Calculations Subneting
After you read the article and understand the concept Subnetting concept Subnetting well. This time you learn techniques subnetting calculation. Subnetting calculation can be done in two ways, ways that binary relatively slow and how quickly the more special. At the bottom of all the questions about subnetting will be in the range of four problems: The number of subnet of the Host per subnet, subnet Block, and address-Host Broadcast.Writing the IP address is 192.168.1.2 with. However, sometimes written with 192.168.1.2/24, what this means? This means that the IP address of 192.168.1.2 subnet mask 255.255.255.0. Did can be such? Yes, / 24 taken from the calculation of the 24 bits that subnet mask diselubung with binary 1. Or in other words, subnet masknya is: 11111111.11111111.11111111.00000000 (255.255.255.0). This concept is called CIDR (Classless Inter-Domain Routing), which was first introduced in 1992 by IEFT.
The next question is how subnet mask that can be used to make subnetting? This is missed by the table below:
Subnet mask value CIDR
255.128.0.0 / 9
255.192.0.0 / 10
255.224.0.0 / 11
255.240.0.0 / 12
255.248.0.0 / 13
255.252.0.0 / 14
255.254.0.0 / 15
255.255.0.0 / 16
255.255.128.0 / 17
255.255.192.0 / 18
255.255.224.0 / 19
Subnet mask value CIDR
255.255.240.0 / 20
255.255.248.0 / 21
255.255.252.0 / 22
255.255.254.0 / 23
255.255.255.0 / 24
255,255,255,128 / 25
255,255,255,192 / 26
255,255,255,224 / 27
255,255,255,240 / 28
255,255,255,248 / 29
255,255,255,252 / 30
SUBNETTING IP ADDRESS ON CLASS C
Ok, now let alone direct training. Subnetting such as what happened with a 192.168.1.0/26 NETWORK ADDRESS?
Analysis: 192.168.1.0 means that the class C with a subnet mask / 26 means 11111111.11111111.11111111.11000000 (255,255,255,192).
Calculations: As I have mentioned before all the questions about subnetting will be headquartered in 4 cases, the number of subnet, the number of hosts per subnet, subnet blocks, host and broadcast address valid. So we finish with a sequence like that:
1. The number of subnet = 2x, where x is the number of binary 1 at oktet last subnet mask (2 oktet for the last class B, and 3 oktet for the last class A). So number of subnet is 22 = 4 subnet
2. Number of Hosts per subnet = 2y - 2, where y is the inverse of x the number of binary 0 on oktet last subnet. So the number of hosts per subnet is 26 - 2 = 62 host
3. Block subnet = 256 - 192 (value oktet last subnet mask) = 64. Subnet is 64 64 = 128, and 128 64 = 192. So the total subnetnya is 0, 64, 128, 192.
4. How to address a host and broadcast a valid? We immediately made the table. As a note, host of the first digit is 1 after the subnet, and broadcast a number is 1 before the next subnet.
Subnet 192.168.1.0 192.168.1.64 192.168.1.128 192.168.1.192
First host 192.168.1.1 192.168.1.65 192.168.1.129 192.168.1.193
Last host 192.168.1.62 192.168.1.126 192.168.1.190 192.168.1.254
Broadcast 192.168.1.63 192.168.1.127 192.168.1.191 192.168.1.255
We have completed the IP address for subnetting Class C. And we can continue for another subnet mask, with the concepts and techniques are the same. Subnet mask that can be used for subnetting class C is:
Subnet mask value CIDR
255,255,255,128 / 25
255,255,255,192 / 26
255,255,255,224 / 27
255,255,255,240 / 28
255,255,255,248 / 29
255,255,255,252 / 30
SUBNETTING IP ADDRESS TO CLASS B
Next we will try to do subnetting IP address class B. First, the subnet mask that can be used for subnetting class B are:
Subnet mask value CIDR
255.255.128.0 / 17
255.255.192.0 / 18
255.255.224.0 / 19
255.255.240.0 / 20
255.255.248.0 / 21
255.255.252.0 / 22
255.255.254.0 / 23
Subnet mask value CIDR
255.255.255.0 / 24
255,255,255,128 / 25
255,255,255,192 / 26
255,255,255,224 / 27
255,255,255,240 / 28
255,255,255,248 / 29
255,255,255,252 / 30
Ok, one question we try to Class B with a network address 172.16.0.0/18.
