GATE CS 2015 Set 3 — Question 46
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Computer Networks → Network Layer: Addressing → Subnetting & CIDR
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Question
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
158 to 158
Solution
1.Identify the Subnet:
- IP Address: 200.10.11.144
- Subnet Mask: /27
- A /27 mask corresponds to . The block size in the last octet is (or ).
- The multiples of 32 are: 0, 32, 64, 96, 128, 160, ...
- The given fourth octet is 144. Since , the network address is 200.10.11.128.
- The next network starts at 200.10.11.160.
- Therefore, the broadcast address for this subnet is (i.e., 200.10.11.159).
- First usable host: Network + 1 = 200.10.11.129
- Last usable host: Broadcast - 1 = 200.10.11.158
- The fourth octet of the last assignable host IP is 158.
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