# Answers to Appendix C Extra Practice Problems

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1 APPENDIX C ANSWERS Answers to Appendix C Extra Practice Problems This document contains the answers and explanations for the problems posed in Appendix C, Extra Practice, of the book. The problems in Appendix C are organized based on the chapters in the book that explain the processes and math formulas used to solve the problems. This document uses that same organization, for easier reference. Answers to Chapter 1 Practice Problems Like most types of math problems, practicing the process can help. This section lists several practice problems for converting numbers between different formats. The four types of problems related to Chapter 1, and posed in this appendix, are as follows: Converting an 8-bit binary number to a decimal number Converting a decimal number to an 8-bit binary number Converting an 8-bit binary number to a hexadecimal number Converting a hexadecimal number to an 8-bit binary number Answers to Converting an 8-bit Binary Number to a Decimal Number Table W-1 lists the answers to the binary-to-decimal conversion problems listed in Table C-2 of Appendix C of the book. Following that, the answers are explained in more detail. Table W-1 Answers to Binary-to-Decimal Conversion Problems Problem Number Binary Decimal Value

2 2 Networking Basics CCNA 1 Companion Guide The process for converting from an 8-bit binary number to decimal, as covered in Chapter 1 of the book, is relatively straightforward. It is repeated here: How To Step 1 Write the powers of 2, in decimal, in the top row of a table, similar to Table W-2. Step 2 On the second line, write the binary number that is to be converted, lining up each binary digit under the powers of 2. Step 3 Multiply each pair of numbers (the numbers in the same column) together. Step 4 Add the eight products from Step 3. Table W-2 shows the details of the process for Problem 1. Table W-2 Answer to Problem 1: Converting to Decimal Value Associated with That Digit or Column The number itself Product of two numbers in same column Sum of all products 85 The first step in the four-step conversion process is always to write the powers of 2, as shown in the top row of Table W-2. Then, in Step 2, the binary number is recorded in the second row. At that point, the last two steps are simple arithmetic: you multiply the two numbers in each column together (Step 3) and then add those eight numbers together (Step 4). In this case, the answer is 85. Tables W-3 through W-7 show the details of the conversion for binary-to-decimal Problems 2 through 6. Table W-3 Answer to Problem 2: Converting to Decimal Value Associated with That Digit or Column The number itself Product of two numbers in same column Sum of all products 170

3 Appendix C Answers 3 Table W-4 Answer to Problem 3: Converting to Decimal Value Associated with That Digit or Column The number itself Product of two numbers in same column Sum of all products 227 Table W-5 Answer to Problem 4: Converting to Decimal Value Associated with That Digit or Column The number itself Product of two numbers in same column Sum of all products 240 Table W-6 Answer to Problem 5: Converting to Decimal Value Associated with That Digit or Column The number itself Product of two numbers in same column Sum of all products 15 Table W-7 Answer to Problem 6: Converting to Decimal Value Associated with That Digit or Column The number itself Product of two numbers in same column Sum of all products 192

4 4 Networking Basics CCNA 1 Companion Guide Answers to Converting a Decimal Number to an 8-bit Binary Number Table W-8 lists the answers to the decimal-to-binary conversion problems listed in Table C-3 of Appendix C of the book. Following that, the answers are explained in more detail. Table W-8 Answers to Decimal-to-Binary Conversion Problems Problem Number Decimal Value Binary Chapter 1 of the book covers the process of converting from decimal to 8-bit binary numbers. The process listed in Chapter 1 is repeated here for reference, after which the answers are provided. Begin by writing down eight powers of 2, starting with 128 on the left, and ending with 1 on the right, similar to Table W-9. The blank lines below the powers of 2 are placeholders in which you will record the binary digits. Table W-9 Table for Converting Decimal to 8-bit Binary Power of Binary Digits Beginning with 128 and moving toward 1, follow these steps eight times, once for each power of 2: Step 1 How To If the decimal number is greater than or equal to the power of 2, do the following: A. Record a 1 as the binary digit under the power of 2. B. Subtract the power of 2 from the decimal number, which results in a number called the remainder. C. Use the remainder for the next step/power of 2. Step 2 If the decimal number is less than the power of 2, do the following: A. Record a 0 as the binary digit under the power of 2. B. Move to the next power of 2 and use the same remainder as in this step.

