This chapter covers the following Cisco-specific objectives for the “Implement an IP addressing scheme and IP services to meet network requirements for a small branch office” section of the of the 640-822 ICND1 exam:
<objective>Describe the need and role of addressing in a network
</objective> <objective>Describe the operation and benefits of using private and public IP addressing
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List the characteristics of IPv4.
Make sure you know how to convert binary to decimal.
Make sure you know how to convert decimal to binary.
Make sure you know how to convert decimal to hexadecimal.
Identify IP address classes, including their networks and hosts, and the IP range value of the first octet of each class.
Define subnet masks and the IP subnet mask format.
Describe CIDR notations and how to determine the CIDR notation based on the subnet mask.
Define RFC 1918, NAT, and PAT.
Make sure you know how to calculate hosts and networks in a subnet.
Determine the Network ID, Broadcast IP, and valid IP range of a subnet.
Name the devices that are used at the Network layer and important traits of each device.
This chapter elaborates on the fundamental concepts that you learned about the Network layer or Layer 3 of the OSI model in Chapter 1, “Standard Internetworking Models.” It reviews the primary functions of the Network layer and then moves right into IP addressing and formats. After you are familiar with address formats and how to convert between them, it reviews subnetting. No matter whether you decide to take the one- or two-test approach to the CCENT certification, you have to know IP address formats and subnetting. They are integral concepts to any Cisco certification. Finally, this chapter discusses network devices that are used at the Network layer, which are routers and Layer 3 switches.
The Network layer of the OSI model serves two primary functions:
Determines the best path selection for a packet based on a logical or virtual address on the network (routing)
Handles ICMP, ARP, and Proxy ARP requests
First, best path determination is made at the Network layer for packet delivery across the network. Routed protocols such as IP are used to define logical addressing, which can identify the destination of a packet or datagram. Logical addresses used for routing consist of network and host bits. Routers also must determine the path through the internetwork for packet transmission. This is similar to how switches use a MAC address and interface for frame delivery. Routers also use an interface along with the logical or IP address.
Second, the Network layer also handles ICMP, ARP, and Proxy ARP requests on the internetwork. Remember the function of each protocol for the ICND1 exam.
Internet Control Messaging Protocol (ICMP) is used by ping and traceroute utilities. Packet Internet Groper (ping) enables you to validate that an IP address exists and can accept requests.
Ping is an echo and the response is an echo response.
Routers send Destination Unreachable messages when they can’t reach the destination network and they are forced to drop the packet. The router that drops the packet sends the ICMP DU message.
A traceroute traces the route or path taken from a client to a remote host. Traceroute also reports the IP addresses of the routers at each next hop on the way to the destination. This is especially useful when you suspect that a router on the route to an unreachable network is responsible for dropping the packet.
Address Resolution Protocol (ARP) maps a known IP address to a MAC address by sending a broadcast ARP. When the destination IP address is on another subnet, the sender broadcasts ARP for the router’s Ethernet port or default gateway, so the MAC address sent back will be that of the router’s Ethernet port.
Reverse ARP (RARP) maps a known MAC address to an IP address.
Proxy ARP enables a router to respond to an ARP request that has been sent to a remote host. Some Unix machines (especially Solaris) rely on Proxy ARP versus default gateways.
Objective:
Describe the need and role of addressing in a network
Internet Protocol (IP) uses logical or virtual addressing to get a packet from a source to its destination. At the Network layer, routers use IP addresses to make best path forwarding decisions. Public IP addresses are used for packets destined for the outside world, whereas private addresses can be used if the packet needs to traverse only an internal network. The ICND1 course focuses on IP version 4 (IPv4). The addresses themselves are assigned by the Internet Assigned Numbers Authority (IANA) to individual organizations based on a request for IP address space. Because the total number of IPv4 addresses is not infinite, strict guidelines are placed on IP space requests to ensure that they are justifiable.
IPv4 addresses
Consist of 32 bits.
Are broken into four octets (8 bits each).
Use dotted-decimal format, such as 172.16.122.204.
Have a minimum value (per octet) of 0 and a maximum value of 255.
Have a Network ID of 0.0.0.0.
Have a Broadcast IP of 255.255.255.255.
Another IP version was created in the event that the IP space from IPv4 is exhausted. That version is called IP version 6 (IPv6). IPv6 has emerged in the Cisco professional-level exams and may appear on a future ICND1 exam. For this reason, IPv6 is introduced later in this chapter.
To understand IP addressing, you must first understand binary. Binary is a computer language that is represented by a bit value of 0 or 1. A 32-bit binary address would resemble 10101010101010101010101010101010. Those 32 bits can be grouped into four octets, or 10101010 10101010 10101010 10101010, for conversion to decimal format. When the bit value is 1, the bit is considered to be on and you can calculate its binary value depending on its placement within the binary octet. When the bit value is 0, the bit is off and has no corresponding binary value. Figure 5.1 displays the binary value and the calculated decimal value of each bit within an octet. Notice that the binary value increases exponentially.
By using the value calculated for each bit, you can easily convert to decimal format. Line up the binary octet with the decimal value that was calculated in Figure 5.1. To calculate the total decimal value of each octet, you would add up the binary value of each bit that is on (1).
