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Question

An organization is granted the block 130.56.0.0/16. The administrator wants to create 1024 subnets. Find first and last addresses of last subnet.

1.

130.56.0.0

130.56.254.255

2.

130.56.255.0

130.56.255.255

3.

130.56.0.192

130.255.255.255

4.

130.56.255.192 

130.56.255.255

The correct answer is

4

Analyzing the Subnetting Requirement

The organization has been assigned the IP block 130.56.0.0/16. This is a Class B network address block. The /16 CIDR notation indicates that the first 16 bits are used for the network portion, and the remaining 16 bits are available for host addresses or subnetting.

The requirement is to divide this block into 1024 subnets.

Calculating Subnet and Host Bits

To determine the number of bits needed for subnetting, we use the formula \(2^{\text{subnet bits}} \ge \text{number of subnets}\). We need 1024 subnets.

\[2^{\text{subnet bits}} = 1024\]

Solving for subnet bits:

\[\text{subnet bits} = \log_2(1024) = 10\]

So, 10 bits must be borrowed from the host portion of the original /16 block to create 1024 subnets.

The original mask was /16. After borrowing 10 bits for subnetting, the new subnet mask will be:

\[\text{New mask} = \text{Original network bits} + \text{Subnet bits} = 16 + 10 = /26\]

A /26 mask means the first 26 bits are used for the combined network and subnet ID, leaving the remaining bits for host addresses.

The number of bits available for host addresses is:

\[\text{Host bits} = \text{Total bits} - \text{New mask bits} = 32 - 26 = 6\]

With 6 host bits, the number of addresses in each subnet is \(2^{\text{host bits}} = 2^6 = 64\). This is also the block size or increment value for the subnets.

The number of usable host addresses per subnet is \(2^{\text{host bits}} - 2 = 2^6 - 2 = 64 - 2 = 62\) (subtracting the network address and the broadcast address).

Understanding the Subnetting Scheme

The original /16 block 130.56.0.0 ranges from 130.56.0.0 to 130.56.255.255.

The new /26 mask means the network and subnet ID occupy the first 26 bits. These 26 bits fall across the first, second, third, and fourth octets.

  • Octet 1: 8 bits (Network)
  • Octet 2: 8 bits (Network)
  • Octet 3: 8 bits (Subnet)
  • Octet 4: 2 bits (Subnet) + 6 bits (Host)

The 10 subnet bits comprise all 8 bits of the 3rd octet and the first 2 bits of the 4th octet. The subnets are formed by incrementing the value represented by these 10 bits, starting from 0.

The 10 subnet bits can represent values from 0 (\(0000000000_2\)) to 1023 (\(1111111111_2\)). This gives us 1024 possible subnet IDs, from 0 to 1023.

Finding the Last Subnet

We need to find the first and last addresses of the last subnet. The last subnet corresponds to the subnet ID 1023.

Let's represent the subnet ID 1023 in binary using 10 bits:

\[1023_{10} = 1111111111_2\]

These 10 bits are composed of the 8 bits for the 3rd octet and the first 2 bits for the 4th octet:

  • 3rd Octet bits: \(11111111_2 = 255_{10}\)
  • 4th Octet (first 2 bits): \(11_2\)

First Address of the Last Subnet (Network Address)

The network address of any subnet has the network and subnet bits set according to the subnet ID, and all host bits set to 0.

For the last subnet (ID 1023), the first 26 bits are the network (16 bits) + subnet ID 1023 (10 bits). In binary:

  • Network (130.56): \(10000010_2 . 00111000_2\)
  • Subnet ID 1023 (10 bits): \(1111111111_2\)
  • Host bits (6 bits): \(000000_2\)

Combining these, the full 32-bit address in binary is:

\[10000010 . 00111000 . 11111111 . 11000000\]

Converting to decimal:

  • Octet 1: \(10000010_2 = 128 + 2 = 130\)
  • Octet 2: \(00111000_2 = 32 + 16 + 8 = 56\)
  • Octet 3: \(11111111_2 = 255\)
  • Octet 4: \(11000000_2 = 128 + 64 = 192\)

So, the first address of the last subnet (network address) is 130.56.255.192.

Last Address of the Last Subnet (Broadcast Address)

The broadcast address of any subnet has the network and subnet bits set according to the subnet ID, and all host bits set to 1.

For the last subnet (ID 1023), the first 26 bits are the network (16 bits) + subnet ID 1023 (10 bits). The host bits are all 1s.

