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GATE CSE 2015 | Set 3 | Question: 38

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73 73 votes
In the network $200.10.11.144/27$, the $\text{fourth}$ octet (in decimal) of the last $\text{IP}$ address of the network which can be assigned to a host is _______.

6 Answers

Best answer
117 117 votes
Answer $= 158$

$144$ in binary $= 100 10000$

out of this $3\;\text{bits}$ in left are subnet bits. $(27\;\text{bits}$ are used for subnet, which means top $3\;\text{bytes}$ and leftmost $3\;\text{bits}$ from the last byte)

So, the $4^{\text{th}}$ octet in the last $\text{IP}$ address of the network which can be assigned to a host is $100 11110$. (its not $100 11111$ because its network broadcast address)

So, $10011110$ is $158$ in decimal.
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38 38 votes

last octate will be 100 11111 = 1 + 2 + 4+ 8+ 16 + 128 = 159

The question seems to by asking about host address. 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 ( 159 - 1 = 158 ) . 

7 7 votes
NID = 27 bits

HOST ID = 5 bits


144 in binary = 10010000

so, 200 . 10 . 11 . 100 00000 ( Reserved for Network id)

      200 . 10 . 11 . 100 00001

      200 . 10 . 11 . 100 00010

      ---------------------------------

      ----------------------------------

     200 . 10 . 11 .100 11110 = 200 . 10 . 11 .158 (Last IP address of the network which  is assigned to host)

     200 . 10 . 11 .  100 11111 = 200 . 10 .11 . 159 (Reserved for DBA)
5 5 votes
Reusing Todd Lammle's method (CCNA) here, which I used in another answer.

$/27$ bits turned on is $255.255.255.224$. Thus $256 - 224 = 32$ is the block size for the last octet. So subnet addresses follow the pattern $200.10.11.0$, $200.10.11.32$, $200.10.11.64$, $200.10.11.96$, $200.10.11.128$, $200.10.11.160$ and so on.

Thus we can easily see that the given IP address is in the $200.10.11.128$ subnet, whose last valid host address is $200.10.11.158$. (as $200.10.11.159$ is the broadcast address).
2 2 votes

The below image explains why 158 should be answer.

As last and first value cant be used to assign to a host

we always assign $2^{hid}$-2 

 

The last bit is 0 as if it were 1, it would have been directed broadcast address of the Network

 

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