The example in Table 5.1 uses binary octet 00000000.
In this case, all the bit values are off (0), so there is no corresponding decimal value. The IP address octet value is also 0.
Table 5.2 uses binary octet 00010001.
In this example, the fourth and last bit values are 1. Add the decimal values to get the total decimal value of that octet. That is, the total decimal value = 17 (16 + 1).
Table 5.3 uses binary octet 11111111.
In Table 5.3, the total decimal value = 255 (128 + 64 + 32 + 16 + 8 + 4 + 2 + 1).
In this case, all the bit values are on (1), so all the decimal values are added together to calculate the IP address octet. The IP address octet value is 255.
Now, you can convert a 32-bit binary address into a dotted-decimal address. In this example, the binary address is 10101010 01010101 11000011 00111100. Start with the first octet 10110000. Table 5.4 shows the conversion of 10101010 from binary to decimal value.
In Table 5.4, the total decimal value = 170 (128 + 32 + 16).
The second octet is 01010101. Table 5.5 shows the conversion of 01010101 from binary to decimal value.
In Table 5.5, the IP octet value = 85 (64 + 16 + 4 + 1).
The third octet is 11000011. Table 5.6 shows the conversion of 11000011 from binary to decimal value.
In Table 5.6, the total decimal value = 195 (128 + 64 + 2 + 1)
The fourth and final octet is 00111100. Table 5.7 shows the conversion of 00111100 from binary to decimal value.
In Table 5.7, the total decimal value = 60 (32 + 16 + 8 + 4).
Based on these calculations, the IP address in dotted-decimal format is 176.85.195.60.
You must also be able to convert an IP address from dotted-decimal format into binary. It helps to work from left to right when converting to binary.
Example IP address = 206.110.28.62
The first octet of 206 can be broken down as follows:
128 | 64 | 32 | 16 | 8 | 4 | 2 | 1 |
1 | 1 | 0 | 0 | 1 | 1 | 1 | 0 |
The octet value is greater than 128, so the first bit is on. Subtract 128 from 206.
206 – 128 = 78
The remainder 78 is greater than 64, so the second bit is also on.
78 – 64 = 14
The remainder 14 is less than 32 and 16, so the third and fourth bits are off. However, 14 is greater than 8, so the fifth bit is on.
14 – 8 = 6
The remainder 6 is greater than 4, so the sixth bit is on.
6 – 4 = 2
The remainder 2 is equal to the seventh bit value, so that bit is also on.
2 – 2 = 0
The last bit value is off because the remainder is 0. Remember that it is an even number, so the last bit will always be 0!
Hexadecimal is a numbering system with a base of 16. Numbers 0 through 9 represent the first 10 decimal digits and the next 6 digits are the letters A through F. Each hexadecimal character is equal to four bits. Hexadecimal format was first introduced in Chapter 1, during the discussion of MAC addresses at the Data Link layer. Figure 5.2 shows the decimal values 0 through 15 and their equivalent hexadecimal values.
There are two ways to calculate hexadecimal from decimal format. With the first method, the decimal value should first be converted to binary format.
Decimal value = 141
128 | 64 | 32 | 16 | 8 | 4 | 2 | 1 |
1 | 0 | 0 | 0 | 1 | 1 | 0 | 1 |
Binary value = 10001101
Now break the binary value into two groups of 4 bits each, which is 1000 and 1101. Then, line up the 4 bits with the last four decimal values that were calculated in Figure 5.1. Again, add up the binary value of each bit that is on (1).
1 | 0 | 0 | 0 | |
8 | 4 | 2 | 1 | |
8 | The combined value is 8 so the hexadecimal character = 8 | |||
1 | 1 | 0 | 1 | |
8 | 4 | 2 | 1 | |
8 | 4 | 1 | The combined value is 13 so the hexadecimal character = D |
The combined hexadecimal address is 0x8D.
The second method for calculating a hexadecimal address is to divide the decimal number by 16 first. So, 141 divided by 16 equals 8 with a remainder of 13, which matches the results from the binary conversion in the first method.
As you know, IP addresses are 32 bits long, represented by dotted-decimal notation. Each address can be divided into two parts:
Network
Host
The number of network octets and host octets determines the IP address class. Table 5.8 shows the three IP defined network classes.
TCP/IP defines two additional address classes:
Class D: Used for multicast addresses
Class E: Used for research purposes
Table 5.9 lists the possible values each class network can have in the first octet. With these values, you can easily identify what class network is being referenced on the exam.
The network portion of an address maintains the same value for all the IP addresses that are assigned from a Class A, B, or C network. Remember that one octet is equal to 8 bits or 1 byte. The Class A network portion is 1 byte, and the host portion takes up the remaining 3 bytes. The Class B network portion is 2 bytes, with the remaining 2 bytes making up the host portion. The Class C network portion is 3 bytes, whereas the host portion is 1 byte. It stands to reason that if fewer bytes are devoted to the network portion of an address, fewer networks are possible for that class of network. With that said, the same is true for the host portion of an address. The fewer host bytes, the fewer total hosts that are available for that class of network.
