ip classes and subnetting.

1. ASSIGNMENT #1
SUBMITTED TO:
MISS SARA SARWAR
SUBMITTED BY:
RABIA ZAFAR
17581556-045
BS IT(5TH
SEMESTER)
SECTION A

2. Question 1:
Write the default Masks for the Class A Class B and Class C IP addresses.
Answer:
 Class A default subnet mask (255.0.0.0).
 Class B default subnet mask (255.255.0.0).
 Class C default subnet mask (255.255.255.0).
Question 2:
How we can distinguish Class A, Class B, Class C, Class D and Class E IP addresses from
each other. Write the range of first octet in decimal and Binary for all the 5 IP address
classes.
Answer:
IP addresses are 32-bit binary numbers that are separated by a dot. The number is
generally represented as 4 octets of numbers from the ranges 0-255.
Range:
Classes Range
Class A 0-126
Class B 128-191
Class C 192-223
Class D 224-239
Class E 240-255
Class A:
Decimal Binary
1 00000001
Class B:
Decimal Binary
128 10000000
Class C:
Decimal Binary
192 11000000
Class D:

3. Decimal Binary
224 11100000
Class E:
Decimal Binary
240 11110000
Question 3:
Write the default subnet Masks for the following IP addresses:
a. 179.65.225.4
b. 222.35.20.18
c. 111.7.80.0?
Answer:
a. 179.65.225.4, this IP belongs to class B. The default subnet mask of class B is
255.255.0.0.
b. 222.35.20.18, this IP belongs to class D. the default subnet mask of class D is
255.255.255.0.
c. 111.7.80.0, this IP belongs to class A. The default subnet mask of class A is
255.0.0.0.
Question 4:
Write down the three available ranges for assigning Private IP addresses recommended
by IANA(Internet Assigned Number Authority).
Answer:
The Internet Assigned Numbers Authority (IANA) has assigned several address ranges
to be used by private networks.
Address ranges to be use by private networks are:
 Class A: 10.0.0.0 to 10.255.255.255
 Class B: 172.16.0.0 to 172.31.255.255
 Class C: 192.168.0.0 to 192.168.255.255
Question 5:
A broadcast address is the one that addresses to all the hosts in any network. State that to
create a broadcast address, all the bits of network ID portion or all the bits of host ID
portion are set to 1? Write down the broadcast addresses of the networks to which the

4. following IP addresses belong, write network addresses and ranges of their valid IP
address too. (No subnetting)
a. 129.65.225.4 b. 211.35.20.18
c. 180.47.115.6
Answer:
(a)
192.65.225.4 Class B
129.65.0.0 Network ID
129.65.255.255 Broadcast ID
(b)
211.35.20.18 Class C
211.35.20.0 Network ID
211.35.20.255 Broadcast ID
(c)
180.47.115.6 Class B
180.47.0.0 Network ID
180.47.255.255 Broadcast ID
Question 6:
Subnet Mask or Custom mask tells us that how many bits are used for Subnet ID portion
and how many for host ID portion. Identify how many bits are used for subnetting in the
following IP address using its subnet mask:
a. IP Address: 135.65.225.4 Subnet Mask: 255.255.240.0
b. IP Address: 210.35.20.18 Subnet Mask: 255.255.255.248
c. IP Address: 190.47.115.6 Subnet Mask: 255.255.254.0
Answer:
(a)
IP address:
135.65.255.4 from class B
Subnet mask:
255.255.240.0

5. 11111111.11111111.11110000.00000000
Bits for subnet ID portion=20
Bits for host ID portion=12
(b)
IP address:
210.35.20.18 from class C
Subnet mask:
255.255.255.248
11111111.11111111.11111111.11111000
Bits for subnet ID portion=29
Bits for host ID portion=3
(c)
IP address:
190.47.115.6 from class B
Subnet mask:
255.255.254.0
11111111.11111111.11111110.00000000
Bits for subnet ID portion=23
Bits for host ID portion=9
Question 7:
Extract the Network Addresses of the given IP addresses in question number 8, using the
subnet masks given with them. (Remember that ANDing the IP address with the Mask
extracts the network address from the given IP address).
Answer:
(a)
IP address: 135.65.225.4

