GATE CSE 2015 Set 3 | Question: 38

Question

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 _______.

Comments

What is meant by fourth octet ?

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@hem chandra joshi ji

fourth octet (in decimal) of the last IP address of the network means

__ __ __ __ __ __ __ __($1^{st}$ octet) . __ __ __ __ __ __ __ __($2^{nd}$ octet) . __ __ __ __ __ __ __ __ ($3^{rd}$ octet). X X X X X X X X ($4^{th}$ octet)

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Remark:

The address with all 1s in host part is broadcast address

The address with all 0s in host part is network address

Answer 1

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.

Comments

@pC

Also, see the list of possible subnet masks for CIDR notation here. The subnet mask of non-contiguous 1’s is not allowed in CIDR notation as IP addresses in the subnets will not be contiguous. It is theoretically possible to use subnet masks of non-contiguous 1’s in classful addressing but it is not used practically.

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Hi 
I am getting question 4^th octact of last host, i have issue with fourth octact like

You know tha subnect bits is 3 , 10010000 which is 100
According to my knowledge ,

          000 is 1^st octact net id 

          001 second

         010 third

        011 fourth 

then why answer using  fourth octact as 100  
Can anyone  please clear my doubt
Again i am repeating i don’t have doubt in host or DBA  
I have dounbt in net id 
Can anyone plz help in this

 

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Who said to subnet the network?

Answer 2

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 ) . 

Comments

why did you subtract 1

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@Jeetmoni saikia Because $10011111 (159)$ is network broadcast address and can’t be assigned to any host. 

Answer 3

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)

Answer 4

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).

Comments

This cleared my doubt about “last IP address” thanks a lot