Bab 1. Perencanaan Desain Jaringan Komputer fileDi dalam jaringan TCP/IP setiap terminal...

IP Address

Fakultas Rekayasa Industri

Institut Teknologi Telkom

Objectives

In this lesson, you will learn about:

A history of the TCP/IP protocol suite

Understand about the tcp/ip address

Type of ip address

Converting from Binary to Decimal

Converting from Decimal to Binary

TCP/IP class

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IP Addressing: introduction TCP/IP was originally developed to enable ARPANET

sites to communicate. ARPANET sites used different computers manufactured

by different vendors, running different operating systems.

The Transmission Control Protocol/Internet Protocol (TCP/IP) protocol suite is the foundation of today's Internet and the foundation of many private computer networks.

To successfully administer and troubleshoot IP internetworks, it is important to understand all aspects of IP addressing.

One of the most important aspects of TCP/IP network administration is the assignment of unique and proper IP addresses to all the nodes of an IP internetwork.

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Types of IP Addresses

An IP address is a 32-bit logical address that can be one of the following types:

Unicast A unicast IP address is assigned to a single network interface attached to an IP internetwork. Unicast IP addresses are used in one-to-one communications.

Broadcast A broadcast IP address is designed to be processed by every IP node on the same network segment. Broadcast IP addresses are used in one-to-everyone communications.

Multicast An IP multicast address is an address on which one or multiple nodes can be listening on the same or different network segments. IP multicast addresses are used in one-to-many communications.

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Expressing IP Addresses

IP address: 32-bit identifier for host,router interface .

The 32-bit IP address is divided from the high-order bit to the low-order bit into four 8-bit quantities called octets.

IP addresses are normally written as four separate decimal octets delimited by a period (a dot). This is known as dotted decimal notation.

For example, the IP address 00001010000000011111000101000011 is subdivided into four octets:

00001010 00000001 11110001 01000011

Each octet is converted to a base 10 number and separated from the others by periods:

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Pengalamatan IP

Di dalam jaringan TCP/IP setiap

terminal diidentifikasi dengan sebuah

alamat IP unik.

Kecuali Router dapat memiliki lebih

dari sebuah alamat IP, karena itu

disebut sebagai Multihomed Device.

TCP/IP

Ilustrasi Pengalamatan IP

Source: www.tcpipguide.com

Badan Internasional

Pengelola IP

Di Asia Pasific pengelolaan IP dilakukan

oleh Asia Pacific Network Information

Center (APNIC).

APNIC bertugas sebagai pembagi blok

nomor IP dan nomor Autonomous System

(AS) kepada para ISP di kawasan Asia

Pasific, selain itu juga mengelola

authoritative resgistration server (whois)

dan reverse domains (in-addr.arpa).

Badan Internasional

Pengelola IP

Selain APNIC badan-badan lain yang bertugas melakukan manajemen IP iniantara lain :

- America Rregistry for Internet Number

(ARIN)

- Reseaux IP Europeens (RIPE)

- African Regional Internet Registry Network

Information Center (AFRINIC)

Koordinasi Internasional dari ke-empatbadan tersebut dipegang oleh International Assigned Number Authority (IANA).

Converting from Binary to Decimal

To convert a binary number to its decimal equivalent, add the numbers represented by the bit positions that are set to 1.

.

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Decimal and Binary Conversion

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Binernya berapa?

Desimalnya berapa?

Konversi Biner –

HexaDesimal - Biner

Angka Hexadesimal mengandung: 0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,F

Contoh:

11000011.10001101

C 3 . 8 D

Note: Format HexaDesimal dipakai untuk pengalamatan IPv6.

Contoh:

Source: www.tcpipguide.com

Unicast IP Addresses

Each network interface on which TCP/IP is active must be identified by a unique, logical, unicast IP address.

The unicast IP address is an internetwork address for IP nodes that contains a network ID and a host ID.

The network ID, or network address, identifies the nodes that are located on the same logical network.

