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Subnetting Question

Posted on 2011-02-13
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Last Modified: 2012-05-11
I have this question in my CCNA book that I understand somewhat but not
all the way:

Question:
Create a minimal configuration enabling IP on each interface on a 2600
series router (two serial, one Ethernet).  The Network Information
Center (NIC) assigns you network 192.168.1.0.  Your boss says that you
need, at most, 60 hosts per LAN subnet.  You also have point-to-point
links attached to the serial interfaces.  When choosing the IP address
values and subnet numbers, you decide to start with the lowest
numerical values.  Assume that point-to-point serial links will be
attached to this router and that EIGRP is the routing protocol.

Answer:
192.168.1.0 is a Class C network number, so it will have 24 network
bits and 8 host bits (by default). This examples calls for at most 60
hosts per LAN subnet.  This means that 2^6 - 2 = 62... so there must be
6 host bits.  That means that we need 24 network bits, 2 subnet bits,
and 6 host bits.  The mask that works for this need is 255.255.255.192,
because it has exactly 6 host bits.  For point-to-point links, we use
mask 255.255.255.252, because it only allows 3 different IPs.

This is the router config:

--------------------------------------------------------------------------------

interface ethernet 0/0
ip address 192.168.1.65 255.255.255.192

interface serial 0/0
ip address 192.168.1.129 255.255.255.252

interface serial 0/1
ip address 192.168.1.133 255.255.255.252

router eigrp 1
network 192.168.1.0

--------------------------------------------------------------------------------

My question is... why is 192.168.1.65 the lowest available IP for the
LAN, given network 192.168.1.0 & subnet 255.255.255.192.  Also, why
does the answer specify 192.168.1.129/133 with network 192.168.1.0 &
subnet 255.255.255.252 as the lowest values for the IP addresses of the
serial interfaces?
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Question by:MarkSal
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Expert Comment

by:gplana
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192.168.1.65 is the lowest available IP because 65 is 01 000001, so subnet 01, host 000001
Remember 0 value for net and for host is reserved (0 and the maximum -all 1s- are the reserved for net address and broadcast respectively).

Hope it helps.
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Accepted Solution

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kdearing earned 500 total points
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First of all, there is no one correct answer, it could be done many different ways.

Using a .192 subnet gives you the possibility of 4 networks within 192.168.1.0
1 - 192.168.1.0 (first IP available is 192.168.1.1)
2 - 192.168.1.64 (first IP available is 192.168.1.65)
3 - 192.168.1.128 (first IP available is 192.168.1.129)
4 - 192.168.1.192 (first IP available is 192.168.1.193)

You can choose any of the 4 networks.
The answer given has chosen the second network to use for the 60 LAN hosts.

Given that you only need one LAN network, the rest of the IPs are available for whatever is left.

As far as the serial interfaces...
A typical serial interface only requires 2 IP addresses.
A .252 subnet give exactly 2 available IPs.
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Expert Comment

by:edster9999
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It is valid but a bit strange.
The first 64 addresses are the same as the next set they have picked.
Maybe they kept the first set for network things.
Maybe it is just unusual to make you think more :)
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Author Comment

by:MarkSal
Comment Utility
kdearing,

Yes thats exactly as I thought....ok not going crazy and thanks that really helped confirm my answer compaired to the answer the the book.

Mark
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Author Closing Comment

by:MarkSal
Comment Utility
Awesome assistance
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Expert Comment

by:rfc1180
Comment Utility
>My question is... why is 192.168.1.65 the lowest available IP for the
LAN, given network 192.168.1.0 & subnet 255.255.255.192

Because you cannot use 192.168.1.0/26 (Technically you can with IP subnet zero, but not in the CCNA, at least when I took it.)

http://www.cisco.com/en/US/tech/tk648/tk361/technologies_tech_note09186a0080093f18.shtml

>Also, why
does the answer specify 192.168.1.129/133 with network 192.168.1.0 &
subnet 255.255.255.252 as the lowest values for the IP addresses of the
serial interfaces?

192.168.0.64 + network incremental (2^6=64)
.64 +64 = 128
They just used the next available network 192.168.1.128/26 and just Subnetted that network into /30s to limit the exposure to wastefulness of IP space.

192.168.1.128/26
-192.168.1.128/30
192.168.1.129 - 192.168.1.130 (broadcast 192.168.1.131)

Next network: 192.168.1.132/30
192.168.1.133 - 192.168.1.134 (broadcast 192.168.1.135)

Billy
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