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IP Address Subneting Jurusan Teknik Informatika Fakultas Teknologi Industri PTA 2020/2021

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Page 1: IP Address Subnetingayu_ws.staff.gunadarma.ac.id/Downloads/files/83644/...IP Address Subneting ﹡Memahami mengenai IP Address, kelas jaringan dalam alamat IP, subnet, subnet mask

IP Address SubnetingJurusan Teknik Informatika

Fakultas Teknologi Industri

PTA 2020/2021

Page 2: IP Address Subnetingayu_ws.staff.gunadarma.ac.id/Downloads/files/83644/...IP Address Subneting ﹡Memahami mengenai IP Address, kelas jaringan dalam alamat IP, subnet, subnet mask

IP Address Subneting

﹡ Memahami mengenai

IP Address, kelas

jaringan dalam alamat

IP, subnet, subnet

mask

﹡ Melakukan subneting

terhadap IP Address

dan VLSM

﹡ IP Address

﹡ Kelas Jaringan

﹡ Subnet dan Subnet

Mask

﹡ Proses Subneting

﹡ VLSM

2

TujuanMateri

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IP Address

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IP Address

Identifikasi Terminal

IP Address 32 bits

IP Address

1. Network ID

2. Host ID

4

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Badan Internasional Pengelola IP

Asia Pasific APNIC (Asia Pacific

Network Information Center

Tugas :

Pembagi blok nomor IP

Pembagi nomor AS (Autonomous

System) ISP Asia Pasific

5

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Badan Internasional Pengelola IP

ARIN (America Registry for Internet

Number)

RIPE (Reseaux IP Europeens)

AFRINIC (African Regional Internet

Registry Network Information Center)

IANA (International Assigned Number

Authority)

6

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IP Address

7

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IP Address

Network ID (Net ID/Nomor Network)

Host ID (Nomor Host/IP Address)

8

Satu Jaringan Net ID sama tetapi nomor

IP tidak sama

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Host Addressing

9

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IP Address

10

Network ID

Host ID

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IP Address

11

Alamat dalam jaringan

(Network Address)

Network ID

Alamat dari peralatan

jaringan unik

Host ID

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IP Addressing

12

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IP Addressing

IP Address kelas A, B dan C Jaringan

besar, medium dan kecil

13

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IP Addressing

IP Address Kelas D Multicasting

IP Address Kelas E IETF (Research)

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Kelas Jaringan

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Kelas IP Address

16

Class A Class B Class C Class D Class E

IP Address

Range :

1-126

00000001-

01111110

IP Address

Range :

128-191

10000000-

10111111

IP Address

Range :

192-223

11000000-

11011111

IP Address

Range :

224-239

11100000-

11101111

IP Address

Range :

240-255

11110000-

11111111

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Kelas IP Addres

17

Valid Host

ID

0 dan 127

(Loopback

Address)

Class

A

Valid Host

ID

128 dan

191

(Loopback

Address)

Class

B

Valid Host

ID

191 dan

224

(Loopback

Address)

Class

C

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Kelas IP Address

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Klasifikasi Jaringan

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IP Address

20

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Subnet dan Subnet

Mask

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Sub Netmask / Netmask

Netmask = IP Address 32 bits

255.0.0.0

255.255.0.0

255.255.255.0

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Sub Netmask / Netmask

Class A

Class B

Class C

IP Subnet mask 255.0.0.0

IP Subnet mask 255.255.0.0

IP Subnet mask 255.255.255.0

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IP dan Netmask

Pengalamatan Logik IP dan Netmask

Perhitungan IP dan Netmask Network

ID

Contoh :

IP : 202.95.151.129

Netmask : 255.255.255.0

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Network ID (Porsi Network)

Host dalam jaringan komunikasi

dengan perangkat dalam satu jaringan

Net ID sama

Subnet Mask menentukan Network ID

25

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Penggunaan Subnet Mask

Subnet Mask IP Address (Net ID dan

Host ID)

Net ID dalam Jaringan Operasi AND

antara Netmask dan IP Address

26

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Contoh Subnet Mask

Misal :

IP Address 10.252.240.6

Netmask 255.255.255.0

Maka :

Net ID ?