5 Appendix C Answers 5 Answer to Problem 1: Converting 255 to Binary Table W-10 shows the results of the preceding process for the first problem, which is to convert decimal 255 to 8-bit binary. 255 is a popular number with IP addressing, so many people simply memorize that 255 is in binary. Table W-10 Answer to Problem 1: Converting 255 to Binary Power of Binary Digits _ _ The decision points at each step are as follows: Step 1 Step 2 Step 3 Step 4 Step 5 Step 6 Step 7 Step 8 The decimal number (255) is greater than 128, so a 1 was recorded in the 128 column, and = 127. The remaining decimal number (127) is greater than or equal to 64, so a 1 was recorded in the 64 column, and = 63. The remaining decimal number (63) is greater than or equal to 32, so a 1 was recorded in the 32 column, and = 31. The remaining decimal number (31) is greater than or equal to 16, so a 1 was recorded in the 16 column, and = 15. The remaining decimal number (15) is greater than or equal to 8, so a 1 was recorded in the 8 column, and 15 8 = 7. The remaining decimal number (7) is greater than or equal to 4, so a 1 was recorded in the 4 column, and 7 4 = 3. The remaining decimal number (3) is greater than or equal to 2, so a 1 was recorded in the 2 column, and 3 2 = 1. The remaining decimal number (1) is greater than or equal to 1, so a 1 was recorded in the 1 column. (The final bit of math, 1 1 = 0, isn t really needed at this point.) Answer to Problem 2: Converting 128 to Binary Table W-11 shows the answer for decimal-to-binary conversion Problem 2. Table W-11 Answer to Problem 2: Converting 128 to Binary Power of Binary Digits _ _

6 6 Networking Basics CCNA 1 Companion Guide The decision points at each step are as follows: Step 1 Step 2 Step 3 Step 4 Step 5 Step 6 Step 7 Step 8 The decimal number (128) is greater than or equal to 128, so a 1 was recorded in the 128 column, and = 0. The remaining decimal number (0) is not greater than or equal to 64, so a 0 was recorded in the 64 column. The remaining decimal number (0) is not greater than or equal to 32, so a 0 was recorded in the 32 column. The remaining decimal number (0) is not greater than or equal to 16, so a 0 was recorded in the 16 column. The remaining decimal number (0) is not greater than or equal to 8, so a 0 was recorded in the 8 column. The remaining decimal number (0) is not greater than or equal to 4, so a 0 was recorded in the 4 column. The remaining decimal number (0) is not greater than or equal to 2, so a 0 was recorded in the 2 column. The remaining decimal number (0) is not greater than or equal to 1, so a 0 was recorded in the 1 column. Answer to Problem 3: Converting 200 to Binary Table W-12 shows the answer for decimal-to-binary conversion Problem 3. Table W-12 Answer to Problem 3: Converting 200 to Binary Power of Binary Digits _ _ The decision points at each step are as follows: Step 1 Step 2 Step 3 Step 4 The decimal number (200) is greater than or equal to 128, so a 1 was recorded in the 128 column, and = 72. The remaining decimal number (72) is greater than or equal to 64, so a 1 was recorded in the 64 column, and = 8. The remaining decimal number (8) is not greater than or equal to 32, so a 0 was recorded in the 32 column. The remaining decimal number (8) is not greater than or equal to 16, so a 0 was recorded in the 16 column.

7 Appendix C Answers 7 Step 5 Step 6 Step 7 Step 8 The remaining decimal number (8) is greater than or equal to 8, so a 1 was recorded in the 8 column, and 8 8 = 0. The remaining decimal number (0) is not greater than or equal to 4, so a 0 was recorded in the 4 column. The remaining decimal number (0) is not greater than or equal to 2, so a 0 was recorded in the 2 column. The remaining decimal number (0) is not greater than or equal to 1, so a 0 was recorded in the 1 column. Answer to Problem 4: Converting 9 to Binary Table W-13 shows the answer for decimal-to-binary conversion Problem 4. Table W-13 Answer to Problem 4: Converting 9 to Binary Power of Binary Digits _ _ The decision points at each step are as follows: Step 1 Step 2 Step 3 Step 4 Step 5 Step 6 Step 7 Step 8 The decimal number (9) is not greater than or equal to 128, so a 0 was recorded in the 128 column. The remaining decimal number (9) is not greater than or equal to 64, so a 0 was recorded in the 64 column. The remaining decimal number (9) is not greater than or equal to 32, so a 0 was recorded in the 32 column. The remaining decimal number (9) is not greater than or equal to 16, so a 0 was recorded in the 16 column. The remaining decimal number (9) is greater than or equal to 8, so a 1 was recorded in the 8 column, and 9 8 = 1. The remaining decimal number (1) is not greater than or equal to 4, so a 0 was recorded in the 4 column. The remaining decimal number (1) is not greater than or equal to 2, so a 0 was recorded in the 2 column. The remaining decimal number (1) is greater than or equal to 1, so a 1 was recorded in the 1 column. (The final bit of math, 1 1 = 0, isn t really needed at this point.)