When calculating the total number of Class A, B, or C networks available, you must subtract 2 from the total. This is a Cisco standard implemented for the ICND1 exam. A host address must be unique for each device or interface on a network. For each Class A, B, or C network, there is always a network identifier (ID) and a broadcast IP address. For this reason, you must also subtract two to calculate the total number of valid hosts per network.
A network ID is the first IP address in a network. This may also be referred to as a subnet ID. Every host bit for the network ID address is turned off (or all 0s). An example of a Class A network ID is 16.0.0.0.
A broadcast IP is the last IP address in a network. Every host bit for the broadcast IP address is turned on (or all 1s). An example of a Class A broadcast IP is 16.255.255.255.
Exam Alert
Power of 2 When you enter the exam room, it’s helpful to write down some things on the paper or white board that is supplied. I would suggest writing down the powers of 2 for quick reference when calculating networks and hosts.
Here’s a calculation for the number of networks for each class:
27 – 2 = 126 total Class A networks
214 – 2 = 16,382 total Class B networks
221 – 2 = 2,097,150 total Class C networks
Note
When calculating the total number of Class A, B, or C networks, the exponent is a multiple of 7.
Now you can calculate the number of hosts per network:
For any Class A network,
Network = 1 byte (8 bits)
Host = 3 bytes (24 bits)
224 – 2 = 16,777,214 total hosts per network
For any Class B network,
Network = 2 bytes (16 bits)
Host = 2 bytes (16 bits)
216 – 2 = 65,534 total hosts per network
For any Class C network,
Network = 3 bytes (24 bits)
Host = 1 byte (8 bits)
28 – 2 = 254 total hosts per network
Sub-networks (subnets) enable you to break a large network of IP addresses down into smaller, manageable address ranges. A smaller address range means fewer hosts on a network. Each subnet becomes a separate broadcast domain. All the devices that are in the same broadcast domain receive all broadcasts. Think if it were possible to have all 16,777,214 Class A network hosts sharing a broadcast domain and receiving all broadcasts. That would be a huge amount of traffic. Subnets enable you to break this large network into smaller address ranges. In this case, smaller is better.
A subnet mask is used to identify which part of an IP address is the network portion. Like the IP address itself, a subnet mask consists of 32 bits. The network portion is represented by all 1s.
The default subnet masks for Class A, Class B, and Class C networks are as follows:
Class A: 255.0.0.0 (11111111 00000000 000000000 000000000)
Class B: 255.255.0.0 (111111111 1111111111 000000000 000000000)
Class C: 255.255.255.0 (111111111 1111111111 111111111 0000000000)
Now that you know what an IP address and subnet mask are, there is a mathematical operation called Boolean AND that helps to identify some important aspects of an IP network. With Boolean AND, you can determine the network ID and broadcast IP given an IP address and subnet mask.
Boolean AND works as follows:
Determines the binary value of the IP address.
Determines the binary value of the subnet mask.
Lines up both binary values one on top of the other.
If the lined-up bit values in both addresses equal 1, the Boolean bit is also 1.
If the lined-up bit values in both addresses do not equal 1, the Boolean bit is 0.
Table 5.10 provides a Boolean example. The decimal IP address value = 124.0.0.0, and the subnet mask = 255.0.0.0.
The network ID in this example is 124.0.0.0. Using Boolean, you can see that the host bits in the last three octets are 0 bits, which identifies the network ID. I emphasized the last host octet in Table 5.10. If you turn all of those host bits on, you will get the broadcast IP, which in this case is 124.255.255.255.
For the next example (see Table 5.11), the decimal IP address value = 135.252.4.0, and the subnet mask = 255.255.0.0.
The network ID in this example is 135.252.0.0. Using Boolean, you can see that the host bits in the last two octets are 0 bits, which identify the network ID. I emphasized both host octets in the example. If you were to turn all those host bits on, you would get the broadcast IP, which in this case is 135.252.255.255.
Classless Interdomain Routing (CIDR) notation may also be used to identify the subnet mask. The mask is written in slash notation as follows:
Class A: /8
Class B: /16
Class C: /24
Describe the operation and benefits of using private and public addressing
The previously listed Class A, B, and C addresses are all IANA assigned public IP addresses. Although it originally seemed that there was sufficient public IPv4 address space available, resources began being consumed quickly. I mentioned that IPv6 was developed in the event that IPv4 address space became exhausted. Other measures were also implemented to alleviate the shortage of IPv4 public IP address space. These measures include RFC1918, Network Address Translation (NAT), and Port Address Translation (PAT). RFC 1918 defines private IP address space. Private address space can be used for traffic that does not need to leave the internal network. Because this traffic is internal to the network, it does not matter if other organizations are using the same address space. Private IP addresses are not routable on the Internet.
IANA Private Address Space Allocations include the following IP address ranges for Class A, Class B, and Class C networks:
Class A: 10.0.0.0–10.255.255.255
Class B: 172.16.0.0–172.31.255.255
Class C: 192.168.0.0–192.168.255.255
NAT translates one IP address to another. Typically this is done between private and public IP addresses. For example, a private IP address can be translated with NAT to a public IP address for outbound transmission to the Internet. NAT can also translate a public IP address to a private IP address for inbound transmission on the internal network. PAT can translate multiple addresses on an internal network to a single public IP address, which is called one-to-many address translation. PAT is available as NAT overloading on Cisco routers.
