EIGRP Redistribution of Different AS(Autonomous System) configuration on GNS3:

https://www.youtube.com/channel/UCQECTKWf3phX5McZIyPM0Mw/videos

EIGRP Redistribution of Different AS(Autonomous System) configuration on GNS3:

In this tutorial i will show you that how to communicate the two different AS(Autonomous System) in gns3. EIGRP Redistribution of Different AS are need to performed on edge router having different AS configure on it. For this i have create a lab in gns3, in our topology R3 is the router on which we shall perform redistribution. For IP and EIGRP configuration you can visit EIGRP configuration on GNS3. But you need to make some changes on R3, R5 and R6 i.e. run "router rigrp 200" instead of "router rigrp 100".  

EIGRP Redistribution Commands:

R3(config)#config t

R3(config)#router eigrp 200

R3(config-router)#network 3.3.3.0 0.0.0.255

R3(config-router)#network 4.4.4.0 0.0.0.255

R3(config)#exit

R3(config)#router eigrp 100

R3(config-router)#network 2.2.2.0 0.0.0.255

R3(config)#router eigrp 100

R3(config-router)#redistribute eigrp 200

R3(config-router)#ex

R3(config)#router eigrp 200

R3(config-router)#redistribute eigrp 100

Verification:

R2#show ip route

You will find the other EIGRP AS routes with EX tag.

EIGRP – Stub Routing

https://www.youtube.com/channel/UCQECTKWf3phX5McZIyPM0Mw/videos

EIGRP – Stub Routing

In hub-and-spoke environments, spoke routers (branch routers) often do not need to maintain complete routing tables of the network, as the paths to other portions of the network are always through the hub routers (head quarter routers). Additionally, the massive query-reply processes between the hub and spoke routers upon a loss route would affect the stability of the network.

A typical connection between a hub router and a spoke router has much less bandwidth than a connection in the network core. Therefore attempting to use such connections as transit paths would typically results in excessive congestion. EIGRP stub routing can prevent this problem by restricting the spoke router from advertising the routes from a hub router to another hub router. As a result, the hub routers would not notice there are alternative paths through the spoke routers. However, stub routing is unable to stop the hub router from advertising routes to spoke router. Manual summarization is required on the hub to advertise only a default route to the spokes.
Note: A stub router would not advertise routes received from a neighbor to another neighbor.

Without stub routing implemented on dual-homed spoke routers, route filtering configuration is required on the spoke routers to prevent them from appearing as transit paths to the hub routers.

When the remote sites are not acting as a transit sites between the regional sites, the regional routers (hubs) should be configured to advertise only a default route to the remote routers. The remote routers (spokes) should be configured to advertise only their directly connected stub networks back to the regional routers to reduce the complexity of EIGRP convergence (the query-reply process) and the sizes of the EIGRP topology table and IP routing table.


EIGRP stub routing an efficient method for limiting the query range, therefore conserves bandwidth due to unnecessary queries, prevents SIA events, and improves network stability.

An EIGRP stub router would inform its stub status to upstream routers via the Hello packets. Any neighboring upstream router that receives such Hello packets will not query the stub router for lost routes, as the stub router has no downstream EIGRP neighbors and hence would not have alternative paths for a lost route. The upstream routers which connected to the stub router would answer any query on behalf of the stub router, which results in improved convergence time.

The eigrp stub [receive-only | connected , static , summary , redistributed] router subcommand configures an EIGRP router as an EIGRP stub router. An EIGRP stub router would advertise its connected and summary routes to other neighboring routers by default. Below describes the optional keywords that can be used to modify this behavior:

receive-only
Restricts an EIGRP stub router from advertising any route to other routers. This option will not allow other options to be specified as it prevents the advertisement of any type of route, which is not really useful; the other options can be configured in any combination. Use this option when the stub router has only a single interface.

connected
Allows an EIGRP stub router to advertise its connected routes. The network statement is required to include the connected interfaces into the EIGRP routing process. The redistribute connected router subcommand can also be used to redistribute connected subnets into the EIGRP routing process. This option is enabled by default and is the most widely used option.

static
Allows an EIGRP stub router to advertise its static routes. The redistribute static router subcommand is required to redistribute static routes.

summary
Allows an EIGRP stub router to advertise its summary routes. Summary routes can be configured manually with the ip summary-address eigrp interface subcommand or automatically with the auto-summary router subcommand. This option is enabled by default.

redistributed
Allows an EIGRP stub router to advertise external EIGRP routes learnt from other routing protocols or other EIGRP autonomous systems.

