network layer -...
TRANSCRIPT
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NETWORK LAYER
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PROLOG
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FUNGSI UTAMA
Addressing
• Fisik
• Logika
•Port
Routing
•MembawaData
•PemilihanRute
Segmentation
• Subjaringan
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ADDRESSING
Physical Logical Port
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PHYSICAL ADDRESSING
Media Access Control
Terdiri bilangan heksadesimal 48 bit yang dibagi dalam 6 oktet
• 08:00:09:3A:20:1B
Alamat ethernet
• Kode Multicast (1 bit pertama)
• Vendor Code (23 bit)
• Globally Administered Address (24 bit)
Sistem penomoran dan pendaftaran diatur oleh IEEE
Kode Vendor didokumentasikan dalam RFC 1700
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LOGICAL ADDRESSING
Identifikasi setiap host dan router secara unik
Penerimaan paket dari jaringan ke jaringanlain
Setiap computer yang terhubung ke jaringanharus tahu alamat IP nya
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PORT ADDRESSING
Titik akhir komunikasi pada Sistem Operasi
Identifikasi tipe service
Disusun oleh bilangan 16-bit
Contoh : http://www.example.com:8080/path/
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COMMON PORT NUMBERS
NUMBER SERVICE
21 File Transfer Protocol
22 Secure Shell
23 Telnet
25 Simple Mail Transfer Protocol
53 Domain Name Services
80 Hyper Text Transfer Protocol
110 Post Office Protocol (POP3)
119 Network News Transfer Protocol (NNTP)
123 Network Time Protocol (NTP)
143 Internet Message Access Protocol (IMAP)
161 Simple Network Management Protocol (SNMP)
194 Internet Relay Chat (IRC)
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ROUTER
Menghubungkan beberapanetwork
• Sama atau berbeda topologi
Membuat subnetwork
Routing Capable
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FORMAT ALAMAT IP (IPV4)
11000000 10101000 00000000 00101000
32 bit
192 . 168 . 0 . 40
Alamat Jaringan
Alamat Host
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KLASIFIKASI IP ADDRESS
IP Address Class A
0 . . . . . . .
32 bit
. . . . . . . . . . . . . . . . . . . . . . . .
Network ID Host ID
Range : 1 . 0 . 0 . 0 sampai 127 . 255 . 255 . 255
Netmask : 255.0.0.0
Jumlah Host : 16.777.214
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KLASIFIKASI IP ADDRESS
IP Address Class B
10. . . . . . .
32 bit
. . . . . . . . . . . . . . . . . . . . . . . .
Network ID Host ID
Range : 128 . 0 . 0 . 0 sampai 191 . 255 . 255 . 255
Netmask : 255.255.0.0
Jumlah Host : 65.532
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KLASIFIKASI IP ADDRESS
IP Address Class C
110. . . . .
32 bit
. . . . . . . . . . . . . . . . . . . . . . . .
Network ID Host ID
Range : 192 . 0 . 0 . 0 sampai 223 . 255 . 255 . 255
Netmask : 255.255.255.0
Jumlah Host : 254
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KLASIFIKASI IP ADDRESS
IP Address Class D
1110
32 bit
. . . . . . . . . . . . . . . . . . . . . . . .
Alamat
Multicast
Range: 224 . 0 . 0 . 0 sampai 239 . 255 . 255 . 255
Multicast: Pengiriman data sekaligus ke sejumlah besar jaringan penerima,
yang dilakukan oleh router multicast khusus
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KLASIFIKASI IP ADDRESS
IP Address Class E
11110
32 bit
. . . . . . . . . . . . . . . . . . . . . . . .
Alamat Khusus Untuk
Experimen dan Cadangan
Range: 240 . 0 . 0 . 0 sampai 247 . 255 . 255 . 255
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KATEGORI ALAMAT IP
Statis
• Permanen Dimiliki
Dinamis
• Diatur DHCP server
Private
• 10 . 0 . 0 . 0
• 172. 16 . 0 . 0
• 192 . 168 . 0 . 0
Publik
• Internet
Khusus
• 0.0.0.0
• 127.0.0.1
• 255.255.255.255
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SUBNETTING
Pemecahan network menjadi beberapa network
yang lebih kecil
Meminjam sebagian bit-bit host sebagai bit-bit
subnet
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NEED FOR SUBNETTING
• Classes A and B have a large number of hosts corresponding to each network ID
• It may be desirable to subdivide the hosts in Class C subnets
• Often, there is a limitation on the number of hosts that could be hosted on a single network segment• The limitation may be imposed by concerns related to the management of
hardware
• Smaller broadcast domains are more efficient and easy to manage
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CONTOH
Diketahui network 192.168.1.0 yang akan
dipecah menjadi beberapa subnet
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CONTOH
1. Konversikan network address 192.168.1.0 menjadi bentuk biner :
11000000 . 10101000 . 00000001 . 00000000
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CONTOH
1. Konversikan network address 192.168.1.0 menjadi bentuk biner :
11000000 10101000 00000001 00000000
Network ID Host ID
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CONTOH
2. Meminjam 2 Buah Bit Host
11000000 10101000 00000001 00000000
Network ID Host ID
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CONTOH
2. Meminjam 2 Buah Bit Host
11000000 10101000 0000000100 000000
Network ID Host ID
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CONTOH
3. Buat Kombinasi Subnet
11000000 10101000 0000000100 000000
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CONTOH
3. Buat Kombinasi Subnet
11000000 10101000 0000000100 000000
11000000 10101000 0000000101 000000
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CONTOH
3. Buat Kombinasi Subnet
11000000 10101000 0000000100 000000
11000000 10101000 0000000101 000000
11000000 10101000 0000000110 000000
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CONTOH
3. Buat Kombinasi Subnet
11000000 10101000 0000000100 000000
11000000 10101000 0000000101 000000
11000000 10101000 0000000110 000000
11000000 10101000 0000000111 000000
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CONTOH
4. Hasil Subnetwork
11000000 10101000 00000001 00000000
11000000 10101000 00000001 01000000
11000000 10101000 00000001 10000000
11000000 10101000 00000001 11000000
192 168 1 64
192 168 1 0
192 168 1 192
192 168 1 128
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CONTOH
5. Hasil Broadcast dari Subnetwork
11000000 10101000 00000001 00111111
11000000 10101000 00000001 01111111
11000000 10101000 00000001 10111111
11000000 10101000 00000001 11111111
192 168 1 127
192 168 1 63
192 168 1 255
192 168 1 191
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192.168.1.0
Network
192.168.1.0 –192.168.1.63
Subnetwork 1
192.168.1.64 –192.168.1.127
Subnetwork 2
192.168.1.128 -192.168.1.191
Subnetwork 3
192.168.1.192 -192.168.1.255
Subnetwork 4
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EXCERCISE
• Consider the case of a class C address 195. 175.25.0 assigned to an organization
• Subnets can be constructed by allocating part of the higher-order bits of the host ID
• Assume that three of the higher-order bits of the host ID are to be reserved for that purpose