+1 vote
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Each of the department requires a separate subnet and respective requirement of address being 28 , 14 and 53 .Also organization got a class C network address 216.85.102.0 .
Which of the following could be possible broadcast address for each subnet ?

A) A- 216.85.102.31 , B-216.85.102.95 ,C- 216.85.102.127

B ) A-216.85.102.63 ,B-216.85.102.31 ,C- 216.85.102.127

C) None of these .

According to me answer should be None of these because in both the options A and B we are subnetting the first network with subnet bit 0 , to 00 and 01 which would not form a different subnet for the department ,please clarify .
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B) ???
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why not B)?
A-216.85.102.00111111 ,
B-216.85.102.00011111 ,
C- 216.85.102.01111111
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I too think it should be B) because address allocation are as follow:-

$216.85.102.16$ to $216.85.102.31$ for Department B.

$216.85.102.32$ to $216.85.102.63$ for Department A.

$216.85.102.64$ to $216.85.102.127$ for Department C.
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How did you approach this ,Please explain .
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@Radha Gogia,Radha Gogia, As per the requirements, B requires 14 addresses, so for B 216.85.102.16 to 216.85.102.31,there are 16 subnetwork addresses, the count which satisfies the requirement. In a similar manner for A as well, 216.85.102.32 to 216.85.102.63,there are 32 subnetwork addresses which satisfies the requirement for network A,for Network C as well 216.85.102.64 to 216.85.102.127,there are 64 subnetwork addresses.This satisfies the requirements for each of the networks A,B and C respectively. I don't know how this range is arrived at,but I've worked it out as per the answer given by @ Jason.

@Jason,Could you please elaborate a bit?How you've got d above range of Network addresses?
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@Jason,Thank you so much. :)

+1 vote

Solving through Options, I am directly taking option (B) based on that you can also verify why option (A) is wrong.

B ) A-216.85.102.63 ,B-216.85.102.31 ,C- 216.85.102.127

A->$216.85.102.63$

Since requirement of Dept A is 28 so rounding off to power of $2$ which is $2^5 = 32$. So, subnet for 32 IPs will be $255.255.255.224$. Means 3 Subnet bits and 5 host bits.

Given A->$216.85.102.63$ expanding its fourth octet in binary 0011 1111. Now here bold letters are SN bits and remaining is host bit. Throught its fourth octet it is clear that its is DBA of first subnet. Seems GOOD DBA for our Dept A.

And Address range is $216.85.102.32$ to $216.85.102.63$

Similarly

B->$216.85.102.31$

Since requirement of Dept B is 14 so rounding off to power of $2$ which is $2^4 = 16$. So, subnet for 16 IPs will be $255.255.255.240$. Means 4 Subnet bits and 4 host bits.

Given B->$216.85.102.31$ expanding its fourth octet in binary 0001 1111. Now here bold letters are SN bits and remaining is host bit. Through its fourth octet it is clear that its is DBA of first subnet. Seems GOOD DBA for our Dept B.

And Address range is $216.85.102.16$ to $216.85.102.31$

Similarly

C->$216.85.102.127$

Since requirement of Dept C is 53 so rounding off to power of $2$ which is $2^6 = 64$. So, subnet for 64 IPs will be $255.255.255.192$. Means 4 Subnet bits and 4 host bits.

Given C->$216.85.102.127$ expanding its fourth octet in binary 0111 1111. Now here bold letters are SN bits and remaining is host bit. Through its fourth octet it is clear that it is DBA of first subnet. Seems GOOD DBA for our Dept C.

And Address range is $216.85.102.64$ to $216.85.102.127$

Note:- To find Address range make all possible combination from host bit, you will get address range for it.

Hint for Option A) Address ranges are:-

A-> $216.85.102.0$ to $216.85.102.31$

B-> $216.85.102.80$ to $216.85.102.95$

C-> $216.85.102.64$ to $216.85.102.127$

Here you can find that there is Address Overlapping between Dept B and Dept C. Hence it can't be the answer.

So, final answer is Option (B)

selected
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Thanks a lot for the precise solution but I think that you approached through the options available , is there any way out through which we can get the options directly ?
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@Jason,Thanks for this lucid descp. :)