Subnet partitioning instance analysis (2)

The previous section mainly focused on understanding the theory of IP addresses and subnet division. This section is used to analyze specific instances. This exercise is also found on the Internet and I feel pretty good. Okay, now.

Instance analysis

1. 192.168.1.0/24 use the subnet mask 255.255.255.240 to divide subnets. The number of available subnets is (), and the number of available host addresses in each subnet is ()
A. 14 14
B. 16 14
C. 254 6
D. 14 62
Analysis: 192.168.1.0/24 is a class C address, and the default network number is 24 bits (255.255.255.0 ). Now we can see that the network number is 28 BITs, the host bit is 4 bits, and the conversion is binary:

1111 1111.1111 1111.1111 1111.1111 0000

Convert 192.168.1.0 to binary: 1100 0000.1010 1000.0000 0001.0000 0000

After the original 192.168.1.0/24 is moved to four bits, the four bits are called subnet bits. The four bits binary can represent the 4 power of 2 and the 16 count. At the same time, because the host space is the network number of all 0 and all 1 subnets, that is, the network segments of 192.168.1.0/28 and 192.168.1.240/28 are prone to ambiguity and cannot be used (note: in some documents, Cisco routes support all 0 and 1 CIDR blocks ). There are four remaining host bits. In total, the first host bit in each subnet is 0 and the last host bit is 1. In summary, the number of available subnets is 16-2 = 14, and the number of available hosts in each subnet is 14. The answer is.

2. The subnet mask is 255.255.0.0. Which of the following IP addresses is not in the same network segment ()
A. 172.25.15.201
B. 172.25.16.15
C. 172.16.25.16
D. 172.25.201.15
Analysis: we can see from the Four answers that the four IP addresses belong to Class B addresses. Subnet Mask 255.255.0.0, which is converted to binary. A total of 16 bits are 1. You can determine that there are 16 bits. Convert the following IP addresses into binary values. Take the first 16 digits. We can see that a, B, and D are in the same network segment. Therefore, only C and others are not in the same network segment. The answer is C.

3. If the subnet mask of Class B address is 255.255.255.255.248, the number of available host addresses in each subnet is ()
A. 10
B. 8
C. 6
D. 4
Analysis: the default subnet mask of the B-type address is 255.255.0. The conversion to binary has 16 1, that is, 16 network bits and 16 host bits. Here, the 255.255.255.248 is converted to binary with 29 1 s, so 13 BITs are removed, that is, the network bits are extended to 29 BITs. The remaining three locations are the primary locations, which have a total of 2 Power 3 and 8 except for all 0 and 1 addresses, the number of available host addresses is 8-2 = 6. The answer is C.

4. For Class c ip addresses, if the subnet mask is 255.255.255.248, the number of subnets (B) can be provided)
A. 16
B. 32
C. 30
D. 128.
Analysis: the default subnet mask of class c ip addresses is 255.255.255.0, and the network bit is a 24-bit binary bit. Here we provide 255.255.255.255.248, 29 network bits, and 5 more as subnet bits. There are 2 power 5, 32 subnets, except all 0 and all 1 subnets, available subnet 32-2 = 30. the answer is C.

5. The three CIDR blocks 192.168.1.0/24,192.168 .2.0/24,192.168 .3.0/24 can be aggregated into which of the following CIDR blocks ()
A. 192.168.1.0/22
B. 192.168.2.0/22
C. 192.168.3.0/22
D. 192.168.0.0/22
Analysis: Convert the third byte of the three CIDR blocks to binary,

192.168.0000 0001.0

192.168.0000 0010.0

192.168.0000 0011.0

It can be easily seen that if the network bit is determined to be 22 binary bits, all three addresses can be mapped to 192.168.0.0/22. The answer is D.

6. What are the default subnet masks of IP address 219.25.23.56?
A.8
B .16
C.24
D.32
Analysis: 219.25.23.56 is a class C address. The default subnet mask is 255.255.255.0 and the default subnet mask is 24 bits. The answer is C.

