modul 2 ip address
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IP ADDRESSPraktikum Jaringan Komputer
2011
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Apaan tuu??Internet Protocol (IP) adalah
protokol yang memberikan alamat atau identitas logika untuk peralatan di jaringan
IP Address adalah alamat yang diberikan ke jaringan dan peralatan jaringan yang menggunakan protokol TCP/IP
Tugas IP : Menyediakan pengiriman
datagram secara connectionless
Menyediakan fragmentasi & reassembly yang mendukung kerja dari data link dengan menggunakan MTU
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FORMAT IP
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Physical Address vs Logical Address12 digit bilangan
hexadecimalPemisah :, -, 00:05:9A:3C:78:0000:05:9A adalah ID
Perusahaan dari IEEE3C:78:00 adalah ID
Produk
32 digit bilangan binerPemisah dot (.)192.168.10.5192.168.10.0 Network
ID.5 Host IDTerbagi menjadi
Network ID dan Host ID
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Internet Protocol Address
IP Versi 4 IP Versi 6
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IPv4 VS IPv632 bit (4.294.967.296)Performance menurun
seiring tabel routingMobilitas terbatasIPSec sbg fitur
tambahanHeader 20 oktetAda header checksumFragmentasi di tiap hop
bila melewati MTUKonfigurasi manualMekanisme best effort
128 bit (3,4 x 10^38)Mampu mengelola
tabel routing besarMobilitas tinggiIPSec fitur wajibHeader 40 oktetChecksum end-to-endMTU DiscoveryKonfigurasi otomatisMekanisme best level
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12 Apr 2023S Ward Abingdon and Witney College CCNA Exploration Semester 1
IPv4 address192. 168. 21. 17
11000000 10101000 00010101 00010001
octetoctet octet octet
network part host part
255. 255. 255. 0
11111111 11111111 11111111 00000000
Prefix /24 Subnet mask:
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IP Address v4
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IP Address
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12 Apr 2023S Ward Abingdon and Witney College CCNA Exploration Semester 1
248 to binary128 64 32 16 8 4 2 1
1 1 1 1 1 0 0 0
248-128120
24-16
8
56-3224
120-6456
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12 Apr 2023S Ward Abingdon and Witney College CCNA Exploration Semester 1
187 to binary128 64 32 16 8 4 2 1
1 0 1 1 1 0 1 1
187-128
59
27-1611
11-83
59-3227
3-21
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12 Apr 2023S Ward Abingdon and Witney College CCNA Exploration Semester 1
89 to binary128 64 32 16 8 4 2 1
0 1 0 1 1 0 0 1
89-6425
25-16
9
9-81
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12 Apr 2023S Ward Abingdon and Witney College CCNA Exploration Semester 1
00110100 to decimal128 64 32 16 8 4 2 1
0 0 1 1 0 1 0 0
32 16 4
32 +16+
452
52
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12 Apr 2023S Ward Abingdon and Witney College CCNA Exploration Semester 1
01010101 to decimal128 64 32 16 8 4 2 1
0 1 0 1 0 1 0 1
64 16 4 1
64 +16+
4+ 185
85
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12 Apr 2023S Ward Abingdon and Witney College CCNA Exploration Semester 1
11001111 to decimal128 64 32 16 8 4 2 1
1 1 0 0 1 1 1 1
128 64 8 4 2 1
128+ 64 + 8+ 4+ 2+ 1207
207
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IP SubnetPembagian suatu kelompok alamat IP menjadi
bagian bagian yang lebih kecil lagi.Contoh :
0 netID subnet number Host IDClass ASubnet
01 816
24
31
10 netID subnet number Host IDClass BSubnet
110 netID
subnet
number
Host IDClass CSubnet
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Contoh Topologi Jaringan
A40 client
B300 client
C1000 client
192.168.0.0255.255.255.192
172.16.0.0255.255.253.0
172.17.0.0255.255.251.0
1
3
2
1 2
3
R1
R2
192.168.1.0255.255.255.251
202.154.62.244/28
INTERNET
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Metode PengirimanAda 4 macam metode pengiriman :
UnicastBroadcastMulticastAnycast
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UnicastPengiriman data dilakukan hanya ke 1
host saja.
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BroadcastPengiriman data disebarkan langsung ke
banyak host.
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MulticastData yang dikirim akan disebarkan di dalam
group tersebut.
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AnycastData yang dikirim dari 1 host langsung
disebarkan ke host tertentu saja.
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IP PUBLIC & IP PRIVATEIP public : IP address yang digunakan pada
jaringan lokal oleh suatu organisasi/perusahaan, organisasi lain dari luar dapat melakukan komunikasi langsung dengan jaringan lokal
Contoh: IP yang didapat dari ISP
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IP PUBLIC & IP PRIVATE (cont.)IP Private : IP Address yang digunakan oleh suatu
organisasi yang diperuntukkan untuk jaringan lokal. Organisasi lain dari luar tidak dapat melakukan komunikasi dengan jaringan lokal
Contoh : Lihat IP masing-masing
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IP PUBLIC & IP PRIVATE (cont.)Untuk
mengkoneksikan suatu jaringan yang menggunakan IP private ke internet, dibutuhkan suatu translasi dari IP private ke IP public, yakni menggunakan Network Address Translation (NAT)
Host Host Host
IP Private
INTERNET
IP Public
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LATIHAN
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1. Ubahlah IP berikut menjadi binary dan Tentukan net id dan host id dari ip berikut :a. 10.14.17.5 b. 172.168.7.3c. 200.11.2.4d. 191.10.3.4e. 10.10.10.10f. 223.224.209.2012. Buatlah ad-hoc dari ip di atas (dipilih saja)?