8 8 Networking Basics CCNA 1 Companion Guide Answer to Problem 5: Converting 100 to Binary Table W-14 shows the answer for decimal-to-binary conversion Problem 5. Table W-14 Answer to Problem 5: Converting 100 to Binary Power of Binary Digits _ _ The decision points at each step are as follows: Step 1 Step 2 Step 3 Step 4 Step 5 Step 6 Step 7 Step 8 The decimal number (100) is not greater than or equal to 128, so a 0 was recorded in the 128 column. The remaining decimal number (100) is greater than or equal to 64, so a 1 was recorded in the 64 column, and = 36. The remaining decimal number (36) is greater than or equal to 32, so a 1 was recorded in the 32 column, and = 4. The remaining decimal number (4) is not greater than or equal to 16, so a 0 was recorded in the 16 column. The remaining decimal number (4) is not greater than or equal to 8, so a 0 was recorded in the 8 column. The remaining decimal number (4) is greater than or equal to 4, so a 1 was recorded in the 4 column, and 4 4 = 0. The remaining decimal number (0) is not greater than or equal to 2, so a 0 was recorded in the 2 column. The remaining decimal number (0) is not greater than or equal to 1, so a 0 was recorded in the 1 column. Answer to Problem 6: Converting 248 to Binary Table W-15 shows the answer for decimal-to-binary conversion Problem 6. Table W-15 Answer to Problem 6: Converting 248 to Binary Power of Binary Digits _ _ The decision points at each step are as follows: Step 1 The decimal number (248) is greater than 128, so a 1 was recorded in the 128 column, and = 120.

9 Appendix C Answers 9 Step 2 Step 3 Step 4 Step 5 Step 6 Step 7 Step 8 The remaining decimal number (120) is greater than or equal to 64, so a 1 was recorded in the 64 column, and = 56. The remaining decimal number (56) is greater than or equal to 32, so a 1 was recorded in the 32 column, and = 24. The remaining decimal number (24) is greater than or equal to 16, so a 1 was recorded in the 16 column, and = 8. The remaining decimal number (8) is greater than or equal to 8, so a 1 was recorded in the 8 column, and 8 8 = 0. The remaining decimal number (0) is not greater than or equal to 4, so a 0 was recorded in the 4 column. The remaining decimal number (0) is not greater than or equal to 2, so a 0 was recorded in the 2 column. The remaining decimal number (0) is not greater than or equal to 1, so a 0 was recorded in the 1 column. Answers to Converting an 8-bit Binary Number to a Hexadecimal Number Table W-16 lists the answers to the binary-to-hexadecimal conversion problems listed in Table C-4 of Appendix C of the book. Following that, the answers are explained in more detail. Table W-16 Answers to Binary-to-Hexadecimal Conversion Problems Problem 8-bit Binary 2-digit Hex Number Number Number AA E F F C0 The process to convert between binary and hex, and vice versa, is relatively simple if you use a conversion chart, as presented in Chapter 1. Table W-17 lists all 16 hexadecimal digits and the 4-digit binary numbers they represent.

10 10 Networking Basics CCNA 1 Companion Guide Table W-17 Binary and Hexadecimal Conversion Chart Hexadecimal Binary Decimal Digit Equivalent Equivalent A B C D E F To perform the conversion, take each set of 4 binary digits and look up the corresponding hexadecimal values. For example, has two sets of 4 binary digits: 0101 and Each set of 4 bits corresponds to hex digit 5, so the hex value is 55. Table W-18 breaks down the logic used to find the answers to all six binary-to-hexadecimal conversion problems listed in Table C-4 of Appendix C.

