Archive for the Subnetting the Easy way!! Category

Subnetting the Easy Way! 1-2-3 Done!

Posted in Subnetting the Easy way!! on April 22, 2008 by itdaddy

Practice Subnetting daily at your job! _ Easy with this site!

 

 http://www.subnettingquestions.com/default_int.asp

 

 

I practice every day with this and man can you get good at subnetting!

Every day use this site for 1 hour at work and watch your successes!

 

freaking  awesome !!!!!!!!!!!!!!!!

 

 

 *******************************************
* IPSubnetter v1.2 great tool
* to verify you answers subnetting
* Click here to download
http://bosondownload.com/utils/bos_sub.exe
*********************************************

Practice Problems: Finding Subnets  

Practice Problems: Finding Hosts      

      http://nuggetlab.com/cisco/ccent_1.zip      

 

                                                  

Memory Cold items you must know:

 

2^n-2=Hosts  cannot use first host (network) and last host (broadcast)

2^n=Subnets (ip subnet zero no -2)  can have 000 and all 111s subnet

 

Note: Prior to 12.0 IOS by default, you could not use the all 0s or all 1s address because they were the Network ID and Broadcast Address. But 12.0 IOS and beyond the command ip subnet-zero is enabled by default. So no longer do you subtract a “2” if they specifiy it. In the Cisco exams Daved Davis says they will give you enough information to know when to use the 2^n-2=hosts or when to use the 2^n. They will give you enough information to know.  

 

Memorize these:

 

2^10=1024

2^6=64

2^5=32

2^3=8………………..number doubles as add 1 to N

2^2=4

 

 

Classes A, B, C, D
 

A – 1-126…………………127 loopback reserved.

B – 128-191

C – 192-223  

D – 224                    multicast 

 

Non-Routable addresses:

 

10.0.0.0 – 10.255.255.255          255.0.0.0  /8

172.16.0.0-172.31.255.255         255.240.0.0 /12

192.168.0.0-192.168.255.255     255.255.0.0  /16

 

 

APIPA (automatic private ip address)

 

169.254.0.0  255.255.0.0 /16 

your machine assignes you this if it cannot find a DHCP server or no STATIC ip is found.

Windows Machines are known for this.

 

 

Loopback Address:(on Computers)

 

127.0.0.1 or localhost DNS name

127.123.3.240 can be loopback adapter as well.

this is used to test your NIC card.

 

 

Network ID and Broadcast in a subnet range:

 

(Network ID)  (Broadcast)……………what is left in the middle is usable IPs

192.168.1.0 – 192.168.1.255…………except where “ip subnet-zero” is used.

192.168.2.0 – 192.168.2.255

192.168.3.0 – 192.168.3.255

192.168.4.0 – 192.168.4.255

 

 

Subnetting TO

 

 

1. Find Number of Subnets/Networks (words used interchangablely)

2. Find Number Hosts 

 

******************************************************************************************** 

Find Number of Subnets 

********************************************************************************************

 2^n=Subnets

 

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Question 1:

*********************************************

 

216.21.5.0  /24
CLASS C
SUBNETS NEEDED = 5

 

1. # OF NETWORK BITS?

1 6 3 1 8 4 2 1
2 4 2 6
8
——————
0 0 0 0 0 1 0 1  = 3 BITS FOR 5 SUBNETS    

2.  BITS TO MAKE MASK-INCREMENT
           
            1 6 3
            2 4 2
            8
255.255.255.1 1 1 00000

Inrement= 32

*Note: with no ip subnet-zero you have
30 usable hosts in below example.

 

3. USE INCREMENT TO FIND NETWORK-RANGE
(network id –   broadcast )     
216.21.5.0   –  216.21.5.31     1
216.21.5.32  –  216.21.5.63     2
216.21.5.64  –  216.21.5.95     3
216.21.5.96  –  216.21.5.127    4
216.21.5.128 –  216.21.5.159    5  Total subnets

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Question 2:

 *********************************************

195.5.20.0
class c
need 50 networks/subnets

 

1. binary-bits

 

1 6 3 1 8 4 2 1
2 4 2 6
8
—————–
0 0 1  -> 6    *quick counts bits to 1
               50 = 6 bits done!

