# BSCI - Multicasting Question

Member Posts: 933
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?

• Member Posts: 460

According to the IPv4 Multicast MAC calculator found here Network Stuff the MAC should be 01-00-5e-11-01-01.
• Member Posts: 933
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.
• Member Posts: 460
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.
• Member Posts: 933
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!!