1

Router Basics and�IP Addressing

CSC499: Special Topic in Computer Networks

Rangsit University

Different Network Standard

• Different Physical Layer
• Voltage/Current
• Signal/Coding
• Connector
• Etc.

• Different Data-Link Layer
• Framing
• Addressing
• MAC
• Etc.

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A

A

B

1

2

2

?

B

Routers

• Creates a single virtual network on top of different kinds of data link Network.

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Network

Network

Network

1.1

A

1.2

B

2.2

1

2.1

2

B

2.1

2

2.1

2.1

2.1

2.1

To 1.1, 1.2 → Filter (Blocked)

To 2.1,2.2 → Switch to port 2

Broadcast Domain

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CD2

CD3

CD4

Switch

MAC Broadcast Address = ffff.ffff.ffff

CD1

Too large Broadcast Domain cause problems

• Performance
• Security
• etc

Switch

Router used to Interconnect BD

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CD2

CD3

CD4

Switch

CD1

BD1

BD2

BD3

Router

IP header details

• Source Address - 32 bit IP source address
• Destination Address - 32 bit IP destination address
• 32 bit address: “dotted-decimal” notation, e.g.,

10011110. 01101100.00000010. 01000111

158. 108. 2. 71

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IP Addressing

• An IP address is partitioned into 2 fields
• Network address (used to determined path)
• Host address (used to specified a device on the net)

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1.0.0.2

1.0.0.3

2.0.0.2

2.0.0.3

Port1

1.0.0.1

Port2

2.0.0.1

Port3

3.0.0.1

network Host ID

1. 0.0.1

0.0.2

0.0.3

2. 0.0.1

3. 0.0.1

• Each interface has its own IP Address
• Machine with >1 Interface, called multi-homed

• Router is multi-homed machine
• Multi-homed need not to be router

Network-Number (Prefix)

Host

32 bits

Routing Table and Default

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1. local (Port 1)

2. local (Port 2)

Default local (Port 3)

Dest. Next

network hop

1. local

Default 1.0.0.1

Dest. Next

network hop

2. local

Default 2.0.0.1

Dest. Next

network hop

Destination 2.2

Destination 1.3

IP with the same Network Number is local can be accessed directly.

1.0.0.2

1.0.0.3

2.0.0.2

2.0.0.3

Port1

1.0.0.1

Port2

2.0.0.1

Port3

3.0.0.1

IP Address

• IP address associated with interface (not machine)
• A Point-to-Point Link is also a Network !!!

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1.0.0.2

1.0.0.3

2.0.0.2

2.0.0.3

4.0.0.2

4.0.0.3

1.0.0.1

2.0.0.1

3.0.0.1

3.0.0.2

4.0.0.1

L2/L3 Addressing Concepts

• IP addresses must be translated in to MAC addresses to be used in Layer 2 (MAC layer)

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Network

IP:1.0.0.2

MAC:0000:aaaa:2222

1.0.0.3

MAC:0000:bbbb:3333

0000:bbbb:3333

1.0.0.3

Network

1.0.0.3

1.0.0.3

?

L2/L3 Addressing Concepts

• IP addresses must be translated in to MAC addresses to be used in Layer 2 (MAC layer)

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IP Header data

Datagram

1.0.0.2

1.0.0.3

IP address

Source →OK

Destination → ?

0000:aaaa:2222

0000:bbbb:3333

MAC address

MAC Header

ARP (Address Resolution Protocol)

• Host IP=1.0.0.2 want to resolve MAC address of IP=1.0.0.3
• 1.0.0.2 send broadcast ARP request
• get unicast ARP reply from 1.0.0.3

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whose IP is

1.0.0.3 ?

No, not me!

Ignore

1.0.0.2

1.0.0.4

Port1

1.0.0.1

1.0.0.3

Number of Hosts

• In network “ 1.” there are 2²⁴ different host IDs

1.0.0.0

1.0.0.1

…

1.255.255.255

• Too large?