So far, we have focused on the network classes and key characteristics of each one. As I mentioned, each network class can also be broken down into smaller groups of IP address ranges or subnets, which may also be referred to as variable-length subnet masking (VLSM). Subnetting is the process of breaking down those larger IP networks into smaller sub-networks. At first, subnetting IP might seem like a daunting task, but it’s not that bad after you get the hang of it. In this section, you need to pull together all the knowledge that you have learned so far about binary, decimal, and subnets.
First, let’s get the easy subnetting out of the way.
For an IP address that has a 255.255.0.0 or 255.255.255.0 subnet mask, you can copy the octets that have a subnet mask value of 255 from the original IP address. For the remaining octets, you will put down a 0. Here’s an example:
IP address = 139.42.6.0
Subnet Mask = 255.255.0.0
The Network ID is 139.42.0.0.
To determine the Broadcast IP of this IP address and subnet mask, just replace the 0 octets from the Network ID with 255.
The Broadcast IP is 139.42.255.255.
Note
When a subnet mask has a value of 255.255.0.0 or 255.255.255.0, you can copy the original IP octets that match the 255 value subnet octets and then use 0 for any remaining octets to determine your Network ID. The Broadcast IP is the same original IP octets that match the 255 value subnet octets and the number 255 rather than 0 for the remaining octet(s).
To understand more difficult subnetting, you need to break down an IP address into network bits, host bits, and subnet bits. The network bits are determined by the network class. Class A has 8 network bits, Class B has 16 network bits, and Class C has 24 network bits. The network bits value is a constant. The host bits, on the other hand, must share space with the subnet bits. To determine the subnet bits for a network, you need to look at the subnet mask in binary. For example,
IP address = 176.85.195.60/22
Subnet Mask = 255.255.252.0
Subnet Mask in Binary = 11111111 11111111 11111100 00000000
Network bits = 16
Host bits = 10
Subnet bits = 6
The subnet mask in binary has 22 bits with a value of 1, which means the CIDR notation is /22. Based on the first octet of the IP address, you know that this is a Class B network. Class B networks have 16 network bits. That leaves 16 bits in the address. The bits that have a value of 0 determine the number of host bits. In this case, there are 10 host bits. The rest of the bits are the subnet bits, so there are 6 subnet bits.
Table 5.12 is a conversion table of decimal to binary values that will help you convert addresses more quickly when taking the exam.
You can use the chart to figure out more IP address values:
IP address = 100.15.209.0/23
Subnet Mask = 255.255.254.0
Subnet Mask in Binary = 11111111 11111111 11111110 00000000
Network bits = 8
Host bits = 9
Subnet bits = 15
IP address = 128.216.55.0/24
Subnet Mask = 255.255.255.0
Subnet Mask in Binary = 11111111 11111111 11111111 00000000
Network bits = 16
Host bits = 8
Subnet bits = 8
IP address = 222.110.8.61/28
Subnet Mask = 255.255.255.240
Subnet Mask in Binary = 11111111 11111111 11111111 11110000
Network bits = 24
Host bits = 4
Subnet bits = 4
To calculate the hosts in a subnet, we can use the formula 2H – 2. The exponent H represents the number of host bits in a network. If you use the subnetting examples, you can determine the hosts in each subnet:
IP address = 176.85.195.60/22
Subnet Mask = 255.255.252.0
Network bits = 16
Host bits = 10
Subnet bits = 6
210 – 2 = 1022 Hosts
IP address = 100.15.209.0/23
Subnet Mask = 255.255.254.0
Network bits = 8
Host bits = 9
Subnet bits = 15
29 – 2 = 510 Hosts
IP address = 128.216.55.0/24
Subnet Mask = 255.255.255.0
Network bits = 16
Host bits = 8
Subnet bits = 8
28 – 2 = 254 Hosts
IP address = 222.110.8.61/28
Subnet Mask = 255.255.255.240
Network bits = 24
Host bits = 4
Subnet bits = 4
24 – 2 = 14 Hosts
To calculate the networks in a subnet, you can use the formula 2N – 2. The exponent N represents the number of subnet bits in a network. You can figure out the number of networks in each subnet with the formula.
IP address = 176.85.195.60/22
Subnet Mask = 255.255.252.0
Network bits = 16
Host bits = 10
Subnet bits = 6
26 – 2 = 62 Networks
IP address = 100.15.209.0/23
Subnet Mask = 255.255.254.0
Network bits = 8
Host bits = 9
Subnet bits = 15
215 – 2 = 32,766 Networks
IP address = 128.216.55.0/24
Subnet Mask = 255.255.255.0
Network bits = 16
Host bits = 8
Subnet bits = 8
28 – 2 = 254 Networks
IP address = 222.110.8.61/28
Subnet Mask = 255.255.255.240
Network bits = 24
Host bits = 4
Subnet bits = 4
24 – 2 = 14 Networks
Exam Alert
The formula to calculate the number of networks or subnets created is 2N – 2. The N represents the subnet bits in a network.
