The New Subnet Mask For Each Of Them I Need The Following
The New Subnet Mask For Each Of Them I Need The Following Information
For each of the subnets, I need the following information: the new subnet mask after subnetting (10%), the network address of the subnet, the broadcast address, the range of available IP addresses, and detailed calculations demonstrating how these answers are derived (50%). The objective is to subnet with as few subnets as possible while still meeting the requirement, thereby maximizing the number of hosts per subnet. The specific scenario involves subnetting the network 47.0.0.0/8 into 7,344 subnets, and providing detailed details for subnets #1, #3665, #3723, and #7344.
Paper For Above instruction
The task of subnetting a large Class A network such as 47.0.0.0/8 into 7,344 subnets demands careful calculation to optimize host availability while meeting subnetting requirements. In this context, the key is to understand the fundamentals of subnet masking, binary manipulation, and how to transition from the original network mask to the new one that yields the desired number of subnets.
Understanding the Original Network and Requirements
The network 47.0.0.0/8 indicates that the default subnet mask for class A addresses is 255.0.0.0, which allows for approximately 16 million hosts per network. However, subnetting enables partitioning such a single large network into smaller, manageable segments—subnets—each with a distinct network address and a limited host capacity.
The requirement to subnet into exactly 7,344 subnets suggests computing the appropriate subnet bits to borrow from the host portion of the address. Each subnetting decision influences the number of available hosts per subnet, as more subnet bits reduce the number of host bits, and vice versa.
Calculating the Number of Subnet Bits Needed
To determine the number of subnet bits required, we use the formula where 2^n ≥ number of subnets, with n being the number of bits borrowed for subnetting. Here, 2^n ≥ 7,344.
Calculating n:
2^10 = 1024 (not sufficient)
2^12 = 4096 (still insufficient)
2^13 = 8192 ≥ 7,344 (suitable)
Therefore, 13 bits are needed for subnetting.
Determining the New Subnet Mask
Since the original mask is /8 (or 255.0.0.0), borrowing 13 bits for subnets modifies the mask as follows:
- Default network bits: 8
- Additional subnet bits: 13
- Total network bits: 8 + 13 = 21
Hence, the new subnet mask becomes /21, equating to 255.255.248.0. The breakdown of the mask is:
- 8 bits for the first octet: 255
- 8 bits for the second octet: 255
- 5 bits for the third octet: 248 (11111000 in binary)
- Remaining 3 bits for hosts: 0 (since they are part of the network bits)
Calculating Host Capacity per Subnet
Given 21 bits for network, the remaining 11 bits are for hosts, which provide 2^11 - 2 = 2048 - 2 = 2046 hosts per subnet (subtracting network and broadcast addresses).
Identifying Specific Subnets: #1, #3665, #3723, and #7344
With a /21 subnet mask, subnet addresses increment in blocks of 8 in the third octet. Here's the step-by-step calculation:
Subnet #1
- Subnet number = 0 (first subnet)
- Network address: 47.0.0.0 (base address + (subnet number × block size))
- Block size in third octet = 8 (from mask /21, 255.255.248.0)
- Network address: 47.0.0.0 + (0 × 8) = 47.0.0.0
- Broadcast address: 47.0.7.255 (network address + block size - 1) in the third octet
- Range of usable addresses: 47.0.0.1 to 47.0.7.254
Subnet #3665
- Subnet number = 3664 (since subnet numbering starts at 0, for the 3665th subnet)
- Calculate third octet offset: 3664 × 8 = 29312
- Find the corresponding third octet: 29312 / 256 = 114 (since 256 × 114 = 29184)
- Remaining in third octet: 29312 - (114 × 256) = 128
- Network address: 47.114.128.0
- Broadcast address: 47.114.135.255
- Host range: 47.114.128.1 to 47.114.135.254
Subnet #3723
- Subnet number = 3722
- Third octet offset: 3722 × 8 = 29776
- Third octet: 29776 / 256 = 116; remainder: 0
- Network address: 47.116.0.0
- Broadcast address: 47.116.7.255
- Host range: 47.116.0.1 to 47.116.7.254
Subnet #7344
- Subnet number = 7343
- Third octet offset: 7343 × 8 = 58744
- Third octet: 58744 / 256 = 229; remainder: 0
- Network address: 47.229.0.0
- Broadcast address: 47.229.7.255
- Host range: 47.229.0.1 to 47.229.7.254
Summary of Calculations and Final Answer
By starting from a base /8 network and borrowing 13 bits for subnetting, the new mask is /21 (255.255.248.0). This configuration provides 2046 hosts per subnet and allows for subnet addresses in increments of 8 in the third octet. The specific subnets requested—#1, #3665, #3723, and #7344—are derived by multiplying the subnet index (minus one) by the block size (8) to find their respective network, broadcast, and host ranges.
Conclusion
This subnetting strategy provides an optimal balance between maximizing host capacity within each subnet and accommodating the total number of required subnets. It demonstrates the critical role of binary calculations in network design and management, ensuring scalable, efficient IP address utilization in large networks like 47.0.0.0/8.
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