subnetting
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STMIK TEKNOKRAT Bandarlampung Manajemen Jaringan TI 12 A
Feri Saputra - 12312431
Nama : Feri Saputra
NPM : 12312431
Kelas : TI 12 A
A. Metode CIDR
1) 172.16.20.0 /25 Class B
Subnet Mask = 255.255.255.128
Bit = 11111111.11111111.11111111.10000000
Jumlah Subnet = 2x = 29 = 512
Host/Subnet = 2y 2 = 27 2 = 126
Blok/Subnet = 256 -128 = 128
Subnet 172.16.20.0 172.16.20.128 172.16.21.0 172.16.21.128
IP Awal 172.16.20.1 172.16.20.129 172.16.21.1 172.16.21.129
IP Akhir 172.16.20.126 172.16.20.254 172.16.21.126 172.16.21.254
Broadcast 172.16.20.127 172.16.20.255 172.16.21.127 172.16.21.255
Dan seterusnya...
2) 192.252.61.80 /26 Class C
Subnet Mask = 255.255.255.192
Bit = 11111111.11111111.11111111.11000000
Jumlah Subnet = 2x = 24 = 4
Host/Subnet = 2y 2 = 26 2 = 62
Blok/Subnet = 256 -192 = 64
Subnet
192.252.61.0 192.252.61.64 192.252.61.128 192.252.61.192
IP Awal 192.252.61.1 192.252.61.65 192.252.61.129 192.252.61.193
IP Akhir 192.252.61.62 192.252.61.126 192.252.61.190 192.252.61.254
Broadcast 192.252.61.63 192.252.61.127 192.252.61.191 192.252.61.255
3) 145.229.179.96 Class B
Subnet Mask = 255.255.252.0
Bit = 11111111.11111111.11111100.00000000
Jumlah Subnet = 2x = 26 = 64
Host/Subnet = 2y 2 = 210 2 = 1022
Blok/Subnet = 256 -252 = 4
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STMIK TEKNOKRAT Bandarlampung Manajemen Jaringan TI 12 A
Feri Saputra - 12312431
Subnet 145.229.179.0
IP Awal 145.229.179.1
IP Akhir 145.229.179.254
Broadcast 145.229.179.255
Dan Seterusnya
B. Metode VLSM
1) 192.168.1.0 /24 Class C
50 HOST?
Gunakan 192.168.1.0 /26 karna /26 dapat menampung 64 HOST
Subnet Mask = 255.255.255.192
Bit = 11111111.11111111.11111111.11000000
Jumlah Subnet = 2x = 22 = 4
Host/Subnet = 2y 2 = 26 2 = 62
Blok/Subnet = 256 -192 = 64
Subnet 192.168.1.0 192.168.1.64 192.168.1.128 192.168.1.192
IP Awal 192.168.1.0 192.168.1.65 192.168.1.129 192.168.1.193
IP Akhir 192.168.1.62 192.168.1.126 192.168.1.190 192.168.1.204
Broadcast 192.168.1.63 192.168.1.127 192.168.1.191 192.168.1.255
2) 192.168.1.0 /24 Class C
100 HOST?
Gunakan 192.168.1.0 /28 karna /25 dapat menampung 128 HOST
Subnet Mask = 255.255.255.128
Bit = 11111111.11111111.11111111.10000000
Jumlah Subnet = 2x = 21 = 2
Host/Subnet = 2y 2 = 27 2 = 126
Blok/Subnet = 256 -128 = 128
Subnet 192.168.1.0 192.168.1.128
IP Awal 192.168.1.1 192.168.1.129
IP Akhir 192.168.1.126 192.168.1.254
Broadcast 192.168.1.127 192.168.1.255
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STMIK TEKNOKRAT Bandarlampung Manajemen Jaringan TI 12 A
Feri Saputra - 12312431
3) 192.168.1.0 /24 Class C
20 HOST?
Gunakan 192.168.1.0 /27 karna /27 dapat menampung 32 HOST
Subnet Mask = 255.255.255.224
Bit = 11111111.11111111.11111111.11100000
Jumlah Subnet = 2x = 23 = 8
Host/Subnet = 2y 2 = 25 2 = 30
Blok/Subnet = 256 -224 = 32
Subnet 192.168.1.0 192.168.1.32
IP Awal 192.168.1.1 192.168.1.33
IP Akhir 192.168.1.30 192.168.1.62
Broadcast 192.168.1.31 192.168.1.63
Dan Seterusnya
4) 192.168.1.0 /24 Class C
6 HOST?
Gunakan 192.168.1.0 /29 karna /29 dapat menampung 8 HOST
Subnet Mask = 255.255.255.248
Bit = 11111111.11111111.11111111.11111000
Jumlah Subnet = 2x = 25 = 32
Host/Subnet = 2y 2 = 23 2 = 6
Blok/Subnet = 256 -248 = 8
Subnet 192.168.1.0 192.168.1.8
IP Awal 192.168.1.1 192.168.1.9
IP Akhir 192.168.1.6 192.168.1.14
Broadcast 192.168.1.7 192.168.1.15
Dan Seterusnya