ISRO CSE 2014 | Question: 57

Question

A supernet has a first address of $205.16.32.0$ and a supernet mask of $255.255.248.0.$  A router receives $4$ packets with the following destination addresses. Which packet belongs to this supernet?

• A. $205.16.42.56$
• B. $205.17.32.76$
• C. $205.16.31.10$
• D. $205.16.39.44$

Comments

Just check the longest prefix match.

Answer 1

AND ip address and supernet mask whichever option gives address of supernet is the ans

Here is d the ans

Comments

       205.16.      00100111
       255.255.    11111000
     -------------------------------
       205.16.32.0   matches perfectly     Hence (D)

U may had doubt regarding option 2, but see 2nd octet is different there too.

Hence, only (D) matches perfectly.

Answer 2

1 _ _ _ _ _ _ _ =2⁷

_ _ _ _ _ _ 0 0 =2⁶

1 _ _ _ _ _ 0 0 =2⁵

2⁷+2⁶ -2⁵=128+64-32 =192-32 =160

Option C  ??

Comments

Sir, so what is the answer?

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160

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Sir,

thank you!

Answer 3

first address of 205.16.32.0 i.e. Class C.
Default Subnet will be 255.255.255.0

supernet mask of 255.255.248.0. i.e. 3 bit for  Supernet and 21 bit for Network ID and rest 8 bit fot Host id.

205.16.39.44 will be correct option.

 

Answer 4

AND of subnet mask 255.255.248.0 & ip address 205.16.39.44 is 205.16.32.0 which is same as address of 205.16.32.0 and 255.255.248.0

So answer is D

Answer 5

Supernet Mask is 255.255.255.248.0  

which is 11111111.11111111.11111000.00000000.

NID= 21 and HID=11

Given first IP address is 205.16.32.0/x here x=21 so we have 11 Hid bits.

So to get range of IP address with a given starting IP is by considering the IP address into two parts fixed and variable. In this case the 21bits are fixed and 11 bits are variable so in first IP address the variable bits are all 0's and in the last IP of range all variable bits are 1's

So  205.16. 00100 000. 00000000    => 205.16.32.0

                    to 

      205.16. 00100 111. 11111111         => 205.16.39.255

So from the given options 205.16.39.44 falls in the above range