60 votes 60 votes The address of a class $\text{B}$ host is to be split into subnets with a $6\;\text{-bit}$ subnet number. What is the maximum number of subnets and the maximum number of hosts in each subnet? $62$ subnets and $262142$ hosts. $64$ subnets and $262142$ hosts. $62$ subnets and $1022$ hosts. $64$ subnets and $1024$ hosts. Computer Networks gatecse-2007 computer-networks subnetting easy isro2016 + – Kathleen asked Sep 21, 2014 edited Dec 4, 2022 by Lakshman Bhaiya Kathleen 35.4k views answer comment Share Follow See all 11 Comments See all 11 11 Comments reply Show 8 previous comments Venkat Sai commented Dec 27, 2017 reply Follow Share at any cost the no of subnets will not be 62 they should be 64 and the no of hosts can be made 2^n in present routers of cisco i have read it some where so 64 ,1024 can be right 1 votes 1 votes smsubham commented Dec 30, 2017 reply Follow Share Similar question: https://gateoverflow.in/480/gate2008-57 1 votes 1 votes habedo007 commented Apr 8, 2020 reply Follow Share Suppose a class B network: 128.16.0.0 Now if we make $2^6$ subnets first one will be: $1000000.00010000. 000000|00.00000000\rightarrow$ mind the separator after 6 zeroes in third octet. The first valid configurable address will be $10000000.00010000.000000|00.00000001$ But people here are removing $2$ subnets (all $6$ zeroes and all $6$ ones) to remove base address and broadcast address. Doing so will render the above completely valid address unavailable. Consequently, we'll not have any network from $10000000.00010000.000000|00.00000001$ to $10000000.00010000.000000|11.11111110$, and $10000000.00010000.111111|00.00000001$ to $10000000.00010000.111111.11.11111110$. And we certainly do not want to do that. So the subnet must remain $64$, and each of them will have $2^{10} - 2=1022$ hosts. 0 votes 0 votes Please log in or register to add a comment.
0 votes 0 votes option B should be correct if it comes in GATE now From wikipedia Refer this then everything becomes a piece of cake shashankrustagi answered Jan 9, 2021 shashankrustagi comment Share Follow See all 0 reply Please log in or register to add a comment.
–1 votes –1 votes ans c) Aditi Dan answered Dec 23, 2014 reshown Jan 9, 2017 by Bikram Aditi Dan comment Share Follow See all 2 Comments See all 2 2 Comments reply amarVashishth commented Nov 11, 2015 reply Follow Share after creating a subnet, inside it 2 IP addresses are excluded one to rep. Subnet ID and other for subnet's DBA, not before, so answer is D 1 votes 1 votes cse23 commented Jul 7, 2016 i reshown by Bikram Jan 9, 2017 reply Follow Share Answer should be (C) we know in CLASS B: HOST ID = 16 we always borrow bits from host id when we do subnetting here 6 bits are used for subnetting so, for hosts, 10 bits are used Total no. of host possible or IP addresses = 2^10 = 1024, one is used for subnet address and one as the broadcast so,no. of hosts =1024-2 =1022 now, with 6 bits we can have 2^6 = 64 subnets and here also 2 will be subtracted for subnet address and broadcast address...so 62 subnets will be the answer. Note: If in question 1022 is not the option then we can also have 1024 as answer. similarly in case of subnet also, 62 is best choice but if not given then we can choose 64 also a/c to modern subnetting. 1 votes 1 votes Please log in or register to add a comment.
–1 votes –1 votes C is the correct answer Pinky Dafouti answered Aug 7, 2016 reshown Jan 9, 2017 by Bikram Pinky Dafouti comment Share Follow See all 0 reply Please log in or register to add a comment.
–1 votes –1 votes Answer is in the link : http://www.techrepublic.com/article/subnetting-a-class-b-network-address/ jaiganeshcse94 answered Aug 13, 2016 reshown Jan 9, 2017 by Bikram jaiganeshcse94 comment Share Follow See all 0 reply Please log in or register to add a comment.