reecriture de subnetting part 3
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shango-wk
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# Subnetting (Part 3 - VLSM)
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# Subnetting (Part 3 - VLSM)
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## Class B Reference Table
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class B Refference table
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| Prefix | Subnets | Hosts |
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| Prefix | Subnets | Hosts |
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| ------ | ------- | ----- |
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|--------|---------|-------|
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| /17 | 2 | 32766 |
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| /17 | 2 | 32766 |
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| /18 | 4 | 16382 |
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| /18 | 4 | 16382 |
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| /19 | 8 | 8190 |
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| /19 | 8 | 8190 |
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@@ -23,10 +22,10 @@ class B Refference table
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---
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---
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class C Refference table
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## Class C Reference Table
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| Prefix | Subnets | Hosts |
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| Prefix | Subnets | Hosts |
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| ------ | ------- | ----- |
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|--------|---------|-------|
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| /25 | 2 | 126 |
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| /25 | 2 | 126 |
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| /26 | 4 | 62 |
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| /26 | 4 | 62 |
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| /27 | 8 | 30 |
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| /27 | 8 | 30 |
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@@ -34,134 +33,129 @@ class C Refference table
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| /29 | 32 | 6 |
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| /29 | 32 | 6 |
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| /30 | 64 | 2 |
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| /30 | 64 | 2 |
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| /31 | | (0) |
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| /31 | | (0) |
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| /32 | | (0)-2 |
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| /32 | | (0–2) |
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## QUIZ Question 1
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---
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You have been given the 172.30.0.0/16. your company requires
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## Quiz Question 1
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100 subnets with at least 500 hosts per subnet. What prefix
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length should you use?
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I should use 172.168.0.0/23
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You have been given the network **172.30.0.0/16**. Your company requires
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because it compatible with 100 subnet and 500 hosts
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**100 subnets with at least 500 hosts per subnet**.
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subnet mask -> 255.255.254.0
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**Answer:**
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- Prefix: `/23`
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- Network: `172.30.0.0/23`
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- Subnet mask: `255.255.254.0`
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---
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## Quiz Question 2
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## Quiz Question 2
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What subnet does host 172.21.111.201/20 belong to ?
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What subnet does host **172.21.111.201/20** belong to?
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10101100.00010101.*0110*1111.11001001
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```
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172.21.111.201
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10101100.00010101.*0110*0000.00000000
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172.21.96.0
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## Quiz question 3
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10101100.00010101.01101111.11001001 (IP)
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10101100.00010101.01100000.00000000 (Network)
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What is the broadcast address of the network 192.168.91.78/26
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```
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11000000.10101000.10110011.*01*001110
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**Answer:** `172.21.96.0/20`
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192.168.91.78
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11000000.10101000.10110011.*01*111111
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192.168.91.127
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## Quiz question 4
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---
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You divide the 172.16.0.0/16 network into 4 subnets of equal size.
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## Quiz Question 3
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Identify the network and broadcast addresses of the second subnet
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Subnet 1 :
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What is the broadcast address of the network **192.168.91.78/26**?
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Network address
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172.16.0.0/18
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10101100.00011010.*00*000000.00000000
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broadcast:
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172.16.0.0/18
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10101100.00011010.*00*111111.00000000
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172.16.63.255/18
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Subnet 2 :
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```
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Network address
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172.16.64.0/18
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10101100.00011010.*01*000000.00000000
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broadcast:
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172.16.65.0/18
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10101100.00011010.*01*111111.00000000
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172.16.127.255/18
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Subnet 3 :
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11000000.10101000.10110011.01001110 (IP)
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Network address
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11000000.10101000.10110011.01111111 (Broadcast)
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172.16.128.0/18
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10101100.00011010.*10*000000.00000000
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broadcast:
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172.16.128.0/18
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10101100.00011010.*10*111111.00000000
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172.16.191.255/18
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Subnet 4 :
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```
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Network address
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172.16.192.0/18
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10101100.00011010.*11*000000.00000000
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broadcast:
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172.16.192.0
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10101100.00011010.*11*111111.00000000
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172.16.255.255/18
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## Quiz question 5
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**Answer:** `192.168.91.127`
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You divide the 172.30.0.0/16 network into subnets of 1000 hosts each. How many subnets
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are you able to make
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I'm able to make 64 subnets
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## Quiz Question 4
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## Subentting class A Networks
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You divide the **172.16.0.0/16** network into 4 equal subnets.
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Identify the network and broadcast addresses.
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The process of subnetting Class A, Class B and Class C networks is Exactly the same !
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### Subnet 1
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- Network: `172.16.0.0/18`
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- Broadcast: `172.16.63.255`
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### Example 2
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### Subnet 2
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- Network: `172.16.64.0/18`
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- Broadcast: `172.16.127.255`
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You hace been given the 10.0.0.0/8 network. You must create 2000 subnets which
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### Subnet 3
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will be distributed to various enterprises.
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- Network: `172.16.128.0/18`
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What prefix length must you use ?
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- Broadcast: `172.16.191.255`
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How many host addresses usable addresses will be in each subnet?
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10.0.0.0
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### Subnet 4
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00001010.00000000.0000000.0000000
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- Network: `172.16.192.0/18`
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2, 4, 8, 16,32, 64,128,256, 512, 1024, 2048
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- Broadcast: `172.16.255.255`
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00001010.*00000000.000*00000.0000000
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Answer:
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10.0.0.0/19
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8192 - 2 = 8190 hosts
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### Example 2
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## Quiz Question 5
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PC1 has an IP address of 10.217.182.223/11.
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You divide the **172.30.0.0/16** network into subnets of **1000 hosts each**.
