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Consider an $IPV4$ addressing system, where at a same time two multicast group are ongoing, by choosing their multicast group address at random.Then the probability they interfere each other is _________________$\times 10^{9}$


how to solve multicast address?

in Computer Networks by Veteran (119k points) | 86 views
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Ma'am is it there in the syllabus?
0

yes, multicast address is in 

See here some points https://gateoverflow.in/241089/icmp-messages-doubt

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@arjun Sir please guide with this

 

Multicast range: 224 - 239, in between this each address represent a different group.

so we have 2^28 groups maximum

"the probability they interfere each other" What I concluded is that two people interfere each other when they want same thing

similarly when two groups of having same multicast address'

1- [probability[ two group of having different multicast address]]

1st group having 2^28 choice

2nd group have 2^28 -1 choice

[probability[ two group of having different multicast address]= (2^28 / 2^28 ) ( 2^ 28 -1  ) / 2^ 28 )  = ( 2^ 28 -1 )  / 2^ 28 )

1- [probability[ two group of having different multicast address]]

1- ( 2^ 28 -1  ) / 2^ 28 ) =  1 / 2^28 = 3.725 * 10^(-9 )

 

can i deduce this q in simple terms:

6 different color balls, prop of two person choosing same ball = 1- prop of two person choosing different ball

1-[(6/6)(5/6)] = 1- (5/6) =  1/6
by Active (2.7k points)
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Is two groups  has

$2^{28}$ and $2^{28}-1$ addresses?
+1
Ma'am First group has 2^28 choices

and the other group has 2^28-1 choices

in multicast, each address represent one group,

it working is different,
0
ok, thanks

is there any range for broadcast address too?
0
ma'am i don't know, bz of this q i studied about multicasting,thanks a lot
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