SUBNETING IP ADDRESS KELAS C
kelas c = 3 oktek pertama adalah network, 1 oktek terakhir adalah host
yang perlu diingat =
Disini saya mempunyai beberapa soal tentang subneting kelas c
SOAL :
Carilah subnet mask,jumlah subnet,jumlah host,blok subnet, dan alamat host yang valid pada kelas C
dengan ip sebagai berikut :
- 192.168.1.0/27
- 192.168.4.0/28
- 192.168.6.0/29
- 192.168.8.0/30
- 192.168.19.0/25
penyelesaian :
yang perlu diingat =
10000000 = 128 ( /25 )
11000000 = 192 ( /26 )
11100000 = 224 ( /27 )
11110000 = 240 ( /28 )
11111000 = 248 ( /29 )
11111100 = 252 ( /30 )
11111110 = 254 ( /31 )
21 =
2 211 = 2048 221 = 2097152
22 =
4 212 = 4096 222 = 4194304
24 =
16 214 = 16384
25 =
32 215 = 32768
26 =
64 216 = 65536
27 =
128 217 = 131072
28 =
256 218 = 262144
29 =
512 219 = 524288
210 =
1024 220 = 1048576
Disini saya mempunyai beberapa soal tentang subneting kelas c
SOAL :
Carilah subnet mask,jumlah subnet,jumlah host,blok subnet, dan alamat host yang valid pada kelas C
dengan ip sebagai berikut :
- 192.168.1.0/27
- 192.168.4.0/28
- 192.168.6.0/29
- 192.168.8.0/30
- 192.168.19.0/25
penyelesaian :
*192.168.1.0/27
Subnet
mask: 11111111.11111111.11111111.11100000
( 255.255.255.224 )
Kelas
C
1. Jumlah subnet =
23 =
8 Subnet
2. Jumlah host =
25 – 2 = 30
Host
3. Blok subnet =
256 – 224 = 32 ( 0,32,64,96,
128,160,192,224 )
4. Alamat host yang valid :
NO
|
SUBNET
|
HOST
AWAL
|
HOST
AKHIR
|
BROADCAST
|
1.
|
192.168.1.0
|
192.168.1.1
|
192.168.1.30
|
192.168.1.31
|
2.
|
192.168.1.32
|
192.168.1.33
|
192.168.1.62
|
192.168.1.63
|
3.
|
192.168.1.64
|
192.168.1.65
|
192.168.1.94
|
192.168.1.95
|
4.
|
192.168.1.96
|
192.168.1.97
|
192.168.1.126
|
192.168.1.127
|
5.
|
192.168.1.128
|
192.168.1.129
|
192.168.1.158
|
192.168.1.159
|
6.
|
192.168.1.160
|
192.168.1.161
|
192.168.1.190
|
192.168.1.191
|
7.
|
192.168.1.192
|
192.168.1.193
|
192.168.1.222
|
192.168.1.223
|
8.
|
192.168.1.224
|
192.168.1.225
|
192.168.1.254
|
192.168.1.255
|
*192.168.4.0/28
Subnet
mask: 11111111.11111111.11111111.11110000
( 255.255.255.240 )
Kelas
C
1. Jumlah subnet =
24 = 16 Subnet
2. Jumlah host =
24 – 2 = 14 Host
3. Blok subnet =
256 – 240 = 16
( 0,16,32,48,64,80,96,112,128,144,160,176,192,208,224,240)
( 0,16,32,48,64,80,96,112,128,144,160,176,192,208,224,240)
4. Alamat host yang valid :
NO
|
SUBNET
|
HOST
AWAL
|
HOST
AKHIR
|
BROADCAST
|
1.
|
192.168.4.0
|
192.168.4.1
|
192.168.4.14
|
192.168.1.15
|
2.
|
192.168.4.16
|
192.168.4.17
|
192.168.4.30
|
192.168.1.31
|
3.
|
192.168.4.32
|
192.168.4.31
|
192.168.4.46
|
192.168.1.47
|
4.
|
192.168.4.48
|
192.168.4.49
|
192.168.4.62
|
192.168.1.63
|
5.
|
192.168.4.64
|
192.168.4.65
|
192.168.4.78
|
192.168.1.79
|
6.
|
192.168.4.80
|
192.168.4.81
|
192.168.4.94
|
192.168.1.95
|
7.
|
192.168.4.96
|
192.168.4.97
|
192.168.4.110
|
192.168.1.111
|
….
|
……….
