Question

A person comes to you and says they need a network with 128 hosts. You currently have the following network:

192.168.0.0/24.

How would you subnet this network to satisfy the requirement?

Group of answer choices

You wouldn't subnet it. It cannot be broken down to satisfy the requirement.

You would borrow one bit and have the subnet mask of 255.255.255.128.

You would borrow two bits and have the subnet mask of 255.255.255.192.

You would borrow 3 bits to come up with the subnet mask of 255.255.224.0.
There are no human answers yet.
Answered by GPT-4o mini
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:
1. **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.

2. **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 \(2^8\).

3. **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.**