexadecimal arithmetic method (no 16-bit binary number is converted to 4 hexadecimal digits, and then calculated according to the hexadecimal addition rule). Compare these two methods.
01000101 00000000 00000000-00011100 4 5 0 0 0 0 1 C
00000000 00000001 00000000-00000000 0 0 0 1 0 0 0 0 00000100 000010001 xxxxxxxx xxxxxxxx 0 4 1 1 0 0 0 0
00001010 00001100 00001110 00000101 0 A 0 C 0 E 0 5
00001100 00000110 00000111 00001001 0 C 0 6 0 7 0 9
01011111 00100100 00010101 00101010 5 F 2 4 1 5 2 A
5 F 2 4 1 5 2 A 7 4 4 E-à8 B B 1
15. What is the maximum transmission unit MTU? It is related to which field in the header of the IP datagram?
Answer: The maximum length of the data field in the frame format defined by the layer in the data chain below the IP layer is related to the total length field in the IP datagram header
16. In the Internet, the datagram transmitted by IP datagram fragments is assembled on the final destination host. There can also be another approach, that is, datagram fragments are assembled once through a network. It is to compare the pros and cons of these two methods.
The reason for assembly at the destination station instead of the router in the middle is:
(1) The router is easier to process datagrams; it has high efficiency and low delay.
(2) Each fragment of the datagram may go through its own path. Therefore, several datagram fragments may always be missing in the assembly of each intermediate router;
(3) Maybe a network will pass after the packet, and it will divide these datagram fragments into smaller fragments. If the router is assembled in the middle, it may be assembled multiple times.
(In order to adapt to the different fragment sizes allowed by different link segments on the path, it may need to be fragmented or assembled again)
17. A 3200-bit TCP message is transmitted to the IP layer, and a 160-bit header becomes a datagram. The Internet below is connected by two LANs through routers. But the data part of the longest data frame that the second LAN can transmit is only 1200 bits. Therefore, the datagram must be fragmented in the router. How many bits of data does the second LAN send to its upper layer (the "data" here of course refers to the data seen by the LAN)?
Answer: The data part of the longest data frame that can be transmitted by the second LAN is only 1200bit, that is, the data part of each IP data slice is less than 1200-160 (bit), because the slice offset is 8 bytes, that is, 64 bits. Unit, so the data part of the IP data piece does not exceed 1024 bits at most, so the 3200-bit message needs to be divided into 4 data pieces, so the number of bits transmitted upwards by the second LAN is equal to (3200+4×160), a total of 3840bit.
18. (1) Some people think: "The ARP protocol provides the network layer with address conversion services, so ARP should belong to the data link layer." Why is this statement wrong?
Because ARP itself is part of the network layer, the ARP protocol provides the service of address conversion for the IP protocol, and the data link layer
The hardware address is used instead of the IP address, and the ARP protocol data link layer itself can operate normally. Therefore ARP is no longer at the data link layer.
(2) Try to explain why the ARP cache has to set a timeout timer of 10-20 minutes every time an item is stored. What will happen if this time is set too large or too small?
Answer: Considering that the IP address and Mac address may be changed (replace the network card, or dynamic host configuration)
It is reasonable to replace a network card in 10-20 minutes. If the timeout period is too short, the traffic of ARP request and response packets will be too frequent. If the timeout period is too long, the host after replacing the network card will be unable to communicate with other hosts on the network.
(3) Cite at least two situations where there is no need to send ARP request packets (that is, there is no need to request to resolve a certain destination IP address to the corresponding hardware address) .
There is an item of the destination IP address in the ARP cache of the source host; the source host sends broadcast packets; the source host and the destination host use a point-to-point link. 19. Host A sends an IP datagram to host B, passing 5 routers on the way. How many times are ARP used in the process of sending IP datagrams? 6 times, once for the host and once for each router.
20. Suppose a router has established the following routing table:
Destination network subnet mask next hop
128.96.39.0 255.255.255.128 interface m0
128.96.39.128 255.255.255.128 interface m1
128.96.40.0 255.255.255.128 R2
192.4.153.0 255.255.255.192 R3
* (Default) —— R4
A total of 5 packets have been received, and their destination addresses are:
（1）128.96.39.10
（2）128.96.40.12
（3）128.96.40.151
（4）192.153.17
（5）192.4.153.90
(1) The IP address of the destination station of the packet is: 128.96.39.10. First and the subnet mask 255.255.255.128 and get 128.96.39.0, it can be seen that the packet is forwarded through interface 0.
(2) The destination IP address of the packet is: 128.96.40.12.
① And the subnet mask 255.255.255.128 is 128.96.40.0, which is not equal to 128.96.39.0.
② With the subnet mask 255.255.255.128, it is 128.96.40.0. After checking the routing table, it can be known that the packet is forwarded by R2.
