## Do I Know This Already Quiz

1. Which of the following is the result of a Boolean AND between IP address 150.150.4.100, mask 255.255.192.0?

2. If mask 255.255.255.128 were used with a Class B network, how many subnets could exist, with how many hosts per subnet, respectively?

Answer: E. Class B networks imply 16 network bits; the mask implies 7 host bits (7 binary 0s in the mask), leaving 9 subnet bits. 29 - 2 yields 510 subnets, and 27 - 2 yields 126 hosts per subnet.

3. If mask 255.255.255.240 were used with a Class C network, how many subnets could exist, with how many hosts per subnet, respectively?

Answer: B. Class C networks imply 24 network bits; the mask implies 4 host bits (4 binary 0s in the mask), leaving 4 subnet bits. 24 - 2 yields 14 subnets, and 24 - 2 yields 14 hosts per subnet.

4. Which of the following IP addresses would not be in the same subnet as 190.4.80.80, mask 255.255.255.0?

5. Which of the following IP addresses would not be in the same subnet as 190.4.80.80, mask 255.255.240.0?

6. Which of the following IP addresses would not be in the same subnet as 190.4.80.80, mask 255.255.255.128?

7. Which of the following subnet masks would allow a Class B network to allow subnets to have up to 150 hosts and allow for up to 164 subnets?

Answer: C. You need 8 bits to number up to 150 hosts because 27 - 2 is less than 150, but 28 - 2 is greater than 150. Similarly, you need 8 subnet bits. The only valid Class B subnet mask with 8 host and 8 subnet bits is 255.255.255.0.

8. Which of the following subnet masks would allow a Class A network to allow subnets to have up to 150 hosts and would allow for up to 164 subnets?

Answer: B. You need 8 host bits and 8 subnet bits. Because the mask is used with a class A network, any mask with the entire second octet as part of the subnet field and with the entire fourth octet as part of the host field meets the requirement.

9. Which of the following are valid subnet numbers in network 180.1.0.0, when using mask 255.255.248.0?

Answer: C, D, E, and F. In this case, the subnet numbers begin with 180.1.0.0 (subnet zero), and then 180.1.8.0, 180.1.16.0, 180.1.24.0, and so on, increasing by 8 in the third octet, up to 180.1.240.0 (last valid subnet) and 180.1.248.0 (broadcast subnet).

10. Which of the following are valid subnet numbers in network 180.1.0.0, when using mask 255.255.255.0?

Answer: A, B, C, D, E, and F. In this case, the subnet numbers begin with 180.1.0.0 (subnet zero), and then 180.1.1.0, 180.1.2.0, 180.1.3.0, and so on, increasing by 1 in the third octet, up to 180.1.254.0 (last valid subnet) and 180.1.255.0 (broadcast subnet).

11. Which of the following best describes a feature of CIDR? Answer: A

12. The phrase "to represent hundreds or thousands of client TCP or UDP connections from different hosts as that same number of connections, but making it appear as if all connections are from one host" best describes which of the following tools?

13. The phrase "grouping a large number of Class C networks into a single group, and putting a single entry for that group in an Internet router, to reduce the overall size of the IP routing table " best describes which of the following tools?

14. The phrase "the use network 10.0.0.0 in an enterprise network" best describes which of the following tools?

Q&A

### 1. Name the parts of an IP address.

Answer: Network, subnet, and host are the three parts of an IP address. However, many people commonly treat the network and subnet parts of an address as a single part, leaving only two parts, the subnet and host parts. On the exam, the multiple-choice format should provide extra clues as to which terminology is used.

2. Define the term subnet mask. What do the bits in the mask whose values are binary 0 tell you about the corresponding IP address(es)?

Answer: A subnet mask defines the number of host bits in an address. The bits of value 0 define which bits in the address are host bits. The mask is an important ingredient in the formula to dissect an IP address; along with knowledge of the number of network bits implied for Class A, B, and C networks, the mask provides a clear definition of the size of the network, subnet, and host parts of an address.