7. A company applies for a class c ip address, but wants to connect six subsidiaries. The largest subsidiary has 26 computers, each of which is in a CIDR block, then the subnet mask should be set?
A.255.255.255.0
B .20.00000000128
C.20.00000000192
D.w.w.w.gov.cn
Analysis: the largest subsidiary has 26 computers, and the number of hosts can be confirmed. 2 ^ 4 <26 <2 ^ 5. The estimated host bit must have at least 5 bits and the network bit should be 27 BITs. The default network number of class C is 24 bits, 27-24 = 3 bits are subnet numbers, 2 ^ 3 = 8, except for all 1 and all 0 subnet numbers, there are 6 Available subnets that meet the conditions. Therefore, the subnet mask should be set to dynamic subnet mask. The answer is D.

8. What is the broadcast IP address issued by the host at startup when the IP address is 10.110.9.113/21?
A.10.110.9.255
B .10.110.15.255
C.10.110.255.255
D.10.20.255. 255
Analysis: first, determine the subnet to which the IP address belongs. The IP address is 10.110.9.113 and the subnet mask is 255.255.248.0. It belongs to the subnet 10.110.8.0/21.

In the previous article "IP address and subnet division (I)", we mentioned that the broadcast address with full host space 1 is the IP address of this network segment, therefore, the broadcast address of 10.110.8.0/21 is the host location of the last 32-21 = 11 BITs, which is 1, that is, 10.110.15.255. the answer is B.

9. To plan a class C network, you need to divide the network into nine subnets. Each subnet can have up to 15 hosts. Which of the following is a suitable subnet mask?
A.20.0000224.0
B .w.w.w.gov.cn
C.20.00000000240
D. No suitable Subnet Mask
Analysis: Class C network is the first, and a can be excluded first. 2 ^ 3 <9 <2 ^ 4, so we need to divide nine subnets. The subnet bit must be at least four. We know that the default subnet mask of the class C address is 255.255.255.0, with 24 network BITs by default. If there are nine subnets, the network bit needs to be expanded by four digits, that is, the current network bit is 24 + 4 = 28 digits. The remaining 32-28 = 4 host bits, and 4 host bits have 2 ^ 4 = 16 hosts, but all 1 and 0 host bits are not available, it is estimated that only 16-2 = 14 hosts are available for each subnet, with less than 15 hosts. Therefore, Class C alone cannot meet the requirements. It is estimated that there is no suitable subnet mask. The answer is D.

10. What is the IP address of the host that is in the same network segment as 10.110.12.29mask?
A.10.110.12.0
B .10.110.12.30
C.10.110.12.31
D.10.110.12.32
Analysis: First, you can confirm the CIDR block of 10.110.12.29. Convert all data to binary for calculation. The network number is 10.110.12.0/27, and the IP address range is 10.110.12.0 ~ 10.110.12.31, all 1 and all 0 addresses cannot be allocated, so the available IP address range is 10.110.12.1 ~ 10.110.12.30, only 10.110.12.30 and 10.110.12.29 belong to the same network segment. The answer is B.

11. What is the network address of the IP address 190.233.27.13/16?
A.190.0.0.0
B .190.233.0.0
C.190.233.27.0
D.190.233.27.1
Analysis: 190.233.27.13/16 adopts the CIDR technology, and the following 16 is the 16-bit network bit. What we usually see is 190.233.27.13, And the subnet mask is 255.255.0.0. To determine some network addresses, you only need to convert the addresses and masks into binary values, and then perform operations on them. The result is 190.233.0.0. The answer is B.

12. What is the IP address range of 125.3.54.56 without any subnet division?
A.125.0.0.0
B .125.3.0.0
C.125.3.54.0
D.125.3.54.32
Analysis: 125.3.54.56 is a Class B address. If no subnet is divided, the default subnet mask 255.0.0.0 is used to convert the address and subnet mask to binary. Then, the calculation result is 125.0.0.0. The answer is.

