To determine how to subnet the network 192.168.0.0/24 to accommodate 128 hosts, we need to consider how many subnets we can create and how many hosts each subnet can support.
Calculation:
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Current Network: 192.168.0.0/24
- This means the subnet mask is 255.255.255.0, allowing for \(2^{(32-24)} = 2^8 = 256\) IP addresses in total (0-255).
- Out of these 256 addresses, we will have 2 addresses reserved (the network address and the broadcast address), leaving us with \(256 - 2 = 254\) usable addresses.
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Hosts Requirement: 128 hosts
- To find out how many bits we need for the hosts, we need to find the smallest power of 2 that is greater than or equal to 128.
- The closest power of 2 that can accommodate 128 hosts is \(2^7 = 128\), but we need to account for the 2 reserved addresses, thus requiring \(2^8 = 256\) addresses in total.
- Therefore, we need 8 bits for hosts which means we can fit 128 hosts perfectly by using \(2^8\).
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Subnetting:
- Since we are starting with a /24 network (which has 8 bits for hosts), we need to borrow some bits from the host part to create subnets.
- Borrowing 1 bit from the host part (from /24 to /25) gives us \(2^1 = 2\) subnets, where each subnet would have \(2^{(32-25)} = 2^7 = 128\) total addresses (126 usable after excluding network and broadcast).
- Therefore, with a subnet mask of 255.255.255.128 (/25), each subnet would have exactly 128 usable host addresses.
Conclusion:
To satisfy the requirement for 128 hosts, you would borrow one bit and have the subnet mask of 255.255.255.128.
So, the correct answer is: You would borrow one bit and have the subnet mask of 255.255.255.128.
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