(3) The destination IP address of the packet is: 128.96.40.151, and the subnet mask 255.255.255.128 will get 128.96.40.128, and the subnet mask 255.255.255.192 will get 128.96.40.128, after checking the routing table It is known that the packet forwarding selects the default route and forwards via R4.
(4) The destination IP address of the packet is: 192.4.153.17. And the subnet mask 255.255.255.128 will be 192.4.153.0. After matching with the subnet mask 255.255.255.192, 192.4.153.0 is obtained. After checking the routing table, it is known that the packet is forwarded via R3.
(5) The destination IP address of the packet is: 192.4.153.90, which is 192.4.153.0 after the subnet mask 255.255.255.128. After matching with the subnet mask 255.255.255.192, 192.4.153.64 is obtained. After checking the routing table, it is known that the packet forwarding selects the default route and is forwarded via R4.
21 An organization is assigned a Class B IP address with a net-id of 129.250.0.0. The organization has 4000 machines distributed in 16 different locations. If you choose a subnet mask of 255.255.255.0, try to assign a subnet mask number to each location, and calculate the minimum and maximum host numbers for each location 4000/16=250, with an average of 250 machines per location . If 255.255.255.0 is selected as the mask, the number of hosts connected to each network=28-2=254>250, and the total number of subnets=28-2=254>16, which can meet actual needs. The following subnet numbers can be assigned to each location
Location: Subnet ID (subnet-id) The minimum and maximum host IP of the subnet network number
1： 00000001 129.250.1.0 129.250.1.1—129.250.1.254
2： 00000010 129.250.2.0 129.250.2.1—129.250.2.254
3： 00000011 129.250.3.0 129.250.3.1—129.250.3.254
4： 00000100 129.250.4.0 129.250.4.1—129.250.4.254
5： 00000101 129.250.5.0 129.250.5.1—129.250.5.254
6： 00000110 129.250.6.0 129.250.6.1—129.250.6.254
7： 00000111 129.250.7.0 129.250.7.1—129.250.7.254
8： 00001000 129.250.8.0 129.250.8.1—129.250.8.254
9： 00001001 129.250.9.0 129.250.9.1—129.250.9.254
10： 00001010 129.250.10.0 129.250.10.1—129.250.10.254
11： 00001011 129.250.11.0 129.250.11.1—129.250.11.254
12： 00001100 129.250.12.0 129.250.12.1—129.250.12.254
13： 00001101 129.250.13.0 129.250.13.1—129.250.13.254
14： 00001110 129.250.14.0 129.250.14.1—129.250.14.254
15： 00001111 129.250.15.0 129.250.15.1—129.250.15.254
16： 00010000 129.250.16.0 129.250.16.1—129.250.16.254
22... The length of a datagram is 4000 bytes (fixed header length). It is now transmitted through a network, but the maximum data length that this network can transmit is 1500 bytes. How can I divide it into several shorter datagram pieces? What is the value of the data field length, fragment offset field and MF flag of each datagram fragment? The fixed header length of the IP datagram is 20 bytes
Total length (bytes) Data length (bytes) MF slice offset
Original data report 4000 3980 0 0
Datagram fragment 1 1500 1480 1 0
Datagram 2 1500 1480 1 185
Datagram 3 1040 1020 0 370
23 There are two cases (using the subnet mask and using CIDR) to write out the Internet IP as a routing algorithm. See textbook P134, P139
24. Try to find the subnet mask (using consecutive masks) that can produce the following number of Class A subnets.
（1）2，（2）6，（3）30，（4）62，（5）122，（6）250.
（1）255.192.0.0，（2）255.224.0.0，（3）255.248.0.0，（4）255.252.0.0，（5）255.254.0.0，（6）255.255.0.0
25. There are 4 subnet masks below. Which ones are not recommended? why?
（1）176.0.0.0，（2）96.0.0.0，（3）127.192.0.0，（4）255.128.0.0。
Only (4) is the mask of continuous 1 and continuous 0, which is recommended
26. There are the following 4 /24 address blocks, try the most possible gathering.
212.56.132.0/24
212.56.133.0/24
212.56.134.0/24
212.56.135.0/24
212=（11010100）2，56=（00111000）2
132=（10000100）2，
133=（10000101）2
134=（10000110）2，
135=（10000111）2
So the common prefix has 22 bits, that is 11010100 00111000 100001, the aggregated CIDR address block is: 212.56.132.0/22
27. There are two CIDR address blocks 208.128/11 and 208.130.28/22. Does any address block contain another address? If so, please point out and explain why.
The prefix of 208.128/11 is: 11010000 100
The prefix of 208.130.28/22 is 11010000 10000010 000101, and its first 11 bits are consistent with the prefix of 208.128/11, so the 208.128/11 address block includes the address block 208.130.28/22.