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Identify the following for PC1's subnet:
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**Answer:** 64 subnets
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1) Network address:10.192.0.0/11
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2) Broadcast address: 10.223.255.255
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3) First usable address: 10.192.0.1
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3) Last usable address:10.223.255.254
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3) Nuber of host (usable) address: 2,097,150
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10.217.182.223
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---
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00001010.*110*11001.10110110.11011111
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## [[VLSM]]
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## Subnetting Class A Networks
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Variable-Length Subnet Masks
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The process of subnetting Class A, B, and C networks is exactly the same.
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- Until now, we have practiced subentting used FLSM (Fixed-Length Subnet Masks).
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- This means that all of the subents use the same prefix length (ie. subnetting a class C
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into 4 subenets using /26).
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- VLSM (Variable-Length Subnet Masks) is the process of creating subnets of different sizes
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to make your use of network addresses more eficient
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- VLSM is more complicated than FLSM, but it's easy if you folow the steps correctly.
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### Exaple 1
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## Example 1
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We must divide 192.168.1.0/24 must divided between 5 subnets
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You are given the network **10.0.0.0/8**.
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You must create **2000 subnets**.
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### Solution
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**Answer:**
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- Prefix: `/19`
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- Usable hosts: `8192 - 2 = 8190`
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---
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## Example 2
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PC1 has IP address **10.217.182.223/11**
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### Subnet Details
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- Network: `10.192.0.0/11`
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- Broadcast: `10.223.255.255`
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- First usable: `10.192.0.1`
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- Last usable: `10.223.255.254`
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- Usable hosts: `2,097,150`
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---
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## VLSM (Variable-Length Subnet Masks)
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- Previously, we used **FLSM (Fixed-Length Subnet Masks)**
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- All subnets had the same prefix length
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- VLSM allows **different subnet sizes** for better efficiency
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- More flexible, slightly more complex
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---
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## Example (VLSM Design)
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```
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Tokyo Lan A = 110 Hosts --- Tokyo Lan B = 8 Hosts
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Tokyo Lan A = 110 Hosts --- Tokyo Lan B = 8 Hosts
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Router
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Router
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@@ -169,61 +163,72 @@ We must divide 192.168.1.0/24 must divided between 5 subnets
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Router
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Router
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Toronto Lan A = 29 Hosts --- Toronto Lan B = 45 Hosts
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Toronto Lan A = 29 Hosts --- Toronto Lan B = 45 Hosts
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```
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#### VLSM Steps
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We must divide **192.168.1.0/24** into 5 subnets:
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1) Assign the largest subnet at the start of the address space.
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- Tokyo LAN A → 110 hosts
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2) Assign the second- largest subnet after it.
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- Tokyo LAN B → 8 hosts
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3) Repeat the process until all subnets have been assigned
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- Toronto LAN A → 29 hosts
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- Toronto LAN B → 45 hosts
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- Point-to-point link
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we will do this in this order
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### Steps
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Tokyo LAN A -> Toronto LAN B -> Toronto LAN A -> Tokyo LAN B -> Point to point
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1. Assign largest subnet first
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2. Continue in descending order
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3. Repeat until done
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##### Tokyo LAN A
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### Allocation Order
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Network address: 192.168.1.0/25
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Tokyo A → Toronto B → Toronto A → Tokyo B → Point-to-point
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Broadcast address: 192.168.1.127/25
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First usable address: 192.168.1.1/25
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Last usable address: 192.168.1.126/25
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Total Number of usable host address: 126
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##### Toronto LAN B
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if we add 1 to the Tokyo Broadcast address we get the next Network address
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Network address: 192.168.1.128/26
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### Tokyo LAN A
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Broadcast address: 192.168.1.191/26
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- Network: `192.168.1.0/25`
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First usable address: 192.168.1.129/26
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- Broadcast: `192.168.1.127`
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Last usable address: 192.168.1.190/26
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- Usable: `192.168.1.1 – 192.168.1.126`
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Total Number of usable host address: 62
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- Hosts: 126
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##### Toronto LAN B
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Network address: 192.168.1.192/27
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### Toronto LAN B
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Broadcast address: 192.168.1.223/27
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- Network: `192.168.1.128/26`
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First usable address:192.168.1.193/27
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- Broadcast: `192.168.1.191`
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Last usable address:192.168.1.222/27
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- Usable: `192.168.1.129 – 192.168.1.190`
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Total Number of usable host address:30
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- Hosts: 62
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##### Tokyo LAN A
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Network address: 192.168.1.224/28
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### Toronto LAN A
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Broadcast address: 192.168.1.239/28
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- Network: `192.168.1.192/27`
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First usable address: 192.168.1.225
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- Broadcast: `192.168.1.223`
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Last usable address: 192.168.1.239
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- Usable: `192.168.1.193 – 192.168.1.222`
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Total Number of usable host address: 14
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- Hosts: 30
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##### Point to Point
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Network address: 192.168.1.240/30
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### Tokyo LAN B
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Broadcast address: 192.168.1.243/30
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- Network: `192.168.1.224/28`
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First usable address: 192.168.1.241/30
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- Broadcast: `192.168.1.239`
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Last usable address: 192.168.1.242/3
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- Usable: `192.168.1.225 – 192.168.1.238`
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Total Number of usable host address: 2
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- Hosts: 14
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---
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### Point-to-Point
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- Network: `192.168.1.240/30`
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- Broadcast: `192.168.1.243`
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- Usable: `192.168.1.241 – 192.168.1.242`
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- Hosts: 2
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---
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## Additional Practice
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## Additional Practice
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- https://www.subnettingquestions.com
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- https://www.subnettingquestions.com
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- https://www.subnetting.org
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- https://www.subnetting.org
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- https://www.subnettingpractice.com
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- https://www.subnettingpractice.com
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```
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