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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 :
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 :
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 =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
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 =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 , Need to take 3 Location net number , The number of available hosts in each subnet is
=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
Because 8 Isomolecular network ,, 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 = 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 = 62 < 120+1 <
-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 = 62 < 62+1 <
-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 = 14 < 23+1 <
-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 < 151 <
-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 < 92 <
-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 < 16 <
-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 < 4 <
-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 < 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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