Note: Configuring EIGRP stub routing would teardown existing EIGRP neighbor relationships!

For you case, implement the EIGRP STUB (and not STUCK as in EIGRP Stuck in Active ) on R4, R6, and R7. Just simply conf t, router eigrp, and the "eigrp stub".

After that, verify the stub configuration on R5 using the show ip eigrp neighbors detail EXEC command on R5. Below is the excerpt from my test lab.

HQ1#sh ip eigrp neighbors detail
IP-EIGRP neighbors for process 100
H   Address                 Interface       Hold Uptime   SRTT   RTO  Q  Seq
                                            (sec)         (ms)       Cnt Num
2   10.10.10.10             Se1/1             14 00:02:28    1  4500  0  22
   Version 12.3/1.2, Retrans: 2, Retries: 0
   Stub Peer Advertising ( CONNECTED ) Routes
   Suppressing queries
1   10.10.10.6              Se1/0             13 00:02:29    1  4500  0  22
   Version 12.3/1.2, Retrans: 2, Retries: 0
   Stub Peer Advertising ( CONNECTED ) Routes
   Suppressing queries
0   10.10.10.2              Fa0/0             12 00:02:30  424  3816  0  25
   Version 12.3/1.2, Retrans: 1, Retries: 0
HQ1#

Subnetting a Class B network address

https://www.youtube.com/channel/UCQECTKWf3phX5McZIyPM0Mw/videos
Class B subnets
In my last Daily Drill Down, we borrowed bits from the host portion of a Class C network address. Class C network addresses only have eight bits to manipulate into subnets. However, a Class B has 16 bits to play with. This will allow more subnets with more hosts per subnet than a Class C network ever could.

Table 1 lists all of the possible Class B subnets:

Table 1

Mask	Binary	Subnets	Hosts per subnet
255.255.128.0	10000000.00000000	2	32,766
255.255.192.0	11000000.00000000	2	16,382
255.255.224.0	11100000.00000000	6	8,190
255.255.240.0	11110000.00000000	14	4,094
255.255.248.0	11111000.00000000	30	2,046
255.255.252.0	11111100.00000000	62	1,022
255.255.254.0	11111110.00000000	126	510
255.255.255.0	11111111.00000000	254	254
255.255.255.128	11111111.10000000	510	126
255.255.255.192	11111111.11000000	1022	62
255.255.255.224	11111111.11100000	2,046	30
255.255.255.240	11111111.11110000	4,094	14
255.255.255.248	11111111.11111000	8,190	6
255.255.255.252	11111111.11111100	16,382	2

All possible Class B subnets

There are quite a few more masks we can use with a Class B network address than we can with a Class C network address. Remember that this is not harder than subnetting with Class C, but it can get confusing if you don’t pay attention to where the subnet bits and host bits are in a mask. This takes practice!

In this Daily Drill Down, I will use the same techniques I used in the Class C article to subnet a network. We’ll start with the Class B subnet mask of 255.255.192.0 and figure out the subnets, broadcast address, and valid host range. We will answer the same five questions we answered for the Class C subnet masks:

1. How many subnets does this mask provide?
2. How many hosts per subnet does this mask provide?
3. What are the valid subnets?
4. What is the broadcast address for each subnet?
5. What is the host range of each subnet?

Before we answer these questions, there is one difference you need to be aware of when subnetting a Class B network address. When subnetting in the third octet, you need to add the fourth octet. For example, on the 255.255.192.0 mask, the subnetting will be done in the third octet. To create a valid subnet, you must add the fourth octet of all 0s and all 1s for the network and broadcast address (0 for all 0s and 255 for all 1s).