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1.0.0.2

1.0.0.4

Port1

1.0.0.1

1.0.0.3

Classful IP Addressing: Default Prefixes

• Use the first 3 bits to identify the length of Network Prefix.
• 32 bits address length, contains 2 parts
• Network Identifier (Net Prefix)
• Host Identifier (Host ID)

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0 Net Prefix Host ID

class A

class B

class C

class D

class E

10 Net Prefix Host ID

110 Net Prefix Host ID

1110 Multicast address

1111 Research

0 8 16 24 32

Default Network Prefixes

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Net

Class prefix start bit host Range

class A /8 0 24 0.0.0.0 ~ 127.255.255.255

class B /16 10 16 128.0.0.0 ~ 191.255.255.255

class C /24 110 8 192.0.0.0 ~ 223.255.255.255

class D /32 1110 - 224.0.0.0 ~ 239.255.255.255

class E /32 11110 - 240.0.0.0 ~ 247.255.255.255

0 Net Prefix Host ID

class A

class B

class C

class D

class E

10 Net Prefix Host ID

110 Net Prefix Host ID

1110 Multicast address

1111 Research

0 8 16 24 32

Network Number

• IP address with “all 0s” host ID is reserved to refer to a network number
• 10.0.0.0
• 172.16.0.0
• 192.168.1.0

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192.168.1.2

192.168.1.3

192.168.1.1

Net work Number

192.168.1.0

172.16.0.2

Net work Number

172.16.0.0

10.0.0.1

Net work Number

10.0.0.0

172.16.0.3

172.16.0.1

Default Network Prefix: Route Selection

Routing Table

Network prefix Next

10.0.0.0 8 s0

172.16.0.0 16 s0

192.169.16.0 24 s1

203.148.240.0 24 f0

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110 01011

10010100

11110000

01100100

24 bits

s0

s1

f0

D:203.148.240.101

Broadcast Address

• IP address with “all 1s” host ID is reserved to refer to broadcast to all hosts in the network
• 10.255.255.255
• 172.16.255.255
• 192.168.1.255

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192.168.1.2

192.168.1.3

192.168.1.1

Broadcast address

192.168.1.255

172.16.0.2

Broadcast address

172.16.255.255

10.0.0.1

Broadcast address

10.255.255.255

172.16.0.3

172.16.0.1

Network Number and Broadcast Domain

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CD2

CD3

CD4

Switch

CD1

BD1

BD2

BD3

Router

192.168.3.0

192.168.3.255

192.168.1.0

192.168.1.255

192.168.2.0

192.168.2.255

• You have to specify network in which to be broadcast.
• That is, routers can be used to separate different BDs.

Subnet mask

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What is subnet mask?

• หมายเลขที่ใช้ในการระบุว่า IP Address ใดๆมีส่วนของ Network ID เป็นอย่างไร โดยการระบุหมายเลข Network ID จะมีการกำหนดหมายเลขของ subnet mask ให้มีค่าของ bit เป็น 1
• ส่วน bit ที่เป็น 0 ในหมายเลข subnet mask จะแสดงถึงส่วนที่เป็น Host ID
• และเมื่อนำ IP Address มาทำการ AND กับ Subnet mask จะได้ Network Number

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เช่น Class A ; subnet mask = /8 หรือเขียนได้เป็น 255.0.0.0

Class B ; subnet mask = /16 หรือเขียนได้เป็น 255.255.0.0

Class C ; subnet mask = /24 หรือเขียนได้เป็น 255.255.255.0

Class A : 10.0.0.0

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00001010.00000000.00000000.00000000

11111111.00000000.00000000.00000000

255 . 0 . 0 . 0

10 . 0 . 0 . 0

00001010.00000000.00000000.00000000

10 . 0 . 0 . 0

Class A : 10.192.0.0

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00001010.11000000.00000000.00000000

