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A user has been assigned an IP address 110.10.32.68/27. Which of the following IP addresses can be assigned for directed broadcasting in the same network.
  • a)
    110.10.32.255
  • b)
    10.10.32.127
  • c)
    110.10.32.95
  • d)
    110.10.32.0
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
A user has been assigned an IP address 110.10.32.68/27. Which of the ...
IP : 110.10.32.68
Subnet mask : 255.255.255.2 24
Net ID : 110.10.32.64
We have 5 bits for host ID. When all bits of host Id is 1 then it is a direct broadcasting address.
So,
\begin{tabular}{cc}
110.10 .32 .010 & 11111 \
\hline Network & Host \
bit & bit \
=110.10.32.95 &
\end{tabular}
So, option (c) is correct.
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Most Upvoted Answer
A user has been assigned an IP address 110.10.32.68/27. Which of the ...
Explanation:


To determine the possible IP addresses that can be assigned for directed broadcasting in the given network, we need to understand the concept of directed broadcasting and calculate the network address and broadcast address.

Given:

IP address: 110.10.32.68/27

Calculating Network Address:

To calculate the network address, we need to perform a bitwise logical AND operation between the IP address and the subnet mask.

Subnet mask: /27 (27 bits set to 1, and the remaining 5 bits set to 0)
Binary representation of the subnet mask: 11111111.11111111.11111111.11100000

Performing bitwise AND operation:
```
IP address: 01101110.00001010.00100000.01000100
Subnet mask: 11111111.11111111.11111111.11100000
Network address:01101110.00001010.00100000.01000000
```
Network address: 110.10.32.64

Calculating Broadcast Address:

To calculate the broadcast address, we need to determine the maximum value that can be represented in the host portion of the IP address.

Given subnet mask: /27, which means we have 27 bits fixed for the network portion and the remaining 5 bits for the host portion.

Number of bits in the host portion: 2^5 = 32
Maximum value in the host portion: 32 - 1 = 31

Performing bitwise OR operation between the network address and the maximum host value:
```
Network address: 01101110.00001010.00100000.01000000
Maximum host value: 00000000.00000000.00000000.00011111
Broadcast address: 01101110.00001010.00100000.01011111
```
Broadcast address: 110.10.32.95

Identifying Possible Directed Broadcasting Addresses:

To determine the possible directed broadcasting addresses, we need to consider the network address and the broadcast address.

In the given options:
a) 110.10.32.255 - This IP address is not within the range of the network address and the broadcast address, so it cannot be assigned for directed broadcasting.
b) 10.10.32.127 - This IP address is not within the range of the network address and the broadcast address, so it cannot be assigned for directed broadcasting.
c) 110.10.32.95 - This IP address is within the range of the network address and the broadcast address, so it can be assigned for directed broadcasting.
d) 110.10.32.0 - This IP address is the network address, so it cannot be assigned for directed broadcasting.

Therefore, the correct answer is option 'C' (110.10.32.95), as it falls within the range of the network address and the broadcast address.
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A user has been assigned an IP address 110.10.32.68/27. Which of the following IP addresses can be assigned for directed broadcasting in the same network.a)110.10.32.255b)10.10.32.127c)110.10.32.95d)110.10.32.0Correct answer is option 'C'. Can you explain this answer?
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