Example 1: Answers for the 255.255.192.0 mask

1. 2-2=2 subnets
2. 2-2=16,382 hosts per subnet
3. 256-192=64.0, 128.0
4. Broadcast for the 64.0 subnet is 127.255. Broadcast for the 128.0 subnet is 191.255.
5. The valid hosts are:

Subnet	64.0	128.0
first host	64.1	128.1
last host	127.254	191.254
broadcast	127.255	191.255

Notice that the numbers in the third octet are the same numbers we used in the fourth octet when subnetting the 192 mask. The only difference is that we add 0 and 255 in the fourth octet.

For the 64.0 subnet, all the hosts between 64.1 and 127.254 are in the 64 subnet. In the 128.0 subnet, the hosts are 128.1 through 191.254.

I know this is confusing, but I promise if you read this complete Daily Drill Down carefully, I can make it easier for you.

Work through a few more with me, and it should start to become clearer.

Example 2: 255.255.240.0

1. 2-2=14 subnets
2. 2-2=4094 hosts per subnet
3. 256-240=16.0, 32.0, 48.0, 64.0, etc.
4. Broadcast for the 16.0 subnet is 31.255. Broadcast for the 32.0 subnet is 47.255, etc.
5. The valid hosts are:

Subnet	16.0	32.0	48.0	64.0
first host	16.1	32.1	48.1	64.1
last host	31.254	47.254	63.254	79.254
broadcast	31.255	47.255	63.255	79.255

Example 3: 255.255.248.0

1. 2-2=30 subnets
2. 2-2=2,046 hosts per subnet
3. 256-248=8.0, 16.0, 24.0, 32.0, 40.0, 48.0, 56.0, 64.0, etc.
4. Broadcast for the 8.0 subnet is 15.255. Broadcast for the 16.0 subnet is 23.255, etc.
5. The valid hosts are:

Subnet	8.0	16.0	24.0	32.0	40.0	48.0	56.0	64.0
first host	8.1	16.1	24.1	32.1	40.1	48.1	56.1	64.1
last host	15.254	23.254	31.254	39.254	47.254	55.254	63.254	71.254
broadcast	15.255	23.255	31.255	39.255	47.255	55.255	63.255	71.255

Example 4: 255.255.252.0

1. 2-2=62 subnets
2. 2-2=1,022 hosts per subnet
3. 256-252=4.0, 8.0, 12.0, 16.0, 20.0, 24.0, 28.0, 32.0, etc.
4. Broadcast for the 4.0 subnet is 7.255. Broadcast for the 8.0 subnet is 11.255, etc.
5. The valid hosts are:

Subnet	4.0	8.0	12.0	16.0	20.0	24.0	28.0	32.0
first host	4.1	8.1	12.1	16.1	20.1	24.1	28.1	32.1
last host	7.254	11.254	15.254	19.254	23.254	27.254	31.254	35.254
broadcast	7.255	11.255	15.255	19.255	23.255	27.255	31.255	35.255

Example 5: 255.255.255.0

1. 2-2=254 subnets
2. 2-2=254 hosts per subnet
3. 256-255=1.0, 2.0, 3.0, 4.0, 5.0, 6.0, 7.0, 8.0, etc.
4. Broadcast for the 1.0 subnet is 1.255. Broadcast for the 2.0 subnet is 2.255, etc.
5. The valid hosts are:

Subnet	1.0	2.0	3.0	4.0	5.0	6.0	7.0	8.0
first host	1.1	2.1	3.1	4.1	5.1	6.1	7.1	8.1
last host	1.254	21.254	3.254	4.254	5.254	6.254	7.254	8.254
broadcast	1.255	2.255	3.255	4.255	5.255	6.255	7.255	8.255

That last example was pretty simple. I hope you notice a pattern now. All the numbers were basically the same except we added the fourth octet into the address.

The more difficult process of subnetting a Class B network address is when you start using bits in the fourth octet for subnetting. For example, what happens when you use this mask with a Class B network address: 255.255.255.128? Is that valid? Absolutely! There are nine bits for subnetting and seven bits for hosts. That is 510 subnets, each with 126 hosts. However, it is the most difficult mask to figure out the valid hosts for.