|
|||
15
|
192.168.4.224
|
192.168.4.225
|
192.168.4.238
|
192.168.4.239
|
16
|
192.168.4.240
|
192.168.4.241
|
192.168.4.254
|
192.168.4.255
|
*192.168.6.0/29
Subnet
mask: 11111111.11111111.11111111.11111000
( 255.255.255.248 )
Kelas
C
1. Jumlah subnet =
25 = 32 Subnet
2. Jumlah host =
23 – 2 = 6 Host
3. Blok subnet =
256 – 248 = 8
( 0,8,16,24,32,40,48,56,64,72,80,88,96,104,112,120,128,136,144,
( 0,8,16,24,32,40,48,56,64,72,80,88,96,104,112,120,128,136,144,
152,160,168,176,184,192,200,208,216,224,232,240,248)
4. Alamat host yang valid :
NO
|
SUBNET
|
HOST
AWAL
|
HOST
AKHIR
|
BROADCAST
|
1.
|
192.168.6.0
|
192.168.6.1
|
192.168.6.6
|
192.168.6.7
|
2.
|
192.168.6.8
|
192.168.6.9
|
192.168.6.14
|
192.168.6.15
|
3.
|
192.168.6.16
|
192.168.6.17
|
192.168.6.22
|
192.168.6.23
|
4.
|
192.168.6.24
|
192.168.6.25
|
192.168.6.30
|
192.168.6.31
|
5.
|
192.168.6.32
|
192.168.6.33
|
192.168.6.38
|
192.168.6.39
|
6.
|
192.168.6.40
|
192.168.6.41
|
192.168.6.46
|
192.168.6.47
|
7.
|
192.168.6.48
|
192.168.6.49
|
192.168.6.54
|
192.168.6.55
|
….
|
||||
31
|
192.168.6.240
|
192.168.6.240
|
192.168.6.240
|
192.168.6.240
|
32
|
192.168.6.248
|
192.168.6.248
|
192.168.6.248
|
192.168.6.248
|
*192.168.8.0/30
Subnet
mask: 11111111.11111111.11111111.11111100
( 255.255.255.252 )
Kelas
C
1. Jumlah subnet =
26 = 64 Subnet
2. Jumlah host =
22 – 2 = 2 Host
3. Blok subnet =
256 – 252 = 4 (0,4,8,12,16,20,24,28,32,36,40,44,48,52,56,60,64,68,72,76,80,84,88,92,96,100,104,108,
112116,120,124,128,132,136,140,144,148,152,156,160,164,168,172,176,180,184,188,192,
196,200,204, 208,212,216,220,224,228,232,236,240,244,248,252)
112116,120,124,128,132,136,140,144,148,152,156,160,164,168,172,176,180,184,188,192,
196,200,204, 208,212,216,220,224,228,232,236,240,244,248,252)
4. Alamat host yang valid :
NO
|
SUBNET
|
HOST
AWAL
|
HOST
AKHIR
|
BROADCAST
|
1.
|
192.168.8.0
|
192.168.8.1
|
192.168.8.2
|
192.168.8.3
|
2.
|
192.168.8.4
|
192.168.8.5
|
192.168.8.6
|
192.168.8.7
|
3.
|
192.168.8.8
|
192.168.8.9
|
192.168.8.10
|
192.168.8.11
|
4.
|
192.168.8.12
|
192.168.8.13
|
192.168.8.14
|
192.168.8.15
|
5.
|
192.168.8.16
|
192.168.8.17
|
192.168.8.18
|
192.168.8.19
|
6.
|
192.168.8.20
|
192.168.8.21
|
192.168.8.22
|
192.168.8.23
|
7.
|
192.168.8.24
|
192.168.8.25
|
192.168.8.26
|
192.168.8.27
|
….
|
||||
63
|
192.168.8.248
|
192.168.8.249
|
192.168.8.250
|
192.168.8.251
|
64
|
192.168.8.252
|
192.168.8.253
|
192.168.8.254
|
192.168.8.255
|
*192.168.19.0/25
Subnet
mask: 11111111.11111111.11111111.10000000
( 255.255.255.128 )
Kelas
C
1. Jumlah subnet =
21 = 2 Subnet
2. Jumlah host =
27 – 2 = 126 Host
3. Blok subnet =
256 – 128 = 128 ( 0,128)
4. Alamat host yang valid :
NO
|
SUBNET
|
HOST
AWAL
|
HOST
AKHIR
|
BROADCAST
|
1.
|
192.168.19.0
|
192.168.19.1
|
192.168.19.126
|
192.168.19.127
|
2.
|
192.168.19.128
|
192.168.19.129
|
192.168.19.254
|
192.168.19.255
|
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