11 Appendix C Answers 11 Table W-18 Answers to Binary-to-Hexadecimal Conversion Problems Problem 8-bit Binary First Set of 4 Hex Second Set Hex Combined Number Number Binary Digits Value of 4 Binary Value 2-digit Digits Hex Value A 1010 A AA E E F F F 0F C C0 Answers to Converting a Hexadecimal Number to an 8-bit Binary Number Table W-19 lists the answers to the hexadecimal-to-binary conversion problems listed in Table C-5 of Appendix C of the book. Following that, the answers are explained in more detail. Table W-19 Answers to Hexadecimal-to-Binary Conversion Problems Problem 2-digit Hex 8-bit Binary Number Number Number 1 F E AB A AA The process to convert between hex and binary is relatively simple if you use a conversion chart, as presented in Chapter 1. Table W-20 provides a chart with all 16 hexadecimal digits and the 4-digit binary numbers they represent.

12 12 Networking Basics CCNA 1 Companion Guide Table W-20 Binary and Hexadecimal Conversion Chart Hexadecimal Binary Decimal Digit Equivalent Equivalent A B C D E F To perform the conversion, take each hexadecimal digit and substitute the 4-bit binary code from the conversion chart. For example, F0 has 2 hex digits. The F corresponds to binary 1111, and the 0 corresponds to binary Table W-21 breaks down the logic used to find the answers to all six hexadecimal-to-binary conversion problems listed in Appendix C.

13 Appendix C Answers 13 Table W-21 Answers to Hexadecimal-to-Binary Conversion Problems Problem Hex First Binary Second Binary Combined 8-Digit Number Value Hex Digit Value Hex Digit Value Binary Value 1 F0 F E E AB A 1010 B A A AA A 1010 A Answers to Chapter 2 Practice Problems You can determine the answers to the data transfer time problems listed in Appendix C in the book with some simple math. Setting up the problem is the only tricky part. To set up the problem, you need to do the following: Find the constraining link s bandwidth. Figure C-1 in Appendix C shows three WAN links, with one of those links being the slowest (constraining) link bandwidth in each problem. Convert the file sizes from megabytes to megabits. The problems list the file sizes in megabytes, and the formula T = S/BW requires the unit of bits instead of bytes for the file size (S). Table W-22 shows the answers to the three data transfer time problems, pointing out the constraining link, that link s bandwidth, the file size in bits, and the result of the T = S/BW calculation. Table W-22 Answers to Data Transfer Time Problems Theoretical Problem Constraining Constraining File Size (Bits) Transfer Time Number Link Bandwidth (bps) (Seconds) 1 C 512,000 80,000, A or C 256,000 40,000, A 64,000 1,600,000,000 25,000 To find the estimated data transfer time using a link s effective throughput, you follow the same rationale as when applying the theoretical maximum. Although the logic works the same as with the previous calculation, in this case, the constraining link s bandwidth is now reduced

14 14 Networking Basics CCNA 1 Companion Guide based on the assumption that other traffic will be competing with this one particular file transfer. Table W-23 lists the answers. Table W-23 Answers to Data Transfer Time Problems Using Estimated Throughput Problem Constraining Constraining Throughput File Size Transfer Time Number Link Bandwidth (bps) (Bits) (Seconds) 1 C 512, ,000 80,000, A or C 256,000 85,333 40,000, A 64,000 16,000 1,600,000, ,000 Answers to Chapter 8 Practice Problems The key to determining the devices in a collision domain and a broadcast domain is to remember what devices break LANs into separate collision and broadcast domains, and which devices do not. Use the following as a guideline: Devices that use Layer 1 forwarding processes repeaters and hubs do not separate LANs into different collision domains or broadcast domains. Devices that use Layer 2 forwarding processes bridges and switches do separate LANs into different collision domains, but do not separate LANs into separate broadcast domains. Devices that use Layer 3 forwarding processes routers separate LANS into different collision domains and different broadcast domains. As for determining which links support full duplex, when LANs are built solely with UTP cabling and fiber-optic cabling, full duplex can be used as long as the cable does not connect to a hub or repeater. Figures W-1 and W-2 show the same network topologies as shown in Figures C-2 and C-3, respectively, in Appendix C. In this case, the collision domains are enclosed with a circle that uses a dotted line style, and the broadcast domains are enclosed by circles that use a solid line style. Also, all links that can use full duplex are shown simply as thicker lines. Figure W-1 Answers to Problem 1 for Identifying Collision Domains, Broadcast Domains, and Full-Duplex Links

15 Appendix C Answers 15 Figure W-2 Answers to Problem 2 for Identifying Collision Domains, Broadcast Domains, and Full-Duplex Links Answers to Chapter 9 Practice Problems Appendix C of the book includes three main types of practice problems related to Chapter 9: Dissecting the structure of an unsubnetted IP address Determining whether an IP address is a private address and whether it can legally be assigned to a host as an IP address, assuming no subnetting is used Dissecting the structure of a subnetted address that uses a simple mask Answers to Dissecting Unsubnetted IP Addresses Table W-24 lists the answers to the problems posed in Table C-8 of Appendix C in the book. Following Table W-24, the text explains the answers.