Zero subnet may also be referred to as subnet zero. The zero subnet is the first subnet in a network and has all binary 0s in the subnet field. For the purpose of taking the ICND1 exam, you should not include the first subnet when calculating the number of networks in a larger subnet. This is one of the two reserved subnet numbers on a network and one of the reasons why you subtract from the total number of networks to get the correct answer for the test. The other network is the broadcast subnet, which has all 1s in the subnet field.
We have been working with the IP address subnet 222.110.8.61/28. After you know how many subnets and hosts are in a subnet, you can determine the network ID for that subnet. So far, you know that 222.110.8.61/28 has 14 hosts and 14 subnets. Before subtracting 2 for the valid number of hosts/networks, your calculations were for 16 hosts and 16 networks. This means that a subnet with a 255.255.255.240 mask is part of a larger subnet with a 16-host increment. The variable part of this subnet is the last octet. So you can automatically write down the first three octets as follows:
222.110.8.x (where the x is variable and has a 16-host increment)
Octet values range from 0 to 255. So the first subnet in the larger network is 222.110.8.0. Now you want to add increments of 16 to the last octet, so you get the following networks:
222.110.8.0 (zero subnet—not valid for the ICND1 exam)
222.110.8.16
222.110.8.32
222.110.8.48
222.110.8.64
222.110.8.80
222.110.8.96
222.110.8.112
222.110.8.128
222.110.8.144
222.110.8.160
222.110.8.176
222.110.8.192
222.110.8.208
222.110.8.224
222.110.8.240 (broadcast subnet—not valid for the ICND1 exam)
This is a list of the Network IDs in that Class C network with a subnet of 255.255.255.240. The Network ID is always an even number. There are 16 total subnets. According to the zero subnet and broadcast subnet rule, the first and last subnet cannot be used. The IP address 222.110.8.61 is greater than 48 and less than 64, so the Network ID or subnet number for 222.110.8.61/28 is 222.110.8.48 (which is highlighted in the list of networks). To get the broadcast IP, subtract 1 from the next Network ID in your list. In this example, the broadcast IP is 222.110.8.63.
There is another math shortcut that can be used to identify the Network ID, which then helps you determine the Broadcast IP. Take another look at 222.110.8.61/28.
IP address = 222.110.8.61
Subnet Mask = 255.255.255.240
Look at the first subnet mask octet from the left that is not a value of 255 and subtract it from 256.
256 – 240 = 16
You want to find the closest multiple of 16 that is less than the last octet in the IP address, which equals 61. You are using the last octet because that is the same octet used from the subnet mask.
16×3 = 48 and 16×4 = 64
Based on the calculations, the Network ID increments are as follows:
222.110.8.48
222.110.8.64
So you use 48 because it is less than 61, and 64 is the Network ID of the next subnet. You come up with the same answer as before.
Network ID = 222.110.8.48
Broadcast IP = 222.110.8.63 (one less than the next Network ID of the next subnet)
Here’s another example:
IP address = 100.15.209.0
Subnet Mask = 255.255.254.0
The first two octets in the subnet mask equal 255, so you need to use the first octet that is not equal to 255, or in this case the third octet from the left. Now you can subtract 254 from 256.
256 – 254 = 2
2 × 104 = 208
2 × 105 = 210
You can see that the valid network ID less than 209 is 208. The next network ID equals 210, so you can fill out the third octet with each value to obtain the following network ID increments:
100.15.208.0
100.15.210.0
For IP address 100.15.209.0 with a subnet mask of 255.255.254.0, you now know the Network ID is 100.15.208.0. The next network is 100.15.210.0, so you want to find the last IP address before that network ID to get the Broadcast IP. Because the value of an octet can range from 0 to 255, the last possible IP before 100.15.210.0 is 100.15.209.255.
Network ID = 100.15.208.0
Broadcast IP = 100.15.209.255
The range of valid IP addresses in a subnet is the first IP address after the Network ID and the last IP address before the Broadcast IP address. If you are given the following IP address and subnet mask, you can determine the range of valid IP addresses:
IP Address = 210.189.16.0
Subnet Mask = 255.255.255.0
First, identify the Network ID, which in this case is 210.189.16.0. Then determine the Broadcast address, which is 210.189.16.255. In this case, the valid IP range is 210.189.16.1 to 210.189.16.254.
Here are some more examples where the Network ID and Broadcast IP have already been determined:
IP address = 100.15.209.0
Subnet Mask = 255.255.254.0
CIDR = /23
Network ID = 100.15.208.0
Broadcast IP = 100.15.209.255
Valid IP range = 100.15.208.1 to 100.15.209.254
IP address = 222.110.8.61
Subnet Mask = 255.255.255.240
CIDR = /28
Network ID = 222.110.8.48
Broadcast IP = 222.110.8.63
Valid IP range = 222.110.8.49 to 222.110.8.62
Exam Alert
The valid range of IP addresses always starts with an odd number and ends with an even number.
