Networking Classroom Series- 01/Feb/2020

Networking Basics

Basic Rules

  • A system will be able to communicate with other system directly if it is in the same network
  • Every system in the network has to have unique id.

Concepts/Terms to be understood

  • TCP
  • HTTP
  • HTTPS
  • ICMP
  • UDP
  • Subnetting
  • CIDR
  • DHCP Server
  • DNS Server
  • DNS Records (CNAME/C-Record and ANAME)
  • Proxy Server
  • Layer – 4 load balancing
  • Layer -7 Load balancing
  • Reverse Proxy
  • Subnets
  • Routing
  • Gateway
  • NAT Servers
  • NAT/Bridge Connections
  • VPN

Network

  • Collection of Systems (hosts)
  • Every network is identified by network id

Note

  • Packets can be sent from one network to other network using router.

IP Address

  • Uniquely identify a host/system on the network
  • IP Address = Network id + host id Preview
  • now enter the following
ipconfig

IPv4 Address. . . . . . . . . . . : 192.168.0.117
Subnet Mask . . . . . . . . . . . : 255.255.255.0
Default Gateway . . . . . . . . . : fe80::f68c:ebff:feab:37ff%20
                                    192.168.0.1


# linux
ifconfig
  • Subnet Mask:

    • Helps in finding network id and host id
    • Also helps in determining size of the network.
  • Default Gateway: ipaddress of your router

  • IP Formats

    • IpV4
    • IPV6

IPV4

  • IPV4 is a 4 byte (32 bit) representation
  • Example:
192.168.0.10
10.40.100.9
77.89.34.8
  • 1 byte = 8 bits => each bit => 0 or 1
    • => Total combinations => 2^8
    • Number range in byte => 0 to 255

Size of network

  • We need ip address and subnet mask
ip: 192.168.10.11
sm: 255.255.255.0

nid: 192.168.10
hid: 11
size of hid: 8 bits => 2^8-1-1 => 254


ip: 172.16.0.8
sm: 255.255.0.0


nid: 172.16
hid: 0.8
size of network => 2^16-2

ip: 10.0.0.7
sm: 255.0.0.0

nid: 10
hid: 0.0.7
size of network => 2^24-2
  • Size of network: 2^n-2
    • n => number of bits for host id
    • one ip address (all zeros) cannot be used bcoz of network id and other ip (all 1’s) bcoz of broadcast ip

Scenario-1: Create a network which should connect 200 devices

  • Subnet mask will be 255.255.255.0 ~> 254

Scenario-2: Create a network which should connect 400 devices

  • Subnet mask will be 255.255.0.0 ~> 65000

Solution for ip address

  • Go for Class Based Ranges. Refer Here
    • Class A
    • Class B
    • Class C
    • Class D
    • Class E Preview
  • CIDR:
    • Lets relook how we calculated network sizes
ip: 192.168.10.11
sm: 255.255.255.0
* SM is considered as decimal number
ip: 192.168.10.11
sm: 255.255.255.0

nid: 192.168.10
hid: 11
size = 2^8-2
* consider SM as binary, all sequence of 1's is network id and all sequence of zero's is host id
ip: 192.168.10.10
SM: 11111111.11111111.11111111.00000000
bits of zeros => 8 => 2^8-2 = 254

ip: 192.168.10.10
SM: 11111111.11111111.11111110.00000000
bits of zeros => 9 => 2^9-2 ~ 500
  • Each subnet mask has following combinations
11111111 => 255
11111110 => 254
11111100 => 252
11111000 => 248
11110000 => 240
11100000 => 224
11000000 => 192
10000000 => 128
  • Same Subnet masks can be represented in CIDR format
ip: 192.168.0.0
sm: 255.255.254.0
sm: 11111111.11111111.11111110.00000000

cidr => 192.168.0.0/23
  • another example
ip 10.0.0.8
sm: 255.255.128.0
sm: 11111111.11111111.10000000.00000000

10.0.0.8/17

Scenario-1: Create a network which should connect 200 devices (Revisited)

  • CIDR => 192.168.0.0/24

Scenario-2: Create a network which should connect 400 devices (Revisited)

  • CIDR => 192.168.0.0/23

Scenario -3 Create a network which should connect to 600 devices

  • Try to find a least value of x so that
(2)^x >= 600

value of x is 10
cidr => 192.168.0.0/22

Scenario -4 Find a cidr which can connect to 2000 devices

2^x >= 2000

x=> 11

cidr => 192.168.0.0/21

CIDR Ranges reserved for private usage

  • 10.0.0.0/8 => 10.0.0.0 – 10.255.255.255
  • 172.16.0.0/12 => 172.16.0.0 – 172.31.255.255
  • 192.168.0.0/16 => 192.168.0.0 – 192.168.255.255

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