3. Given the IP address 134.141.7.11 and the mask 255.255.255.0, what is the subnet number?

Answer: The subnet is 134.141.7.0. The binary algorithm is shown in the table that follows.

 Address 134.141.7.11 1000 0110 1000 1101 0000 0111 0000 1011 Mask 255.255.255.0 1111 1111 1111 1111 1111 1111 0000 0000 Result 134.141.7.0 1000 0110 1000 1101 0000 0111 0000 0000

4. Given the IP address 193.193.7.7 and the mask 255.255.255.0, what is the subnet number?

Answer: The network number is 193.193.7.0. Because this is a Class C address and the mask used is 255.255.255.0 (the default), no subnetting is in use. The binary algorithm is shown in the table that follows.

 Address 193.193.7.7 1100 0001 1100 0001 0000 0111 0000 0111 Mask 255.255.255.0 1111 1111 1111 1111 1111 1111 0000 0000 Result 193.193.7.0 1100 0001 1100 0001 0000 0111 0000 0000

5. Given the IP address 200.1.1.130 and the mask 255.255.255.224, what is the subnet number?

Answer: The answer is 200.1.1.128. The table that follows shows the subnet chart to help you learn the way to calculate the subnet number without binary math. The magic number is 256 - 224 = 32.

 Octet 1 2 3 4 Comments Address 200 1 1 130 — Mask 255 255 255 224 Interesting octet is the fourth octet (magic = 256 - 224 = 32). Subnet number 200 1 1 128 128 is the closest multiple of the magic number not greater than 130. First address 200 1 1 129 Add 1 to the last octet of the subnet number. Broadcast 200 1 1 159 Subnet + magic - 1. Last address 200 1 1 158 Subtract 1 from broadcast.

6. Given the IP address 220.8.7.100 and the mask 255.255.255.240, what is the subnet number?

Answer: The answer is 220.8.7.96. The table that follows shows the subnet chart to help you learn the way to calculate the subnet number without binary math. The magic number is 256-240=16.

 Octet 1 2 3 4 Comments Address 220 8 7 100 — Mask 255 255 255 240 Interesting octet is the fourth octet. Subnet number 220 8 7 96 96 is the closest multiple of the magic number not greater than 100. First address 220 8 7 97 Add 1 to the last octet. Broadcast 220 8 7 111 Subnet + magic - 1. Last address 220 8 7 110 Subtract 1 from broadcast.

 Address 134.141.7.11 1000 0110 1000 1101 0000 0111 0000 1011 Mask 255.255.255.0 1111 1111 1111 1111 1111 1111 0000 0000 Result 134.141.7.0 1000 0110 1000 1101 0000 0111 0000 0000 Broadcast address 134.141.7.255 1000 0110 1000 1101 0000 0111 1111 1111

Answer: The broadcast address is 193.193.7.255. Because this is a Class C address and the mask used is 255.255.255.0 (the default), no subnetting is in use. The binary algorithm is shown in the table that follows.

 Address 193.193.7.7 1100 0001 1100 0001 0000 0111 0000 0111 Mask 255.255.255.0 1111 1111 1111 1111 1111 1111 0000 0000 Result 193.193.7.0 1100 0001 1100 0001 0000 0111 0000 0000 Broadcast address 193.193.7.255 1100 0001 1100 0001 0000 0111 1111 1111

Answer: The broadcast address is 200.1.1.159. The binary algorithm math is shown in the table that follows. The easy decimal algorithm is shown in the answer to an earlier question.

 Address 200.1.1.130 1100 1000 0000 0001 0000 0001 1000 0010 Mask 255.255.255.224 1111 1111 1111 1111 1111 1111 1110 0000 Result 200.1.1.128 1100 1000 0000 0001 0000 0001 1000 0000 Broadcast address 200.1.1.159 1100 1000 0000 0001 0000 0001 1001 1111

 Address 220.8.7.100 1101 1100 0000 1000 0000 0111 0110 0100 Mask 255.255.255.240 1111 1111 1111 1111 1111 1111 1111 0000 Result 220.8.7.96 1101 1100 0000 1000 0000 0111 0110 0000 Broadcast address 220.8.7.111 1101 1100 0000 1000 0000 0111 0110 1111

