Archive for the Subnet VLSM (largest 1st) Category

Subnetting with VLSM (largest 1st)

Posted in Subnet VLSM (largest 1st) on May 2, 2008 by itdaddy

 

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 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

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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.
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Done with VLSM
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 www.routemyworld.ent    great site check it out!

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