Q1: Given an IP address 94.0.0.
0/20, answer the following
questions:
1. What is the IP address class?
Answer:
Identify the first octet of the IP address (94).
The IP address class is determined as follows:
o Class A: 1-126
o Class B: 128-191
o Class C: 192-223
Since 94 falls in the range of Class A, the IP address is Class A.
2. What are the Netid and Hostid?
Answer:
For a Class A address:
o Netid: First octet (94).
o Hostid: Remaining three octets (0.0.0).
Therefore:
o Netid: 94
o Hostid: 0.0.0
3. How many bits were borrowed for subnetting?
Answer:
The default subnet mask for Class A is /8.
The given subnet mask is /20.
Bits borrowed = 20 - 8 = 12 bits.
4. How many total subnets are created?
Answer:
Number of subnets = 2^12 = 4096 subnets.
5. How many bits are left for hosts?
Answer:
Total bits in an IPv4 address = 32.
Bits used for subnetting = 20.
Bits left for hosts = 32 - 20 = 12 bits.
6. How many hosts per subnet?
Answer:
Number of hosts = 2^12 - 2 = 4094 hosts per subnet.
7. How many total hosts are available across all subnets?
Answer:
Total hosts = 4096 x 4094 = 16,769,024 hosts.
8. What is the new subnet mask after subnetting?
Answer:
The given subnet mask is /20.
Convert /20 to dotted decimal:
o 255.255.240.0
New subnet mask: 255.255.240.0
9. Show the subnet address, host range, and broadcast address of the first
and last subnet.
Answer:
First Subnet:
o Subnet Address: 94.0.0.0
o Host Range: 94.0.0.1 to 94.0.15.254
o Broadcast Address: 94.0.15.255
Last Subnet:
o Subnet Address: 94.255.240.0
o Host Range: 94.255.240.1 to 94.255.255.254
o Broadcast Address: 94.255.255.255
Q2: Given the Class C IP address 192.168.2.24 and subnet mask
255.255.255.192, answer the following questions:
1. What is the default subnet mask?
Answer:
The default subnet mask for Class C is 255.255.255.0 (/24).
2. How many total addresses are granted by the given subnet mask?
Answer:
The given subnet mask is 255.255.255.192 (/26).
Number of total addresses = 2^6 = 64 addresses.
3. How many subnets and valid hosts per subnet are created?
Answer:
Default mask for Class C: /24
Given mask: /26
Bits borrowed: 26 - 24 = 2 bits
Number of subnets: 2^2 = 4 subnets
Number of valid hosts per subnet: 2^6 - 2 = 62 hosts
4. What is the valid host range for the first subnet?
Answer:
First Subnet:
o Subnet Address: 192.168.2.0
o Host Range: 192.168.2.1 to 192.168.2.62
o Broadcast Address: 192.168.2.63
Q3: Given the IP address 100.70.52.70/19, answer the following
questions:
1. What is the network address?
Answer:
Subnet mask for /19: 255.255.224.0
Network address: 100.70.32.0
2. What is the broadcast address?
Answer:
Broadcast address: 100.70.63.255
3. What is the first valid host address?
Answer:
First valid host address: 100.70.32.1
4. What is the last valid host address?
Answer:
Last valid host address: 100.70.63.254
5. How many total hosts are available?
Answer:
Number of host bits: 13
Total host addresses: 2^13 - 2 = 8190
6. How many subnets can be created if each subnet must support at least 30
hosts?
Answer:
Host bits required: 5 (2^5 - 2 = 30)
Subnet bits: 32 - 19 - 5 = 8 bits
Maximum number of subnets: 2^8 = 256
Q4: Given a network block with IP address 222.86.4.0/22, answer
the following questions:
1. How many subnets can be created to support 46 hosts per subnet?
Answer:
Host bits required: 6
Subnet bits: 10 - 6 = 4 bits
Number of subnets: 2^4 = 16
2. What is the maximum possible number of hosts per subnet?
Answer:
2^6 - 2 = 62
3. What is the new subnet mask after subnetting?
Answer:
/22 + 4 = /26
255.255.255.192
4. What is the host range and broadcast address for subnet #4
(222.86.4.192/26)?
Answer:
Host Range: 222.86.4.193 to 222.86.4.254
Broadcast Address: 222.86.4.255
Q5: What is VLSM (Variable Length Subnet Mask)?
Answer:
VLSM allows subnets of varying sizes within the same network,
enabling efficient IP address utilization.
Example using 222.86.4.0/22:
o Subnet 1: 50 hosts (/26)
o Subnet 2: 100 hosts (/25)
o Subnet 3: 10 hosts (/28)
This method minimizes IP wastage by allocating appropriate subnet
sizes based on requirements.
