# You are given an IP address for a host 172.168.35.10/20 What is/are the

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24 Οκτ 2013 (πριν από 4 χρόνια και 8 μήνες)

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1

You are given an IP address for a host 172.168.35.10/20

What is/are the

The number of useable hosts available for this subnet?

The number of useable subnets available for this
network?

The assignable address range for this subnet?

2

Given an IP address for a host 172.168.35.10/20

The address is of class B.

/20

11111111.11111111.11110000.00000000 =
255.255.240.0

256
-
240 = 16

Multiples of 16: 16, 32, 48, 64, 80, ….

The third octet is 35 which is between 32 and 48, so we
choose the minimum value: 32

172.168.32.0

3

172.168.32.0

and the next subnet
172.168.48.0

. We know that the broadcast

172.168.48.0

-

1

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

172.168.47.255

4

The number of useable hosts available for this subnet?

The address is of class B and
11111111.11111111.11110000.00000000, so we borrowed 4
bits to make subnetting and 12 bits are left for hosts

x = 12

the number of useable hosts = 2
x
-
2 = 2
12
-
2 =
4094

5

The number of useable subnets available for this
network?

so we borrowed 4 bits to make subnetting

x = 4

the number of useable subnets = 2
x
-
2 = 2
4
-
2 =
14

6

The assignable address range for this subnet?

172.168.32.1

172.168.32.2

174.168.47.254

7

of 130.10.0.0

Work out the

Subnet mask required for this subnet

useable subnets

The number of hosts for each subnet

The assignable address range of the first 5 useable
subnets

8

of 130.10.0.0

Number of subnets = 2
x
-
2, so we have to find the value of
x so as to get a value closer or equal to 2000.

If we choose x = 11, we get 2
11

-

2 = 2048

2 = 2046.

9

Subnet mask required for this subnet

255.255.0.0. But, we need to borrow 11 (eleven) bits for
subnetting

11111111.11111111.
11111111
.
111
00000

255.255.255.224

10

useable subnets:

256

224 = 32

Multiples of 32: 32, 64, 96, 128, 160, 192, …..

1
st

subnet:
130.10.0.32

130.10.0.63

2
nd

subnet:
130.10.0.64

130.10.0.95

3
rd

subnet:
130.10.0.96

130.10.0.127

4
th

subnet:
130.10.0.128

130.10.0.159

5
th

subnet:
130.10.0.160

130.10.0.191

11

The number of hosts for each subnet

We borrowed 11 (eleven) bits for subnetting, so we are
left with 5 bits for host.

x = 5

number of hosts in each subnets = 2
x
-
2 = 2
5
-
2 =
30

12

The assignable address range of the first 5 useable
subnets:

1
st

subnet:
130.10.0.33

.....

130.10.0.62

2
nd

subnet:
130.10.0.65

…..

130.10.0.94

3
rd

subnet:
130.10.0.97

…..

130.10.0.126

4
th

subnet:
130.10.0.129

…..

130.10.0.158

5
th

subnet:
130.10.0.161

..…

130.10.0.190