13. If a subnet CIDR block is 2.0.0.0 and the mask is 255.255.224.0, what is the valid subnet CIDR block address?
A.2.1.16.0
B .2.2.32.0
C.2.3.48.0
D.2.4.172.0
Analysis: First, you can determine that 2.0.0.0 is a type-a network, and the mask is 255.255.224.0. Then, you can determine that the network has 19 network bits and the host bits have 32-19 = 13 BITs.

The default subnet mask of A-class network is 255.0.0, which has eight network bits. It can be concluded that the subnet bits are 19-8 = 11 bits, the host space is 32-a total of 2 ^ 11 = 2048 subnets. Except for subnet numbers that are full 1 and full 0, they cannot be allocated. There are 2042 valid subnets.

In sequence: 2.0.32.0/19, 2.0.64.0/19, 2.0.96.0/19 ....

2.1.0.0/19, 2.1.32.0/19, 2.1.64.0/19 .....
2.2.0.0/19, 2.2.32.0/19, 2.2.64.0/19 .....

2.3.0.0/19, 2.3.32.0/19, and 2.3.64.0/19 .....

.....

The valid CIDR Block of the four network addresses is 2.2.32.0. The answer is B.

14. If the IP address range of a subnet is 5.32.0.0 and the mask is 255.224.0.0, what is the maximum host address allowed?
A.5.32.254.254
B .5.32.0000254
C.5.63.20.254
D.5.63.255.255
Analysis: first, we can find that the network bit is 11 bits and the host bit is 21 bits. The maximum address is 5.63.20.254 because the host space of all 1 cannot be allocated. The answer is C.

15. In a network with a subnet mask of 255.255.255.240.0, what are valid network segment addresses?
A.150.150.0.0
B .150.150.0.8
C.150.150.8.0
D.150.150.16.0
Analysis: first, we can determine from the four CIDR blocks that the IP addresses are for the B-class network. The default subnet mask is 255.255.0.0, with 16-bit network bits; it can be inferred from 255.255.255.240.0 that there are 20 network bits. Therefore, the subnet has 20-16 = 4, and the subnet bit, that is, the number from left to right after being converted to binary, 17th ~ 20. Four digits cannot be 1 or 0. In summary, only 150.150.16.0 is a valid CIDR block address. The answer is D.

16. If the mask of the class C subnet is too many then the number of subnet digits included. Number of subnets. Which of the following is the correct number of hosts in each subnet?
A.2, 2,62
B .3, 6,30
C.4, 14,14
D.5, 30,6
Analysis: the default subnet mask of the class C network is 255.255.255.0. You can determine that there are 24-bit network bits. Here, the subnet mask is too large. You can determine that there are 27 network bits, then the subnet BITs have 27-24 = 3 bits; the number of subnets is 2 ^ 3-2 = 6 (except for the full subnet 1 and full 0 ). The remaining 32-27 = 5 primary locations can be inferred from the given subnet mask, then, the number of hosts in each subnet is 2 ^ 5-2 = 30 (except for all 1 and 0 ). The answer is B.

17. Network Address: 172.16.0.0. If the subnet mask 255.255.192.0 is used, which of the following statements is true ()
A. Two Valid subnets are divided;
B. Four valid subnets are divided;
C. The broadcast address of one of the subnets is 172.16.191.255;
D. The broadcast address of one of the subnets is 172.16.128.255.
Analysis: The network address is 172.16.0.0, which can be regarded as a Class B address. The default subnet mask is 255.255.0.0, and the default host bit is 16 bits;

It is inferred from subnet mask 255.255.192.0 that it has 18-bit network bits. It is estimated that there are two subnet bits and there are 2 ^ 2 = 4 subnets, except that the subnet bits are all 1 and all 0, there are two valid subnets, 172.16.64.0/18, 172.16.128.0/18, and 172.16.128.0/18 respectively. The broadcast addresses of the two subnets (that is, all the digits after the network bit are 1) are 172.16.127.255 and 172.16.191.255 respectively. The answer is AC.