11111111.00000000.00000000.00000000

255 . 0 . 0 . 0

10 . 192 . 0 . 0

00001010.00000000.00000000.00000000

10 . 0 . 0 . 0

ดังนั้น10.0.0.0 และ 10.192.0.0 จึงอยู่ในเน็ตเวิร์คเดียวกัน

• 10.0.1.0 /8
• 10.123.5.0 /8
• 10.222.0.176 /8
• 10.?.?.? /8

​

​

• subnet mask จึงเป็นหมายเลขที่ใช้ในการดูว่า ip address ใดๆ อยู่ในเน็ตเวิร์คเดียวกันหรือไม่

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Class B : 172.16.0.0

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10101100.00010000.00000000.00000000

11111111.11111111.00000000.00000000

255 . 255 . 0 . 0

172 . 16 . 0 . 0

10101100.00010000.00000000.00000000

172 . 16 . 0 . 0

Class B : 172.16.15.255

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10101100.00010000.00001111.11111111

11111111.11111111.00000000.00000000

255 . 255 . 0 . 0

172 . 16 . 15 . 255

10101100.00010000.00000000.00000000

172 . 16 . 0 . 0

ดังนั้น172.16.0.0 และ 172.16.15.0 จึงอยู่ในเน็ตเวิร์คเดียวกัน

• 172.16.1.0 /16
• 172.16.157.25 /16
• 172.16.228.176 /16
• 172.16.?.? /16

​

​

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Class C : 192.177.143.0

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11000000.10110001.10001111.00000000

11111111.11111111.11111111.00000000

255 . 255 . 255 . 0

192 . 177 . 143 . 0

11000000.10110001.10001111.00000000

192 . 177 . 143 . 0

Class C : 192.177.143.240

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11000000.10110001.10001111.11110000

11111111.11111111.11111111.00000000

255 . 255 . 255 . 0

192 . 177 . 143 . 0

11000000.10110001.10001111.00000000

192 . 177 . 143 . 0

ดังนั้น192.177.143.0 และ 192.177.143.240 จึงอยู่ในเน็ตเวิร์คเดียวกัน

• 192.177.143.10 /24
• 192.177.143.25 /24
• 192.177.143.176 /24
• 192.177.143.? /24

​

​

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1

IP Addressing�Subnetting

CSC499: Special Topic in Computer Networks

Rangsit University

พิจารณา Network address space

Class Host ID address spaces usable

A 24 2²⁴ =16,677,216 16,677,214

B 16 2¹⁶ = 65,536 65,534

C 8 2⁸ = 256 254

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0 Net Prefix Host ID

class A

class B

class C

10 Net Prefix Host ID

110 Net Prefix Host ID

0 8 16 24 32

Class A และ B จึงมีหมายเลข IP สำหรับใช้งาน จำนวนมากไม่เหมาะแก่การดูแลและจัดการ จึงควรมีการแบ่งเน็ตเวิร์ค Class A, B หรือแม้แต่ C ให้มีขนาดที่เล็กลง ( subnetting )

Q: Give the example of Class A, B, C Network Number?

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• Class A (0.0.0.0 ~ 127.255.255.255)
• 1.0.0.0
• 54.0.0.0
• 121.0.0.0
• Class B (128.0.0.0 ~ 191.255.255.255)
• 132.52.0.0
• 171.9.0.0
• 178.235.0.0
• Class C (192.0.0.0 ~ 223.255.255.255)
• 192.168.2.0
• 203.148.5.0
• 222.234.225.0

Problems with Classful 2-level hierarchy

• Class B “Flat network” more than 65,000 hosts
• How to manage?
• Single Admin.?
• Who-Where?
• Performance?
• Broadcast
• Class B “subdivided network” to smaller groups with router ?

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How to assign subnet

• RFC 950 define standard to divide a single Classes A, B or C host ID into smaller pieces.