16 16 Networking Basics CCNA 1 Companion Guide Table W-24 Answers to Dissecting Unsubnetted IP Addresses Problem IP Address Class Network Number of Number of Number Number Octets in the Octets in the Network Part Host Part A C B B D A C B To dissect an unsubnetted IP address, you first need to be able to quickly determine whether an address is a Class A, B, or C address. When no subnetting is used, these addresses have a set size for the network and host parts of the address, as shown in Table W-25, and as covered in Chapter 9 in the book. Table W-25 Classful IP Address Classes Based on the First Octet Range of Values in Class Length of Network Part Length of Host Part First Octet (Inclusive) A 1 octet 3 octets B 2 octets 2 octets C 3 octets 1 octet The only other fact needed to solve the problems in Table C-8 (repeated in this document, with answers, as Table W-24) is to remember that a network number has all decimal 0s in the host octets. For example, Problem 1 lists IP address From the first octet s value (10), and Table W-25, it is clear that is in a Class A network. By definition, then, with no subnetting, the address has a 1-octet network part and a 3-octet host part. The network number would be the same as the IP address ( ), except that all the host octets would instead be 0s. In this case, the host octets are the last 3 octets, so the network number is The following list explains how the answers for Problems 2 through 7 were found: Problem 2 IP address is a Class C address, because the first octet (200) is between 192 and 223. That means it has 3 octets in the network part and 1 octet in the host part. So, the network number is the same as the IP address, except that the last octet is 0, making the network number

21 Appendix C Answers 21 Problem 3 Problem is a Class B address, so it has 2 network octets. With mask , it has 1 host octet (the last octet), leaving 1 subnet octet. The subnet number would be , which is the same number as , but with the 1 host octet changed to be is a Class B address, so it has 2 network octets. With mask , it has 1 host octet (the last octet), leaving 1 subnet octet. The subnet number would be , which is the same number as , but with the 1 host octet changed to be 0. Problem is a Class A address, so it has 1 network octet. With mask , it has 1 host octet (the last octet), leaving 2 subnet octets. The subnet number would be , which is the same number as , but with the 1 host octet changed to be 0. Problem is a Class A address, so it has 1 network octet. With mask , it has 2 host octets (the last 2 octets), leaving 1 subnet octet. The subnet number would be , which is the same number as , but with the 2 host octets changed to be 0. Problem 7 Problem is a Class C address, so it has 3 network octets. With mask , it has 1 host octet (the last octet), leaving no subnet octets. In effect, no subnetting is used for this problem. The network number would be , which is the same number as , but with the 1 host octet changed to be is a Class B address, so it has 2 network octets. With mask , it has 1 host octet (the last octet), leaving 1 subnet octet. The subnet number would be , which is the same number as , but with the 1 host octet changed to be 0. Answers to Chapter 10 Practice Problems Appendix C lists six different types of problems that can be solved based on Chapter 10 of the book. This section provides the answers to those problems. The types of problems are as follows: Determining the number of required subnets for an internetwork, based on a diagram of the internetwork Determining the minimum number of subnet and host bits required in the structure of an IP address to support the maximum number of required subnets and hosts per subnet Determining the subnet mask(s) that support the stated maximum number of subnets and hosts per subnet For a chosen IP network and mask, determining all the subnets of that network

23 Appendix C Answers 23 Figure W-4 Answer to Problem 2 for Determining the Number of Required Subnets Legend: Subnet Figure W-5 Answer to Problem 3 for Determining the Number of Required Subnets Server1 Legend: Subnet Server2 SW6 SW1 Server3 R1 R2 R3 R4 SW3

