Wednesday, January 12, 2011

SUBNETTING CALCULATION

Now it's time we learn the technique of calculating subnetting. Calculation of subnetting can be done in two ways, the way the binary is relatively slow and the special way faster. Virtually all the questions about subnetting will range in four problems: Number of Subnet, Number of Hosts per Subnet, Block Subnet, and Host-Broadcast Address.
Writing an IP address is usually 192.168.1.2. But sometimes written with 192.168.1.2/24, what this means? This means that the IP address 192.168.1.2 with a subnet mask of 255.255.255.0. You know how come like that?
Yes, / 24 taken from counting that 24-bit subnet mask diselubung with binary 1. Or in other
words, the 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.

Subnet Mask The next question is how that could be used to perform subnetting? This is answered 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 IN CLASS C

Okay, now let's just straight training. Subnetting like what happened with a 192.168.1.0/26 NETWORK ADDRESS?

analysis:

Class C IP address 192.168.1.0 with a subnet mask / 26 means 11111111.11111111.11111111.11000000 (255 255 255 192).

calculation:

As I mentioned before all the questions about subnetting will be centered in 4 cases, the number of subnets, the number of hosts per subnet, subnet blocks, host and broadcast address is valid. So we finish up with that order:

1. Number of Subnet = 2x, where x is the number of binary 1 in the last octet subnet mask (2 last octet for class B, and 3 the last octet for class A). So the subnet number is 22 = 4 subnets

2. Number of Hosts per subnet = 2y - 2, where y is the inverse of x is the number of binary 0 in the last octet subnet. So the number of hosts per subnet is 26-2 = 62 hosts

3. Block Subnet = 256-192 (the last octet subnet mask) = 64. The next subnet is 64 + 64 = 128, and 128 +64 = 192. So the total subnet are 0, 64, 128, 192.

How to host and broadcast address is valid? We immediately made ​​the table. For the record, the first host is the first digit after the subnet, and the broadcast is a number 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 subnetting for Class C. And we can go on again foranother subnet mask, with the same concepts and techniques. Subnet mask can be usedfor subnetting class C is
Subnet Mask
CIDR Value
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 THE CLASS B
Next we will try to do subnetting for an IP address class B. First, the subnet mask that can be used for subnetting a class B are:


Subnet Mask
CIDR Value
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
CIDR Value
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, let's try a problem for the Class B network address 172.16.0.0/18.
Analysis :
172.16.0.0 it means B Class Address, With the Subnet Mask /18 it means 11111111.11111111.11000000.00000000 (255.255.192.0).
Calculation :
1. Number of Subnet = 2x, where x is the number of binary 1 on the second last octet. So the subnet number is 22 = 4 subnets
2. Number of Hosts per subnet = 2y - 2, where y is the inverse of x is the number of binary 0 in the last two octets. So the number of hosts per subnet is 214-2 = 16382 hosts
3. Block Subnet = 256-192 = 64. The next subnet is 64 + 64 = 128, and 128 +64 = 192. So the total subnet are 0, 64, 128, 192.
4. Host and broadcast address is valid?

Subnet
172.16.0.0
172.16.64.0
172.16.128.0
172.16.192.0
First Host
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 having trouble? Okay, let's  try once again for Class B. How is network address 172.16.0.0/25.
Analisa:
172.16.0.0 is B Class address, with Subnet Mask /25 it means 11111111.11111111.11111111.10000000 (255.255.255.128).

calculation:

1. Number of Subnet = 29 = 512 subnets
2. Number of Hosts per Subnet = 27-2 = 126 host
3. Block Subnet = 256-128 = 128.
4. Host and broadcast address is valid?

172.16.0.0
172.16.0.128
172.16.1.0
172.16.255.128
First Host
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? Okay, before going into Class A, try again from Class C, and read slowly
SUBNETTING IP ADDRESS ON THE CLASS A
If it is established and understood, we go to Class A. The concept is all the same. The difference is in where we play octet subnet blocks. If Class C in octets to 4 (the last), grade B in octet 3 and 4 (2 last octet), if Class A in octet 2, 3 and 4 (3 last octet). Then the subnet mask that can be used for subnetting class A subnet mask is all of CIDR / 8 to / 30.
We try to exercise for the network address 10.0.0.0/16.
analysis:
10.0.0.0 means the class A, with a subnet mask / 16 means 11111111.11111111.00000000.00000000 (255.255.0.0).
calculation:
1. Number of Subnet = 28 = 256 subnets
2. Number of Hosts per Subnet = 216-2 = 65 534 hosts
3. Block Subnet = 256-255 = 1. So subnet full: 0,1,2,3,4, etc.
4. Host and broadcast address is valid?Subnet
10.0.0.0
10.1.0.0
10.254.0.0
10.255.0.0
First Host
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 read this last paragraph, you already understand subnetting calculations properly. If not get it, you repeat that continues to this article slowly from above. For subnetting memorizing techniques that are faster, waiting in the next article

Note: All of the above assumption that the calculation of subnet IP Subnet-zeroes (and IP Subnet-Ones) are calculated by default. Todd Lamle latest version of the book and also CCNA after 2005 had to accommodate a problem-zeroes subnet IP (and IP Subnet-Ones) this. CCNA pre-2005 did not include it by default (although in reality we can activate it with the command ip subnet-zeroes), so maybe in a few books on CCNA and CNAP test questions, you still find a formula calculating number of Subnet = 2x - 2

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