It is important to understand the general concepts related to IP addressing and subnetting. Figure 5.3 is a chart or quick sheet that will help you check your answers and provide you a guideline for the ICND1 exam. I suggested that you memorize the powers of 2. I also suggest that you memorize this chart and write it down on your scrap paper or white board when you start the exam. I found it to be extremely helpful!
The most common network device found at the Network layer is a router; however, Layer 3 switches may also be implemented to create a WAN.
Both routers and Layer 3 switches can carry out these functions:
Suppress broadcasts or multicasts
Determine the best path for data transfer (routing)
Strip down and add to Data Link layer frames
Implement access lists for packet filtering (permit/deny statements)
Set up quality of service (QoS) qualifiers to measure network performance
It is important to know that both these devices can be used at the Network layer. However, for the purpose of the ICND1 exam, routers are more widely recognized and, therefore, are referred to when discussing Layer 3 functions.
Routers join a minimum of two networks together to create an internetwork or WAN. So far, we have discussed devices that are used at the Physical layer (hubs and repeaters) and the Data Link layer (Layer 2 switches and bridges). Layer 2 switches and bridges create a separate collision domain for each segment of the LAN. Routers and Layer 3 switches create a separate broadcast domain for each segment of a WAN. A broadcast domain is a group of nodes that can receive one another’s broadcast messages. Figure 5.4 demonstrates how a router creates broadcast domains whereas the connected switches create collision domains.
Figure 5.5 demonstrates a simple LAN with one router and two segments. In this network, any traffic that is generated by Matt’s PC has the source MAC and source IP address of that PC. If Matt is sending a frame to the server on the other segment of that WAN, the destination IP address will be that of the server he is trying to reach. Because the server is not on the same segment as Matt, the destination MAC address is that of the router, which is the default gateway. The router takes a look at the frame and at its own routing table. It then decides what interface to use to forward the frame based on the network portion of the IP address. The router attaches its own MAC address as the source MAC address of the frame before sending the frame to the server.
A routing table on a router contains the following information:
Network Address
Interface: Exit interface used to forward packets
Metric: Distance to reach a remote network
Exam Alert
For the exam, you should understand how a packet traverses the network and determine the source and destination IP and MAC address as the packet moves from device to device.
Figure 5.6 exemplifies a WAN with two routers. Each router has a separate routing table to make best path decisions.
Routers provide packet switching between networks and can provide packet filtering based on a network address or application layer port level.
Two packet types are used at Layer 3:
Data packets: Transport data across the internetwork and are supported by routed protocols such as IP and IPX.
Route update packets: Send updates to neighbor routers about all networks connected to that internetwork and are supported by routing protocols such as RIP, EIGRP, and OSPF.
Note
Specific Cisco router models are reviewed in Chapter 6, “Introduction to Cisco Routers and Switches.”
IP configurations are covered in Chapter 8, “Foundation Cisco Configurations.”
Chapter 10, “Introduction to Routing and Routing Protocols,” details routing terminology.
Chapter 11, “Distance Vector Routing Protocols,” reviews RIP and IGRP routing protocols.
Layer 3 switches are typically called multilayer switches. I already listed the commonalities between routers and Layer 3 switches. There are also a couple of differences worth mentioning. The number one difference between a router and a Layer 3 switch is packet switching throughput. Whereas a router has evolved over the years to process more than one million packets per second (pps), a Layer 3 switch can process millions of pps. That said, Layer 3 switches process more traffic in a shorter time.
Whereas routers use microprocessor-based engines, Layer 3 switches use ASIC hardware to perform packet switching. Layer 2 switches use ASIC hardware to forward frames.
Take a deep breath. That was a lot to cover! IP addressing and IP subnetting are vital to studying for the ICND1 exam. In fact, they are integral to understanding networking in general. This chapter touched on many of the concepts that are elaborated later in this book. A lot of device configurations are in upcoming chapters. Knowing your IP format and subnet masks is the first big hurdle when working with IP routing protocols such as OSPF and EIGRP. The next chapter introduces Cisco routers and switches as well as other important Cisco network components. You will also be introduced to Cisco IOS Software, which you will need to understand to configure a Cisco device.
Link-local address
Unique/site-local address
Global address
Practice makes perfect! Address conversion is pretty much guaranteed to be on the ICND1 exam. I promise you that it is worth your while to go over these exercises until you have the system down pat. I will give you the bit values for four octets. Convert each octet to decimal and fill in the corresponding dotted-decimal IP address value.
Keep in mind that the bit values for each octet are:
128 | 64 | 32 | 16 | 8 | 4 | 2 | 1 |
Estimated Time: 10 minutes
First Octet: | |||||||
Binary1 | 1 | 0 | 1 | 1 | 0 | 0 | 0 |
Decimal_____________ | |||||||
Second Octet: | |||||||
Binary0 | 1 | 0 | 1 | 1 | 1 | 1 | 1 |
Decimal_____________ | |||||||
Third Octet: | |||||||
Binary1 | 0 | 1 | 1 | 1 | 1 | 1 | 1 |
Decimal_____________ | |||||||
Fourth Octet: | |||||||
Binary1 | 1 | 1 | 0 | 0 | 0 | 0 | 0 |
Decimal_____________ | |||||||
IP Address = _____________. _____________. _____________. _____________ | |||||||
Conversions go both ways. Now, I will give you the decimal values for four octets. Convert each octet to binary and fill in the corresponding binary address value.