Analysis: 172.16.0.0 means the class B, with a subnet mask / 18 means 11111111.11111111.11000000.00000000 (255.255.192.0).
Calculations:
1. The number of subnet = 2x, where x is the number of binary 1 at 2 oktet last. So number of subnet is 22 = 4 subnet
2. Number of Hosts per subnet = 2y - 2, where y is the inverse of x the number of binary 0 on the last 2 oktet. So the number of hosts per subnet is 214 - 2 = 16,382 hosts
3. Block subnet = 256 - 192 = 64. Subnet is 64 64 = 128, and 128 64 = 192. So the total subnetnya is 0, 64, 128, 192.
4. Host and broadcast address valid?
Subnet 172.16.0.0 172.16.64.0 172.16.128.0 172.16.192.0
Host First 172.16.0.1 172.16.64.1 172.16.128.1 172.16.192.1
Last host 172.16.63.254 172.16.127.254 172.16.191.254 172.16.255.254
Broadcast 172.16.63.255 172.16.127.255 172.16.191.255 172.16 .. 255,255
Still confused? Ok we try one more for Class B. What with the network address 172.16.0.0/25.
Analysis: 172.16.0.0 means the class B, with a subnet mask / 25 means 11111111.11111111.11111111.10000000 (255,255,255,128).
Calculations:
1. The number of subnet = 29 = 512 subnet
2. Number of Hosts per subnet = 27 - 2 = 126 hosts
3. Block subnet = 256 - 128 = 128.
4. Host and broadcast address valid?
Subnet 172.16.0.0 172.16.0.128 172.16.1.0 ... 172.16.255.128
Host First 172.16.0.1 172.16.0.129 172.16.1.1 ... 172.16.255.129
Last host 172.16.0.126 172.16.0.254 172.16.1.126 ... 172.16.255.254
Broadcast 172.16.0.127 172.16.0.255 172.16.1.127 ... 172.16.255.255
Still confused as well? Ok prior to the Class A, try to repeat again from the Class C, and read slowly
SUBNETTING IP ADDRESS TO A CLASS
If you have mantab and understand, we go to Class A. The concept all the same. The difference is in where we play OKTET subnet blocks. If the Class C oktet to 4 (last), class B in Oktet 3 and 4 (2 oktet last), a Class A oktet 2, 3 and 4 (3 oktet last). Then the subnet mask that can be used for subnetting is all class A subnet mask of a CIDR / 8 / 30.
We try to exercise 10.0.0.0/16 network address.
Analysis: 10.0.0.0 means the class A, with a subnet mask / 16 means 11111111.11111111.00000000.00000000 (255.255.0.0).
Calculations:
1. The number of subnet = 28 = 256 subnet
2. Number of Hosts per subnet = 216 - 2 = 65534 host
3. Block subnet = 256 to 255 = 1. So the full subnet: 0,1,2,3,4, etc.
4. Host and broadcast address valid?
Subnet 10.0.0.0 10.1.0.0 ... 10.254.0.0 10.255.0.0
Host First 10.0.0.1 10.1.0.1 ... 10.254.0.1 10.255.0.1
Last host 10.0.255.254 10.1.255.254 ... 10,254,255,254 10,255,255,254
Broadcast 10.0.255.255 10.1.255.255 ... 10,254,255,255 10,255,255,255
Hopefully after you have read the last paragraph of this, you have to understand subnetting calculation with the good. Even if not also understand, you continue to repeat this article slowly from the top. Hapalan subnetting techniques for faster, waiting in the next article
Note: All the above calculation of the subnet berasumsikan that the IP subnet-Zeroes (and IP subnet-Ones) is calculated by default. The latest version of Buku Todd Lamle and CCNA also after 2005 to accommodate the problem is the IP subnet-Zeroes (and IP subnet-Ones) this. CCNA pre-2005 does not include default (although in reality we can activate the command ip-subnet zeroes), so it may in some books on CCNA and questions CNAP test, you still find a formula calculation of the number of subnet = 2x - 2 (taken Satrio articles from the facts taken from this site wahono)
