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Given IP = 210.15.131.141 and subnet mask = 255.255.255.224. Calculate the last host of last subnet?

a) 210.15.131.222

b)210.15.131.162

c)210.15.131.202

the answer given is A, but i have a doubt  since this is a class C address.. and 224=11100000

we have 3 bits for subnets that is , we will have 8 subnets from 000 to 111 .. so the last subnet will be 111.

for the host id bits .. in the last octet.. we will have 11110 .. since 11111 is for DBA .. so acc to me .. the answer should be 210.15.31.254 .

where am i going wrong ? someone pls clarify it ..
asked in Computer Networks by Active (2.2k points) | 50 views

i think they have calculated by taking 110 as last subnet 210.15.131.11011110 -> 210.15.31.222

yes but , how can 110 possibly be the last subnet ?

any references or examples?
Ya, I have came across this kind of  problem

Read Bikram Sir's comment in this question

It will clear everything.

https://gateoverflow.in/1265/gate2007-67-isro2016-72
Got it , thanks a lot bro.

1 Answer

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Given Subnet Mask = 255.255.255.224

It is a class C network, so 24 MSB belongs to Network, rest 8 belongs to Host.

Subnet Mask is all 1s for subnet and network ID part. 0's for host ID part.

So we get 3 bit for subnet, Total number of subnet = 23 = 8

Now comes the main part.

--

According to RFC 950 specification, we have to subtract the subnet with all 0's and all 1s

Therefore, last subnet's network ID will be  = 210.15.131. (110 00000)

Last subnet's Directed Broadcast Address = 210.15.131. (110 11111)

Therefore, last host of last subnet  = 210.15.131. (110 11110) = 210.15.131.222

--

According to RFC 1878 specification, we DON'T have to subtract the subnet with all 0's and all 1s

Therefore, last subnet's network ID will be  = 210.15.131. (111 00000)

Last subnet's Directed Broadcast Address = 210.15.131. (111 11111)

Therefore, last host of last subnet  = 210.15.131. (111 11110) = 210.15.131.254

 

Which standard to follow for GATE ?

The new standard, i.e. RFC 1878

answered by Active (1.8k points)
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