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Typical problems of subnet division and super network construction

2022-07-07 01:20:00 【CUMT Deyi】

Subnet division of classified address ：

1. Known one C The class network address is 192.189.25.0, Now it is divided into different subnets , The requirement is ： The number of hosts in each subnet should not exceed 25 platform , At least how many subnets can be divided ？ The subnet mask is ？

Explain ：

$2^{4}< 25< 2^{5}$

The host number part of the subnet only needs 5 position , Subnet number needs 8-5=3 position , Can be divided into 6 Subnet ( If 000 and 111 no way , The topic says at least ） Subnet mask ： 255.255.255.224

2. A unit applied for a C Class network address ：200.165.68.0, Due to business needs, the internal must be divided into 5 Separate subnets , The number of hosts owned by each subnet is 24、28、16、16、20 platform , Please use subnet partition , Establish these five subnets , Write the network address of each subnet 、 You can use IP Address range 、 Direct broadcast address and subnet mask .

Explain ：

The topic requires division 5 Subnet , First verify whether it can be divided into equal lengths , Determine the division method ：

$2^{2}< 5< 2^{3}$

borrow 3 It's enough to give the subnet number , Then the host number can occupy 5 position , The maximum number of hosts that can be provided is $2^{5}-2$ =30,30 More than the number of hosts required in each subnet , Therefore, equal length division can be carried out .

It is easy to determine that the subnet mask is ：255.255.255.224

subnet	Subnet network address	You can use IP Address range ( whole 0,1 no need ）	Direct address
1	200.165.68.00100000(32）	200.165.68.33~200.165.68.62	200.165.68.63
2	200.165.68.01000000(64）	200.165.68.65~200.165.68.94	200.165.68.95
3	200.165.68.01100000(96）	200.165.68.97~200.165.68.126	200.165.68.127
4	200.165.68.10000000(128）	200.165.68.129~200.165.68.158	200.165.68.159
5	200.165.68.10100000(160）	200.165.68.161~200.165.68.190	200.165.68.191
6	200.165.68.11000000(192）	200.165.68.193~200.165.68.222	200.165.68.223

3. Known one C The class network address is 192.189.25.0, Now press RFC950 It is stipulated to divide it into multiple subnets , requirement ：
(1) The number of hosts in each subnet should not exceed 25 platform ;
(2) Address utilization is the highest . Please write the subnet mask and reason of the subnet partition scheme ？

Explain ： This is a C Class network , The default subnet mask is 255.255.255.M If you want to accommodate 25 Console host , be

$2^{{4}}-2< 25< 2^{5}-2$

Therefore, the subnet host number should be left at least 5 Only bits can , It can be 5 position 、6 position 、7 Bit or 8 position . Press RFC950 Regulations , The subnet number is all 0 And all 1 Not available , So the longer the subnet number , The smaller the variation range of the number of hosts , The less host addresses are wasted . therefore , In the case of the highest address utilization , The subnet number is 3 position , You can get $2^{3}-2$ =6 Available subnets , 192.189.25.xxx00000 Subnet mask ： 255.255.255.224

4. A unit applied for a C Class network address ：200.165.68.0, Due to business needs, the internal must be divided into 5 Separate subnets , The number of hosts owned by each subnet is 24、28、16、5、14 platform , Please follow RFC1878 The regulations of are divided into equal length subnet and variable length subnet , Establish these five subnets , Write the network address of each subnet 、 You can use IP Address range 、 Broadcast address and subnet mask .

Explain ：

（1) Wait for the long subnet

To divide 5 Subnet , be $2^{2}< 5< 2^{3}$ , Need to take 3 Location net number , The number of available hosts in each subnet is $2^{5}-2$ =30 platform ( The number of hosts should always be reduced 2）, The maximum number of hosts in the subnetwork of the topic can be 28 The requirements of , So the subnet mask is 255.255.255.224

subnet	Subnet network address	You can use IP Address range	Broadcast address
0	200.165.68.0	200.165.68.1~200.165.68.30	200.165.68.31
1	200.165.68.32	200.165.68.33~200.165.68.62	200.165.68.63
2	200.165.68.64	200.165.68.65~200.165.68.94	200.165.68.95
3	200.165.68.96	200.165.68.97~200.165.68.126	200.165.68.127
4	200.165.68.128	200.165.68.129~200.165.68.158	200.165.68.159
5	200.165.68.160	200.165.68.161~200.165.68.190	200.165.68.191
6	200.165.68.192	200.165.68.193~200.165.68.222	200.165.68.223
7	200.165.68.224	200.165.68.225~200.165.68.254	200.165.68.255

（2） Variable length subnet

Divide the molecular network from high to low according to the number of hosts in the subnet , And consider that each subnet needs a routing interface IP Address , be

So the host number 5 position , Subnet number 3 position Comply with RFC1878 The provisions of the , whole 0 And all 1 Subnet available

available 8 Subnet ：

Pay attention to the difference between the following questions and the above questions , Begin to enter CIDR Subnet division under Technology ：

1. Find the network address block 212.110.96.0/20 The maximum number of hosts included , as well as 8 Wait for the molecular network , Mask and number of hosts of each subnet .

