Question

which subnet address would include address 192.168.1.88/27 as a usable host address give broadcast and subnet mask

Answer 1

To determine the subnet address that includes the address 192.168.1.88 with a subnet mask of /27, we first need to understand how the /27 subnet mask works.

1. **Subnet Mask**: A /27 subnet provides 32 - 27 = 5 bits for host addresses. This means we have \(2^5 = 32\) total addresses in this subnet.

2. **Subnet Address Calculation**: The subnet mask in dotted decimal notation for /27 is 255.255.255.224. The subnet addresses span 32 addresses, starting from a base address. With /27, the base addresses that are divisible by 32 within the range of 192.168.1.x are:

- 192.168.1.0 (0-31)
- 192.168.1.32 (32-63)
- 192.168.1.64 (64-95)
- 192.168.1.96 (96-127)
- 192.168.1.128 (128-159)
- 192.168.1.160 (160-191)
- 192.168.1.192 (192-223)
- 192.168.1.224 (224-255)

Since the address 192.168.1.88 falls in the range of 64 to 95, it belongs to the subnet:

- **Subnet Address**: 192.168.1.64
- **Subnet Mask**: 255.255.255.224 (/27)

3. **Broadcast Address Calculation**: The broadcast address is the last address in the subnet. For the subnet 192.168.1.64/27, the range of usable IP addresses is from 192.168.1.65 to 192.168.1.94.

Thus, the broadcast address for this subnet is:

- **Broadcast Address**: 192.168.1.95

### Summary:
- **Subnet Address**: 192.168.1.64
- **Subnet Mask**: 255.255.255.224 (/27)
- **Broadcast Address**: 192.168.1.95

The usable host addresses are from 192.168.1.65 to 192.168.1.94, and the address 192.168.1.88 falls within this range.