The host ID, or host address, identifies a node within a network. A node is a router or host (a nonrouter interface such as a workstation, server, or other TCP/IP-based system). The host ID must be unique within each network segment.

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IP broadcast

IP broadcast addresses are used for single-packet one-to-every one delivery

IP broadcast addresses can be used only as the destination IP address.

The IP network broadcast address is the address formed by setting all the host bits to 1 for a classful address

example of a network broadcast address for the classfulnetwork ID 131.107.0.0/16 is ?

IP routers do not forward network broadcast packets.

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Kategori Pengalamatan IP

Ada 3 macam kategori pengalamatan

IP, yaitu:

- Classfull Addressing (conventional):

pengalamatan berdasarkan kelas,

tanpa perlu ada subnetting.

- Subnetted Classfull Addressing:

pengalamatan dengan subnetting.

- Classless Addressing: CIDR

Class A, B, C, D, and E IP Addresses

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Rules for Enumerating Network IDs

When enumerating IP network IDs, the following rules apply:

The network ID cannot begin with 127 as the first octet. All 127.x.y.z addresses are reserved as loopback addresses.

All the bits in the network ID cannot be set to 1. Network IDs set to all 1s are reserved for broadcast addresses.

All the bits in the network ID cannot be set to 0. Network IDs set to all 0s are reserved for indicating a host on the local network.

The network ID must be unique to the IP internetwork.

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Address Class Ranges of Network IDs

Address

Class

First

Network ID

Last Network

ID

Number of

Networks

Class A 1.0.0.0 126.0.0.0 126

Class B 128.0.0.0 191.255.0.0 16,384

Class C 192.0.0.0 223.255.255.0 2,097,152

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Rules for Enumerating Host IDs

When enumerating IP host IDs, the following rules apply:

All bits in the host ID cannot be set to 1. Host IDs set to all 1s are reserved for broadcast addresses.

All the bits in the host ID cannot be set to 0. Host IDs set to all 0s are reserved for the expression of IP network IDs.

The host ID must be unique to the network.

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Subnet Mask Address

Determines which part of an IP address is the network field and which part is the host field.

Follow these steps to determine the subnet mask:

1. Express the subnetwork IP address in binary form.

2. Replace the network and subnet portion of the address with all 1s.

3. Replace the host portion of the address with all 0s.

4. Convert the binary expression back to dotted-decimal notation.

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Why do we need Subnets

A network with 16,777,214 host addresses (Class A) , or even one with 65,534 (Class B), is likely to be unwieldy

Class C network with 254 addresses may well be undesirably large for many organizations.

As a result of

traffic patterns and congestion

upper limits on the number of allowable nodes in a network

distance limitations on LANs,

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Class InterDomain Routing (CIDR)

Was developed to allow routing and manipulation of smaller IP blocks that previously possible with natural subnets

Particularly useful for sub netting Class C network Blocks

Was one means for trying to conserve IP space

Supported by almost all modern network devices

RFCs 1517–1520 introduced CIDR.

Many organization have > 256 computers but few have more than several thousand

Instead of giving class B (16384 nets) give sufficient contiguous class C addresses to satisfy needs

< 256 addresses assign 1 class C

…

< 8192 addresses assign 32 contiguous Class C nets

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CIDR and Standard Notation

/16 – 255.255.0.0 /17 – 255.255.128.0 /18 – 255.255.192.0 /19 – 255.255.224.0 /20 – 255.255.240.0 /21 – 255.255.248.0 /22 – 255.255.252.0 /23 – 255.255.254.0 /24 – 255.255.255.0 /25 - 255.255.255.128 /26- 255.255.255.192 /27- 255.255.255.224 /28- 255.255.255.240 /29 - 255.255.255.248 /30 - 255.255.255.252 /31 - 255.255.255.254 /32 - 255.255.255.255

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Broadcast Addressing

Address: 10010010.11100111.01111 011.00001111

Netmask: 11111111.11111111.11111 000.00000000

Network: 10010010.11100111.01111 000.00000000

Broadcast: 10010010.11100111.01111 111.11111111

Broadcast address is the address used tocommunicate with all hosts on the local network.

