Shulou(Shulou.com)06/02 Report--

The boss gives a network segment:

192.168.55.0 / 24, host address: 256; available address: 254

192.168.55.0000 0000 / 30 Murray-> Host IP address: 8 Available address: 6192.168.55.0000 1000 / 29-> Host IP address: 8192.168.55.0001 0000 / 29-> Host IP address: 8192.168.55.0001 1000 / 29-> Host IP address: 8192.168.55.0010 0000 / 29-> Host IP address: 8192.168.55.0010 1000 / 29-> Host IP address: 8192.168.55.0011 0000 / 29-- > Host IP address: 8192.168.55.0011 1000 / 29-> Host IP address: 8192.168.55.0000 0000 / 29-> Host IP address: 8192.168.55.0000 0000 / 29-- -- > Host IP address: 8192.168.55.0000 0000 / 29-> Host IP address: 8192.168.55.0000 0000 / 29-- > Host IP address: 8 available address: 329.6 = 192

Network address: host bits are all 0

Represents an area and does not represent any hosts / devices

Broadcast address: host bits are all 1

Represents "all" hosts / devices in an area

Note:

Later, when describing and configuring the IP address, you must use it with the subnet mask.

The more subnets are divided, the more address waste is likely to occur.

The address of the given network segment: 192.168.55.0 Universe 24

If you change the subnet mask to 30

So how many subnets will be generated?

How many host IP addresses are there in each subnet?

How many host addresses are available per subnet?

Write down the network address and broadcast address of the first five network segments

64 subnets; 42 network segment 1, 192.168.55.0Universe 30; 192.168.55.3Universe segment 2,192.168.55.4Universe 30; 192.168.55.7 Universe 30 subnet mask:-function: distinguish between network bits and host bits in the IP address;-structure: always 1 on the left, 0 on the right; and never cross. 255.255.64.01111111111111111.01000000.00000000-> wildcard bits (wildcard)-IP: network bit + host bit; host bit-N,-> number of hosts to the N power of 2 XRV 0123456789 10 to the power 1 = XX:00 10 11 19 99 10 to the power 2 = 10 "10 XXX: 10 to the power 3 = 10" 10 "10 Y 1 2 to the power of YY:00, 01, 10, 11 2 to the power 2 = YYY: 2 to the third power of 2 = 192.168.100.0 hip 24-> 256254 available 2 = > department: 3, indicating that there are 3 different network segments; 1 = > max:50 hosts / each department minimizes the number of wasted IP addresses in each network segment; determines the address space of the three network segments; calculation ideas: 1. First pay attention to the number of hosts in each network segment, so as to determine the number of host bits in each network segment. The N power of 2 is greater than or equal to 502 N > = 62. Based on the number of host bits, the number of "network bits" of each new network segment is determined, and the "subnet mask" used by each network segment is determined. M=32-N=32-6 subnet 26, that is, the subnet mask is 255.255.255.192 or / 263. Based on the number of subnet masks of the new network segment and the original subnet mask, determine the number of "subnet bits", thus determining the number of subnets. Number of subnet bits = number of subnet masks of the new network-number of subnet masks of the old network 26-24 = 2, so the number of subnets is: 2 to the power of 2 Subnet address: 192.168.100.00 0000Universe 26 = "192.168.100.0Universe 26, network address 192.168.100.63lap26 192.168.100.01 000000lap26 =" 192.168.100.64Compact 26 The network address is 192.168.100.127ax 26 192.168.100.100000swap 26 = "192.168.100.128Universe 26, the network address 192.168.100.191Compare 26 192.168.100.11 000000Compact 26 =" 192.168.100.192 Universe 26. Network address 192.168.100.255Universe 26

Case 2:

Given IP address range: 192.168.65.0 Universe 24

5 departments

22 mainframes per department

Minimize the number of wasted IP addresses in each department

Determine how much IP address space is used by each of the five departments?

1. The number of host bits in the new network segment is: 5

2. The number of network bits in the new network segment is 27.

255.255.255.224

3. The subnet bit is: 3, so the number of subnets is: 8

List the address space for each subnet separately:

192.168.65.000 00000swap 27 = "192.168.65.0 Universe 27, network address

192.168.65.31Universe 27, broadcast address

192.168.65.001 00000ax 27 = "192.168.65.32 Universe 27"

192.168.65.63/27

192.168.65.010000Universe 27 = "192.168.65.64Universe 27"

192.168.65.95/27

192.168.65.011 00000Universe 27 = "192.168.65.96Universe 27"

192.168.65.127/27

192.168.65.10000000Universe 27 = "192.168.65.128Universe 27"

192.168.65.159/27

192.168.65.101 00000Universe 27 = "192.168.65.160Universe 27

192.168.65.191/27

192.168.65.11000000Universe 27 = "192.168.65.192 gamma 27"

192.168.65.223/27

192.168.65.111 00000Universe 27 = "192.168.65.224Universe 27

192.168.65.255/27

VLSM:

Virable length subnet mask-variable length subnet mask

=

Case 2 in PPT:

192.168.22.0/24

When subnetting an original network segment, if the number of hosts per subnet

If it is different, then pay attention to the subnet with the largest number of hosts and calculate it first:

