Transcript IP Addressing

Mr. Mark Welton
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IPv4 address are 32-bit numbers represented in dotted
decimal notation of 8 bit segments
00001010.00001000.01100100.00011000
10.8.100.24
So why 8 bit segments?
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We started with a classful system (Class
A,B,C,etc)
Each class is created by 8-bits of the binary IP
8-bit processing systems where easier and
cheaper to build (RFC 791 published in 1981)
Class A
11000000
Class B
Class C
10101000
00000000
00000000
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We are accustomed to the decimal system a
base 10 system
The number 12410 is 100+20+4 or
◦ 1x102+2x101+4x100
◦ 1x100+2x10+4x1
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The binary number system is a base 2 system
011111002 is
0x27+1x26+1x25+1x24+1x23+1x22+0x21+0
x20
0x128+1x64+1x32+1x16+1x8+1x4+0x2+
0x1
64+32+16+8+4 or 12410
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So each octet (8-bit binary number) goes
from
◦ 000000002 – 1111111112
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So what is the value of 1111111112
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128+64+32+16+8+4+2+1 = 25510
So what is the hexadecimal value?
8+4+2+1 = 15 or F16
27
26
128 64
25
24
32 16
23
22
21
20
8
4
2
1
11111111
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IP address allocation is rarely done properly
First mistake I see is people not understand
what I just covered
Second mistake I see is not understanding
public vs private vs reserved IP addresses
Third mistake I see is not understanding how
to take a large prefix and break it down to
usable network prefixes that allow for growth
Fourth mistake I see is people not
understanding why we do it
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We divide the IP space to create segments
that makes sense to us
Segmentation = routing
Each IP address allocation is a L2 network
which needs a router to move to the next
network
The better we do this the easier routing and
ACLs are to do
The easier the network is to troubleshoot
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RFC 1918 “Address Allocation for Private
Internets”
10.0.0.0 – 10.255.255.255 (10/8 prefix)
172.16.0.0 – 172.31.255.255 (172.16/12
prefix)
192.168.0.0 – 192.168.255.255
(192.168/16)
These are the IP address spaces that can be
used internally in an enterprise
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“link local” block
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reserves lowest Class B
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Also defines loop back space (RFC 1700)
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Also defines multicast address space (RFC 5771)
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So you should never use these IP address spaces!
◦ 169.254.0.0 – 169.254.255.255 (169.254.0.0/16)
◦ To be used when a device can not get an IP address
through DHCP
◦ 128.0.0.0 -128.0.255.255 (128.0.0.0/16)
◦ Not able to be used under old class system but can be
assigned to someone now
◦ 127.0.0.0 – 127.255.255.255 (127.0.0.0/8)
◦ Used for a machine to communicate internally
◦ 224.0.0.0 – 239.255.255.255 (224.0.0.0/4)
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Misuse of Public IP address space can cause
network routing problems for you network
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Prefix 10.0.0.0/8 has what subnet mask?
The 8 says the first 8 bits must be ones
So the first octet would be 255 and all others
would be zero
255.0.0.0
128 64
32 16
8
4
2
1
11111111
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What about 172.16.0.0/16?
192.168.0.0/24?
172.16.0.0/12?
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Redefined how a traditionally Classful IP
network could be used and subnetted (in
equal size block)
With VLSM, subnets can be any size if they
follow the binary rules
VLSM allows networks to be subdivided
192.168.1.192
11000000
10101000
00000001
11000000
10101000
00000001
11001000
11111111
11111000
11111111
11110000
192.168.1.200
11000000
/29 255.255.255.248
11111111
11111111
/28 255.255.255.240
11111111
11111111
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We use it all the time but do you really know
what it is?
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CIDR is sort of the inverse of VLSM
Where VLSM prescibes rules for subdividing
networks, CIDR prescribes rules for
referencing groups of networks with a single
route statement
Why would we want to do this?
Smaller routing tables are more
logical, easier to understand,
easier to troubleshoot, and
require less CPU and memory
for the routers.
