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IP Addressing and Subnetting Fundamentals

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Question 1

In CIDR notation, the network 192.168.5.0 with subnet mask 255.255.255.0 is written as 192.168.5.0/[[blank1]].

Explanation

255.255.255.0 in binary is 11111111.11111111.11111111.00000000 — the first 24 bits are all 1s. In CIDR notation, the prefix length after the slash represents the number of consecutive 1-bits in the subnet mask. So 255.255.255.0 = /24. Common conversions: 255.0.0.0 = /8, 255.255.0.0 = /16, 255.255.255.128 = /25, 255.255.255.192 = /26.

Question 2

Which of the following is the broadcast address for the network 10.1.1.0/28?

Answer choices

  • A. 10.1.1.14

  • B. 10.1.1.15 (Correct)

  • C. 10.1.1.16

  • D. 10.1.1.255

Explanation

A /28 subnet has 4 bits for hosts (32-28=4), giving 16 addresses per subnet (2^4=16). For the network 10.1.1.0/28: the network address is 10.1.1.0 (first address), the broadcast address is 10.1.1.15 (last address, 0+16-1=15), and usable hosts are 10.1.1.1 through 10.1.1.14 (14 usable). The subnet mask is 255.255.255.240. The next subnet starts at 10.1.1.16.

Question 3

You need to subnet the network 192.168.10.0/24 into subnets that each support at least 50 hosts. What is the minimum subnet mask you should use?

Answer choices

  • A. /25 (255.255.255.128)

  • B. /26 (255.255.255.192) (Correct)

  • C. /27 (255.255.255.224)

  • D. /28 (255.255.255.240)

Explanation

To support at least 50 hosts, you need at least 52 addresses (50 + network + broadcast). /26 gives 2^6 = 64 addresses, with 62 usable hosts — enough for 50. /27 gives only 30 usable hosts, which is not enough. /25 gives 126 usable hosts (more than needed but still valid). The question asks for the minimum subnet (fewest host bits while still meeting the 50-host requirement), so /26 is correct.

Question 4

Arrange the following steps in the correct order to manually calculate the usable host range for the subnet 192.168.100.64/26:

Answer choices

  • A. A) Determine the broadcast address (last address in the subnet)

  • B. B) Identify the prefix length: /26 = 6 host bits

  • C. C) The usable range is from .65 to .126

  • D. D) Calculate total addresses: 2^6 = 64

  • E. E) Network address = 192.168.100.64; broadcast = 192.168.100.127

Explanation

Subnet calculation steps: (1) Identify prefix — /26 means 6 host bits. (2) Calculate block size — 2^6 = 64. (3) Identify subnet — 64 is a multiple of 64, so network is 192.168.100.64. (4) Broadcast is network + 63 = 192.168.100.127. (5) Usable hosts: .65 through .126 (64 total - 2 = 62 usable). First usable = network + 1, last usable = broadcast - 1.

Question 5

A network administrator needs to create 6 subnets from the 192.168.50.0/24 network using VLSM. What is the minimum number of subnet bits needed to create at least 6 subnets?

Answer choices

  • A. 2 bits (4 subnets)

  • B. 3 bits (8 subnets) (Correct)

  • C. 4 bits (16 subnets)

  • D. 6 bits (64 subnets)

Explanation

To create subnets, you borrow bits from the host portion. The number of subnets = 2^n where n = borrowed bits. 2^2 = 4 subnets (not enough for 6), 2^3 = 8 subnets (sufficient for 6). Borrowing 3 bits from a /24 creates a /27. Each /27 subnet has 32 addresses with 30 usable hosts. You get 8 subnets of /27 from a /24.