Estimated Time: 10 minutes | |
First Octet: | |
Decimal | 224 |
Binary__________________ | |
Second Octet: | |
Decimal | 137 |
Binary__________________ | |
Third Octet: | |
Decimal | 15 |
Binary__________________ | |
Fourth Octet: | |
Decimal | 253 |
Binary__________________ | |
Binary Address = _____________ _____________ _____________ _____________ |
You may be asked to determine a hexadecimal value given a decimal value. In this challenge, I have given you the decimal value. You need to convert that to binary and then break the binary octet into two 4-bit groups.
Keep in mind that the bit values for each 4-bit group are:
8 | 4 | 2 | 1 |
Also, Figure 5.2 has the decimal-to-hexadecimal values.
Estimated Time: 10 minutes
Decimal | = | 105 | |||||||
Binary | = | ____ | ____ | ____ | ____ | ____ | ____ | ____ | ____ |
First 4 bits of the binary octet: | ||||||
____ | ____ | ____ | ____ | = Decimal _______ Hexadecimal_______ | ||
Last 4 bits of the binary octet: | ||||||
____ | ____ | ____ | ____ | = Decimal _______ Hexadecimal_______ | ||
The combined hexadecimal value is 0x_________.
There are several useful memorization tools for the IP subnetting questions on the ICND1 exam. Fill in the binary equivalent (one octet) for each of the following decimal values.
Estimated Time: 5 minutes
Decimal | Binary |
|---|---|
0 | |
128 | |
192 | |
224 | |
240 | |
248 | |
252 | |
254 | |
255 |
In this exercise, please use Boolean AND to determine the Network ID, given the IP address 134.141.7.130 and subnet mask 255.255.255.0.
Estimated Time: 5 minutes
IP Address Binary |
Subnet Mask Binary |
Boolean AND |
Network ID = _____________. _____________. _____________. _____________
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The first hexadecimal character is 5, which is the same as the decimal value.
The second hexadecimal character is F, which is 15 in decimal.
Now you can combine the two 4-bit groups to get 01011111, which converts to 95 in decimal. | |||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
IP Address Binary = 11001000 00111110 10110111 00011010 Subnet Mask Binary = 11111111 11111111 11111111 00000000 Boolean AND = 11001000 00111110 10110111 00000000 The correct answer is 200.62.183.0. | |||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
This is considered an easy mask. You just write down the original octets with a subnet mask value of 255 and a 0 for each octet with a 0 value in the subnet mask. The correct answer is 32.116.5.0. | |||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
Because this has an easy subnet mask, you can copy the first three octets and then replace the last octet with 255. The correct answer is 213.50.201.255. | |||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
This IP address subnet mask is 255.255.255.0, so the network ID is 220.9.3.0. The broadcast IP is 220.9.3.255. The correct answer is 220.9.3.1 to 220.9.3.254. | |||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
Zero subnet may also be referred to as subnet zero. The zero subnet is the first subnet in a network and has all binary 0s in the subnet field. For the purpose of taking the ICND1 exam, you should not include the first subnet when calculating the number of networks in a larger subnet. This is one of the two reserved subnet numbers on a network and one of the reasons why you subtract from the total number of networks to get the correct answer for the test. The other network is the broadcast subnet, which has all 1s in the subnet field. | |||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
Traffic that is generated by a device has the source MAC and source IP address of that device. If a frame is sent to a server on another segment of a WAN, the destination IP address is that of the server the frame is trying to reach. Because the server is not on the same segment, the destination MAC address is that of the router, which is the default gateway. The router takes a look at the frame and at its own routing table. It then decides what interface to use to forward the frame, based on the network portion of the IP address. The router attaches its own MAC address as the source MAC address of the frame before the frame is sent to the server. | |||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
Suppress broadcasts or multicasts. Determine the best path for data transfer (routing). Strip down and add to Data Link layer frames. Implement access lists for packet filtering (permit/deny statements). Set up quality of service (QoS) qualifiers to measure network performance. | |||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
A. The decimal equivalent of 10010111 00000110 10101100 01110111 is 151.6.172.119. Answers B and D can be eliminated right away because both IP addresses end with an even number. | |
B. The first octet range for Class B addresses is 128 to 191. Answers A, C, and D are incorrect because the range of 1 to 126 is Class A, 192 to 223 is Class C, and 224 to 239 is reserved for multicast. | |
C. Answer C is correct because the first octet range for Class C addresses is 192 to 223. Answers A, B, and D are incorrect because the first octet range for Class A addresses is 1 to 126, for Class B is 128 to 191, and 224 to 239 is reserved for multicast. | |