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IP address allocation is rarely done properly
First mistake I see is people not understand what
I just covered Check
Second mistake I see is not understanding public
vs private vs reserved IP addresses Check
Third mistake I see is not understanding how to
take a large prefix and break it down to usable
network prefixes that allow for growth Not Yet
Fourth mistake I see is people not understanding
why we do it
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Allocate a block of IP addresses that can be
referenced with a single access-list (filter)
entry
Always allocate more IP addresses than
requested
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Need 30 IP addresses for a server farm of
database servers
Should we use a /27 255.255.255.224?
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Need 30 IP addresses for a server farm of
database servers
Should we use a /27 255.255.255.224?
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Allowing for 30 percent growth is a good rule
of thumb
Round up to the next binary boundary
64 IP addresses or a /26 255.255.255.192
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Now let say the server farm subnet was
already allocated using 10.100.100.0/24
There are currently 10 servers in place
.1 for the router and 2-11 for the servers
You need to issue 30 more IP addresses on
this subnet
Now what???
Just give them 12-42 right???
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Allocating groups of devices into subnettable
ranges
◦ allows you to remove them from the network and
place them elsewhere without significant changes
to the IP network design
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You could allocate the range of 32-63
◦ Access-list 101 permit ip any 10.100.100.32 255.255.255.224 eq web
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So we are good right???
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You should think ahead and allocate 64 IP
addresses on a bit boundary
So you should allocate 64-127
Right???
◦ Access-list 101 permit ip any 10.100.100.64
255.255.255.192 eq web
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Meets both rules so we are good???
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IP address allocation is rarely done properly
First mistake I see is people not understand what
I just covered Check
Second mistake I see is not understanding public
vs private vs reserved IP addresses Check
Third mistake I see is not understanding how to
take a large prefix and break it down to usable
network prefixes that allow for growth Not Yet
Fourth mistake I see is people not understanding
why we do it Know why you are allocating the IP
and allow for growth
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There are three methods you can use to allocate IP
addresses and IP subnets
Sequential– assign the first numerical subnet and
then the next and so on, most commonly used. It is
easy to understand
Divide by half - every time a network is allocated, the
smallest available chunk is divided by half for use
while preserving a large portion of IP address space
for additional growth
Reverse binary – subnets are allocated by counting in
binary with the most and least significant bits
reversed. Is the most logical method, but is hard to
understand
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Some of us have been doing this for so long
we remember the rule of all-zeros and allones as it relates to subnets
RFC 1878 states, “This practice
(of excluding all-zeros and all-ones subnets)
is obsolete. Modern software will be able to
utilize all definable networks.”
Sometimes you maybe in an environment
where legacy equipment can not do this
Or the staff still think they have to follow the
rule
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For the Cisco people in the class. You will
need to know Cisco’s way to pass the CCNA
 or know how to get the answer to the question based on
how Cisco or vendor X tests
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Everyone knows (or should now) that two IP
addresses are used in every subnet (one for
the gateway and one for broadcast)
 Unless you have done enough networking to know you
can use a /31 for to routers in a point-to-point
connection. DO NOT ASKING IF YOU ARE NOT GOING TO
TAKE CCDP!!!
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A /24 subnet has 256 host IP addresses –
254 IP are usable by host devices
Everything is based on the subnet masks
which is based on binary
Everything will be powers of 2 and will either
produce 256 or be divisible by 256
The maximum value of an octet is 255 (but
remember we count from 0 so 256 number)
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Subnet masks are, by their nature, inclusive
There are only nine values that are possible
for any octet in a subnet mask
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What the author is trying to say is a /24 or
255.255.255.0 would have 256 host with
16,777,216 possible subets (256*256*256*1)
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I find an expanded for of the horizontal
format very useful
Increment
Number of
hosts
128
64
32
16
8
4
2
1
CIDR
/25
/26
/27
/28
/29
/30
/31
/32
mask
128
192
224
240
248
252
254
255
Usable
Hosts
126
62
30
14
6
2
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