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Frame tagging places a unique user id in the header of each frame that allows that frame to traverse the
switching fabric.
Chapter: 2
52. What are the valid hosts for the following network: 192.168.10.40 255.255.255.224?
A. 192.168.10.32-64 B. 192.168.10.33-64
C. 192.168.10.33-62
D. 192.168.10.34-65
Answer: C
224 is 3 bits for subnets and 5 bits for host addressing. That means we have 6 subnets each with 30 hosts.
The valid subnets are: 256-224=32, 64, 96, 128, 160, 192. The valid hosts are the numbers between the subnets minus all zero's and all 1's. The valid hosts in the 32 subnet are 33-62, 63 is the broadcast for the 32
subnet
Chapter: 2
53. What are the valid hosts for the following network: 192.168.10.56 255.255.255.240?
A. 48-64 B. 48-63
C. 49-62
D. 49-61
Answer: C
A 240 mask gives us 4 -bits for subnets and 4-bits for host addressing. This gives us 14 subnets each with
14 hosts. The subnets are 256-240=16,32,48,64,80,96,112,128,144, 160,176,192,208,224. The valid hosts
are the numbers between the subnets, minus all zeros and all 1's. The valid hosts for the IP address specified in the question is the in the 48 subnet range or 49-62, 63 is the broadcast address.
Chapter: 2
54. What are the valid hosts for the following network: 192.168.10.5 255.255.255.252?
A. 10.4 through 10.7
B. 10.5 through 10.6 C. 10.4 through 10.10
D. 10.4 through 10.8
Answer: B