Subnetting with VLSM (largest 1st)

 

******************************
 Question VLSM Scenario  
*******************************
note: Do the 3 steps like you would do
hosts or subnetting; but do the largest host first!
(end of each router)
60 hosts  = 192.168.1.0  -1.63    /26
20 hosts  = 192.168.1.64 -1.95    /27
20 hosts  = 192.168.1.96 -1.127   /27
(between 2 routers)
serial = 192.168.1.128 – 1.131 /30
serial = 192.168.1.132 – 1.135 /30
serial = 192.168.1.136-  1.139 /30
6 subnets/networks total

******************************
Largets First   60 Hosts needed
*******************************

1. binary->bits

1 6 3 1 8 4 2 1
2 4 2 6
8
—————–
0 0 1->6 bits

 

2.bits->mask->increment

255.255.255.192
/26
255.255.255.11000000
2 bits for subnets
6 bits for hosts

2^2=4 subnets
2^6-2=62 hosts

increment = 64

3.increment->range

increment = 64
(network id   – broadcast address)
192.168.1.0       – 1.63             (take this range)    
192.168.1.64      – 1.127
192.168.1.128     – 1.191
192.168.1.192     – 1.255
192.168.2.0       – 2.63
=================================
 20 host VLSM Question
=================================

20 hosts needed:

1.binary to bit

1 6 3 1 8 4 2 1
2 4 2 6
8
—————–
0 0 0 1-> 5 bits to 20 host

 2. bit to mask to increment

class c is
255.255.255.0
255.255.255.11100000
255.255.255.224
/27

2^5-2 = 30 hosts
2^3= 8 subnets
2^4-2 = 14 hosts
increment = 32

3. increment to ranges

increment = 32
ranges
(network id          – broadcast id)

192.168.1.0      1.31
192.168.1.32     1.63
(above taken with the 60 hosts)

192.168.1.64     1.95      20 hosts here
192.168.1.96     1.127     20 hosts here
192.168.1.128
*************************************
Except for the 3 serial subnets left!
*************************************

need 2 hosts
3 subnets but 2 hosts only needed

2^2-2=2
2 bits in hosts

6 bits in subnets!

1. binary to bits

1 6 3 1 8 4 2 1
2 4 2 6
8
—————–
0 0 0 0 0 0 1 ->

bits = 2
2. bits to mask and increment

class c
255.255.255.0
255.255.255.11111100
/30
255.255.255.252

increment = 2
2^6= 64 subnets
3. increment to range

(network id   – broadcast)
192.168.1.0
192.168.1.2
192.168.1.4
192.168.1.6
192.168.1.8
192.168.1.10
……
192.168.1.128 – 1.131 /30
192.168.1.132 -135    /30
192.168.1.136- 1.139  /30
=================
Routing protocol on each interface
you cannot put 192.168.1.0 on each interface of that
router, it will tell you it overlaps. What you do is this
you put the actual network id examples below
=================
Router Rip
version 2

network 192.168.1.136  255.255.255.252
 network 192.168.1.96  255.255.255.224

make sure you have exact ip subnet and mask correct on each interface that you are going to have the routing protocol configured on. and it will work. Just putting at /24 subnet mask and 192.168.1.0 /24 is wrong for each interface…

note: notice it is not 192.168.1.0
it cannot be for both interfaces to run routing protocol.
==================
Done with VLSM
==================

 www.routemyworld.ent    great site check it out!

Advertisements

3 Responses to “Subnetting with VLSM (largest 1st)”

  1. U have just made these calculations so easy for me..U rock IT Daddy…

    Thank U

    • super yeah be nice if there was some kind of work book we could do but i made my own work book on
      vlsm subnetting and soon ipv6 we need to do it and do it in some kind of a work book there is not enogh
      work books stuff on tasks that we need to drill so you have to make your own. cool man glad i could help 😉

  2. venkatesh.R Says:

    Hi this site is very useful to me. Thank u very much

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: