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	25
Subnet B	75

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

Before proceeding, verify your IP addresses with the instructor.

Answer 1

1.

a) Since the number of hosts = 25, this means that we need log₂25 ~ 5 host bits. This means that subnet mask will have 27 bits set.

b) IP mask is given as: 11111111.11111111.11111111.11100000

c) The new IP mask is given as: /27 (255.255.255.224)

d) Number of usable subnets = 8

e) Number of usable hosts per subnet = 2⁵ - 2(reserved) = 30

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