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
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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.
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).
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.
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.
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:
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:
Combining these, the full 32-bit address in binary is:
\[10000010 . 00111000 . 11111111 . 11000000\]Converting to decimal:
So, the first address of the last subnet (network address) is 130.56.255.192.
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.
Combining these, the full 32-bit address in binary is:
\[10000010 . 00111000 . 11111111 . 11111111\]Converting to decimal:
So, the last address of the last subnet (broadcast address) is 130.56.255.255.
| 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.
| 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. |
Subnetting is the process of dividing a large network into smaller, more manageable sub-networks. Key concepts include:
Understanding how to calculate the number of subnets, hosts per subnet, network addresses, and broadcast addresses is fundamental in network design and administration.
An organization is granted the block 130.56.0.0/16. The administrator wants to create 1024 subnets. Find the subnet prefix.
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 |
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)
