Question

Given an IP address and mask of 192.168.0.0 /24 (address / mask), design an IP addressing scheme that satisfies the following requirements. Network address/mask and the number of hosts for Subnets A and B will be provided by your instructor.

Subnet	Number of Hosts
Subnet A	55
Subnet B	25

The 0^th subnet is used. No subnet calculators may be used. All work must be shown on the other side of this page.

		Subnet A
Specification		Student Input	Points
Number of bits in the subnet			(5 points)
IP mask (binary)
New IP mask (decimal)
Maximum number of usable subnets (including the 0th subnet)
Number of usable hosts per subnet
IP Subnet
First IP Host address
Last IP Host address

		Subnet B
Specification		Student Input	Points
Number of bits in the subnet			(5 points)
IP mask (binary)
New IP mask (decimal)
Maximum number of usable subnets (including the 0th subnet)
Number of usable hosts per subnet
IP Subnet
First IP Host address
Last IP Host address

Host computers will use the first IP address in the subnet. The network router will use the LAST network host address. The switch will use the second to the last network host address.

Write down the IP address information for each device:

Device	IP address	Subnet Mask	Gateway	Points
PC-A				(5 points)
R1-G0/0			N/A
R1-G0/1			N/A
S1			N/A
PC-B

Answer 1

Given IP address is: 192.168.0.0/24

Subnet	Number of Hosts
A	55
B	25

For Subnet A:

Number of hosts needed is => 55

55 is near to 64 (in terms of powers of 2) => 64 = 2⁶

So, 6 bits needed for host.

Network bits are => 32 - 6 = 26

Specification	Answer
Number of bits in the subnet	2 (i.e., 26 - 24 (old mask) = 2)
IP Mask (binary)	11111111.11111111.11111111.11000000
New IP Mask (Decimal)	255.255.255.192
Maximum number of usable subnets (including the 0th subnet)	4 (subnet bits = 2, so 22 = 4)
Number of usable hosts per subnet	62 ( host bits = 6, so 26 - 2 = 62)
IP Subnet	192.168.0.0/26
First IP Host address	192.168.0.1
Last IP Host address	192.168.0.62

For Subnet B:

Number of hosts needed is => 25

25 is near to 32 (in terms of powers of 2) => 32 = 2⁵

So, 5 bits needed for host.

Network bits are => 32 - 5 = 27

IP address information for each device given:

Device	IP address	Subnet Mask	Gateway
PC-A	192.168.0.1	255.255.255.192	192.168.0.63
R1-G0/0	192.168.0.63	255.255.255.192	N/A
R1-G0/1	192.168.0.95	255.255.255.224	N/A
S1	192.168.0.94	255.255.255.224	N/A
PC-B	192.168.0.65	255.255.255.224	192.168.0.95