27

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Contoh Subnet Mask

Misal :

IP Address 10.252.240.6 dan Netmask

255.255.255.0

Konversi menjadi biner dan AND kan

00001010.11111100.11110000.00000110

11111111.11111111.11111111.00000000 AND

00001010.11111100.11110000.00000000

Net ID : 10.252.240.0

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Contoh Subnet Mask

Contoh :

IP : 202.95.151.129

Netmask : 255.255.255.0

Maka :

Net ID ?

Host ID ?

29

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Contoh Subnet Mask

Contoh :

IP : 202.95.151.129

Netmask : 255.255.255.0

Konversi :

11001010.01011111.10010111.10000001

11111111.11111111.11111111.00000000 AND

11001010.01011111.10010111.00000000

Net ID : 202.95.151.0 dan Host ID : 12930

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Pembagian Subnet Mask

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Jangkauan Network ID

Class A Net ID : 8 bit dan Host ID : 24 bit

Host Class A dalam satu jaringan (2^24) - 2 =

16.777.214 host

2 alamat loopback

32

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Jangkauan Network ID

Class B Net ID : 16 bit dan Host ID : 16 bit

Host Class B dalam satu jaringan (2^16) - 2 =

65.534

33

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Jangkauan Network ID

Class C Net ID : 24 bit dan Host ID : 8

bit

Host Class C dalam satu jaringan 2^8 =

256 254 host

34

Broadcast : 1

bit terakhir

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IP Private Address

﹡ Koneksi jaringan dengan IP Private ke

internet NAT

﹡ IP Private Address kekurangan alamat

IP Publik

35

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IP Private Address

36

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IP Private Address

Automatic Private IP Addressing (APIPA)

IP Address dunia

IP Private dengan Range IP Address :

169.254.0.1 – 169.254.255.254

Tidak routing dan deadlock di internet

37

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Contoh Network Address

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Proses SubNetting

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SubNetting / Sub-network

Teknik jaringan yang menggunakan 32 bit

kode pada IP Address.

Cara menciptakan jaringan pada IP Class

A, B dan C

Cara mendesain jaringan mengatur IP

Address dan pembatasan host

Class IP Address Network ID 8 bit

(Class A), Network ID 16 bit (Class B)

dan Network ID 24 bit (Class C)

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Subnetting

IP Address baru Net ID baru

Memecah 1 network network kecil

Jumlah Subnet : (2^n) – 2

n Jumlah bit yang digunakan (1)

Jumlah host per subnet : (2^m) – 2

m Jumlah bit yang belum digunakan (0)

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Mengapa

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Alasan Subneting

Untuk pengalokasian alamat IP yang

terbatas efisien

Membatasi jumlah host performansi

Network ID banyak Host ID sedikit

Memudahkan pengelolaan jaringan

Membantu pengembangan jaringan

43

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Gambaran Subnetting

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Kasus Subnetting

45

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Kasus Subnetting

46

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Subnetting Address

47

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Subnetting Address

48

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Contoh Subnetting (1)

IP Address : 130.200.0.0

Subnet Mask :

255.255.224.0 Class

B Maka :

Kelompok IP ? Atau Range IP ?

49

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Contoh Subnetting (1)

IP Address : 130.200.0.0

Subnet Mask : 255.255.224.0 Class B

Maka :

Subnet Mask :

11111111.11111111.11100000.00000000

Kelompok subnet : 2^5 = 32

130.200.32.0 – 130.200.224.0

broadcast 2 bit

50

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Contoh Subnetting (1)

Subnet : 32 64 96 128 160 192 224

Kelompok IP :

130.200.32.1 - 130.200.63.254 ==> loopback

130.200.64.1 - 130.200.95.254

130.200.96.1 - 130.200.127.254

130.200.128.1 - 130.200.159.254

130.200.160.1 - 130.200.191.254

130.200.192.1 - 130.200.224.254

130.200.223.254 (broadcast)51

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Contoh Subnetting (2)

Network ID : 130.200.0.0

Subnet Mask : 255.255.192.0 Maka :

Kelompok IP ?