Broadcast address is defined as the highest valuethat is on a network

Calculate by replacing all the host address portionbits with 1s

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Perbandingn antara IPv4 dan

IP version 6

28 Computer Network - Industrial Engineering -

Faculty of Industrial Engineering

Background

IPv4’s addresses is only 232

IPv4 has minimum security mechanism

IPv4 has no Quality of Service

IPv4 did not support IP roaming

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Advantage of IPv6 over IPv4

IPv6 has 2128 address space

IPv6 simplify the header easier for

router to process

IPv6 has more security features

IPv6 has quality of service

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IPv6 Addressing Scheme

16 bytes address

Written in 8 group of four hexadecimal

notation separated by colon

8000:0000:0000:0000:0123:4567:89AB:CDEF

First Leading Zero can be ommited

8000::0123:4567:89AB:CDEF

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IPv4 vs IPv6 Header

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Addressing

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Global Unicast Address

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IPv6 Address Allocation

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IPv4 and IPv6 Internetworking

many ways to do internetworking, 2 of

them are:

Tunnel IPv6 in IPv4 network

Address translation NAT PT

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Tunneling IPv6 in IPv4

`

IPv6

Network

IPv6

Network

IPv4

Network

PacketIPv6

HeaderPacket

IPv6

Header

IPv4

HeaderPacket

IPv6

Header

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Address Translation

`

IPv6

Network

IPv6

Network

IPv4

Network

PacketIPv6

Header

PacketIPv4

Header

PacketIPv6

Header

IPv4

Network

PacketIPv4

Header

PacketIPv4

Header38

Subnetting &

Supernetting

Mengapa SubNetting?

SubNetting adalah proses membagi

sebuah network menjadi beberapa

Sub-network.

Sebagai contoh, dalam sebuah

jaringan lokal yang menggunakan

alamat kelas B 172.16.0.0 terdapat

65.534 host address.

Efisiensi pengelolaan jaringan dapat

ditingkatkan dengan cara melakukan

subnetting terhadap network tersebut.

Mengapa SubNetting (Cont.)

Alasan-alasan perlunya dibentuk subnetting antara lain :

- Memudahkan pengelolaan jaringan.

- Mereduksi traffic yang disebabkan oleh broadcast maupun benturan (collision).

- Membantu pengembangan jaringan ke jarak geografis yang lebih jauh (LAN ke MAN).

Subnetting

The increasing number of host connected to the internet

Restrictions on the network size

In subnetting, a network is divided into smaller subnetworks with each subnet having its own subnet address

In supernetting, a organization can combine several class C to create a large range of addresses. In other word, several networks are combined to create a supernetwork

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Without subnetting

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With Subnettiing

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Masking

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Applying bit-wise-and operation to achieve masking

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Ilustrasi sebuah Network

tanpa Subnet

SubNetting

Pembentukan subnet dilakukan

dengan cara mengambil beberapa bit

pada bagian HostId untuk dijadikan

SubnetId. Contoh:

Source: www.tcpipguide.com

Subnet Mask

Source: www.tcpipguide.com

Subnet Mask (Cont.)

Dalam contoh di atas, sebuah

jaringan kelas B dengan Network-Id :

154.71.0.0.

Subnet Mask dalam bentuk desimal

adalah: 255.255.248.0

Dengan demikian 5 bit pertama pada

octet ke 3 adalah Subnet-Id,

sedangkan sisa bit adalah Host-Id.

Default Subnet-Mask

Konversi Subnet-Mask

1 0 0 0 0 0 0 0 = 128

1 1 0 0 0 0 0 0 = 192

1 1 1 0 0 0 0 0 = 224

1 1 1 1 0 0 0 0 = 240

1 1 1 1 1 0 0 0 = 248

1 1 1 1 1 1 0 0 = 252

1 1 1 1 1 1 1 0 = 254

1 1 1 1 1 1 1 1 = 255

Menentukan SubNet-Id

Source: www.tcpipguide.com

Menentukan Subnet-Id

Router menentukan sebuah IP address merupakan anggota dari subnet tertentu melalui proses masking seperti dalam gambar di atas.

