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on 22 June 2014

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B. To answer this, you must be able to determine which Class C mask provides how many hosts and subnets. The 255.255.255.0 mask provides one network with 254 hosts. The 255.255.255.192 provides two subnets each with 62 hosts. The 255.255.255.224 provides 6 subnets, each with 30 hosts, and the 255.255.255.248 mask provides 30 subnets, each with 6 hosts.
A. 255.255.255.0
B. 255.255.255.192
C. 255.255.255.224
D. 255.255.255.248

If you are using a Class C network ID with
two subnets
and need
31 hosts per network
, which of the following masks should you use?
D. This is a Class C network address with 5 bits of subnetting. The valid subnets are 256 – 248 = 8, 16, 24, 32, 40, etc. Since the host ID is 33, we are in the 32 subnet. The next subnet is 40, so our broadcast address is 39
A. 192.168.10.40
B. 192.168.10.255
C. 192.168.255.255
D. 192.168.10.39

E. This is a Class B network address with 12 bits of subnetting—8 in the third octet and 4 in the fourth octet. The subnet in the third octet is 10, and the subnets in the fourth octet are 256 – 240 = 16, 32, 48, etc. Since the fourth octet is using 22, the host is in the 16 subnet, and since the next subnet is 32, the broadcast address for the 16 subnet is 31. The valid host range is the numbers in between, or 17–30.
A. 172.16.10.20 through 172.16.10.22
B. 172.16.10.1 through 172.16.10.255
C. 172.16.10.16 through 172.16.10.23
D. 172.16.10.17 through 172.16.10.31
E. 172.16.10.17 through 172.16.10.30
What valid host range is the IP address 172.16.10.22 255.255.255.240 a part of?
C. Take a look at the answers and see which subnet mask will give you what you need for subnetting.

252 gives you 62 subnets, 248 gives you 30 subnets, 240 gives you 14 subnets, and 255is invalid. Only the third option (240) gives you what you need.
A. 255.255.255.252
B. 255.255.255.248
C. 255.255.255.240
D. 255.255.255.255
If you wanted to have 12 subnets with a Class C network ID, which subnet mask would you use?
C. This is a Class B network address with 10 bits of subnetting—8 in the third octet and 2 in the fourth octet. The subnet in the third octet is 8, and the subnets in the fourth octet are 256 – 192 = 64, and 128. However, as long as all the subnet bits in the third octet are not all on at once, the subnets in the fourth octet really can be 0, 64, 128 and 192. This means that the host is in the 128 subnet and since the next subnet is 192, our broadcast address is 8.191.
A. 172.16.255.255
B. 172.16.8.127
C. 172.16.8.191
D. 172.16.8.255