Example 6: The Class B 255.255.255.128 subnet mask:

1. 2-2=510 subnets
2. 2-2=126 hosts per subnet
3. For the third octet, the mask would be 256-255=1, 2, 3, 4, 5, 6, etc.
4. For the fourth octet, the mask would be 256-128=128, which is one subnet if it is used. However, if you turn the subnet bit off, the value is 0. This means that for every subnet in the third octet, the fourth octet has two subnets: 0 and 128, for example 1.0 and 1.128.
5. Broadcast for the 0.128 subnet is 128.255; the broadcast for the 1.0 subnet is 1.127. Broadcast for the 1.128 subnet is 1.255, etc.
6. The valid hosts are:

Subnet	0.128	1.0	1.128	2.0	2.128	3.0	3.128	4.0
first host	0.129	1.1	1.129	2.1	2.129	3.1	3.129	4.1
last host	0.254	1.126	1.254	2.126	2.254	3.126	3.254	4.126
broadcast	0.255	1.127	1.255	2.127	2.255	3.127	3.255	4.127

The thing to remember is that for every subnet in the third octet, there are two in the fourth octet: 0 and 128. For the 0 subnet, the broadcast address is always 127. For the 128 subnet, the broadcast address is always 255.

Let’s continue with more subnetting into the fourth octet. This is exactly like subnetting a Class C network address, but the third octet is part of the subnet address.

Example 7: Class B network 255.255.255.192

1. 2-2=1022 subnets
2. 2-2=62 hosts per subnet
3. 256-255=1.0, 2.0, 3.0, etc. for the third octet. 256-192=64, 128, 192 for the fourth octet. For every valid subnet in the third octet, we get four subnets in the fourth octet: 0, 64, 128, and 192.
4. Broadcast for the 1.0 subnet is 1.63, since the next subnet is 1.64. Broadcast for the 1.64 subnet is 1.127, since the next subnet is 1.128. Broadcast for the 1.128 subnet is 1.191, since the next subnet is 1.192. Broadcast for the 1.192 subnet is 1.255.
5. The valid hosts are as follows:

Subnet	0.64	0.128	0.192	1.0	1.64	1.128	1.192	2.0
first host	0.65	0.129	0.193	1.1	1.65	1.129	1.193	2.1
last host	0.126	0.190	0.254	1.62	1.126	1.190	1.254	2.62
broadcast	0.127	0.191	0.255	1.63	1.127	1.191	1.255	2.63

On this one, the 0 and 192 subnets are valid, since we are using the third octet as well. The subnet range is 0.64 through 255.128. 0.0 is not valid since no subnet bits are on. 255.192 is not valid because then all subnet bits would be on.

Example 8: Class B network 255.255.255.224

1. 2-2=2046 subnets
2. 2-2=30 hosts per subnet
3. 256-255=1.0, 2.0, 3.0, etc. for the third octet. 256-224=32, 64, 96, 128, 160, 192 for the subnet value. (For every value in the third octet, we get eight subnets in the fourth octet: 0, 32, 64, 96, 128, 160, 192, 224.)
4. Broadcast for the 1.0 subnet is 1.63, since the next subnet is 1.64. Broadcast for the 1.64 subnet is 1.127, since the next subnet is 1.128. Broadcast for the 1.128 subnet is 1.191, since the next subnet is 1.192. Broadcast for the 1.192 subnet is 1.255.
5. The valid hosts are:

Subnet	0.32	0.64	0.96	0.128	0.160	0.192	0.224	1.0
first host	0.33	0.65	0.97	0.129	0.161	0.193	0.225	1.1
last host	0.62	0.94	0.126	0.158	0.190	0.222	0.254	1.30
broadcast	0.63	0.95	0.127	1.159	0.191	0.223	0.255	1.31

For this subnet mask, the 0 and 224 subnets are valid as long as not all subnet bits in the third octet are off or all subnet bits in the fourth octet are on.

When would we use this valuable information? All the time! For example, if you have a host configuration of 172.16.10.33 255.255.255.224, what subnet, broadcast address, and valid host range is this host a member of? (We would solve this question with the information presented above.)
256-224=32, 64

Bingo! In the fourth octet, the host address is 33. That is between 32 and 64, so the host is in the 32 subnet, which has a broadcast address of 63, and the valid host range is 33-62. Easy. Just remember that the subnet is 10.32 because the third octet is part of the subnet address.

Let’s try another one. You have a host configuration of 172.16.10.33 255.255.255.240. What subnet, broadcast address, and valid host range is this host a member of?