IP address: 154.71.150.42 dioperasikan AND dengan subnet-mask. Didapat Subnet-Id: 18.

Sedangkan IP address dari subnet tersebut adalah: 154.71.144.0.

IP Address dari Subnet

Determining the Subnet ID of an IP Address

Through Subnet Masking

Component Octet 1 Octet 2 Octet 3 Octet 4

IP Address10011010

(154)

01000111

(71)

10010110

(150)

00101010

(42)

Subnet Mask11111111

(255)

11111111

(255)

11111000

(248)

00000000

(0)

Result of AND

Masking

10011010

(154)

01000111

(71)

10010000

(144)

00000000

(0)

Dengan CIDR, dapat dituliskan sebagai:

154.71.150.42/21.

Contoh Kasus 1

Sebuah jaringan dengan network-id: 192.16.9.0 akan dibagi ke dalam 3 buah subnet. Tentukan IP address untuk setiap subnet.

No IP 192.16.9.0 adalah Kelas C, dengan host-Id berada pada 8 bit terakhir. Karena itu, subnet-id harus berada pada 8 bit terakhir.

Penyelesaian Kasus 1

Kebutuhan 3 subnet berarti

membutuhkan sebanyak 3 bit.

Karena itu subnet-mask ditentukan:

11111111.11111111.11111111.11100000

255. 255. 255. 224

Penyelesaian Kasus 1

Kombinasi subnet: 000, 001, 010, 011, 100, 101, 110, 111.

Karena itu 3 bit pertama dialokasikan untuk subnet.

192.16.9.b b b b b b b b

subnet

Penyelesaian Kasus 1:

Subnet Host Decimal

000 00000 - 11111 0-31

001 00000 – 11111 32 – 63

010 00000 – 11111 64 – 95

011 00000 – 11111 96 - 127

100 00000 – 11111 128 - 159

101 00000 – 11111 160 – 191

110 00000 – 11111 192 – 223

111 00000 - 11111 224 - 255

Kesimpulan Kasus 1

Jumlah subnet yang terbentuk ada 23=8. Tetapi subnet 000 dan 111 tidak dapat digunakan. Karena itu jumlah subnet yang dapat digunakan adalah: (23-2=6).

Jumlah host yang terbentuk untuk masing-masing subnet 25=32. Sedang host yang dapat digunakan sebanyak 25-2=30. Host-Id: 00000 dan 11111 tidak dapat digunakan.

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Addressing example

The example given in the curriculum

shows subnetting without VLSM using

172.16.0.0/22. (172.16.0.0 –

172.16.3.255)

They produce 4 subnets each with

510 addresses.

This is impossible. It will be

corrected.

You can do it if you start with

172.16.0.0/21 (172.16.0.0 –

172.16.7.255)

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Addressing example no

VLSM

172.16.0.0/21

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What we have and need

Given IP address 172.16.0.0/21

That’s 172.16.0.0 to 172.16.7.255

4 subnets needed:

Student LAN has 481 hosts

Instructor LAN has 69 hosts

Administrator LAN has 23 hosts

WAN has 2 hosts

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Without VLSM – same size subnets

Biggest subnet has 481 hosts.

Formula for hosts is 2n – 2

n = 9 gives 510 hosts (n = 8 gives

only 254)

So 9 host bits needed.

That means 32 – 9 = 23 network bits

/23 or subnet mask 255.255.254.065

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Network addresses

/23 so subnet mask in binary is

11111111 11111111 11111110.00000000

Octet 3 is the interesting one.