 

2. bits mask-increment

 

class c
255.255.255.11111100 
2 host bits left
6 subnet bits used
255.255.255.252 mask used
255-3=252

increment = 4

 

3. increment-range

 

2 usable
2 not used
4 usable if “ip subnet-zero” used
(network id  –  broadcast address)
195.5.20.0   –  195.5.20.3
195.5.20.4   –  195.5.20.7
195.5.20.8   –  195.5.20.11
195.5.20.12  –  195.5.20.15
195.5.20.16  –  195.5.20.(20-1)

 

 ********************************************* 
Question 3:     

 *********************************************

150.5.0.0
Class B
255.255.0.0 /16
Need 100 Subnets

 

1. binary->bits


1 6 3 1 8 4 2 1
2 4 2 6
8
——————
  1 1 -> 7 bits needed

 

2. bits->mask->increment

255.255.11111110.00000000
255.255.254.0 /23
increment= 2

 

3. increment->range

 

(network id – broadcast)
150.5.0.0  – 150.5.1.255
150.5.2.0  – 150.5.3.255
150.5.4.0  – 150.5.5.255
150.5.6.0  – 150.5.x.y

*********************************************

Question 4:

*********************************************

 

need 500 networks/subnets
10.0.0.0
class A    /8
255.0.0.0

 

1. binary->bits

5 2.1 6 3 1 8 4 2 1
1 5 2 4 2 6
2 6 8
—————-
  1 1 ->9 bits = 500 networks 
 
  2^9-2=510

 

2.bits->mask->increment

 

255.0.0.0
/8    –> /17
255.255.128.0
255.11111111.10000000.00000000
place = 128 increment = 128

 

3.increment->ranges

 

10.0.0.0       – 10.0.127.255
10.0.128.0   – 10.0.255.255
10.1.0.0       – 10.1.127.255
10.1.128.0   – 10.1.255.255
10.2.0.0       – 10.2.127.255
10.2.128.0   – 10.2.255.255

 

 ************************************************************************************************

Find Number of Hosts 

 ************************************************************************************************

2^n-2=Hosts

2^n=Host (ip subnet-zero used) 

 

 ******************************
 Question 1:  
*******************************
216.21.5.0 /24
class c
need 30 hosts per subnet

 

1. binary->bits

1 6 3 1 8 4 2 1
2 4 2 6
8
—————–
0 0 0 1 1-> 5 bits = 30 host binary

means in this case
5 bits or 0s for host.

 

2.bits->mask->increment

 

5 bits to make subnet mask
class c = 255.255.255.0
255.255.255.11100000  (5bits for hosts)
2^5-2= 32 hosts = 30 host met!
2^3=8 subnets

3.increment->range

(network id   – broadcast address)
           30 hosts
216.21.5.0    –   216.21.5.31
216.21.5.32   –   216.21.5.63
216.21.5.64   –   216.21.5.95
216.21.5.96   –   216.21.5.127
216.21.5.128  –   216.21.5.159
216.21.5.160

 **************************************************

Question 2:

***************************************************

 

hosts->50

195.5.20.0
class c
255.255.255.0

1. binary->bits

1 6 3 1 8 4 2 1
2 4 2 6
8
—————–
0 0 1 1->6bits to make 50 hosts

 

2.bits->mask->increment

255.255.255.11000000
increment = 64
255.255.255.192
/26

 

3.increment->range

(network id   – broadcast address)
   
195.5.20.0      – 195.5.83.255
195.5.84.0      – 195.5.147.255
195.5.148.0     – 195.5.211.255
195.5.212.0     – 195.5.x.y

(2^6-2)=62   hosts = 60
2^2= subnets

 

********************************************

Question 3:

********************************************

150.5.0.0
class B
255.255.0.0

hosts->500 per subnet

1. binary->bits

 

2 1 6 3 1 8 4 2 1
5 2 4 2 6
6 8
—————–
1 ->9 BITS TO = 500
2^9-2=510

2.bits->mask->increment

255.255.11111110.00000000
255.255.254.0
/23
increment = 2

3.increment->range

(network id   – broadcast address)
150.5.0.0  – 150.5.1.255
150.5.2.0  – 150.5.3.255
150.5.4.0  – 150.5.5.255
150.5.6.0  –

2^9-2 = 510 host
2^7   =  128 = subnets/networks

 

******************************************

Question 4:

******************************************

 10.0.0.0
class A
255.0.0.0
hosts->100

1. binary->bits

 

1 6 3 1 8 4 2 1
2 4 2 6
8
—————–
0 1->7 bits to make 100 hosts

 

2.bits->mask->increment

255.11111111.11111111.10000000
255.255.255.128 /25
increment = 128

3.increment->range

(network id   – broadcast address)

10.0.0.0       –   10.0.0.127
10.0.0.128     –   10.0.0.255
10.0.1.0       –   10.0.1.127
10.0.1.128     –   10.0.1.255
10.0.2.0       –   10.0.2.127
10.0.2.128     –   10.0.2.255

2^7-2 = 126 hosts
2^17 =  131072 subnets
*********************************************************************************

Subnetting VLSM The Easy way!!!

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