Since we did not go through this mask in this Daily Drill Down, you’ll have to figure it out on your own. It is done the same way as all the others.
256-240=16, 32, 48

Bingo! The host is in the 10.32 subnet, which has a broadcast address of 10.47 and a valid host range of 10.33 through 10.46.

Let’s keep going: You have a host configuration of 172.16.10.33 255.255.255.248. What subnet, broadcast address, and valid host range is this host a member of?
256-248=8, 16, 24, 32, 40

Bingo! The host is in the 10.32 subnet, which has a broadcast address of 10.39 and valid host range of 10.33 through 10.38. Easy, huh?

One more: You have a host configuration of 172.16.10.17 255.255.255.252. What subnet, broadcast address, and valid host range is this host a member of?
256-252=4, 8, 12, 16, 20

Bingo! You have a subnet of 10.16, with a broadcast of 10.19 and valid host range of 10.17 through 10.18.

thats the class b subnetting

Subnetting a Class C network address

https://www.youtube.com/channel/UCQECTKWf3phX5McZIyPM0Mw/videos

Let’s face it, some day you are going to have to subnet a network. Although IP addressing isn’t a network administrator’s favorite task, it’s a critical skill that you must have. In this Daily Drill Down, I will continue our discussion of IP addressing by exploring the process of taking a large IP address range and dividing it into smaller, more manageable pieces. Since this is a complicated, involved process, I will only discuss the process of breaking up Class C networks. First, we need to discuss why we want to subnet a network and the advantages of doing so.

Why subnetting?

By creating smaller IP networks (instead of having one large network), we can obtain better security, smaller collision and broadcast domains, and greater administrative control of each network. Think of a network like streets in a city. Each house on this network is known by the street and by the address. Think of the addresses on the houses as the hardware addresses of a host. For IP to communicate with a host, the IP address must be known, and the router connected to the network on which this host is located must also know the hardware address of the house.

What if a city didn’t have many blocks but just one long street? The mailman would go crazy trying to get the mail delivered to each house correctly because he would have to know the address of every house. It’s the same scenario with IP. By creating smaller networks, we can more effectively get data to each host.

Subnetting a Class C network

So you understand why you want to subnet, but how do you do it? Your goal is to look at an IP address and subnet mask of a host and then determine three things quickly:

1. The subnet the host is located in

2. The broadcast address of the subnet

3. The valid host range of the subnet used to configure hosts

Once the subnet is determined, the broadcast address must be found. Why? Because these are not valid host addresses and cannot be assigned to host configurations. Also, by determining the subnet and broadcast addresses, we can easily determine the host addresses because the valid host range is always the numbers between the subnet address and the broadcast address.

If we use the default subnet mask with a Class C network address, then we already know that three bytes are used to define the network and only one byte is used to define the hosts on each network.

The default Class C mask is: 255.255.255.0. To make smaller networks, called subnetworks, we will borrow bits from the host portion of the mask. Since the Class C mask only uses the last octet for host addressing, we only have 8 bits at our disposal. Therefore, only the following masks can be used with Class C networks (Table A).

---

Subset zero
Take note that in the table below I do not assume subnet zero. Cisco does teach a subnet zero assumption but they do not test that way. I have chosen to follow the exam.

---

Table A

Mask	Binary	# Subnet bits	# Host bits	Subnets	Hosts
255.255.255.128	10000000	1	7	2	126
255.255.255.192	11000000	2	6	2	62
255.255.255.224	11100000	3	5	6	30
255.255.255.240	11110000	4	4	14	14
255.255.255.248	11111000	5	3	30	6
255.255.255.252	11111100	6	2	62	2

Class C masks

You can see in Table A that the bits that are turned on (1s) are used for subnetting, while the bits that are turned off (0s) are used for addressing of hosts. You can use some easy math to determine the number of subnets and hosts per subnet for each different mask.

To determine the number of subnets, use the 2^x-2, where the x exponent is the number of subnet bits in the mask.

To determine the number of hosts, use the 2^x-2, where the x exponent is the number of host bits in the mask.

To determine the mask you need for your network, you must first determine your business requirements. Count the number of networks and the number of hosts per network that you need. Then determine the mask by using the equations shown above—and don’t forget to factor for growth.