6. Subnet mask: 255.255.0.0
Binary of IP address: 10000111.010000011.11100001.00000100
Binary of subnet mask: 11111111.11111111.00000000.00000000
---------------------------------------------------------------------
10000111.01000001.00000000.00000000
Network address: 135.65.0.0
(b)
IP address: 210.35.20.18
Subnet mask: 255.255.255.0
Binary of IP address: 11010010.00100011.00010100.00010010
Binary of subnet mask: 11111111.11111111.11111111.00000000
-------------------------------------------------------------------
11010010.00100011.00010100.00000000
Network address: 210.34.20.0
(c)
IP address: 190.47.115.6
Subnet mask: 255.255.0.0
Binary of IP address: 10111110.00101111.01110011.00000110
Binary of subnet mask: 11111111.11111111.00000000.00000000
---------------------------------------------------------------------
10111110.00101111.00000000.00000000
Network address: 190.47.0.0
Question 8:
Which of the following Subnet masks would allow a class A network to allow subnets to
have up to 150 hosts and allow for up to 164 subnets?
a. 255.0.0.0

7. b. 255.255.255.0
c. 255.255.192.0
d. 255.255.240.0
e.255.255.252.0
Answer:
(a)
255.0.0.0 from class A
Binary: 11111111.00000000.00000000.00000000
Subnet: 2^s =2^0 =1
Host: 2^h – 2 = 2^24 -2 =16,777,216 -2 =16,777,214
(b)
255.255.255.0 from class C
Binary: 11111111.11111111.11111111.00000000
Subnet: 2^s= 2^0 = 1
Host: 2^h-2 =2^8-2=256-2 =254
(c)
255.255.192.0 from class B
Binary: 11111111.11111111.11000000.00000000
Subnet: 2^s=2^2=4
Host: 2^s-2=2^14-2=16384-2=16382
(d)
255.255.240.0 from class B
Binary: 11111111.11111111.11110000.0000000
Subnet: 2^s =2^4 =16
Host: 2^h-2 =2^12 -2=4096-2=4094
(e)

8. 255.255.252.0 from class B
Binary: 11111111.11111111.11111100.00000000
Subnet: 2^s = 2^6 =64
Host: 2^h-2 = 2^10-2=1024-2=1022
Question 9:
Suppose you have a class C Network 208.94.115.0. Your task is to design a subnet scheme
so that we can create 16 Network segments (subnets) within this Network. Each subnet
should support 10-14 hosts.
2n
>= 14
24
>= 14
16 > 14
So, n = 4.
a. How many bits would you use for the subnet ID?
28 bits are used for subnet ID.
b. How many bits would you use for the Host ID?
4 bits are used for Host ID
c. How many maximum possible subnets will be there?
24
subnets are possible because we have 8 bits for Host ID. So,
16 maximum subnets are possible.
d. How many maximum possible hosts will be there in each subnet?
24
subnets are possible because we have 8 bits for Host ID. So,
16 maximum hosts are possible per subnet.
e. Write down the Subnet Mask of your scheme.
208.94.115.11110000
255.255.255.240
f. Write down All the valid IP addresses
Valid IP’s
 208.94.115.0/28
 208.94.115.16/28
 208.94.115.32/28
 208.94.115..48/28
 208.94.115.64/28
 208.94.115.80/28
 208.94.115.96/28
 208.94.115.112/28
 208.94.115.128/28

9.  208.94.115.160/28
 208.94.115.176/28
 208.94.115.192/28
 208.94.115.208/28
 208.94.115.224/28
 208.94.115.240/28
g. The broadcast address of the First subnet of your scheme.
208.94.115.15 /28
Question 10:
Suppose you have a class C Network 220.94.115.0. Your task is to design a subnet
scheme so that we can create 28 Network segments (subnets) within this Network. Each
subnet should support hosts as given below.
a. 2 Network Segment support 30 Hosts
b. 4 Network Segment support 14 Hosts
c. 8 Network Segment support 6 Hosts
d. 14 Network Segment support 2 Hosts
Steps
1- IP
220.94.115.0
2- Convert to binary
220.94.115.00000000
3- Find n
2n
>= max host value
2n
>= 30
25
>= 30
N=5
4- Give bits to network
220.94.115.11100000
5- IP of this network
220.94.115.224
6- Subnet Mask of IP
255.255.255.224/27
7- Subnets
 220.94.115.0/27
 220.94.115.32/027
 220.94.115.64/027

10.  220.94.115.96/027
 220.94.115.128/027
 220.94.115.160/027
 220.94.115.192/027
 220.94.115.224/027
These are the network segments that can be created. Each subnet will support
up to 32 hosts.
To make the 28 segments we can use Class A or B for more segments.