Upload
vothu
View
228
Download
0
Embed Size (px)
Citation preview
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
2
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.
3
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.
4
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:
6
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.
.
12
Decimal and Binary Conversion
13
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.
16
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.
17
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
19
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.
20
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
21
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.
22
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.
23
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,
24
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
25
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
26
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
27
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
29
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
30
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
31
IPv4 vs IPv6 Header
32
Addressing
33
Global Unicast Address
34
IPv6 Address Allocation
35
IPv4 and IPv6 Internetworking
many ways to do internetworking, 2 of
them are:
Tunnel IPv6 in IPv4 network
Address translation NAT PT
36
Tunneling IPv6 in IPv4
`
IPv6
Network
IPv6
Network
IPv4
Network
PacketIPv6
HeaderPacket
IPv6
Header
IPv4
HeaderPacket
IPv6
Header
37
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
42
Without subnetting
43
With Subnettiing
44
Masking
45
Applying bit-wise-and operation to achieve masking
46
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.
61
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)
62
Com
pute
r Netw
ork
-In
dustria
l Engin
eerin
g -
Faculty
of In
dustria
l Engin
eerin
g
Addressing example no
VLSM
172.16.0.0/21
63
Com
pute
r Netw
ork
-In
dustria
l Engin
eerin
g -
Faculty
of In
dustria
l Engin
eerin
g
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
64
Com
pute
r Netw
ork
-In
dustria
l Engin
eerin
g -
Faculty
of In
dustria
l Engin
eerin
g
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
Com
pute
r Netw
ork
-In
dustria
l Engin
eerin
g -
Faculty
of In
dustria
l Engin
eerin
g
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
66
Com
pute
r Netw
ork
-In
dustria
l Engin
eerin
g -
Faculty
of In
dustria
l Engin
eerin
g
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
67
Com
pute
r Netw
ork
-In
dustria
l Engin
eerin
g -
Faculty
of In
dustria
l Engin
eerin
g
Addressing example with VLSM
172.16.0.0/22 is OK
68
Com
pute
r Netw
ork
-In
dustria
l Engin
eerin
g -
Faculty
of In
dustria
l Engin
eerin
g
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
69
Com
pute
r Netw
ork
-In
dustria
l Engin
eerin
g -
Faculty
of In
dustria
l Engin
eerin
g
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
70
Com
pute
r Netw
ork
-In
dustria
l Engin
eerin
g -
Faculty
of In
dustria
l Engin
eerin
g
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
71
Com
pute
r Netw
ork
-In
dustria
l Engin
eerin
g -
Faculty
of In
dustria
l Engin
eerin
g
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
72
Com
pute
r Netw
ork
-In
dustria
l Engin
eerin
g -
Faculty
of In
dustria
l Engin
eerin
g
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
Com
pute
r Netw
ork
-In
dustria
l Engin
eerin
g -
Faculty
of In
dustria
l Engin
eerin
g
Visually with VLSM
Student
Instructor
Admin
WAN
172.16.0.0 172.16.1.0 172.16.2.0 172.16.3.0
74
Com
pute
r Netw
ork
-In
dustria
l Engin
eerin
g -
Faculty
of In
dustria
l Engin
eerin
g
Case 2. Given
192.168.1.0/24
75
Com
pute
r Netw
ork
-In
dustria
l Engin
eerin
g -
Faculty
of In
dustria
l Engin
eerin
g
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
Com
pute
r Netw
ork
-In
dustria
l Engin
eerin
g -
Faculty
of In
dustria
l Engin
eerin
g
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
Com
pute
r Netw
ork
-In
dustria
l Engin
eerin
g -
Faculty
of In
dustria
l Engin
eerin
g
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
Com
pute
r Netw
ork
-In
dustria
l Engin
eerin
g -
Faculty
of In
dustria
l Engin
eerin
g
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
Com
pute
r Netw
ork
-In
dustria
l Engin
eerin
g -
Faculty
of In
dustria
l Engin
eerin
g
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
Com
pute
r Netw
ork
-In
dustria
l Engin
eerin
g -
Faculty
of In
dustria
l Engin
eerin
g
VisualB
E
A
D
C
One octet
available
81
Com
pute
r Netw
ork
-In
dustria
l Engin
eerin
g -
Faculty
of In
dustria
l Engin
eerin
g
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
82
Com
pute
r Netw
ork
-In
dustria
l Engin
eerin
g -
Faculty
of In
dustria
l Engin
eerin
g
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
83
Com
pute
r Netw
ork
-In
dustria
l Engin
eerin
g -
Faculty
of In
dustria
l Engin
eerin
g
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
84
Net id Host id
Net id SubNet id Host id
Com
pute
r Netw
ork
-In
dustria
l Engin
eerin
g -
Faculty
of In
dustria
l Engin
eerin
g
Theres is 1024 subnets Each subnet can have 16,384 hosts/computer
85
Com
pute
r Netw
ork
-In
dustria
l Engin
eerin
g -
Faculty
of In
dustria
l Engin
eerin
g
86
Com
pute
r Netw
ork
-In
dustria
l Engin
eerin
g -
Faculty
of In
dustria
l Engin
eerin
g
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.