  • Network (130.56): \(10000010_2 . 00111000_2\)
  • Subnet ID 1023 (10 bits): \(1111111111_2\)
  • Host bits (6 bits): \(111111_2\)

Combining these, the full 32-bit address in binary is:

\[10000010 . 00111000 . 11111111 . 11111111\]

Converting to decimal:

  • Octet 1: \(10000010_2 = 130\)
  • Octet 2: \(00111000_2 = 56\)
  • Octet 3: \(11111111_2 = 255\)
  • Octet 4: \(11111111_2 = 255\)

So, the last address of the last subnet (broadcast address) is 130.56.255.255.

Summary of the Last Subnet

Subnet Detail Address
Network Address (First Address) 130.56.255.192
Broadcast Address (Last Address) 130.56.255.255
Subnet Mask 255.255.255.192 (/26)

Comparing these results with the provided options, the first and last addresses of the last subnet are 130.56.255.192 and 130.56.255.255.

Revision Table - Subnetting Calculations

Parameter Calculation/Value Explanation
Original Block 130.56.0.0/16 Given network block.
Needed Subnets 1024 Requirement for division.
Subnet Bits 10 \(2^{10} = 1024\). Bits borrowed for subnets.
New CIDR Mask /26 Original /16 + 10 subnet bits.
New Subnet Mask 255.255.255.192 Decimal representation of /26 (11111111.11111111.11111111.11000000).
Host Bits 6 32 total bits - 26 network/subnet bits.
Addresses per Subnet 64 \(2^6\). Block size.
Usable Hosts per Subnet 62 \(2^6 - 2\). Total addresses minus network and broadcast.
Last Subnet ID 1023 Subnet IDs run from 0 to 1023 for 1024 subnets.
Last Subnet Network Address 130.56.255.192 Network + Subnet bits (1023) with Host bits 0s.
Last Subnet Broadcast Address 130.56.255.255 Network + Subnet bits (1023) with Host bits 1s.

Additional Information - IP Subnetting Concepts

Subnetting is the process of dividing a large network into smaller, more manageable sub-networks. Key concepts include:

  • IP Address Classes: Historically, IP addresses were grouped into classes (A, B, C, D, E) determining the default network/host split. 130.56.0.0 is a Class B address.
  • CIDR (Classless Inter-Domain Routing): Replaced classful addressing, allowing more flexible network divisions using a prefix length (e.g., /16, /24, /26).
  • Subnet Mask: A 32-bit number used to distinguish the network portion of an IP address from the host portion. In binary, 1s represent the network/subnet part and 0s represent the host part. For /26, the mask is 255.255.255.192.
  • Network Address: The first address in a subnet, where all host bits are 0. It is used to identify the subnet itself.
  • Broadcast Address: The last address in a subnet, where all host bits are 1s. Packets sent to this address are delivered to all devices on that subnet.
  • Usable IP Range: The IP addresses between the network address and the broadcast address. These can be assigned to hosts.
  • Variable Length Subnet Masking (VLSM): Allows using different subnet masks for different subnets within the same network block, optimizing IP address usage.

Understanding how to calculate the number of subnets, hosts per subnet, network addresses, and broadcast addresses is fundamental in network design and administration.

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Important Questions from IPv4

  1. An organization is granted the block 130.56.0.0/16. The administrator wants to create 1024 subnets. Find the subnet prefix.

  2. An organization is granted the block 130.56.0.0/16. The administrator wants to create 1024 subnets.

    Find first and last addresses of First subnet.

    1.

    130.56.0.0

    130.56.254.254

    2.

    130.56.0.0

    130.56.0.63

    3.

    130.0.0.0

    130.255.255.255

    4.

    130.56.0.0

    130.56.255.63

  3. Suppose a message of size 15000 bytes is transmitted from a source to a destination using IPv4 protocol via two routers as shown in the figure. Each router has a defined maximum transmission unit (MTU) as shown in the figure, including IP header. The number of fragments that will be delivered to the destination is ________ . (Answer in integer)

  4. A machine receives an IPv4 datagram. The protocol field of the IPv4 header has the protocol number of a protocol X.
    Which ONE of the following is NOT a possible candidate for X?
  5. Consider a network that uses Ethernet and IPv4. Assume that IPv4 headers do not use any options field. Each Ethernet frame can carry a maximum of 1500 bytes in its data field. A UDP segment is transmitted. The payload (data) in the UDP segment is 7488 bytes.
    Which ONE of the following choices has the CORRECT total number of fragments transmitted and the size of the last fragment including IPv4 header?
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