24 24 Networking Basics CCNA 1 Companion Guide Figure W-6 has three routers with two LAN interfaces plus a WAN interface. R2, R3, and R4 each have two LAN interfaces connected to two different LAN switches. So, each of these three routers connects to two LAN broadcast domains, requiring six subnets. R1 on the left connects to a single LAN broadcast domain, requiring a single subnet. Finally, the three WAN point-to-point links each require a subnet, so for this problem, ten subnets are required. Figure W-6 Answer to Problem 4 for Determining the Number of Required Subnets R2 R1 R3 Legend: Subnet R4 Answers to Determining the Required Number of Subnet and Host Bits Table W-30 lists the answers for the problems stated in Table C-11 of Appendix C. Following Table W-30, the text explains the answers. Table W-30 Answers to Determining the Number of Subnet and Host Bits Required Problem Classful Maximum Maximum Minimum Number Minimum Number Network Number of Number of Required Number of Subnets of Hosts per Subnet Subnet Bits Required Being Needed Needed Host Bits Subnetted

25 Appendix C Answers 25 The minimum number of subnet bits required is s, where s is the smallest number that makes the following formula true: 2 s 2 => the number of required subnets Similarly, the minimum number of host bits required is h, where h is the smallest number that makes the following formula true: 2 h 2 => the number of required hosts per subnet As a result, determining the answer to each problem relates specifically to either calculating or memorizing the powers of 2. Table W-31 lists the powers of 2 through Table W-31 The Powers of 2, 2 0 Through ,384 32,768 With the information in Table W-31, all you need to do is look at the requirements for the number of subnets and hosts and do some quick comparisons to find the number of required subnet and host bits. The following list describes the logic used to determine the answers in Table W-30. Problem 1 To create 4000 subnets, 11 subnet bits are not enough, because is only However, 12 subnet bits would be enough, because is 4094, which is larger than To support 4000 hosts per subnet, the same math applies, with 11 host bits not being enough, but 12 host bits supporting 4094 hosts per subnet, because is Problem 2 To create 200 subnets, 7 subnet bits are not enough, because is only 126. However, 8 subnet bits would be enough, because is 254, which is larger than 200. The same math applies to the process of deciding the number of host bits needed to support 200 hosts per subnet. Problem 3 To create 10 subnets, 3 subnet bits are not enough, because is only 6. However, 4 subnet bits would be enough, because is 14, which is larger than 10. The same math applies to the process of deciding the number of host bits needed to support 10 hosts per subnet. Problem 4 To create 200 subnets, 7 subnet bits are not enough, because is only 126. However, 8 subnet bits would be enough, because is 254, which is larger than 200. To support 100 hosts per subnet, 6 host bits are not enough, because is only 62. However, 7 host bits would be enough, because is 126, which is larger than 100. Problem 5 To create 50 subnets, 5 subnet bits are not enough, because is only 30. However, 6 subnet bits would be enough, because is 62, which is larger than 50. To support 2 hosts per subnet, 1 host bit is not enough, because is 0. However, 2 host bits would be enough, because is 2, which provides the exact number of hosts needed per subnet.

26 26 Networking Basics CCNA 1 Companion Guide Problem 6 To create 255 subnets, 8 subnet bits are not enough, because is only 254. However, 9 subnet bits would be enough, because is 510, which is larger than 255. To create 200 hosts per subnet, 7 host bits are not enough, because is only 126. However, 8 host bits would be enough, because is 254, which is larger than 200. In this case, with a Class B network, the requirements cannot be met, because the network part is 16 bits long per the Class B network rules, and the subnet and host parts need a total of 17 bits, making the network, subnet, and host parts 33 bits long and the address is only 32 bits long. Answers to Determining the Subnet Mask Table W-32 shows the answers to the problems posed in Table C-12 of Appendix C. Following Table W-32, the text explains the answers. Table W-32 Answers to Determining the Subnet Mask, Given a Set of Requirements Problem Classful Number of Number of Only Possible Mask That Number Network Required Required Mask, or Mask Supports the Being Subnet Host Bits That Supports Maximum Subnetted Bits the Maximum Number of Number of Subnets Hosts per Subnet (as Needed) Chapter 10 describes two processes to find the correct subnet masks: one using binary math and one that works in some cases but uses decimal math. The next section explains the answers to all six problems using binary math, and the section following that explains four of the problems using decimal math. Determining the Answer Using Binary Math The network, subnet, and host parts of an IPv4 address must total 32 bits. In some cases, the design requirements may exceed 32 bits and thus may not be allowed. For instance, in a Class B network, the network part is always 16 bits long. In that case, if the number of subnets and hosts per subnet specified in the design would require more than 16 combined subnet and host

