READ Free Dumps For CISCO- 100-105
| Question ID 14479 | What is the subnet address for the IP address 172.19.20.23/28?
|
| Option A | 172.19.20.0
|
| Option B | 172.19.20.15
|
| Option C | 172.19.20.16
|
| Option D | 172.19.20.20
|
| Option E | 172.19.20.32
|
| Correct Answer | C |
Explanation Explanation: From the /28 we can get the following: Increment: 16 (/28 = 11111111.11111111.11111111.11110000) Network address: 172.19.20.16 (because 16 < 23) Broadcast address: 172.16.20.31 (because 31 = 16 + 16 1)
| Question ID 14480 | If an Ethernet port on a router was assigned an IP address of 172.16.112.1/20, what is the
maximum number of hosts allowed on this subnet?
|
| Option A | 1024
|
| Option B | 2046
|
| Option C | 4094
|
| Option D | 4096
|
| Option E | 8190
|
| Correct Answer | C |
Explanation Explanation: Each octet represents eight bits. The bits, in turn, represent (from left to right): 128, 64, 32 , 16 , 8, 4, 2, 1 Add them up and you get 255. Add one for the all zeros option, and the total is 256. Now, take away one of these for the network address (all zeros) and another for the broadcast address (all ones). Each octet represents 254 possible hosts. Or 254 possible networks. Unless you have subnet zero set on your network gear, in which case you could conceivably have 255. The CIDR addressing format (/20) tells us that 20 bits are used for the network portion, so the maximum number of networks are 2^20 minus one if you have subnet zero enabled, or minus 2 if not. You asked about the number of hosts. That will be 32 minus the number of network bits, minus two. So calculate it as (2^(32-20))-2, or (2^12)-2 = 4094