11. Given the IP address 134.141.7.11 and the mask 255.255.255.0, what are the assignable IP addresses in this subnet?

11. Given the IP address 134.141.7.11 and the mask 255.255.255.0, what are the assignable IP addresses in this subnet?

12. Given the IP address 193.193.7.7 and the mask 255.255.255.0, what are the assignable IP addresses in this subnet?

13. Given the IP address 200.1.1.130 and the mask 255.255.255.224, what are the assignable IP addresses in this subnet?

14. Given the IP address 220.8.7.100 and the mask 255.255.255.240, what are the assignable IP addresses in this subnet?

15. Given the IP address 134.141.7.7 and the mask 255.255.255.0, what are all the subnet numbers if the same (static) mask is used for all subnets in this network?

Answer: The answer is 134.141.1.0, 134.141.2.0, 134.141.3.0, and so on, up to 134.141.254.0. 134.141.0.0 is the zero subnet, and 134.141.255.0 is the broadcast subnet.

16. Given the IP address 220.8.7.100 and the mask 255.255.255.240, what are all the subnet numbers if the same (static) mask is used for all subnets in this network?

Answer: The answer is not as obvious in this question. The Class C network number is 220.8.7.0. The mask implies that bits 25 through 28, which are the first 4 bits in the fourth octet, comprise the subnet field. The answer is 220.8.7.16, 220.8.7.32, 220.8.7.48, and so on, through 220.8.7.224. 220.8.7.0 is the zero subnet, and 220.8.7.240 is the broadcast subnet. The following table outlines the easy decimal algorithm to figure out the subnet numbers.

 Octet 1 2 3 4 Comments Network number 220 8 7 0 — Mask 255 255 255 240 The last octet is interesting; the magic number is 256 - 240 = 16. Subnet zero 220 8 7 0 Copy the network number; it's the zero subnet. First subnet 220 8 7 16 Add magic to the last subnet number's interesting octet. Next subnet 220 8 7 32 Add magic to the previous one. Last subnet 220 8 7 224 You eventually get her... Broadcast subnet 220 8 7 240 ...and then here, the broadcast subnet, because the next one is 256, which is invalid.

17. How many IP addresses could be assigned in each subnet of 134.141.0.0, assuming that a mask of 255.255.255.0 is used? If the same (static) mask is used for all subnets, how many subnets are there?

Answer: There will be 2hostbits, or 28 hosts per subnet, minus two special cases. The number of subnets will be 2subnetblts, or 28, minus two special cases.

 Number of Number of Network and Network Number of Number of Hosts per Number of Mask Bits Host Bits Subnet Bits Subnet Subnets 134.141.0.0, 16 8 8 254 254 255.255.255.0

18. How many IP addresses could be assigned in each subnet of 220.8.7.0, assuming that a mask of 255.255.255.240 is used? If the same (static) mask is used for all subnets, how many subnets are there?

Answer: There will be 2hostbits, or 24 hosts per subnet, minus two special cases. The number of subnets will be 2subnetblts, or 24, minus two special cases.

 Network Bits Subnet Bits Number of Hosts per Subnet Number of Subnets 220.8.7.0, 255.255.255. 240 24 4 4 14 14

19. You design a network for a customer, and the customer insists that you use the same subnet mask on every subnet. The customer will use network 10.0.0.0 and needs 200 subnets, each with 200 hosts maximum. What subnet mask would you use to allow the largest amount of growth in subnets? Which mask would work and would allow for the most growth in the number of hosts per subnet?

Answer: Network 10.0.0.0 is a Class A network, so you have 24 host bits with no subnetting. To number 200 subnets, you will need at least 8 subnet bits because 28 is 256. Likewise, to number 200 hosts per subnet, you will need 8 host bits. So, you need to pick a mask with at least 8 subnet bits and 8 host bits. 255.255.0.0 is a mask with 8 subnet bits and 16 host bits. That would allow for the 200 subnets and 200 hosts, while allowing the number of hosts per subnet to grow to 216 - 2, quite a large number. Similarly, a mask of 255.255.255.0 gives you 16 subnet bits, allowing 216 - 2 subnets, each with 28 - 2 hosts per subnet.

