IPV4 Multicast addressing

Question

Consider an IPV4 addressing system,where at same time two multicast groups are ongoing,by choosing their multicast group addresses at random. The probability they interfere each other i.e. choosing same multicast address is ?

Answer 1

i think it should be

(1/2^28)*(1/2^28)

 

but it has given (1/2^28) in made easy solutions

Comments

Initially, I also had the same impression, but when you look at it carefully,

Number of choices for 1st multicast group address: $2^{28}$

Number of choices for 2nd multicast group address: 1(we choose only the address chosen by group 1)

Calculating the probability: ($2^{28}$ / $2^{28}$) * (1 / $2^{28}$)

Which gives us: 1/($2^{28}$)

Still confused?

According to you the answer should be: (1/$2^{28}$)*(1/$2^{28}$)

This unfortunately is the probability of both the multicast groups selecting the same(only a particular, say, 224.1.1.1) group address.

But, we have $2^{28}$ choices: Therefore multiplying your answer with $2^{28}$ gives us the correct result.

Answer 2

Datagrams with multicast address are simultaneously transmitted to one or more multicast host groups or networked computers. 

Multicast addresses range from 224.0.0.0 to 239.255.255.255. Examples for IPV4-reserved addresses for multicasting are as follows: 

• 224.0.0.0: Base address reserved
• 224.0.0.1: Used for all multicasting host groups
• 224.0.0.2: Used for all subnet routers
• 224.0.0.5 and 224.0.0.6: Used by Open Shortest Path First, an interior gateway protocol for all network segment routing information

Multicast addresses in IPV4 are defined using leading address bits of 1110, which originate from the classful network design of the early Internet when this group of addresses was designated as Class D. Multicast addresses in IPV6 have the prefix ff00::/8. IPv6 multicast addresses are generally formed from four-bit groups.

 (1/2^28)

3.72

Comments

just correct final answer it is 3.72 * 10^(-9)