Non classification system is adopted CIDR, slash notation , Introduced “ network prefix ” The concept .

Address block ： 212.110.01100000.0 /20 （ front 20 Bits make up the network prefix ）

Minimum address 212.110.01100000.00000000 Indicates the address block

Maximum address 212.110.01101111.11111111 Address block broadcast address

Maximum number of hosts $2^{32-20}-2=2^{12}-2$

Because 8 Isomolecular network , $8=2^{3}$ , So you need to borrow it from the host location 3 Bit to network prefix , It becomes 23 position

212.110.01100000.00000000 /23

212.110.01100010.00000000 /23

212.110.01100100.00000000 /23

212.110.01100110.00000000 /23

212.110.01101000.00000000 /23

212.110.01101010.00000000 /23

212.110.01101100.00000000 /23

212.110.01101110.00000000 /23

Subnet mask /23 or 255.255.254.0 Host number $2^{9}-2$ = 510

2. A company has 3 subsidiaries , The number of hosts in each subsidiary is 120、62、23. hold 202.16.134.0/23 The network segment of is allocated to each subsidiary . It is required to allocate from the minimum address of the address block , Give the allocation with the highest address utilization programme , Include each address block assigned 、 Minimum address 、 Maximum address and mask , List the remaining address blocks . Ask for a solution Question process , The address is expressed in dotted decimal .

Explain ：

CIDR The subnet division under the address is the division of address blocks , As long as it is in accordance with 2 To divide , And each address block can meet the requirements of the number of hosts , It can be divided many times , And there is no need to exclude that the subnet number is all 0 And all 1 Of situation .

202.16.134.0/23 In binary 202.16.10000110.00000000/23

To improve address utilization , Start with the smallest address , First assign address blocks to the maximum number of hosts

The first division ：

$2^{6}$ – 2 = 62 < 120+1 < $2^{7}$ -2 = 126

The host number needs 7 position , The prefix is 25 position , So put the address block 202.16.134.0/23 Divide into four 202.16.10000110.00000000/25 【202.16.134.0/25】 Assigned to 120 A subsidiary of the host

The remaining address block is ：

202.16.10000110.10000000/25 【202.16.134.128/25】

202.16.10000111.00000000/25 【202.16.135.0/25】

202.16.10000111.10000000/25 【202.16.135.128/25】

The second division ：

$2^{6}$ -2 = 62 < 62+1 < $2^{7}$ -2 = 128 The host number needs 7 position , The prefix is 25 position

202.16.10000110.10000000/25 【202.16.134.128/25】 Assigned to 62 A subsidiary of the host

The remaining address block is

202.16.10000111.00000000/25 【202.16.135.0/25】

202.16.10000111.10000000/25 【202.16.135.128/25】

The third division ：

$2^{4}$ -2 = 14 < 23+1 < $2^{5}$ -2 = 30 The host number needs 5 position , The prefix is 27 position , So put the address block 202.16.135.0/25 Divide into four

202.16.10000111.00000000/27 【202.16.135.0/27】 Assigned to 23 A subsidiary of the host

202.16.10000111.00100000/27 【202.16.135.32/27】

202.16.10000111.01000000/27 【202.16.135.64/27】

202.16.10000111.01100000/27 【202.16.135.96/27】 The last two address blocks can be merged

Final address allocation scheme

Assigned to 120 Address block of host ：

202.16.10000110.00000000/25 【202.16.134.0/25】

Minimum address ： 202.16.134.0 Maximum address ： 202.16.134.127 Mask ： /25 perhaps 255.255.255.128

Assigned to 62 Address block of host ：

202.16.10000110.10000000/25 【202.16.134.128/25】

Minimum address ： 202.16.134.128 Maximum address ： 202.16.134.255 Mask ： /25 perhaps 255.255.255.128

Assigned to 23 A subsidiary of the host

202.16.10000111.00000000/27 【202.16.135.0/27】

Minimum address ： 202.16.135.0 Maximum address ： 202.16.135.31 Mask ： /27 perhaps 255.255.255.224

The remaining address block is

202.16.10000111.00100000/27 【202.16.135.32/27】

202.16.10000111.01000000/26 【202.16.135.64/26】

202.16.10000111.10000000/25 【202.16.135.128/25】

3. There are the following 4 individual /24 Address block , Try to polymerize as much as possible .
212.56.132.0/24
212.56.133.0/24
212.56.134.0/24
212.56.135.0/24

Explain ：

Route aggregation , Also known as the construction of hypernets .

