A network administrator assigns a multicast addressof 239.255.8.5 to an application
running on a device with an Ethernet MAC address of 01.b2.7d.05.f1.80. Which Layer
2 multicast address will this device use?
A.
01.00.5e.7F.08.05
B.
01.b2.7d.05.f1.80
C.
01.b2.7d.0a.08.05
D.
01.00.5e.05.f1.80
E.
ff.ff.ff.ff.ff.ff
As you know class D is multicast and the first four bits are 1110. This means that we have 28 bits left for multicast addresses. However at layer 2 the 01-00-5e MAC range has only reserved 01-00-5e-00-00-00 to 01-00-5e-7f-ff-ff. This means that we can only map 23 bits of the IP to the MAC address. 5 bits of the IP is lost which means that 32 multicast IP addresses will map to the same multicast MAC.
Now for the conversion itself:
239.255.8.5
1110 1111 1111 1111 0000 1000 0000 0101
We will use the last 23 bits of this IP. The first 25 bits of the MAC are always the same:
0000 0001 0000 0000 0101 0111 0
Then we add the final 23 bits of the IP to this:
0000 0001 0000 0000 0101 0111 0111 1111 0000 1000 0000 0101
This corresponds to:
01-00-5e-7f-08-05
0
0