1. Default Prefik:
- Class A with Subnet mask 255.0.0.0; or prefik = / 8; or binary = 11111111.00000000.0000000.0000000
- Class B Subnet mask 255.255.0.0; or prefik = / 16; or binary = 11111111.11111111.0000000.0000000
- Class C Subnet mask 255.255.255.0; or prefik = / 24; or binary = 11111111.11111111.1111111.0000000
2. Number of Subnet:
2x where x is the number of binary 1 in the last octet or can use a formula Prefik - Default_Prefik.
3. Number of Hosts per Subnet:
2y - 2 where y is the number of binary 0 in the last octet or can use a formula max_prefik - Prefik.
4. Block Subnet:
If Prefik> 24
Use the formula = 32 - Prefik
If Prefik> 16 <24 Use the formula = 24 - Prefik If Prefik> 8 <16 Use the formula = 16 - Prefik 5. Subnet Mask: is the last block of the block Subnet. Note prefiknya, if prefik> of 24 = 255.255.255.sm *
if prefik> of 16 = 255.255.sm *. 0
if prefik> from 255.sm *. 8 = 0.0
* sm = Last Block of Block Subnet
(Case Study) The calculation of IP
1. 202.95.8.0/25
2. 191.200.10.0/19
3. 180.20.10.0/25
answer:
1. IP Address 202.95.8.0/25
o Number of Subnet:
X = 25 - 24 = 1 -> 2X = 21 = 2 Subnet
o The number of hosts per subnet:
2y -2
Y = 32-25 = 7 -> 27-2 = 128 -2 = 126 Host
o Block Subnet:
* remember the key no. 4
32-25 = 7 -> 27 = 128 (block subnet multiples of 128)
Block Subnet = 0.128
o Subnetmask:
* remember the key no. 5
Subnetmask = 255 255 255 128
o Table IP:
Subnet 202.95.8.0 - 202.95.8.128 * *2. IP Address 191.200.10.0/19
First host 202.95.8.1 - 202.95.8.129
Last Host 202.95.8.126 - 202.95.8.254
Broadcast Address 202.95.8.127 - 202.95.8.255
* is a subnet block
o Number of Subnet:
X = 19-16 = 3 -> 2X = 23 = 8 Subnet
o The number of hosts per subnet:
2y -2
Y = 32-19 = 13 -> 213-2 = 8192 2 = 8190 Host
o Block Subnet:
* remember the key no. 4
24-19 = 5 -> 25 = 32 (subnet blocks multiple of 32)
Block Subnet = 0, 32, 64, 96, 128, 160, 192, 224
o Subnetmask:
* remember the key no. 5
Subnet mask = 255.255.128.0
o Table IP:
Subnet 191.200.0*.0 191.200.32*.0 191.200.64*.0 ... 191.200.224*.0
First Host 191.200.0.1 191.200.32.1 191.200.64.1 191.200.224.1
Last Host 191.200.31.254 191.200.63.254 191.200.127.254 191.200.255.254
Broadcast Address 191.200.31.255 191.200.63.255 191.200.127.255 191.200.255.255
3. IP Address 180.20.10.0/25
o Number of Subnet:
X = 25-16 = 9 -> 2X = 29 = 512 Subnet
o The number of hosts per subnet:
2y -2
Y = 32-25 = 7 -> 27-2 = 128 -2 = 126 Host
o Block Subnet:
* remember the key no. 4
32-25 = 7 -> 27 = 128
Block Subnet = 0, 128
o Subnetmask:
* remember the key no. 5
Subnetmask = 255 255 255 128
o Table IP:
Subnet 180.20.10.0 180.20.10.128 180.20.11.0 ... 180.20.255.128
First Host 180.20.10.1 180.20.10.129 180.20.11.1 ... 180.20.255.129
Last Host 180.20.10.126 180.20.10.254 180.20.11.126 ... 180.20.255.254
Broadcast Address 180.20.10.127 180.20.10.255 180.20.11.127 ... 180.20.255.255
Source : forummikrotik.com
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