52

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Contoh Subnetting (2)

Network ID : 130.200.0.0

Subnet Mask : 255.255.192.0

Class B Maka :

Subnet mask :

11111111.11111111.11000000.00000000

Kelompok subnet : 2^6 = 6453

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Contoh Subnetting (2)

Subnet : 64 128 192

Kelompok IP :

130.200.64.1 - 130.200.127.254

Loopback

130.200.128.1 - 130.200.192.254

broadcast

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CIDR (Classless Inter

Domain Routing)

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CIDR (Classless Internet Domain

Routing)

Alternatif cara klasifikasi alamat IP

Superneting

Mekanisme routing efisien membagi

alamat IP jaringan dalam class A, B dan

C

Pengembangan alamat IP

penggunaan alamat IP

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CIDR (Classless Inter Domain

Routing)

Metode ISP alokasi jumlah IP Address

ISP ukuran blok (blok size) tertentu

Contoh kelas B subnet mask

255.255.0.0 dengan nilai /16

16 jumlah bit 1 dalam subnet mask =

11111111.11111111.00000000.00000000

255.255.0.0/16

57

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CIDR (Classless Internet Domain

Routing)

Contoh : blok IP 192.168.32/28 28 bit

yang bernilai 1

Nilai maksimum : /32 1 byte = 8 bit (4

byte dalam alamat IP)

Subnet mask terbesar : /30 2 bit

sebagai host

58

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Nilai CIDR

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Contoh CIDR : Subnetting Kelas C

Kelas C host 8 bit

Subnet Kelas C

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Cara Subnetting CIDR

Berapa banyak jumlah subnet yang

dibutuhkan?

Berapa banyak jumlah valid host yang

dibutuhkan persubnet?

Mana saja subnet yang valid?

Apa Alamat Broadcast setiap subnet?

Apa Alamat valid host setiap subnet?

61

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Cara Subnetting CIDR

Berapa banyak jumlah subnet?

Rumus : 2^x = jumlah subnet

x = jumlah bit 1 yang dipinjam untuk subnet

Contoh : 11000000 subnet : 2^2 = 4

Berapa banyak jumlah host persubnet?

Rumus : (2^y) - 2 = jumlah host per -subnet

y = jumlah bit 0 yang terbentuk

Contoh : 11000000 subnet : (2^6) - 2 =

6262

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Cara Subnetting CIDR

Menentukan valid subnet dengan tepat

Rumus : 256 – subnetmask = blok size

Contoh : 256 – 192 = 64

64 subnet pertama

Subnet kedua : 64+64 = 128

Subnet ketiga : 128 + 64 = 192 dst

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Cara Subnetting CIDR

Menentukan Alamat Network

Dengan membuat semua bit Host menjadi bernilai 0

(biner)

Angka berikutnya adalah alamat valid Host pertama

Menentukan Alamat Broadcast

Dengan membuat semua bit Host menjadi bernilai 1

(biner)

Angka berikutnya adalah alamat subnet berikutnya

Alamat Valid Host adalah alamat yang berada di

antara Alamat Network dan Alamat Broadcast

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Contoh CIDR

Alamat Network : 172.16.0.0

Subnet Mask : 255.255.192.0

Maka :

Subnet ?

Host pertama ?

Alamat Broadcast ?

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Contoh CIDR

Alamat Class B

Alamat Network 172.16.0.0

Subnet Mask 255.255.192.0

Maka :

11111111.11111111.11000000.00000000

Subnet 192 = (2^2) – 2 = 2

Host (2^14) – 2 = 16.382

Subnet valid = 256 – 192 = 64; subnet

selanjutnya = 64 + 64 = 128, 128 + 64 = 19266

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Solusi Kasus 1

Subnet 172.16.64.0 - 172.16.128.0

Host pertama 172.16.64.1 - 172.16.128.1

Host terakhir 172.16.127.254 - 172.16.192.254

Broadcast 172.16.127.255 - 172.16.199.255

Subnet 64.0 atau 172.16.64.0 host pertama 64.1 atau

172.16.74.1 - 172.16.127.254 dan alamat broadcast

172.16.127.255

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Latihan

Alamat Network 10.0.0.0

Subnet Mask 255.255.0.0

Tentukan :

Subnet ?

Host pertama ?

Alamat Broadcast ?

