TUGAS PRODUKTIF PERHITUNGAN
SUBNETTING
1. IP KELAS A
a). 20.18.10.0/14
11111111.11111100.00000000.00000000
2555.252.0.0
v Subnet = 2x
= 26
= 64
v Jumlah host = 2y-2
= 218-2
= 262144-2
= 262142
v Blok subnet =
256-252
= 4
(0,4,8,12,16,………………….,52,56,60,64)
|
Subnet
|
Host pertama
|
Host terakhir
|
Broadcast
|
|
20.0.0.0
|
20.0.0.1
|
20.3.255.254
|
20.3.255.255
|
|
20.4.0.0
|
20.4.0.1
|
20.7.255.254
|
20.7.255.255
|
|
20.8.0.0
|
20.8.0.1
|
20.11.255.254
|
20.11.255.255
|
|
20.12.0.0
|
20.12.0.1
|
20.15.255.254
|
20.15.255.255
|
|
-
- -
- -
|
-
- - - -
- -
|
-
- - - - - -
- -
|
-
- - - - - - - -
|
|
20.52.0.0
|
20.52.0.1
|
20.55.255.254
|
20.55.255.255
|
|
20.56.0.0
|
20.56.0.1
|
20.59.255.254
|
20.59.255.255
|
|
20.60.0.0
|
20.60.0.1
|
20.63.255.254
|
20.63.255.255
|
|
20.64.0.0
|
20.64.0.1
|
20.255.255.254
|
20.255.255.255
|
b). 28.50.2.15
/11
11111111.11100000.00000000.00000000
255.224.0.0
v Subnet = 2x
= 23
= 8
v Jumlah Host = 2y
– 2
=221
– 2
=2097152
– 2
=
2097150
v Blok Subnet =
256 – 224
=32
(
0,32,64,96,128,160,192,224 )
|
Subnet
|
Host Pertama
|
Host Terakhir
|
Broadcast
|
|
28.0.0.0
|
28.0.0.1
|
28.31.255.254
|
28.31.255.255
|
|
28.32.0.0
|
28.32.0.1
|
28.63.255.254
|
28.63.255.255
|
|
28.64.0.0
|
28.64.0.1
|
28.95.255.254
|
28.95.255.255
|
|
28.96.0.0
|
28.96.0.1
|
28.127.255.254
|
28.127.255.255
|
|
28.128.0.0
|
28.128.0.1
|
28.159.255.254
|
28.159.255.255
|
|
28.160.0.0
|
28.160.0.1
|
28.191.255.254
|
28.191.255.255
|
|
28.192.0.0
|
28.192.0.1
|
28.223.255.254
|
28.223.255.255
|
|
28.224.0.0
|
28.224.0.1
|
28.255.255.254
|
28.255.255.255
|
2. IP KELAS B
a.) 150.29.100.40
/17
11111111.11111111.10000000.00000000
255.255.128.0
v Subnet = 2x
= 21
= 2
v Jumlah Host = 2y
– 2
= 215 – 2
= 32768 – 2
= 32766
v Blok Subnet =
256 – 128
= 128
( 0,128)
|
Subnet
|
Host
Pertama
|
Host
Terakhir
|
Broadcast
|
|
150.29.0.0
|
150.29.0.1
|
150.29.127.254
|
150.29.127.255
|
|
150.29.128.0
|
150.29.128.1
|
150.29.255.254
|
150.29.255.255
|
b.) 191.22.24.0 /22
11111111.11111111.11111100.00000000
255.255.252.0
v Subnet = 2x
= 26
= 64
v Jumlah Host = 2y
– 2
= 210 - 2
= 1024 – 2
= 1022
v Blok Subnet =
256 – 252
= 4
( 0,4,8,12,16, …………..
,52,56,60,64)
|
Subnet
|
Host
Pertama
|
Host
Terakhir
|
Broadcast
|
|
191.22.0.0
|
191.22.0.1
|
191.22.3.254
|
191.22.3.255
|
|
191.22.4.0
|
191.22.4.1
|
191.22.7.254
|
191.22.7.255
|
|
191.22.8.0
|
191.22.8.1
|
191.22.11.254
|
191.22.11.255
|
|
191.22.12.0
|
191.22.12.1
|
191.22.15.254
|
191.22.15.255
|
|
--- ---
---
|
--- ---
---
|
--- ---
---
|
--- ---
---
|
|
191.22.52.0
|
191.22.52.1
|
191.22.55.254
|
191.22.55.255
|
|
191.22.56.0
|
191.22.56.1
|
191.22.59.254
|
191.22.59.255
|
|
191.22.60.0
|
191.22.60.1
|
191.22.63.254
|
191.22.63.255
|
|
191.22.64.0
|
191.22.64.1
|
191.22.255.254
|
191.22.255.255
|
3. IP KELAS C
a.) 192.168.1.0 /27
11111111.11111111.11111111.11100000
255.255.255.224
v Subnet = 2x
= 23
= 8
v Jumlah Host = 2y
– 2
= 25 - 2
= 32 – 2
= 30
v Blok Subnet =
256 – 224
= 32
(
0,32,64,96,128,160,192,224 )
|
Subnet
|
Host
Pertama
|
Host
Terakhir
|
Broadcast
|
|
192.168.1.0
|
192.168.1.1
|
192.168.1.30
|
192.168.1.31
|
|
192.168.1.32
|
192.168.1.33
|
192.168.1.62
|
192.168.1.63
|
|
192.168.1.64
|
192.168.1.65
|
192.168.1.94
|
192.168.1.95
|
|
192.168.1.96
|
192.168.1.97
|
192.168.1.126
|
192.168.1.127
|
|
192.168.1.128
|
192.168.1.129
|
192.168.1.158
|
192.168.1.159
|
|
192.168.1.160
|
192.168.1.161
|
192.168.1.190
|
192.168.1.191
|
|
192.168.1.192
|
192.168.1.193
|
192.168.1.222
|
192.168.1.223
|
|
192.168.1.224
|
192.168.1.2255
|
192.168.1.254
|
192.168.1.255
|
b.) 192.90.23.130
/26
11111111.11111111.11111111.11000000
255.255.255.192
v Subnet = 2x
= 22 = 4
v Jumlah Host = 2y
– 2
= 26 - 2
= 64 – 2 = 62
v Blok Subnet =
256 – 192
=64
(
0,64,128,192 )
|
Subnet
|
Host Pertama
|
Host Terakhir
|
Broadcast
|
|
192.90.23.0
|
192.90.23.1
|
192.90.23.62
|
192.90.23.63
|
|
192.90.23.64
|
192.90.23.65
|
192.90.23.126
|
192.90.23.127
|
|
192.90.23.128
|
192.90.23.129
|
192.90.23.190
|
192.90.23.191
|
|
192.90.23.192
|
192.90.23.193
|
192.90.23.254
|
192.90.23.255
|