28. The routing table of the known router R1 is shown in Table 4-12. Table 4-12 The routing table address mask of router R1 in Exercise 4-28 Destination network address Next hop address Router interface
/26 140.5.12.64 180.15.2.5 m2
/ 24 130.5.8.0 190.16.6.2 m1
/ 16 110.71.0.0 …… m0
/16 180.15.0.0 …… m2
/ 16 196.16.0.0 …… m1
Default 110.71.4.5 m0
Try to draw a connection topology between the network and the necessary routers, and mark the necessary IP addresses and interfaces. It should be pointed out if it is uncertain. See the answer after class P380 for graphics
29. An autonomous system has 5 LANs, and the connection diagram is shown in Figure 4-55. The numbers of hosts on LAN2 to LAN5 are:
91, 150, 3, and 15. The IP address block allocated by this autonomous system is 30.138.118/23. Try to give the address block (including prefix) of each LAN. 30.138.118/23–à30.138.0111 011 When assigning network prefixes, you should assign more addresses first. The topic does not say that there are several hosts on LAN1, but at least 3 addresses are needed for three routers.
There are many answers to this question. Two different answers are given below:
第一组答案 第二组答案
LAN1 30.138.119.192/29 30.138.118.192/27
LAN2 30.138.119.0/25 30.138.118.0/25
LAN3 30.138.118.0/24 30.138.119.0/24
LAN4 30.138.119.200/29 30.138.118.224/27
LAN5 30.138.119.128/26 30.138.118.128/27
30. A large company has a headquarters and three subordinate departments. The network prefix assigned by the company is 192.77.33/24. The company's network layout is shown in Figure 4-56. The headquarters has five local area networks, among which LAN1-LAN4 are connected to router R1, and R1 is connected to router R5 through LAN5. R5 is connected with the local area networks LAN6～LAN8 of the three remote departments through a wide area network. The number marked next to each LAN is the number of hosts on the LAN. Try to assign an appropriate network prefix to each local area network. See answer after class P380
31. Which of the following addresses matches 86.32/12: Please explain why.
（1）86.33.224．123：（2）86.79.65.216；（3）86.58.119.74; (4) 86.68.206.154。
86.32/12 è 86.00100000 The 12-bit prefix underlined means that the first 4 bits of the second byte are in the prefix.
The first 4 bits of the second byte of the four addresses given are: 0010, 0100, 0011 and 0100. Therefore only (1) is matched.
32. Which of the following addresses 2.52.90.140 matches? Please explain why.
(1) 0/4; (2) 32/4; (3) 4/6 (4) 152.0/11
The prefix (1) matches the address 2.52.90.140
2.52.90.140 is 0000 0010.52.90.140
0/4 is 0000 0000
32/4 is 0010 0000
4/6 is 0000 0100
80/4 is 0101 0000
33. Which of the following prefixes matches the addresses 152.7.77.159 and 152.31.47.252? Please explain why.
(1) 152.40/13; (2) 153.40/9; (3) 152.64/12; (4) 152.0/11.
The prefix (4) matches both addresses
34. How many bits does the network prefix correspond to the following masks?
（1）192.0.0.0；（2）240.0.0.0；（3）255.254.0.0；（4）255.255.255.252。
（1）/2 ; (2) /4 ; (3) /11 ; (4) /30 。
35. One of the addresses in the known address block is 140.120.84.24/20. Try to find the minimum address and maximum address in this address block. What is the address mask? How many addresses are in the address block? How many class C addresses are equivalent to? 140.120.84.24 è 140.120.(0101 0100).24
最小地址是 140.120.(0101 0000).0/20 (80)
最大地址是 140.120.(0101 1111).255/20 (95)
地址数是4096.相当于16个C类地址。
36. One address in the known address block is 190.87.140.202/29. Recalculate the previous question.
190.87.140.202/29 is 190.87.140. (1100 1010) / 29
最小地址是 190.87.140.(1100 1000)/29 200
最大地址是 190.87.140.(1100 1111)/29 207
地址数是8.相当于1/32个C类地址。
37. A certain unit is allocated an address block 136.23.12.64/26. Now it needs to be further divided into 4 equally large subnets. Just ask:
（1）每一个子网的网络前缀有多长？
（2）每一个子网中有多少个地址？
（3）每一个子网的地址是什么？
(4) What are the smallest and largest addresses that can be allocated to the host for each subnet?
(1) The prefix of each subnet is 28 bits.
(2) Four of the addresses of each subnet are reserved for the host, so there are 16 addresses in total.