Value of last network bit in octet 3 is 2

So network numbers go up in 2s

172.16.0.0

172.16.2.0

172.16.4.0

172.16.6.0

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Subnet with no VLSM

Network Subnet

address

Host range Broadcast

address

Student 172.16.0.0/23 172.16.0.1 -

172.16.1.254

172.16.1.255

Instructor 172.16.2.0/23 172.16.2.1 -

172.16.3.254

172.16.3.255

Admin 172.16.4.0/23 172.16.4.1 -

172.16.5.254

172.16.5.255

WAN 172.16.6.0/23 172.16.6.1 -

172.16.7.254

172.16.7.255

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Addressing example with VLSM

172.16.0.0/22 is OK

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What we have and need

Given IP address 172.16.0.0/22

That’s 172.16.0.0 to 172.16.3.255

4 subnets needed:

Student LAN has 481 hosts

Instructor LAN has 69 hosts

Administrator LAN has 23 hosts

WAN has 2 hosts

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With VLSM

Student subnet has 481 hosts.

Formula for hosts is 2n – 2

n = 9 gives 510 hosts (n = 8 gives

only 254)

So 9 host bits needed.

That means 32 – 9 = 23 network bits

/23 or subnet mask 255.255.254.0

Network address 172.16.0.0

Broadcast address 172.16.1.255

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With VLSM

Instructor subnet has 69 hosts.

Formula for hosts is 2n – 2

n = 7 gives 126 hosts (n = 6 gives

only 62)

So 7 host bits needed.

That means 32 – 7 = 25 network bits

/25 or subnet mask 255.255.255.128

Network address 172.16.2.0

Broadcast address 172.16.2.127

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With VLSM

Admin subnet has 23 hosts.

Formula for hosts is 2n – 2

n = 5 gives 30 hosts (n = 4 gives only

14)

So 5 host bits needed.

That means 32 – 5 = 27 network bits

/27 or subnet mask 255.255.255.224

Network address 172.16.2.128

Broadcast address 172.16.2.159

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With VLSM

WAN subnet has 2 hosts.

Formula for hosts is 2n – 2

n = 2 gives 2 hosts

So 2 host bits needed.

That means 32 – 2 = 30 network bits

/30 or subnet mask 255.255.255.252

Network address 172.16.2.160

Broadcast address 172.16.2.16373

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Visually with VLSM

Student

Instructor

Admin

WAN

172.16.0.0 172.16.1.0 172.16.2.0 172.16.3.0

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Case 2. Given

192.168.1.0/24

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Subnet 192.168.1.0/24

2 subnets with 28 hosts each (largest)

5 host bits 25 - 2 = 30 would be just

enough

But allow for expansion: 6 host bits

give 62

Network bits 32 - 6 = 26

so /26 or subnet mask

255.255.255.192

Network Subnet address Host range Broadcast

address

B 192.168.1.0/26 192.168.1.1

- 192.168.1.62

192.168.1.63

E 192.168.1.64/26 192.168.1.65 -

192.168.1.126

192.168.1.12776

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Subnet 192.168.1.1/24

1 subnets with 14 hosts

4 host bits 24 - 2 = 14 would be just

enough

But allow for expansion: 5 host bits

give 30

Network bits 32 - 5 = 27

so /27 or subnet mask

255.255.255.224

0-127 range already used

Network Subnet address Host range Broadcast

address

A 192.168.1.128/27 192.168.1.129 -

192.168.1.158

192.168.1.159 77

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Subnet 192.168.1.1/24

1 subnets with 7 hosts

4 host bits 24 - 2 = 14 is enough

Network bits 32 - 4 = 28

so /28 or subnet mask

255.255.255.240

0-159 range already usedNetwork Subnet address Host range Broadcast

address

D 192.168.1.160/28 192.168.1.161 -

192.168.1.174

192.168.1.175 78

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Subnet 192.168.1.1/24

1 subnets with 2 hosts

2 host bits 22 - 2 = 2 is enough

Network bits 32 - 2 = 30

so /30 or subnet mask

255.255.255.252

0-175 range already usedNetwork Subnet address Host range Broadcast

address

C 192.168.1.176/30 192.168.1.177 -

192.168.1.178

192.168.1.179 79

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Subnet plan with VLSM

Networ

k

Subnet address Host range Broadcast

address

B 192.168.1.0/26 192.168.1.1

- 192.168.1.62

192.168.1.63

E 192.168.1.64/26 192.168.1.65 -

192.168.1.126

192.168.1.127

A 192.168.1.128/27 192.168.1.129 -

192.168.1.158

192.168.1.159

D 192.168.1.160/28 192.168.1.161 -

192.168.1.174

192.168.1.175

C 192.168.1.176/30 192.168.1.177 -

192.168.1.178

192.168.1.17980

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VisualB

E

A

D

C

One octet

available

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Subnetting class A

A class A address:

Is made of a one-byte netid and a three-byte hostid

Can have one single physical network with up to 16.777.214 (224-2)

If we want more physical networks, we can divide this one range into several smaller ranges

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Example :

A organization with a class A needs a least 1000 subnetworks. Find the subnet mask and configuration of each network

Solution:

• We need at least 1002 subnet to allow the all-1s and all-0s subnetids

• This means that the minimum number of bits to be allocated for subnetting should be 10 (29 < 1,002< 1010)

• Fourteen bits are left to define the hostid

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Without subnetting

masking 255.0.0.0

In binarry : 11111111 00000000 00000000 0000000

With subnetting

Masking 255.255.192.0

In binarry : 11111111 11111111 11000000 00000000

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Net id Host id

Net id SubNet id Host id

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Theres is 1024 subnets Each subnet can have 16,384 hosts/computer

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Subnetting class B

A class B:

Is made A two byte netid and two-byte hostid

Can have one single physical network and up to 65,534 hosts on the network.

If we wan more physical network, we can divide this one big range into several smaller ranges.

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Example

An organization with a class B address needs at least 12 subnetwork. Find subnet mask and configuration of each subnetwork

Solution:

There is a need for at least 14 subnetworks, 12 as specified plus 2 reserve as special address. This means that the minimum number of bits should be 4 (23 < 14 < 24)

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Without subnetting

masking 255.0.0.0

In binarry : 11111111 11111111 00000000 0000000

With subnetting

Masking 255.255.240.0

In binarry : 11111111 11111111 11110000 00000000

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Net id Host id

Net id SubNet id Host id

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Each subnet 4094 hosts/computers90

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Subnetting class C

A class C address:

is made of a three byte netid and one-byte hostid

Can have one single physical network and up to 254 (28 –2) host on that network

If we want more physical network, we can divide this one range into several smaller range.

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Example:

An organization with a class C needs at least five subnetworks. Find the subnet mask and configuration of each subnetwork

Solution:

There is a need for at least seven subnetworks, five specifief and two reserved a special address.

This means that minimum number of bits should be 3 ( 22< 7 < 23)

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Without subnetting

masking 255.0.0.0

In binarry : 11111111 11111111 11111111 0000000

With subnetting

Masking 255.255.240.0

In binarry : 11111111 11111111 11111111 11100000

Net id Host id

Net id

SubNet id

Host id

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There is 8 subnet

Each subnet can have 32 hosts 95

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Supernetting

Depend on the need of an organization

One or more classes c can be jointed to make one supernetwork

Example: an organization that needs 1000 address can be granted four class c addresses. The organization can then use these address in one supernetwork, in four network, or in more then four networks.

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Supernetting

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Supernet Mask

Can be assigned to a block of class C network address, if the number of net. Address is a power of two

Default mask for a class C address 255.255.255.0

If some of the 1s are changed to 0s, we can have a mask for a group of class C

Supernet mask is the reverseof the subnet mask

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The beginning address can beX.Y.32.0, but itcan not be X.Y.33.0

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Example:

With supernet 255.255.252.0, we can have four class C combined into one supernetwork

If we choose the first address to be X.Y.32.0, the other three addres X.Y.33.0, X.Y.34.0 and X.Y.35.0

If the router recieves a packet, it applies the supernetmask to the destination address and compare the result to the lowest address. If the result and the lowest address are the same, the packet belong to the supernet

Suppose a packet arrives with destination address X.Y.33.4. After applying the mask, the result is X.Y.32.0 (the lowest address), the packet belong to the supernet

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CIDRClasslessInterdomainRouting

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