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Class B address such as 158.108.0.0/16 might use its third byte to identify subnet within the class, e.g.,

• subnet #1 158.108.0.X x=host id range from 1-254
• subnet #2 158.108.1.X

…

• subnet #2 158.108.255.X

Network ID Subnet address host address

host ID

choose

appropriate size

Network ID host address

Subnet Mask

• 32 bit number, tells router to recognize the subnet field, call subnet mask

subnet rule:

“The bit covering the network ID and subnet part of an address are set to 1”

• Example: class B with 8 bit subnet mask

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1111 1111 1111 1111 1111 1111 0000 0000

network mask= 255.255.255.0

/24

Extended Net-prefix

Class B 16 bits+

subnet 8 bits

Network 158.108.0.0 /16

158.108.1.0 /24

🡪 158.108.2.0 /24

158.108.3.0 /24

Subnet mask bits

• Subnet Mask don’t limit to multiple number byte

#mask

bits 128 64 32 16 8 4 2 1

1 1 0 0 0 0 0 0 0 = 128 (2 subnets)

2 1 1 0 0 0 0 0 0 = 192 (4 subnets)

3 1 1 1 0 0 0 0 0 = 224 (8 subnets)

4 1 1 1 1 0 0 0 0 = 240 (16 subnets)

5 1 1 1 1 1 0 0 0 = 248 (32 subnets)

6 1 1 1 1 1 1 0 0 = 252 (64 subnets)

7 1 1 1 1 1 1 1 0 = 254 (128 subnets)

8 1 1 1 1 1 1 1 1 = 255 (256 subnets)

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Subnet Numbers

For a class C (/24) with 3 subnet bit

extended-network prefix : /27 bit (มาจาก 24+3)

subnet mask: 255.255.255.224

e.g. 193.1.1.0 / 24 , means

Base net:1100 0001.0000 0001.0000 0001.0000 0000 = 193.1.1.0/24

is partitioned into 2³ = 8 subnets

Subnet #0:1100 0001.0000 0001.0000 0001.0000 0000 = 193.1.1.0/27

Subnet #1:1100 0001.0000 0001.0000 0001.0010 0000 = 193.1.1.32/27

Subnet #2:1100 0001.0000 0001.0000 0001.0100 0000 = 193.1.1.64/27

Subnet #3:1100 0001.0000 0001.0000 0001.0110 0000 = 193.1.1.96/27

Subnet #4:1100 0001.0000 0001.0000 0001.1000 0000 = 193.1.1.128/27

Subnet #5:1100 0001.0000 0001.0000 0001.1010 0000 = 193.1.1.160/27

Subnet #6:1100 0001.0000 0001.0000 0001.1100 0000 = 193.1.1.192/27

Subnet #7:1100 0001.0000 0001.0000 0001.1110 0000 = 193.1.1.224/27

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Number of Hosts in each Subnet

• Consider subnet #1

1100 0001.0000 0001.0000 0001.0010 0000 = 193.1.1.32/27

we have 2⁵=32 IP addresses

1100 0001.0000 0001.0000 0001.0010 0000 = 193.1.1.32

1100 0001.0000 0001.0000 0001.0010 0001 = 193.1.1.33

…

1100 0001.0000 0001.0000 0001.0011 1111 = 193.1.1.63

But,

193.1.1.32 (all 0s) is reserved to refer to this network number

193.1.1.63 (all 1s) is reserved to broadcast on this network

The maximum number of hosts can be assigned is

2⁵-2=30.

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Example 1: Class C of No. 203.108.1.0

• Total 3 Subnets
• max. 60 hosts are need

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Department A max. 60 hosts

Department B

…

…

WAN-Link

Example 1: Class C of No. 203.108.1.0

• Subnet Mask 1 bit - 2 Sunbets
• Subnet Mask 2 bit - 4 Subnets
• Net prefix /26 (class C 24 bits + subnet 2 bits)

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64 IP No.