20. Referring to Figure A-1, Fred has been configured with IP address 10.1.1.1, Router A's Ethernet has been configured with 10.1.1.100, Router A's Serial interface uses 10.1.1.101, Router B's serial uses 10.1.1.102, Router B's Ethernet uses 10.1.1.200, and the web server uses 10.1.1.201. Mask 255.255.255.192 is used in all cases. Is anything wrong with this network? What is the easiest thing that could be done to fix it? You can assume any working interior routing protocol.

 A B a. "4 / L-H Fred

Answer: Router A's Ethernet interface and Fred's Ethernet should be in the same subnet, but they are not. Fred's configuration implies a subnet with IP addresses ranging from 10.1.1.1 through 10.1.1.62; Router A's Ethernet configuration implies a subnet with addresses between 10.1.1.65 and 10.1.1.126. Also, Router A's two interfaces must be in different subnets; as configured, they would be in the same subnet. So, the solution is to change Router A's Ethernet IP address to something between 10.1.1.1 and 10.1.1.62, making it be in the same subnet as Fred.

21. Referring to Figure A-1, Fred has been configured with IP address 10.1.1.1, mask 255.255.255.0; Router A's Ethernet has been configured with 10.1.1.100, mask 255.255.255.224; Router A's serial interface uses 10.1.1.129, mask 255.255.255.252; Router B's serial uses 10.1.1.130, mask 255.255.255.252; Router B's Ethernet uses 10.1.1.200, mask 255.255.255.224; and the web server uses 10.1.1.201, mask 255.255.255.224. Is anything wrong with this network? What is the easiest thing that could be done to fix it? You can assume any working interior routing protocol.

Answer: Fred's configuration implies a subnet with a range of addresses from 10.1.1.1 through 10.1.1.254, so he thinks that Router A's Ethernet interface is in the same subnet. However, Router A's configuration implies a subnet with addresses from 10.1.1.97 through 10.1.1.126, so Router A does not think that Fred is on the same subnet as Router A's Ethernet. Several options exist for fixing the problem. You could change the mask used by Fred and Router A's Ethernet to 255.255.255.128, which makes them both reside in the same subnet.

22. Referring to Figure A-1, Fred has been configured with IP address 10.1.1.1, mask 255.255.255.240; Router A's Ethernet has been configured with 10.1.1.2, mask 255.255.255.240; Router A's Serial interface uses 10.1.1.129, mask 255.255.255.252; Router B's serial uses 10.1.1.130, mask 255.255.255.252; Router B's Ethernet uses 10.1.1.200, mask 255.255.255.128; and the web server uses 10.1.1.201, mask 255.255.255.128. Is anything wrong with this network? What is the easiest thing that could be done to fix it? You can assume any working interior routing protocol.

Answer: Router B's configuration implies a subnet with a range of addresses from 10.1.1.129 to 10.1.1.130 on the serial link, and 10.1.1.129 to 10.1.1.254 on the Ethernet. So, the subnets overlap. One solution would be to configure Router B and the web server's masks to 255.255.255.192, which would change the subnet so that the valid addresses would be between 10.1.1.193 and 10.1.1.254.

23. What are the valid private IP network numbers, according to RFC 1918?

Answer: Network 10.0.0.0, Class B networks that from 172.16.0.0 through 172.31.0.0, and Class C networks beginning with 192.168.

25. How does CIDR help reduce the size of Internet routing tables?

Answer: By using a routing protocol that exchanges the mask as well as the subnet/ network number, a classless view of the number can be attained. By advertising many networks as a single route, the routing table can be shortened. For instance, 198.0.0.0/8 (198.0.0.0, mask 255.0.0.0) defines a set of addresses whose first 8 bits are equal to decimal 198. Instead of the more than 65,000 routes needed to list a route for each class C network that starts with 198, CIDR allows those routes to be represented by a single route.