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Jawaban Latihan

Alamat Class A

Alamat Network 10.0.0.0

Subnet Mask 255.255.0.0

Maka :

Subnet 255 = 11111111 (2^8) – 2 = 254

Host (2^16) – 2 = 65.534

Subnet valid = 256 – 255 = 1, 2, 3 dan seterusnya

Subnetnya = 10.1.0.0, 10.2.0.0 sampai 10.254.0.0

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Solusi Kasus 1

Subnet 10.1.0.0 … 10.254.0.0

Host pertama 10.1.0.1 … 10.254.0.1

Host terakhir 10.1.255.254… 10.254.255.254

Broadcast 10.1.255.255… 10.254.255.255

Subnet 10.1.0.0 host pertama 10.1.0.1 – 10.254.0.1

dan alamat broadcast 10.1.255.255

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VLSM (Variabel Length

Subent Mask)

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VLSM (Variabel Length Subnet

Mask)

Memudahkan dalam membuat subnet

dengan panjang prefix yang berbeda-

beda di berbagai tipe design jaringan

Protokol VLSM RIPv2, EIGRP, OSPF,

IS-IS

Mengehmat banyak IP

Subnet mask berbeda setiap interface

router

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VLSM (Variabel Length Subnet

Mask)

Implementasi pengalokasian blok IP yang

dilakukan oleh pemilik network dari blok

IP yang diberikan

Sifatnya local dan tidak dikenal di internet

Bentuk lain teknik subnetting

banyaknya host yang dibuat, banyak

jaringan subnet maupun class

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Keuntungan Subnetting VLSM

Mengurangi lalu lintas jaringan

Optimatisasi kerja jaringan

Pengelolaan yang sederhana

Membantu pengembangan jaringan ke

jarak geografis yang jauh

Menghemat ruang alamat

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Subnetting Tanpa VLSM

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Subnetting dengan VLSM

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Kasus

Jaringan perkantoran IP Address Class C

dengan alamat network 192.168.32.0

Jaringan dibagi 5 subnet per divisi

masing-masing divisi tidak terkait

Rincian host per divisi adalah :

Subnet 1 : 50 host

Subnet 2 : 50 host

Subnet 3 : 50 host

Subnet 4 : 30 host

Subnet 5 : 30 host 77

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Solusi Subnetting

Subnetting 255.255.255.192 4 subnet

dengan tiap subnet 64 host (11000000)

Kasus 5 subnet

Subnet 255.255.255.224 8 subnet tiap

subnet maksimal 32 host (11100000)

Kasus 50 host untuk 3 subnet dan 30

host untuk 2 subnet

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Solusi VLSM

Alamat network 192.168.32.0 Class IP C

Subnet mask default 255.255.255.0

Maka :

192.168.32.0

11000000.10101000.00100000.00000000

255.255.255.0

11111111.11111111.11111111.00000000

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Langkah Subnetting VLSM

1. Netmask untuk 3 divisi yang jumlah client

50 host. Teknik subnetting CIDR

netmask 255.255.255.192 dengan

maksimal host 62 2^6 = 64

192.168.32.0

11000000.10101000.00100000.00000000

255.255.255.192

11111111.11111111.11111111.11000000

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Langkah Subnetting VLSM

Jumlah subnet : 2^2 = 4

Jumlah host tiap subnet : (2^6) – 2 = 62

host

Blok subnet : 256 – 192 = 64 atau 2^6 = 64

0 64 128 192

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Langkah Subnetting VLSM

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Subnet 1 2 3 4

192.168.32.0 192.168.32.64 192.168.32.128 192.168.32.192

Host

Pertama

192.168.32.1 192.168.32.65 192.168.32.129 192.168.32.193

Host

Terakhir

192.168.32.62 192.168.32.126 192.168.32.190 192.168.32.254

Broadcast 192.168.32.63 192.168.32.127 192.168.32.191 192.168.32.255

Maksimal

Host

62 62 62 62

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Langkah VLSM

Teknik Subnetting network ID

192.168.32.192

Maka :

192.168.32.192

11000000.10101000.00100000.11000000

255.255.255.192

11111111.11111111.11111111.11000000

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Langkah VLSM

30 client (2^x) – 2 = 30

2^x = 32

x = 5

30 host (2^5) - 2 = 30

Netmask : 255.255.255.224

11111111.11111111.11111111.11100000

Blok subnet : 256 – 224 = 32

192 224

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Langkah VLSM

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Subnet 4 5

192.168.32.192 192.168.32.224

Host

Pertama

192.168.32.193 192.168.32.225

Host

Terakhit

192.168.32.222 192.168.32.254

Broadcast 192.168.32.223 192.168.32.255

Maksimal

Host

30 30

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Thanks!Any questions?

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