(3) The address blocks of the four subnets are:
The first address block 136.23.12.64/28 can be allocated to the host
Minimum address: 136.23.12.01000001=136.23.12.65/28
Maximum address: 136.23.12.01001110=136.23.12.78/28
The second address block 136.23.12.80/28 can be allocated to the host
Minimum address: 136.23.12.01010001=136.23.12.81/28
Maximum address: 136.23.12.01011110=136.23.12.94/28
The third address block 136.23.12.96/28 can be allocated to the host
Minimum address: 136.23.12.01100001=136.23.12.97/28
Maximum address: 136.23.12.01101110=136.23.12.110/28
The fourth address block 136.23.12.112/28 can be allocated to the host
Minimum address: 136.23.12.01110001=136.23.12.113/28
Maximum address: 136.23.12.01111110=136.23.12.126/28
38. What is the main difference between IGP and EGP?
IGP: The routing protocol used within the autonomous system; strive for the best routing
EGP: A routing protocol that is convenient to use in different autonomous systems; strive for better routing (not going around in circles)
EGP must consider other aspects of the policy and requires multiple routes. Accessibility may be more important in terms of costs.
IGP: Internal Gateway Protocol. It only cares about how to transmit datagrams in this autonomous system, and has nothing to do with what protocols are used by other autonomous systems on the Internet.
EGP: External Gateway Protocol, a protocol for transferring routing information on the border of different ASs, regardless of which protocol is used inside the AS.
Note: IGP mainly considers how to work efficiently within the AS. In most cases, the best route is found. There are many explanations for costs and costs.
39. Briefly describe the main features of RIP, OSPF and BGP routing protocols.
Main features RIP OSPF BGP
Gateway protocol inside and outside
Routing table content destination network, next stop, distance to destination network, next stop, distance to destination network, complete
path
The optimal path depends on multiple strategies of hop count
Algorithm distance vector link state distance vector
Transmission mode Transport layer UDP IP datagram to establish TCP connection
Other simple, low efficiency, 16 unreachable hops, fast transmission of good news, slow transmission of bad news, high efficiency, frequent exchange of information by routers, difficulty in maintaining consistency, large scale, uniform measurement of reachability
40. RIP uses UDP, OSPF uses IP, and BGP uses TCP. What are the advantages of this? Why does RIP periodically exchange router information with neighboring sites, but BGP does not do so?
RIP only exchanges information with neighboring stations. The use of UDP has no reliable guarantee, but the overhead is small and can meet the requirements of RIP; OSPF uses a reliable flooding method and directly uses IP, which is flexible and low overhead;
BGP needs to exchange the entire routing table and update information. TCP provides reliable delivery to reduce bandwidth consumption; RIP uses UDP, which does not guarantee reliable delivery. Therefore, it must constantly (periodically) exchange information with neighbors to update routing information in time. But BGP uses TCP that guarantees reliable delivery, so there is no need to do so.
41. Assume that the routing table of router B in the network has the following items (these three columns represent "destination network", "distance" and "next hop router")
N1 7 A
N2 2 B
N6 8 F
N8 4 E
N9 4 F
Now B receives the routing information from C (the two columns represent the "destination network" and "distance" respectively):
N2 4
N3 8
N6 4
N8 3
N9 5
Try to find the updated routing table of router B (describe each step in detail).
The updated routing table of Router B is as follows:
N1 7 A No new information, no change
N2 5 C Same next hop, update
N3 9 C New items, add in
N6 5 C Different next hop, shorter distance, update
N8 4 E Different next hops, the same distance, no change
N9 4 F Different next hop, greater distance, no change
42. Assume that the routing table of router A in the network has the following items (the format is the same as the above question):
N1 4 B
N2 2 C
N3 1 F
N4 5 G
Now A receives the routing information from C (the format is the same as the above question):
N1 2
N2 1
N3 3
N4 7
Try to find the updated routing table of router A (describe each step in detail).
The updated routing table of Router A is as follows:
N1 3 C Different next hop, shorter distance, change
N2 2 C Different next hops, the same distance, unchanged
N3 1 F Different next hop, greater distance, no change
N4 5 G No new information, no change
43. What are the main points of the IGMP protocol? How is tunnel technology used?
IGMP can be divided into two stages:
Phase 1: When a host joins a new multicast group, the host should send an IGMP message to the multicast address of the multicast group to declare that it wants to become a member of the group. After the local multicast router receives the IGMP message, it forwards the group membership to other multicast routers on the Internet.
The second stage: Because the group membership is dynamic, the local multicast router must periodically inquire the hosts on the local LAN to know whether these hosts are still members of the group. As long as a host responds to a certain group, the multicast router considers this group to be active. However, a group still does not respond to a host after several inquiries, and the membership of the group is not forwarded to other multicast routers. Tunneling technology: The multicast datagram is encapsulated into a unicast IP datagram, which can traverse a network that does not support multicast and reach another network that supports multicast.
44. What is VPN? What are the characteristics, advantages and disadvantages of VPN? How many types of VPN are there?
P171-173
45. What is NAT? What are the characteristics of NAPT? What are the advantages and disadvantages of NAT? What are the advantages and disadvantages of NAT? P173-174