18. Which of the following statements about the host address 192.168.19.125 (subnet mask: 255.255.255.255.248) is true ()
A. subnet address: 192.168.19.120;
B. The subnet address is 192.168.19.121;
C. The broadcast address is 192.168.19.127;
D. The broadcast address is 192.168.19.128.
Analysis: the host address and subnet mask are converted to binary for calculation, and the subnet address is 192.168.19.120;

According to zookeeper 255.248, the network bit size is 29 and the host bit size is 3. The broadcast address of this network segment is 192.168.19.127 when the master location is 1 after the network bit. The answer is AC.

19. A Class C address: 192.168.5.0 for subnet planning. Each subnet must have 10 hosts. Which subnet mask is used for the most reasonable division ()
A. Use subnet mask 255.255.255.255.192;
B. Use the subnet mask;
C. Use the subnet mask 255.255.255.255.240;
D. Use the subnet mask 255.255.255.252.
Analysis: the default subnet mask of class C addresses is 255.255.255.0, with 24 network bits and 8 host bits;

If each subnet requires 10 hosts, then 2 ^ 3 <10 <2 ^ 4, the host bit must be at least 4, and 32-4 = 28 network bits, the subnet mask is 255.255.255.255.255.240. The answer is C.

20. For the Network Address 192.168.1.0/24, select "invalid Subnet Mask". Which of the following statements is true ()
A. Four valid subnets are divided;
B. Six valid subnets are divided;
C. The number of valid hosts in each subnet is 30;
D. The number of valid hosts in each subnet is 31;
E. The number of valid hosts in each subnet is 32.
Analysis: The Network Address 192.168.1.0/24 is a class C network, the default subnet mask is 255.255.255.0, the network bit has 24 bits, and the host bit has 8 bits;

The subnet mask contains 27 BITs, and the subnet bit is 27-24 = 3 bits, and the subnet bit is 1 and all 0 cannot be allocated, valid subnets are 2 ^ 3-2 = 6;

If the number of valid hosts is 2 ^ 5-2 = 30, the number of valid hosts is calculated from the number of valid hosts. The answer is BC.

21. IP Address: 192.168.12.72, subnet mask: 255.255.255.255.192, the network address and broadcast address of the IP address segment are ()
A. 192.168.12.32, 192.168.12.127;
B. 192.168.0.0, 255.255.255.255;
C. 192.168.12.43, 255.255.255.128;
D. 192.168.12.64, 192.168.12.127.
Analysis: It is obtained from subnet mask 255.255.255.255.192 that it has 26-bit network bits and 6-bit host bits.

Convert the IP address and subnet mask to binary format and calculate the network address 192.168.12.64.

The broadcast address of this CIDR block is 192.168.12.127 when the 6-bit host is 1. The answer is D.

22. 172.16.10.32/24 represents ()
A. network address;
B. Host address;
C. multicast address;
D. broadcast address.
Analysis: there are 24 network bits in 172.16.10.32/24. The network address is the network address when the master location is 0 after the network bit is exceeded, that is, 172.16.10.0/24;

The multicast address range is 224.0.0.0 ~ 239.00000000255, excluding C;

The broadcast address is 172.16.10.255, and D is excluded when the host bit is set to 1 after the network bit is exceeded. The answer is B.

23. The IP address of a subnet is 10.32.0.0 and the mask is 255.224.0.0. The maximum host address allowed is ()
A. 10.32.254.254;
B. 10.32.0000254;
C. 10.63.0000254;
D. 10.63.0000255.
Analysis: The subnet mask 255.224.0.0 shows that it has 11-bit network bits and 21-bit host bits. The maximum host address of the network segment is when the host bits are all 1 and then minus 1. The host space is 10.63.0000255, and 10.63.0000254 is subtracted from 1. The answer is C.

This article is from the eagleeye blog, please be sure to keep this source http://eagle2014.blog.51cto.com/7992740/1439252