For example, if you have eight networks and each requires 10 hosts, you would use the Class C mask of 255.255.255.240. Why? Because 240 in binary is 11110000, which means you have four subnet bits and four host bits. Using our math, we’d get the following:
2⁴-2=14 subnets
2⁴-2=14 hosts

Many people find it easy to memorize the Class C information because Class C networks have few bits to manipulate. However, there is an easier way to subnet.

Easy subnetting

Instead of memorizing the entire table (Table A), it’s possible to glance at a host address and quickly determine the necessary information if you’ve memorized key parts of the table. First, you need to know your binary-to-decimal conversion. Memorize the number of bits used with each mask that are shown in Table A. Second, you need to remember the following:
256-192=64
256-224=32
256-240=16
256-248=8
256-252=4

Once you have the two steps memorized, you can begin subnetting. Our first example will use the Class C mask of 255.255.255.192. Ask five simple questions to gather all the facts:

1. How many subnet bits are used in this mask?

2. How many host bits are available per subnet?

3. What are the subnet addresses?

4. What is the broadcast address of each subnet?

5. What is the valid host range of each subnet?

You already know how to answer questions one and two. To answer question three, use the formula 256-subnetmask to get the first subnet and your variable. Keep adding this number to itself until you get to the subnet mask value to determine the valid subnets. Once you verify all of the subnets, you can determine the broadcast address by looking at the next subnet’s value. The broadcast address is the number just before the next subnet number. Once you have the subnet number and broadcast address, the valid hosts are the numbers in between.

Here are the answers using 255.255.255.192:

1. How many subnet bits are used in this mask?
   Answer: 2
   2²-2=2 subnets
2. How many host bits are available per subnet?
   Answer: 6
   2⁶-2=62 hosts per subnet
3. What are the subnet addresses?
   Answer: 256-192=64 (the first subnet)
   64+64=128 (the second subnet)
   64+128=192. However, although 192 is the subnet mask value, it’s not a valid subnet. The valid subnets are 64 and 128.
4. What is the broadcast address of each subnet?
   Answer: 64 is the first subnet and 128 is the second subnet. The broadcast address is always the number before the next subnet. The broadcast address of the 64 subnet is 127. The broadcast address of the 128 subnet is 191.
5. What is the valid host range of each subnet?
   Answer: The valid hosts are the numbers between the subnet number and the mask. For the 64 subnet, the valid host range is 64-126. For the 128 subnet, the valid host range is 129-190.

Let’s do a second example using the Class C mask of 255.255.255.224. Here are the answers:

1. How many subnet bits are used in this mask?
   Answer: 3 bits or 2³-2=6 subnets
2. How many host bits are available per subnet?
   Answer: 5 bits or 2⁵-2=30 hosts per subnet
3. What are the subnet addresses?
   Answer: 256-224 =32, 64, 96, 128, 160 and 192 (Six subnets found by continuing to add 32 to itself.)
4. What is the broadcast address of each subnet?
   Answer: The broadcast address for the 32 subnet is 63. The broadcast address for the 64 subnet is 95. The broadcast address for the 96 subnet is 127. The broadcast address for the 160 subnet is 191. The broadcast address for the 192 subnet is 223 (since 224 is the mask).
5. What is the valid host range of each subnet?
   Answer: The valid hosts are the numbers in between the subnet and broadcast addresses. For example, the 32 subnet valid hosts are 33-62.

Let’s do a third example using the Class C mask of 255.255.255.240. Here are the answers:

1. How many subnet bits are used in this mask?
   Answer: 4 bits or 2⁴-2=14 subnets
2. How many host bits are available per subnet?
   Answer: 4 bits or 2⁴-2=14 hosts per subnet
3. What are the subnet addresses?
   Answer: 256-240 =16, 32, 48, 64, 80, 96, 112, 128, 144. 160, 176, 192, 208 and 224 (14 subnets found by continuing to add 16 to itself.)
4. What is the broadcast address of each subnet?
   Answer: Here are some examples of the broadcast address: The broadcast address for the 16 subnet is 31. The broadcast address for the 32 subnet is 47. The broadcast address for the 64 subnet is 79. The broadcast address for the 96 subnet is 111. The broadcast address for the 160 subnet is 175. The broadcast address for the 192 subnet is 207.
5. What is the valid host range of each subnet?
   Answer: The valid hosts are the numbers in between the subnet and broadcast addresses. The 32 subnet valid hosts are 33-46.