27 Appendix C Answers 27 bits, the design requirements simply cannot be met. For problems like those in Table C-12 of Appendix C, first make sure that the required minimum size of the network, subnet, and host parts of the address add up to 32 bits or less, to ensure that the design requirements can be met. For all six problems in Table C-12, the requirements can be met. In cases in which the number of bits in the network, subnet, and host parts totals 32 bits, one and only one subnet mask meets the requirements. That is true of the first three problems in Table C-12 and Table W-32. For these problems, you can use the following algorithm, as listed in Chapter 10 of the book: How To Step 1 Write down 8, 16, or 24 binary 1s, depending on whether the network being subnetted is a Class A, B, or C network, respectively. These bits represent the network bits. Step 2 Going left to right, write down an additional s binary 1s, with s being the number of subnet bits. Step 3 Write down binary 0s for the rest of the bits; these represent the host bits. Step 4 Convert this 32-bit binary number, 8 bits at a time, back to decimal. Answer to Problem 1: Determining the Subnet Mask For Problem 1, Class A network 10 is used, so Step 1 requires that the mask have an octet of binary 1s. For Step 2, 12 more binary 1s are needed, because the requirements state a need for 12 subnet bits. For Step 3, the rest of the bits are binary 0s. Finally, the resulting 32-bit number must be converted to decimal, 8 bits at a time. The only tricky part of that process is to remember the 8-bits-at-a-time part, especially for the third octet, because it has some subnet bits and some host bits. The resulting mask is The following outlines the first three steps in the process: Step 1 Step 2 Step Answer to Problem 2: Determining the Subnet Mask For Problem 2, Class B network is used, so Step 1 requires that the mask have two octets of binary 1s. For Step 2, 10 more binary 1s are needed, because the requirements state a need for 10 subnet bits. For Step 3, the rest of the bits are binary 0s. Finally, the resulting 32-bit number must be converted to decimal, 8 bits at a time. Again, remember the 8-bits-at-a-time part, especially for the fourth octet, because it has some subnet bits and some host bits. The resulting mask is The following outlines the first three steps in the process: Step 1 Step 2 Step

28 28 Networking Basics CCNA 1 Companion Guide Answer to Problem 3: Determining the Subnet Mask For Problem 3, Class C network is used, so Step 1 requires that the mask have 3 octets of binary 1s. For Step 2, 5 more binary 1s are needed, because the requirements state a need for 5 subnet bits. For Step 3, the rest of the bits are binary 0s. Finally, the resulting 32-bit number must be converted to decimal, 8 bits at a time. The resulting mask is The following outlines the first three steps in the process: Step 1 Step 2 Step Finding Multiple Subnet Masks for Problems 4, 5, and 6 The first three problems had only one valid answer, but Problems 4, 5, and 6 have at least two answers. As indicated in the description of Figure in Chapter 10, if the required number of bits in the combined network, subnet, and host fields adds up to less than 32, multiple masks meet the requirements. In those cases, you can examine the structure of the mask and determine which bits must be network, subnet, and host bits, and then list the remaining bits as wildcard bits. These wildcard bits can be either 0 or 1, within restrictions. Figure W-7 shows the format of the possible masks for Problems 4, 5, and 6, with w representing wildcard bits. Figure W-7 Finding the Subnet Mask: Answers for Problems 4, 5, and 6 Network bits Subnet Host bits bits Problem w0000 Network bits Subnet bits Host bits Problem w w Network bits Subnet bits Host bits Problem wwww wwww0000 W represents wildcard bits To find the possible masks, you simply substitute 0s or 1s for the wildcard bits. The more 0s used in the wildcard bits, the larger the host part of the addresses, and the more hosts supported. The more 1s used in the wildcard bits, the more subnet bits, and the larger the number of

### examines the ideas related to Class A, Class B, and Class C networks (in other words, classful IP networks).

This chapter covers the following subjects: Classful Network Concepts: This section examines the ideas related to Class A, Class B, and Class C networks (in other words, classful IP networks). Practice

### Lab 10.3.5a Basic Subnetting

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### Source net: 200.1.1.0 Destination net: 200.1.2.0 Subnet mask: 255.255.255.0 Subnet mask: 255.255.255.0. Router Hub

then to a router. Remember that with a Class C network address, the first 3 octets, or 24 bits, are assigned as the network address. So, these are two different Class C networks. This leaves one octet,

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