Question

An organization is granted the block 150.36.0.0/16.The administrator wants to create 512 subnets. Find number of addresses in each subnet. Find the first and last addresses in subnet 1.

(A) 128, 150.36.0.1 and 150.36.0.127 (B) 128, 150.36.0.129 and 150.36.0.255
(C) 126, 150.36.0.1 and 150.36.0.126 (D) 126, 150.36.0.129 and 150.36.0.254.

Comments

If they ask second subnet first and last address then it will be 150.36.0.129 to 150.36.0.254.am I right?? Please correct... Thanks in advance

Answer 1

Ans - c

512 subnets => 9 bits from host id are reserved. Therefore, 150.36.NNNNNNNN.NHHHHHHH is our scenario. There the first network id is : 150.36.00000000.00000000 => 150.36.0.0 and the first usable IP address is 150.36.0.1 The broadcast address of this subnet will be 150.36.0.127 => The last usable IP address of first subnet will be : 150.36.0.126

Comments

Answer is not option(a) because First address is reserved for Interface Id and last address is reserved for broadcast Id.

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In option a , 128, 150.36.0.1 is the first usable ip address but 128, 150.36.0.127 is the broadcast address. Either give network and broadcast address OR give first and last usable ip addresses. Don't mix them up.

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But the question asks the first and last address of the first subnet not the first and last usable address of the first subnet

Answer 2

no of bits for host=32-16-9=7

so no of host =2^7-2=126

no of address=128

Comments

Are there questions based on discontinous subnet mask in GATE ?

Answer 3

Number of subnet = 512 i.e 9 bits 
so total bits for subnetting 16+9 = 25 

 

so host bits are 32 -25 = 7 

therefore total number of address = 2^7 = 128 
total host address = 128 -2 = 126

Comments

Its asking address not host address so why you are giving host address

Answer 4

Is b correct

Comments

yes answer is (b)

becoz last address of the subnet is Direct broadcast address.