Vlan 2: 90 available hosts

The n power of 2 is greater than or equal to 90 square 2; so n > = 7, so the final N is 7; therefore, the number of network bits of the new network segment is: 25. Therefore, if the subnet mask is 255.255.255.128, then the number of subnets is 2, which is subnet 1: 192.168.22.0ax 25, subnet 2: 192.168.22.128x25, so we can assign subnet 1 to VLAN 2. Then, we should continue subnetting "subnet 2", which should meet the requirements of VLAN 1: the number of hosts is 50, then the n power of 2 is greater than or equal to 50 square 2; so n > = 6, so the final N is 6; therefore, the number of network bits in the new network segment is: 26 Therefore, the subnet mask is 255.255.255.192, then the number of subnets is 2, namely: subnet 2-1: 192.168.22.128Universe 26 subnet 2-2: 192.168.22.192 Universe 26, so we can change the subnet 2-1. Assigned to VLAN 1 for use.

So, next we should continue subnetting "subnet 2-2", which should satisfy VLAN3/4.

Requirement: the number of hosts is 50, so:

The n-th power of 2 is greater than or equal to 15. 2

The n-th power of 2 is greater than or equal to 26 times 2.

So n > = 5, so the final N is 5.

Therefore, the number of network bits in the new network segment is: 27

Therefore, the subnet mask is 255.255.255.224

Then, the number of subnets is 2, which are:

Subnet 2-2-1: 192.168.22.192 xanth27

Subnet 2-2-2: 192.168.22.224Universe 27

So, we can assign subnet 2-2-1 to VLAN 3.

So, we can assign subnet 2-2-2 to VLAN 4.

VLAN2: 192.168.22.0Universe 25-Network address

192.168.22.1 Compact 25-> PC2 address

192.168.22.126 Compact 25-> PC2 Gateway

192.168.22.127Compact 25-broadcast address

VLAN1: 192.168.22.128swap 26-Network address

192.168.22.129Universe 26-> PC1 address

192.168.22.190amp 26-> PC1 Gateway

192.168.22.191Compact 26-broadcast address

VLAN3: 192.168.22.192 Universe 27-Network address

192.168.22.1931 PC3 27-> address

192.168.22.222Universe 27-> PC3 gateway

192.168.22.223Universe 27-broadcast address

VLAN4: 192.168.22.224swap 27-Network address

192.168.22.225Universe 27-> PC4 address

192.168.22.254Compact 27-> PC4 Gateway

192.168.22.255Compact 27-broadcast address

IP address Summary:

In fact, the essence is the reverse process of "subnetting".

In the process of determining the summary address, we should:

Find the same bit of multiple IP addresses (from left to right)

The same bits, unchanged, written directly, represent the network bits of the summarized IP address.

Different bits, directly represented by 0, represent the host bits of the summarized IP address.

The length of the same bit, indicating the length of the subnet mask of the summarized IP address

Case study:

Case 1: class C address division

1.1 question

The class C address 192.168.55.0 take 24 is divided using the subnet mask / 29 and / 30, respectively. What is the subnet and host range after the division?

1.2 steps

To implement this case, you need to follow these steps.

Step 1: class C address 192.168.55.0 Universe 24 is divided using the subnet mask / 29

Step 2: class C address 192.168.55.0 Universe 24 is divided using the subnet mask / 30

Case 2: subnetting and VLAN configuration

2.1 question

The network segment of the company is 192.168.220.There are 50 mainframes in the VLAN1 of the production department and 90 VLAN2 in the sales department.

There are 15 VLAN3 in Finance Department and 26 VLAN4 in customer Service Department, which requires subnetting to realize network interworking.

2.2 scenario

Use eNSP to build the experimental environment, as shown in figure-1.

2.3 steps

To implement this case, you need to follow these steps.

Step 1: subnetting

The subnetting result is shown in figure-2:

Step 2: VLAN configuration

[SW3] dis cu

#

Sysname SW3

#

Vlan batch 2 to 4

#

Cluster enable

Ntdp enable

Ndp enable

#

Drop illegal-mac alarm

#

Diffserv domain default

#

Drop-profile default

#

Aaa

Authentication-scheme default

Authorization-scheme default

Accounting-scheme default

Domain default

Domain default_admin

Local-user admin password simple admin

Local-user admin service-type http

#

Interface Vlanif1

Ip address 192.168.22.190 255.255.255.192

#

Interface Vlanif2

Ip address 192.168.22.126 255.255.255.128

#

Interface Vlanif3

Ip address 192.168.22.222 255.255.255.224

#

Interface Vlanif4

Ip address 192.168.22.254 255.255.255.224

#

Interface MEth0/0/1

#

Interface GigabitEthernet0/0/1

Port link-type trunk

Port trunk allow-pass vlan 2 to 4094

#

Interface GigabitEthernet0/0/2

Port link-type trunk

Port trunk allow-pass vlan 2 to 4094

#

Interface GigabitEthernet0/0/3

#

Interface GigabitEthernet0/0/4

#

Interface GigabitEthernet0/0/5

#

Interface GigabitEthernet0/0/6

#

Interface GigabitEthernet0/0/7

#

Interface GigabitEthernet0/0/8

#

Interface GigabitEthernet0/0/9

#

Interface GigabitEthernet0/0/10

#

Interface GigabitEthernet0/0/11

#

Interface GigabitEthernet0/0/12

#

Interface GigabitEthernet0/0/13

#

Interface GigabitEthernet0/0/14

#

Interface GigabitEthernet0/0/15

#

Interface GigabitEthernet0/0/16

#

Interface GigabitEthernet0/0/17

#

Interface GigabitEthernet0/0/18

#

Interface GigabitEthernet0/0/19

#

Interface GigabitEthernet0/0/20

#

Interface GigabitEthernet0/0/21

#

Interface GigabitEthernet0/0/22

#

Interface GigabitEthernet0/0/23

#

Interface GigabitEthernet0/0/24

#

Interface NULL0

#

User-interface con 0

User-interface vty 0 4

#

Return

[SW1] dis cu

#

Sysname SW1

#

Vlan batch 2 to 4

#

Cluster enable

Ntdp enable

Ndp enable

#

Drop illegal-mac alarm

#

Diffserv domain default

#

Drop-profile default

#

Aaa

Authentication-scheme default

Authorization-scheme default

Accounting-scheme default

Domain default

Domain default_admin

Local-user admin password simple admin

Local-user admin service-type http

#

Interface Vlanif1

#

Interface MEth0/0/1

#

Interface Ethernet0/0/1

Port link-type access

#

Interface Ethernet0/0/2

Port link-type access

Port default vlan 2

#

Interface Ethernet0/0/3

#

Interface Ethernet0/0/4

#

Interface Ethernet0/0/5

#

Interface Ethernet0/0/6

#

Interface Ethernet0/0/7

#

Interface Ethernet0/0/8

#

Interface Ethernet0/0/9

#

Interface Ethernet0/0/10

#

Interface Ethernet0/0/11

#

Interface Ethernet0/0/12

#

Interface Ethernet0/0/13

#

Interface Ethernet0/0/14

#

Interface Ethernet0/0/15

#

Interface Ethernet0/0/16

#

Interface Ethernet0/0/17

#

Interface Ethernet0/0/18

#

Interface Ethernet0/0/19

#

Interface Ethernet0/0/20

#

Interface Ethernet0/0/21

#

Interface Ethernet0/0/22

#

Interface GigabitEthernet0/0/1

Port link-type trunk

Port trunk allow-pass vlan 2 to 4094

#

Interface GigabitEthernet0/0/2

#

Interface NULL0

#

User-interface con 0

User-interface vty 0 4

#

Return

[SW2] dis cu

#

Sysname SW2

#

Vlan batch 2 to 4

#

Cluster enable

Ntdp enable

Ndp enable

#

Drop illegal-mac alarm

#

Diffserv domain default

#

Drop-profile default

#

Aaa

Authentication-scheme default

Authorization-scheme default

Accounting-scheme default

Domain default

Domain default_admin

Local-user admin password simple admin

Local-user admin service-type http

#

Interface Vlanif1

#

Interface MEth0/0/1

#

Interface Ethernet0/0/1

Port link-type access

Port default vlan 3

#

Interface Ethernet0/0/2

Port link-type access

Port default vlan 4

#

Interface Ethernet0/0/3

#

Interface Ethernet0/0/4

#

Interface Ethernet0/0/5

#

Interface Ethernet0/0/6

#

Interface Ethernet0/0/7

#

Interface Ethernet0/0/8

#

Interface Ethernet0/0/9

#

Interface Ethernet0/0/10

#

Interface Ethernet0/0/11

#

Interface Ethernet0/0/12

#

Interface Ethernet0/0/13

#

Interface Ethernet0/0/14

#

Interface Ethernet0/0/15

#

Interface Ethernet0/0/16

#

Interface Ethernet0/0/17

#

Interface Ethernet0/0/18

#

Interface Ethernet0/0/19

#

Interface Ethernet0/0/20

#

Interface Ethernet0/0/21

#

Interface Ethernet0/0/22

#

Interface GigabitEthernet0/0/1

Port link-type trunk

Port trunk allow-pass vlan 2 to 4094

#

Interface GigabitEthernet0/0/2

#

Interface NULL0

#

User-interface con 0

User-interface vty 0 4

#

Return

The VLAN 1 host configuration is shown in figure-3:

The VLAN 2 host configuration is shown in figure-4:

The VLAN 3 host configuration is shown in figure-5:

The VLAN 4 host configuration is shown in figure-6:

Verify that all four hosts can ping each other.

3 case 3:IP address summary

3.1 question

Summarize the following IP address field into one address field, requiring the summary address to contain exactly all addresses, that is, requiring the longest subnet mask of the summary address field.

1) 10.16.5.0, 10.16.7.0, 10.16.8.0, and 10.16.8.0

2) 192.168.5.8Compact 29, 192.168.5.16swap 28

3.2 steps

To implement this case, you need to follow these steps.

1) 10.16.0.0 Universe 20

2) 192.168.0.0 Universe 27

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