Let’s do a fourth example using the Class C mask of 255.255.255.248. Here are the answers:

1. How many subnet bits are used in this mask?
   Answer: 5 bits or 2⁵-2=30 subnets
2. How many host bits are available per subnet?
   Answer: 3 bits or 2³-2=6 hosts per subnet
3. What are the subnet addresses?
   Answer 256-248 =8, 16, 24, 32, 40, 48, and so forth. The last subnet is 240 (30 subnets found by continuing to add 8 to itself).
4. What is the broadcast address of each subnet?
   Answer: The broadcast address for the 8 subnet is 15. The broadcast address for the 16 subnet is 23. The broadcast address for the 48 subnet is 55.
5. What is the valid host range of each subnet?
   Answer: The valid hosts are the numbers in between the subnet and broadcast addresses. For example, the 32 subnet valid hosts are 33-38.

Let’s do a fifth example using the Class C mask of 255.255.255.252. Here are the answers:

1. How many subnet bits are used in this mask?
   Answer: 6 bits or 2⁶-2=62 subnets
2. How many host bits are available per subnet?
   Answer: 2 bits or 2²-2=2 hosts per subnet
3. What are the subnet addresses?
   Answer: 256-252 =4, 8, 12, 16, 20, and so forth. The last subnet is 248 (62 subnets found by continuing to add 4 to itself).
4. What is the broadcast address of each subnet?
   Answer: The broadcast address for the 4 subnet is 7. The broadcast address for the 8 subnet is 11. The broadcast address for the 12 subnet is 15. The broadcast address for the 20 subnet is 23.
5. What is the valid host range of each subnet?
   Answer: The valid hosts are the numbers in between the subnet and broadcast addresses. For example, the 16 subnet valid hosts are 17 and 18.

How do I use this information?
Let’s take a look at an example that will highlight how the above information is applied.

A host configuration has an IP configuration of 192.168.10.17 255.255.255.248. What are the subnet, broadcast address, and host range that this host is a member of? The answer is: 256-248=8, 16, 24. This host is in the 16 subnet, the broadcast address of the 16 subnet is 23, and the valid host range is 17-22. Pretty easy!

Here is an explanation of this example: First, I used 256-subnetmask to get the variable and first subnet. Then I kept adding this number to itself until I passed the host address. The subnet is the number before the host address, and the broadcast address is the number right before the next subnet. The valid hosts are the numbers in between the subnet and broadcast address.

Let’s examine a second example. A host configuration has an IP configuration of 192.168.10.37 255.255.255.240. What are the subnet, broadcast address, and host range this host is a member of? The answer is: 256-240=16, 32, 48. This host is in the 32 subnet, the broadcast address of the 32 subnet is 47, and the valid host range is 33-46.

Let’s go through a third example: A host configuration has an IP configuration of 192.168.10.44 255.255.255.224. What are the subnet, broadcast address, and host range this host is a member of? The answer is: 256-224=32, 64. This host is in the 32 subnet, the broadcast address of the 32 subnet is 63, and the valid host range is 33-62.

Here’s a fourth example: A host configuration has an IP configuration of 192.168.10.17 255.255.255.252. What are the subnet, broadcast address, and host range this host is a member of? The answer is: 256-252=4, 8, 12, 16, 20. This host is in the 16 subnet, the broadcast address of the 16 subnet is 19, and the valid host range is 17-18.

Let’s go through a final example. A host configuration has an IP configuration of 192.168.10.88 255.255.255.192. What are the subnet, broadcast address and host range this host is a member of? The answer is: 256-192=64, 128. This host is in the 64 subnet, the broadcast address of the 64 subnet is 127, and the valid host range must be 65-126.

Conclusion
It is important to be able to subnet quickly and efficiently. After studying the examples presented in this Daily Drill Down, you should be familiar with this process with Class C addresses. Practice your subnetting as much as possible, and the process will get easier and easier. In my next Daily Drill Down, I’ll take subnetting a step further and discuss subnetting a Class B network address.