To determine the correct subnetting approach for the requirement of 128 hosts within the 192.168.0.0/24 network, let's analyze the options based on the total number of usable hosts each subnet can provide.
A /24 network has:
- Subnet mask: 255.255.255.0
- Total IPs: 2^(32-24) = 256
- Usable IPs: 256 - 2 (network and broadcast addresses) = 254 usable IPs
To find out how many bits we need to borrow for a subnet that can accommodate at least 128 hosts, we can use the formula for calculating the number of usable hosts in a subnet:
Usable Hosts = 2^(number_of_host_bits) - 2
Now, since we need at least 128 usable IPs, we can solve it as follows:
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Calculate the minimum number of host bits required:
- To satisfy 128 usable hosts:
- 2^(n) - 2 >= 128
- 2^(n) >= 130
- n must be at least 8 (since 2^8 = 256, and 256 - 2 = 254, which is sufficient).
- To satisfy 128 usable hosts:
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The original subnet mask is /24, which means there are 8 bits for host addresses.
- If we keep 8 host bits, we can create 1 subnet with 254 usable addresses.
- If we borrow 1 bit from the host portion, we can create additional subnets:
- New mask: /25 (255.255.255.128)
- Usable addresses per subnet: 2^(7) - 2 = 126 (insufficient for 128 hosts).
- If we borrow 2 bits:
- New mask: /26 (255.255.255.192)
- Usable addresses per subnet: 2^(6) - 2 = 62 (insufficient).
- If we borrow 3 bits:
- New mask: /27 (255.255.255.224)
- Usable addresses per subnet: 2^(5) - 2 = 30 (insufficient).
- If we borrow only 1 bit and keep the 8 host bits, we can use /25 which provides 126 usable addresses – thus not sufficient.
So, for 128 hosts, we need to retain the original /24 mask (255.255.255.0) as it provides 254 usable addresses without borrowing bits.
Therefore, the best option from those provided is:
You wouldn't subnet it. It cannot be broken down to satisfy the requirement.
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