Question 5411 – Data-Communication
April 30, 2024Question 11661 – Queues-and-Stacks
April 30, 2024Question 8430 – Subnetting
In the network 200.10.11.144/27, the fourth octet (in decimal) of the last IP address of the network which can be assigned to a host is ____________.
Correct Answer: A
Question 4 Explanation:
No. of bit in HID part = 32-27 = 5 bits
Subnet mask is 255.255.255.224
Do AND with given IP and subnet mask then we get NID 200.10.11.128
In fourth octet first three bit will fixed for subnet and remaining 5 bits is for HID, so maximum value as 11111.
The address with all 1s in host part is broadcast address and can’t be assigned to a host.
So the maximum possible last octal in a host IP is 10011110 which is 158.
Subnet mask is 255.255.255.224
Do AND with given IP and subnet mask then we get NID 200.10.11.128
In fourth octet first three bit will fixed for subnet and remaining 5 bits is for HID, so maximum value as 11111.
The address with all 1s in host part is broadcast address and can’t be assigned to a host.
So the maximum possible last octal in a host IP is 10011110 which is 158.
158
157
156
155
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