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192.168.10.112.
6.
The broadcast number is 1 less then the next subnet address, so it would be 192.168.10.119.
So, the answers to the question are that 192.0.0.0 addresses are in Class C, the subnet address of
192.168.10.115 is 192.168.10.112, and the broadcast address of that subnet is 192.168.10.119.
A Class B subnet problem
Calculate the valid subnetwork and broadcast addresses for a Class B address of 172.16.1.128 with a subnet
mask of 255.255.255.0. Lab 13-4 details the steps used to calculate the answers to this problem.
Lab 13-4: Calculating the subnet and broadcast addresses for a Class B address
1. First of all, using the information you memorized from Chapter 12, you know that 172.0.0.0
networks are Class B networks.
2.
From the subnet value in the third octet, 255, we know that 8 bits are in use for the mask (out of
a possible 14 available).
3.
Using the subnet formula, we get 28 2 or 254 subnets, which in this case is also the number of
hosts per subnet.
4.
Adding the 254 hosts and the one broadcast address means that the entire fourth octet is used
for each subnet.
5.
To determine the valid subnetworks, subtract 256 255 = 1. The valid subnetwork numbers are
172.16.1.0, 172.16.2.0, ... through 172.16. 254.0.
6.
The answers to this problem are that the subnetwork address is a Class B address of 172.16.1.0
and the broadcast address is 172.16.1.255.
Subnet planning problem
On the exam, you may encounter a question similar to this scenario:
Your company has been assigned a Class C address with a network number of 192.168.250.0. Your boss
wants you to plan for network expansion but doesnÇt want to have more than 20 people on any LAN. Come
up with a networking scheme.
The key to this question is the number of hosts per subnet. A Class C network can have a maximum of 62
hosts on each of two subnets or a minimum of two hosts on each of 62 subnets, depending on the subnet mask
used.
To allow for 20 hosts per subnetwork (LAN) requires a subnet mask of 255.255.255.224. Remember that to
discover the number of host addresses on a subnet, take 2 to the power represented by the number of bits in
the host portion of the subnet mask, and then subtract 2 from this number:
25 2 = 30
Thirty hosts is about as close as you can get to 20 and still allow for 20 hosts. This uses 5 bits in the mask for
hosts, which leaves 3 bits for the subnetwork, or 6 subnets (see ÝClass C subnetsÛ earlier in this chapter). To
figure out the subnet numbers, subtract 256 224 = 32. This would result in the network scheme shown in
Table 13-8.
159
Table 13-8: Networking Scheme for Subnet Planning Problem
Subnet Address Valid Hosts Broadcast
192.168.250.32 .33 - .62 .63
192.168.250.64 .65 - .94 .95
192.168.250.96 .97 - .126 .127
192.168.250.128 .129 - .158 .159
192.168.250.160 .161 - .190 .191
192.168.250.192 .193 - .222 .223
Prep Test
1. Given a Class A address of 120.40.168.13 and a subnet mask of 255.0.0.0, what is the network
address?
A. 120.40.168.0
B. 120.40.0.0
C. 120.0.0.0
D. 0.40.168.13
2.
What is the binary equivalent of the subnet mask 255.255.224.0?
A. 11100000
B. 11111111 11100000 00000000
C. 11111111 11111111 11110000 00000000
D. 11111111 11111111 11100000 00000000
3.
Using a Class B address with a 255.255.255.248 how many bits are being borrowed from the host
portion of the address?
A. 11
B. 10
C. 12
D. 5
E. None of the above
4.
Given a Class C address of 194.16.121.1 and a subnet mask of 255.255.255.224 how many
160
subnetworks can there be and how many hosts can be on each subnetwork?
A. 30 subnets and 6 hosts
B. 16 subnets and 16 hosts
C. 14 subnets and 14 hosts
D. 6 subnets and 30 hosts
5.
Given an IP address of 204.200.106.1 and a subnet mask of 255.255.255.240 what is the broadcast
address for subnet three?
A. 204.200.106.48
B. 204.200.106.63
C. 204.200.106.47
D. 204.200.106.79
6. Using a Class C address of 192.168.1.0 with a subnet mask of 255.255.255.248 how many
subnetwork addresses are available?
A. 64
B. 32
C. 30
D. 16
7.
If you have a need for no more than 12 subnetworks with at least 15 hosts per subnet which address
scheme would be best?
A. Class C with 255.255.255.224 subnet mask
B. Class B with 255.255.224.0 subnet mask
C. Class B with 255.255.255.240 subnet mask
D. Class B with 255.255.240.0 subnet mask
8.
Which series includes only valid subnets of a 192.168.1.0 IP address with a subnet mask of
255.255.255.224?
A. 192.168.32.0, 192.168.128.0, 192.168.228.0
B. 192.168.0.0, 192.168.64.0, 192.168.160
C. 192.168.64.0, 192.168.160.0, 192.168.192.0
D. 192.168.33.0, 192.168.99.0, 192.168.193.0
161
9. What is the dotted decimal equivalent of the binary IP address 10101100 00010000 1100100
11111010?
A. 172.32.100.250
B. 172.16.100.250
C. 172.16.104.249
D. 172.8.50.250
10.
How many hosts are available on a Class B network that uses 27 bits for the network address in its
subnet mask?
A. 30
B. 62
C. 16
D. 126
Answers
1. C. No guessing allowed. You must know how this is computed. If youÇre still having problems, go
back over the Labs in this chapter, substituting different values until you get it. See ÝSubnetting
network IDs.Û
2.
D. This is computed as 255 = 128 + 64 + 32 + 16 + 8 + 4 + 2 + 1, 255 = 128 + 64 + 32 + 16 + 8 + 4 +
2 + 1, and 224 = 128 + 64 +32. Review ÝSubnetting a Class A network.Û
3.
E. Thirteen bits were borrowed: 8 (255) from the third octet plus 5 (248) from the fourth octet. Take a
look at ÝBorrowing bits to grow a subnet.Û
4.
D. There can be 6 subnets and 30 hosts: 3 bits are used for 224 (128 + 64 + 32), (23)-2 = 6, leaving 5
bits for the hosts (25)-2 = 30. Check out ÝSubnetting Class B and Class C networks.Û
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