212.56.132.0/24 212.56.10000100

212.56.133.0/24 212.56.10000101

212.56.134.0/24 212.56.10000110

212.56.135.0/24 212.56.10000111

Before the third byte 6 The bits are the same , Only the last two are different , therefore 4 The common prefix of addresses is before 22 position , namely ： 212.56.100001 Most likely aggregated CIDR Address block is ： 212.56.132.0/22

4. Assigned to an autonomous system IP The address is fast 30.138.118/23, Yes 5 A LAN , The connection diagram is shown in the figure below , The number of hosts in each LAN is shown on the figure , Try to give the address block of each LAN （ Including the prefix ）.

Explain ：

LAN1 Need at least 3 individual IP The address is assigned to three routers

LAN2 need 91 Host computer +1 A router interface =92 An address

LAN3 need 150 Host computer +1 A router interface =151 An address

LAN4 need 3 Host computer +1 A router interface =4 An address

LAN5 need 15 Host computer +1 A router interface =16 An address The answer is not unique

30.138.118/23 118=64+32+16+4+2 Address block range

Minimum address 30.138.01110110.00000000 The Internet

….

Maximum address 30.138.01110111.11111111 Broadcast address

LAN3 Need at least 151 An address $2^{7}$ -2 < 151 < $2^{8}$ -2 Host location is 8 position , Prefix 24 position

LAN3 Assigned address block ： 30.138.118/24 here ,30.138.118/23 Half of the address block has been allocated

Minimum address 30.138.01110110.00000000 The Internet

....

Maximum address 30.138.01110110.11111111 Broadcast address

The range of the remaining address block 30.138.119.0/24

Minimum address 30.138.01110111.00000000

….

Maximum address 30.138.01110111.11111111

LAN2 Need at least 92 An address $2^{6}$ - 2 < 92 < $2^{7}$ -2 Host location is 7 position , Prefix 25 position

LAN2 The assigned address block is ： 30.138.119.0/25

Minimum address 30.138.01110111.00000000 The Internet

….

Maximum address 30.138.01110111.01111111 Broadcast address

here ,30.138.119.0/24 The address block is left 1/2 30.138.118/23 Only 1/4

The range of the remaining address block 30.138.119.128/25

Minimum address 30.138.01110111.10000000

….

Maximum address 30.138.01110111.11111111

LAN5 Need at least 16 An address $2^{4}$ - 2 < 16 < $2^{5}$ -2 Host location is 5 position , Prefix 27 position

LAN5 The assigned address block is ： 30.138.119.128/27

Minimum address 30.138.01110111.10000000 The Internet

….

Maximum address 30.138.01110111.10011111 Broadcast address

The range of the remaining address block A contiguous address block consisting of two different prefixes

30.138.119.160/27

Minimum address 30.138.01110111.10100000

….

Maximum address 30.138.01110111.10111111

30.138.119.192/26

Minimum address 30.138.01110111.11000000

….

Maximum address 30.138.01110111.11111111

LAN4 Need at least 4 An address $2^{2}$ - 2 < 4 < $2^{3}$ -2 Host location is 3 position , Prefix 29 position

Select address block 30.138.119.160/27

Minimum address 30.138.01110111.10100000

….

Maximum address 30.138.01110111.10111111

LAN4 The assigned address block is ：30.138.119.160/29

Minimum address 30.138.01110111.10100000 The Internet

….

Maximum address 30.138.01110111.10100111 Broadcast address

The range of the remaining address block A continuous address block consisting of three different prefixes

30.138.119.168/29

Minimum address 30.138.01110111.10101000

….

Maximum address 30.138.01110111.10101111

30.138.119.176/28

Minimum address 30.138.01110111.10110000

….

Maximum address 30.138.01110111.10111111

30.138.119.192/26

Minimum address 30.138.01110111.11000000

….

Maximum address 30.138.01110111.11111111

LAN1 Need at least 3 An address $2^{2}$ - 2 < 3 < $2^{3}$ -2 Host location is 3 position , Prefix 29 position

Select the appropriate address block

30.138.119.168/29

Minimum address 30.138.01110111.10101000 The Internet

….

Maximum address 30.138.01110111.11111111 Broadcast address

Just right for LAN1 Address block size requirements LAN1 The assigned address block is ： 30.138.119.168/29

The range of the remaining address block A contiguous address block consisting of two different prefixes

30.138.119.176/28

Minimum address 30.138.01110111.10110000

….

Maximum address 30.138.01110111.10111111

30.138.119.192/26

Minimum address 30.138.01110111.11000000

….

Maximum address 30.138.01110111.11111111

CIDR When dividing address blocks , Give priority to dividing large address blocks ; When dividing Small address block , Try to divide from both sides of the address block inward .

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