B. There are 254 hosts per Class C network. For any Class C network, there are 24 network bits and 8 host bits: 28 – 2 = 254. | |
B. The Network ID of 124.199.7.18/28 is 124.199.7.16. If you write out the subnet mask it is 255.255.255.240. Subtract 240 from 256 and you have 16; 16×1 = 16, which is the first valid increment that can be the network ID. The next network ID is 124.199.7.32. | |
C. Subnet mask 255.255.255.240 allows for 14 networks and 14 hosts. Answer A is incorrect because subnet mask 255.255.255.0 allows for only one network with 254 hosts. Answer B is incorrect because subnet mask 255.255.255.224 allows for 6 networks and 30 hosts, and answer D is incorrect because subnet mask 255.255.255.248 allows for 30 networks and 6 hosts. | |
B, C. 208.62.15.1 and 208.62.15.30 are valid host addresses in the 208.62.15.0 network with a 255.255.255.224 subnet mask. Answer A is incorrect because 208.62.15.0 is the network ID and therefore is not a valid host address. Answer D is incorrect because 208.62.15.32 is the network ID of the next network and is not valid. | |
B. The network address 192.131.10.0 and subnet mask 255.255.255.0 has one network with 254 hosts. | |
A. With a network address of 218.41.99.24 and a subnet mask of 255.255.255.252, 2 valid host addresses are available. Answer B is incorrect because subnet mask 255.255.255.248 has 6 available hosts. Answers C and D are incorrect because subnet mask 255.255.255.240 has 14 available hosts, and subnet mask 255.255.255.224 has 30 available hosts. | |
D. Subnet mask 255.255.255.224 would allow for 6 networks with 30 available hosts per network. Answer A is incorrect because subnet mask 255.255.255.252 allows for 62 networks but only 2 hosts per network. Answers B and C are incorrect because subnet mask 255.255.255.240 allows for 14 networks with only 14 hosts per network, and subnet mask 255.255.255.248 allows for 30 networks with only 8 hosts per network. | |
C. There are 65,534 hosts available for a Class B network. Answers A and D are incorrect because a Class C network has 254 possible hosts, and a Class A has 16,777,214 possible hosts. | |
C. The broadcast IP for 196.23.250.32/27 is 196.23.250.63. Answer A is incorrect because 196.23.250.32 is the Network ID of 196.23.250.32/27. Answers B and D are incorrect because 196.23.250.33 is the first valid IP in the /27 network and 196.23.250.64 is the Network ID of the next network. | |
A. The only subnet that allows for 250 networks is 255.255.255.0. Subnet mask 255.255.255.0 can create 254 subnets. Answer B is incorrect because subnet mask 255.255.254.0 can create 126 subnets. Answer C is incorrect because subnet mask 255.255.252.0 can create 62 subnets. Answer D is incorrect because subnet mask 255.255.248.0 can create 30 subnets. | |
B. You can have 14 subnetworks with a Class B network that has a subnet mask of 255.255.240.0. | |
D. Answer D is correct because there are 126 possible hosts for a Class B network with a subnet mask of 255.255.255.128. Answer A is incorrect because a Class B network with a subnet mask of 255.255.255.0 has 254 possible hosts. Answers B and C are incorrect because the Network ID and Broadcast IP addresses were not subtracted from the total number of IPs in each subnet. | |
B, C, D. RFC 1918 defines the following private IP address ranges: 10.0.0.0 to 10.255.255.255, 172.16.0.0–172.31.255.255, and 192.168.0.0 to 192.168.255.255. Answer A is incorrect because the address 1.0.0.0 is a Class A public address. | |
B. 24 is the correct CIDR notation for the 128.250.62.0 network with a subnet mask of 255.255.255.0. Answer A is incorrect because /23 is the CIDR notation for 255.255.254.0. Answer C is incorrect because /25 is the CIDR notation for 255.255.255.128, and answer D is incorrect because /26 is the CIDR notation for 255.255.255.192. | |
C, D. Class D defines multicast address space and Class E defines IP address space reserved for research. Answers A and B are incorrect because Classes B and C are both defined for public use. | |
D. The valid host range for 160.254.101.167/27 is 160.254.101.161 to 160.254.101.190. Answers A and B are incorrect because both 160.254.101.128 to 160.254.101.254 and 160.254.101.129 to 160.254.101.254 are part of a 255.255.255.128 subnet. Answer C is incorrect because with the range 160.254.101.160 to 160.254.101.191, the first IP is the Network ID and it is not a valid host IP address. | |
A. The Milford router forwards the packet through the E0 interface because the network matches the 172.23.0.0 network. Answer B is incorrect because interface E1 is going to the 172.24.0.0 network, and answer C is incorrect because interface S0 is going to the Philadelphia router via the 172.22.0.0 network. Answer D is incorrect because interface S1 does not exist in this scenario. |
The following are some recommended readings on IPv4, private and public IP addressing, subnetting, and routing:
“IPv4,” http://en.wikipedia.org/wiki/IPv4.
“RFC 1918,” http://www.faqs.org/rfcs/rfc1918.html.
“IP Addressing and Subnetting for New Users,” http://www.cisco.com/warp/public/701/3.html.
“IP Address Subnetting Tutorial,” http://www.ralphb.net/IPSubnet/.
“Network Calculators,” http://www.subnetmask.info/.
Andrew Colton. Cisco IOS for IP Routing. Rocket Science Press, Inc., 2002.