Subnet mask-NID

Question

For a class C network if IP address of a computer is 200.99.39.112 and subnet mask is 255.255.255.224 the decimal value of last octet of last host of sixth subnet is ?

shouldn't it be 190 but the answer given is 222? anyone help

Comments

Correct, Please solve it .

I am getting 190 :/

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If it is 222 then it will be the decimal value of last octet of last host of 7th segment..

 I also get 190    . 😒

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LATEST VERSION WE CAN NOT CONSIDER TWO SUBNET 000 ,111 BECAUSE THESE TWO SUBNET MACHING WITH NID AND DBA OF NETWORK..

Ist Subnet 001

2nd subnet 010

3rd subnet 011

4th subnet 100

4th subnet 101

last subnet110

Answer 1

class C default NetWork mask : 255.255.255.0 ..But subnet mask is 255.255.255.224

224 = 11100000 = 128 + 64 + 32

so, 3 1's have been borrowed from host bits which will now become subnet bit.  

for subnets there are 8 subnets ( 2³ = 8 ) possible like 000, 001, 010 , 011 and so on .....  the first subnet is 000, second subnet 001, third 010 and fourth 011, fifth 100 , sixth 101 , seventh 110 , eight subnet 111 .

Hence subnet no of 6th subnet = 101 , 

The address with all 1s ( 111 11111 ) 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 111 11110 , 

here it asked 6th subnet so last octet of last host will be 101 11110 .

The decimal value of last octet of last host of sixth subnet is 101 11110 = 128+32+16+8+4+2 =190.

Comments

GEEKSFORGEEKS...

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@ bad_engineer 

okk i understand :)

made a poll here ..

https://www.facebook.com/groups/gateoverflow/permalink/652856628252884/

lets see what people say ..

i think most people go with old convention ..as it is followed by all coaching institutes and also follow in test series  ..

I can see , most people vote for new convention :)

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@bad also made easy test solution has given wrong answer for this question

Answer 2

last address of sixth subnet will be 209.99.39.11011110   last octet 222

Comments

@reena_kandarl  

I agreed with your answer . I am getting the same answer. But there is a confusion how to choose 6th subnet or 4th subnet ?

@Bikram Sir plz guide

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@ashwina

class C default NetWork mask : 255.255.255.0 ..But subnet mask is 255.255.255.224

224 = 11100000 = 128 + 64 + 32

so, 3 1's have been borrowed from host bits which will now become subnet bit. 

for subnets there are 8 subnets ( 2³ = 8 ) possible like 000, 001, 010 , 011 and so on .....  the first subnet is 000, second subnet 001, third 010 and fourth 011, fifth 100 , sixth 101 , seventh 110 , eigth 111 .

Hence subnet no of 6th subnet = 101 , sixth host will be : 00000   00000110

The decimal value of last octet of last host of sixth subnet is 101 11110 = 128+32+16+8+4+2 =190.

As the number of available subnets is 2ⁿ, where n is the number of bits used for the network portion of the address , here to calculate number of subnets we can not subtract 2 so that's why 000 is subnet 1 

Reference:

[1] https://en.wikipedia.org/wiki/Subnetwork#Subnet_and_host_counts

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Thanks @Bikram Sir

Answer 3

If we divide according to decimal value then 110 part subnet otherwise 101 so according to 110 - 222(ans), 101 - 190(ans).

Comments

@Dlllp

It is fill in type question how we will know which subnet to take 101 or 110?

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fill in type then I will go for  -- 101(190)

click following link (page-9 titled as "ANDING With Custom subnet masks")

https://www.kirkwood.edu/pdf/uploaded/569/ip_addressing_&_subnetting_workbook.pdf

Answer 4

As the given IP belongs to network of class C we know that out of 32 bits the first 24 bits are reserved for the nid . The subnet mask given to us is 255.255.255.224.

The last octet in this gives us the information about the subnet id bits i.e 3 as 224=11100000

So they are asking for the decimal value of the last octet of the last host of the 6th subnet ... So SID =110 As it is the last host rest all bits are 11110(last bit is not one because if it was 1 then it becomes directed broadcast address)

So we get 1101 1110=128+64+16+8+4+2=222.

Note: They are referring to subnet with id =6 if 6th subnet would be there then id would be 101 and ans will be 190 as

000- 1st subnet 001=2nd subnet 010- 3rd subnet 011-4th subnet 100-5th subnet 101-6th subnet 110-7th subnet 111-8thsubnet

Comments

Yeah anjali please elaborate a little why u took 110 instead of 101, Rest i got it :)

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The question is about the sixth subnet not the one with subnet id 6.

6th subnet is actually the one with subnet id 5 (101)

The last host id in subnet 5 is

                                = subnet id of 6 - 2

                                = (1100 0000) - 2

                                = 192 - 2

                                = 190 //

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@Na462 as they had mentioned last host of the 6th subnet so I took it as the one with the id =6 i.e. 110 that is why I took 110 instead of 101..

Answer 5

The question is about the sixth subnet not the one with subnet id 6.

6th subnet is actually the one with subnet id 5 (101)

The last host id in subnet 5 is

                                = subnet id of 6 - 2

                                = (1100 0000) - 2

                                = 192 - 2

                                = 190 //

Answer 6

Answer given is correct cause first subnet starts with001 not with 000 cause the first subnet id and network id will be same so in subnet we don't consider the 2 subnet like here 000 and 111 cause later one will clash with DBA of network and DBA of last subnet .