Computer Networks

Question

Question From PW

Given :8 th answer is d

:   9 th answer is b

(Answer seems to be incorrect (I don’t know exactly)).

 

 

Comments

8. In class C, there are $2^{21}$ networks and $2^8$ IP addresses / Network. Therefore total #IP addresses = $2^{21} * 2^8$ = $2^{29}$
    
9.  Multicast address is represented by $224.0.0.0/4$ in CIDR notation. Their range is from $224.0.0.0$ to $239.255.255.255$
   
   It’s not like we will only use 28 bits of IP address to represent multicast address, I am not getting what they are trying to ask.

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I think I got the answer for 9^th one :

starting 4 bit is used for representing class D ip.

Rest are 28 bits which are used for multi casting.

Answer 1

Ans 8). In class C, the first octet’s first three bits are already fixed.

Total number of bits = 32, Fixed bits in class C = 3, So Remaining bits = Total number of bits – Fixed Bits

                                                                                                                            = 32 – 3 = 29 bits

So, the number of Ip addresses = $2^{^{29}}$ Ip addresses

option D is correct.

 

Ans 9). Here Class D is given, The first octet’s first four bits are already fixed.

Total number of bits = 32, Fixed bits in class D = 4, So Remaining bits = Total number of bits – Fixed Bits

                                                                                                                            = 32 – 4 = 28 bits

And the application of Class D is used for the Multicasting purpose only.

 All 28 bits are used for multicasting only.

Option B is correct.