64 IP No.

64 IP No.

64 IP No.

256 IP No.

203.108.1.0 /26

.00 00 0000

203.108.1.64 /26

.01 00 0000

203.108.1.128 /26

.10 00 0000

203.108.1.192 /26

.11 00 0000

Example 1: Class C of No. 203.108.1.0

• Total 3 Subnets with max. 60 hosts are need

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203.108.1.65 / 26

Department A

Sunet #1: 203.108.1.64

Department B

Subnet #2: 203.108.1.128

203.108.1.66 / 26

203.108.1.67 / 26

203.108.1.126 / 26

…

203.108.1.130 / 26

203.108.1.131 / 26

203.108.1.190 / 26

…

203.108.1.129 / 26

WAN-Link

Subnet#0

203.108.1.0

203.108.1.1 / 26

203.108.1.2 / 26

Given a C class 203.148.240.0

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Network 1

subnet # 1

link 1

subnet # 6

Network 5 subnets

Links 3 subnets

Total 8 subnets

Network 2

subnet # 2

Network 3

subnet # 3

Network 4

subnet # 4

Network 5

subnet # 5

link 2

subnet # 7

link 3

subnet # 8

Solutions

8 subnets are required

• 1 bits mask ->2¹ = 2 subnets
• 2 bits mask ->2² = 4 subnets
• 3 bits mask ->2³ = 8 subnets
• 4 bits mask ->2⁴ = 16 subnets

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Solution 1: 3-Bit Subnet Mask

• Prefix = / 27 (24+3)
• net mask 255.255.255.224
• Class C contain 256 IP addresses
• Each Subnet contain

256/8 - 2= 30 IP addresses (hosts)

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Subnet # 1

203.148.240.0

Subnet # 2

203.148.240.32

32 IP #

Subnet # 3

203.148.240.64

Subnet # 8

203.148.240.224

Subnet # 4

203.148.240.96

Subnet # 5

203.148.240.128

Subnet # 6

203.148.240.160

Subnet # 7

203.148.240.192

Solution 1:3 Bit Subnet Mask

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Subnet # 6

203.148.240.192 / 27

Subnet # 7

203.148.240.160 / 27

Subnet # 8

203.148.240.224 / 27

Subnet # 1

203.148.240.0 / 27

Subnet # 2

203.148.240.32 / 27

Subnet # 3

203.148.240.64 / 27

Subnet # 4

203.148.240.96 / 27

Subnet # 5

203.148.240.128 / 27

Problems

• No spare subnet
• Each link, subnets 6 ~ 8,

require max. 2 IP

waste 30 x 3 = 90 IPs.

• 30 hosts too large for some

nets, too small for some nets

Solution 2: 4-Bit Subnet Mask

• 16 Subnet
• Prefix = / 28 (24+4)
• net mask 255.255.255.240
• Class C contain 256 IP addresses
• Max. host on each subnet

= 256/16 - 2= 14 hosts

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Subnet # 1

203.148.240.0

Subnet # 2

203.148.240.16

16 IP #

Subnet # 3

203.148.240.32

Subnet # 16

203.148.240.240

Subnet # 14

203.148.240.208

Subnet # 15

203.148.240.224

.

.

.

.

.

.

Subnet Number Assignment

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Subnet # 6

203.148.240.128 / 28

Subnet # 7

203.148.240.112 / 28

Subnet # 8

203.148.240.144 / 28

Subnet # 1

203.148.240.16 / 28

Subnet # 2

203.148.240.32 / 28

Subnet # 3

203.148.240.48 / 28

Subnet # 4

203.148.240.64 / 28

Subnet # 5

203.148.240.80 / 28

Problems

• There are 8 unused subnets

waste 128 IPs

• Each link, subnets 6 ~ 8,

require max. 2 IP

waste 36 IPs.

• Subnet 1~ 5, may require more than 14 IPs.