87
Com
pute
r Netw
ork
-In
dustria
l Engin
eerin
g -
Faculty
of In
dustria
l Engin
eerin
g
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)
88
Com
pute
r Netw
ork
-In
dustria
l Engin
eerin
g -
Faculty
of In
dustria
l Engin
eerin
g
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
89
Net id Host id
Net id SubNet id Host id
Com
pute
r Netw
ork
-In
dustria
l Engin
eerin
g -
Faculty
of In
dustria
l Engin
eerin
g
Each subnet 4094 hosts/computers90
Com
pute
r Netw
ork
-In
dustria
l Engin
eerin
g -
Faculty
of In
dustria
l Engin
eerin
g
91
Com
pute
r Netw
ork
-In
dustria
l Engin
eerin
g -
Faculty
of In
dustria
l Engin
eerin
g
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.
92
Com
pute
r Netw
ork
-In
dustria
l Engin
eerin
g -
Faculty
of In
dustria
l Engin
eerin
g
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)
93
Com
pute
r Netw
ork
-In
dustria
l Engin
eerin
g -
Faculty
of In
dustria
l Engin
eerin
g
94
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
Com
pute
r Netw
ork
-In
dustria
l Engin
eerin
g -
Faculty
of In
dustria
l Engin
eerin
g
There is 8 subnet
Each subnet can have 32 hosts 95
Com
pute
r Netw
ork
-In
dustria
l Engin
eerin
g -
Faculty
of In
dustria
l Engin
eerin
g
96
Com
pute
r Netw
ork
-In
dustria
l Engin
eerin
g -
Faculty
of In
dustria
l Engin
eerin
g
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.
97
Com
pute
r Netw
ork
-In
dustria
l Engin
eerin
g -
Faculty
of In
dustria
l Engin
eerin
g
Supernetting
98
Com
pute
r Netw
ork
-In
dustria
l Engin
eerin
g -
Faculty
of In
dustria
l Engin
eerin
g
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
99
Com
pute
r Netw
ork
-In
dustria
l Engin
eerin
g -
Faculty
of In
dustria
l Engin
eerin
g
100
The beginning address can beX.Y.32.0, but itcan not be X.Y.33.0
Com
pute
r Netw
ork
-In
dustria
l Engin
eerin
g -
Faculty
of In
dustria
l Engin
eerin
g
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
101
Com
pute
r Netw
ork
-In
dustria
l Engin
eerin
g -
Faculty
of In
dustria
l Engin
eerin
g
102
Com
pute
r Netw
ork
-In
dustria
l Engin
eerin
g -
Faculty
of In
dustria
l Engin
eerin
g
103
Com
pute
r Netw
ork
-In
dustria
l Engin
eerin
g -
Faculty
of In
dustria
l Engin
eerin
g
104
CIDRClasslessInterdomainRouting
Com
pute
r Netw
ork
-In
dustria
l Engin
eerin
g -
Faculty
of In
dustria
l Engin
eerin
g