rpsc programmer #computer networks

Question

if direct broadcast address is 200.200.200.31. which of the following can be subnet mask?

A. 255.255.255.192

B. 255.255.255.224

C. 255.255.255.248

D. 255.255.255.128

Answer 1

In order to do direct broadcast first we specify the NETWORK ID and then we put all 1's in HOST ID.

Give Direct Broadcast address : 200.200.200.31 

Equivalent Decimal representation : 11001000. 11001000. 11001000. 00011111

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One thing that we need to observe here that in 4th octet 00011111, here we don't have clarity that

out  of these fives 1's, how many of them belongs to NETWORK ID?

One thing that we are sure about is that NETWORK ID is alteast 27-bits. NETWORK ID can be

of 27, 28, 29, 30, 31-bits.

NETWORK ID length can't be 32, because if it is 32-bits then it will no longer be Direct Broadcast

Address, it will become Limited Broadcast Address.

So, Differenet possiblities are : 

11001000. 11001000. 11001000. 00011111

11001000. 11001000. 11001000. 00011111

11001000. 11001000. 11001000. 00011111

11001000. 11001000. 11001000. 00011111

11001000. 11001000. 11001000. 00011111

(Red Portion Represents NETWORK ID and Black Portion Belongs to HOST ID.)

 

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NOTE : In order to find subnet mask we put all 1's in NETWORK ID and all 0's in HOST ID

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Now, Put all 1's in NETWORK ID and all 0's in HOST ID in each possibility to get Subnet Mask

11111111. 11111111. 11111111. 11100000 $\equiv$ 255.255.255.224

11111111. 11111111. 11111111. 11110000 $\equiv$ 255.255.255.240

11111111. 11111111. 11111111. 11111000 $\equiv$ 255.255.255.248

11111111. 11111111. 11111111. 11111100 $\equiv$ 255.255.255.252

11111111. 11111111. 11111111. 11111110 $\equiv$ 255.255.255.254

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So, Option B, C are correct possibilities.

Comments

here we can Apply the rule  of  longest prefixe matching so most strict ans should be option C right .

Answer 2

I think the answer should be B 255.255.255.224
 

Because the 4th octet will be 00011111. Now we can say the the last 5 bits represent the host ID, because all bits in the HOST ID are 1 in the DBA (Direct broadcasting address).

Hence the network ID will be of 27 bits (atleast) if we consider host ID as 5 bits.

Now, in subnet mask all the bits in network ID and subnet ID are set (i.e 1) and all the bits in host ID are 0.

so the subnet mask of the given will be 11111111.11111111.11111111.11100000

which has decimal equivalent 255.255.255.224.

 

As RPSC PROGRAMMER doesn't have MSQs so option B will be the best choice.