Subnetting with VLSM Exercise Answers

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Oct 24, 2013 (3 years and 9 months ago)

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Subnetting with VLSM Exercise

Answers




A network administrator has the address range 192.168.12.0/24 to address the networks
shown.


1

How many bits are in the network part of the address and how many are in the host
part of the address?

24 in the networ
k part (3 octets) and 8 in the host part.


2

How many networks need to be addressed?

5


3

Using older subnetting methods without VLSM, all the subnets need to be the same
size. How many bits would you need to borrow to have enough networks?


3


4

How many bits
would be left in the host part?

5


5

How many hosts could you have on each network?

30


6

Can you address the networks without using VLSM?

No


7

From now on you will use VLSM to address the networks.


How many hosts are in the largest network?

50


8

H
ow many host bits do you need to address these hosts?

6


9

How many bits are available for borrowing?

2


10

What will the subnet mask be when you borrow these bits?

/26 or 255.255.255.192


11

How many addresses are there on a network of this size?
64


1
2

How many of these addresses can be used for hosts?
62


13

How many subnets do you get if you borrow this number of bits?
4


14

What are the network addresses of these subnets?


192.168.12.0, 192.168.12.64, 192.168.12.128, 192.168.12.192


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15

Fill in the

first row of the table on page 2, using the lowest numbered network address
and the name of your largest LAN


16

How many hosts will be needed on your second biggest LAN?

20


17

How many host bits do you need to address this number of hosts?
5


18

How m
any extra bits can you borrow, and how many bits are borrowed altogether?


1 more. 3 borrowed altogether.

19

What will the subnet mask be when you borrow these bits?

/27 or 255.255.255.224


20

How many addresses are there on a network this size?

32


21

Ho
w many of these addresses can be used for hosts?

30


22

Start with the second network address that you listed in step 14.
Enter your second
network into the table.


23

Now deal with your third LAN. Although your third LAN is smaller, consider using the
sa
me subnet mask (and therefore number of hosts) as for your second LAN. This will
allow room for growth.


24

Now deal with the two point to point links. There will never be more than two hosts on
these links so you should use subnets that allow only two ho
sts.



Subnet
name

Subnet mask

Network
address

Broadcast
address

Host address
range

Centre
LAN

255.255.255.192

or /26

192.168.12.0

192.168.12.63

192.168.12.1


192.168.12.62

West LAN

255.255.255.224

or /27

192.168.12.64

192.168.12.95

192.168.12.65


192.
168.12.94

East LAN

255.255.255.224

or /27

192.168.12.96

192.168.12.127

192.168.12.97
-

192.168.12.126

West link

255.255.255.252

or /30

192.168.12.128

192.168.12.131

192.168.12.129
-

192.168.12.131

East link

255.255.255.252

or /30

192.168.12.132

192.168.
12.135

192.168.12.133
-

192.168.12.134


There is no single right answer. There are a lot of possible solutions.