BSCI - Multicasting Question

billscott92787

Hey guys,



I have a question in regards to multicasting, this question is actually in the Exam Certification Guide for the BSCI Cahtper 17, Question #7.



7. Which of the following multicast addresses does not share a multicast MAC Address.


A. 225.145.1.1
B. 227.17.1.1
C. 235.17.1.1
D. 237.117.1.1


The answer shows as D. But, I get different multicast macs for A and D.


Here is what I get:


225.145.1.1

01-00-5e-91-01-01



227.17.1.1

01-00-5e-11-01-01



237.17.1.1

01-00-5e-11-01-1




237.117.1.1

01-00-5e-75-01-01



Am I doing something wrong, why do they say the only right answer is D?

burbankmarc

Your calculations must be wrong.

According to the IPv4 Multicast MAC calculator found here Network Stuff the MAC should be 01-00-5e-11-01-01.

billscott92787

Thank you, I definitely figured that. I'm trying to see what I'm doing wrong.


for 225.145.1.1


I broke the last three into binary

145 -> 10010001 128 + 16 + 1 = 145

and of course we all know the other two so I'm not going to post them here.


If I double check my decimal to hex on http://www.statman.info/conversions/hexadecimal.html


I get 91 like I did. But, if I use the one you did I get 145 as 11 in hex.

burbankmarc

Remember the 25th bit of the MAC is ALWAYS 0. so it goes from

145
10010001

to

00010001 which is 17 in dec. 11 in hex.

billscott92787

burbankmarc wrote: »

Remember the 25th bit of the MAC is ALWAYS 0. so it goes from

145
10010001

to

00010001 which is 17 in dec. 11 in hex.

AH, duh! I forgot the highest bit always gets set to zero, I feel like an idiot lol. Thank you VERY much for reminding me of that. I'm sure that will help me to not forget when my BSCI attempt comes up!!