Question

Assume that we have a machine with IP address 15.107.33.204 with netmask 255.255.255.240.

Assume that you manage a /11 address space, and that we have fully divided it up into into

Child networks the size of the one on which this machine resides. Answer the following questions:

a.

(1 pt) What is the netmask of the parent network in dotted decimal and “slash” notations?

b.

(1 pt) What “class” is the parent network?

c.

(1 pt) How many bits do you manage?

d.

(1 pt) What is the parent’s IP address (network number)?

e.

(1 pt) What is the netmask of the child network in dotted decimal and “slash” notations?

f.

(1 pt) What is the “class” of the child network?

g.

(1 pt) How many bits are there for host number portion for each of these subnets?

h.

(1 pt) What is the child’s IP address (network number)?

i.

(1 pt) How many bits are available for network number portion, within the parent address space, for these subnets?

j.

(1 pt) How many usable subnets of this size can be created?

k.

(1 pt) How many usable addresses will be available on each of these subnets?

l.

(1 pt) List the IP addresses of first three and last three of these usable subnets.

m.

(1 pt) Give the network number (IP address) of the next to the last usable subnet.

n.

(1 pt) Give the broadcast address of the next to the last usable subnet.

o.

(1 pt) Give the

first usable address on the next to the last usable subnet.

p.

(1 pt) Give the last usable address on the next to the last usable subnet.

Answer 1

a.

Given subnet prefix = 11

An IP address of size 32 bits.

Network bits = 11

Host bits = 21

There are 4 octet in IPv4 address.

First 11 bits will be 1 and other will be 0.

Thus netmask in binary: 11111111.11100000.00000000.00000000

Convert the binary to decimal.

Netmask in dotted decimal form: 255.224.0.0

Slash Notation: 11

b.

Class A address range: 0 to 127

First octet has 15 that corresponds to class A.

Parent network belongs to class A.

But if the network is considered with /11 address space then it is a classless network.

c.

Managable bits = host bits = 32-11 = 21

d.

Parent IP address can be calculated as:

Given IP address: 15.107.33.204

IP in binary: 00001111.01101011.00100001.11001100

Set the last 21 bits as 0.

Parent IP address: 00001111.01100000.00000000.00000000

Now convert the IP to decimal:

15.96.0.0

e.

The child network is subnetted to netmask :255.255.255.240

Netmask in slash notation can be calculated as follows:

n = 256-240

n= 16

Number of bits^ 2 = n

Number of bits^ 2 = 16

Number of bits = 4

Subnet mask = 32-4 = 28

Slash notation: 28

Dotted notation: 255.255.255.240

f.

Child network has subnet mask of /28 thus it is classless.

g.

IP address size = 32

Subnet mask = 28

Number of bits for host portion = 32-28 = 4

h.

Given IP address: 15.107.33.204

IP in binary: 00001111.01101011.00100001.11001100

Netmask = /28

Number of bits to be set as 0 = 32-28

Set the last 4 bits as 0.

Child IP address: 00001111.01101011.00100001.11000000

Now convert the IP to decimal:

15.107.33.192

i.

Bits available in parents address for subnet = new subnet prefix – old subnet prefix

= 28-11

Bits available = 17

j.

Number of subnets = 2^Bits Available = 2^17

k.

Two address are reserved for each subnet one for broadcast and one for network.

Number of usable address = (2^(IP Address Size – Subnet Prefix))-2

= (2^(32-28))-2

= (2^4) – 2

= 16 – 2

Number of usable address = 14

l.

Child network address: 15.107.33.192

Address Range: 15.107.33.193 to 15.107.33.206

First three usable IP’s are 15.107.33.193, 15.107.33.194 and 15.107.33.195.

Last three usable IP’s are 15.107.33.204, 15.107.33.205 and 15.107.33.206.

m.

First IP address = Parent IP = 15.96.0.0

Last IP address can be calculated as follows:

Given IP address: 15.107.33.204

IP in binary: 00001111.01101011.00100001.11001100

Set the last 21 bits as 1.

Parent IP address: 00001111.01111111. 11111111.11111111

Now convert the IP to decimal:

15.127.255.255

Number of IP’s = 16/ subnet

Last network address: 15.107.255.240/28

Second last network address: 15.107.255.224/28

n.

Since the network address of second last usable subnet is 15.107.255.224/28

IP’s in subnet are 16.

Thus, Broadcast Address is 15.107.255.239

o.

Since the network address of second last usable subnet is 15.107.255.224/28

Thus, the first usable address is 15.107.255.225           

p.

Since the broadcast address of second last usable subnet is 15.107.255.239/28

Thus, the first usable address is 15.107.255.238