CCNA NOTE BOOK: Management in IP addressing

Classful and Classless Addressing

VLSM

CIDR

Routing Table Structure

Routing Table lookup Process

Routing Behavior

Routing Table Labs

OBJECTIVES

By the end of this module, you should be able to understand:

What is the basic IP address and Subnet Mask of networking all about;

Fundamental idea of IP address and Subnet Mask

Understand the Classfull and Classless Concept.

Understand the Classes of IP address concepts

Understand the Variable Length Subnet Masks (VLSM) concept

Understand the Summarization concept

MIND MAP

IP address and Subnet Mask

In the topic you will learn more about how to find and understand the use of IP address and subnet mask. IP address as you know have two types

Classfull

Classless

Classfull is usually use and always interchange to Classless for example

Classfull : 200.100.10.1 with subnet 255.255.255.0 and can represent into

Classless: 200.100.10.1/24

Before you go deeper into subnetting you all need to know the Classfull IP address. From the table below you will easily identified the correct IP address and its subnet mask

Class	Valid dotted Decimal	Valid number	Subnet Mask ( Default )
A	A.X.X.X	1 – 126	255.0.0.0
B	B.B.X.X	128 – 191	255.255.0.0
C	C.C.C.X	192 - 233	255.255.255.0
Broadcast	255.255.255.255

Some valid subnet mask

Class	Subnet Mask	Bits value	CIDR value
A	255.0.0.0	11111111.00000000.00000000.00000000	/8
	255.128.0.0	11111111.10000000.00000000.00000000	/9
	255.192.0.0	11111111.11000000.00000000.00000000	/10
	255.224.0.0	11111111.11100000.00000000.00000000	/11
	255.240.0.0	11111111.11110000.00000000.00000000	/12
	255.248.0.0	11111111.11111000.00000000.00000000	/13
	255.252.0.0	11111111.11111100.00000000.00000000	/14
	255.254.0.0	11111111.11111110.00000000.00000000	/15
B	255.255.0.0	11111111.11111111.00000000.00000000	/16
	255.255.128.0	11111111.11111111.10000000.00000000	/17
	255.255.192.0	11111111.11111111.11000000.00000000	/18
	255.255.224.0	11111111.11111111.11100000.00000000	/19
	255.255.240.0	11111111.11111111.11110000.00000000	/20
	255.255.248.0	11111111.11111111.11111000.00000000	/21
	255.255.252.0	11111111.11111111.11111100.00000000	/22
	255.255.254.0	11111111.11111111.11111110.00000000	/23

Class	Subnet Mask	Bits Value	Note
C	255.255.255.0	11111111.11111111.11111111.00000000	/24
	255.255.255.128	11111111.11111111.11111111.10000000	/25
	255.255.255.192	11111111.11111111.11111111.11000000	/26
	255.255.255.224	11111111.11111111.11111111.11100000	/27
	255.255.255.240	11111111.11111111.11111111.11110000	/28
	255.255.255.248	11111111.11111111.11111111.11111000	/29
	255.255.255.252	11111111.11111111.11111111.11111100	/30
	255.255.255.254	11111111.11111111.11111111.11111110
	255.255.255.255	11111111.11111111.11111111.11111111	Broadcast

What you all must understand here is when you creating subnetting, always make sure that the bits 1's must be full from left to right. CIDR or classless inter domain routing is the number of 1's in IP address.

As you all know the host ID and Network ID is as followed:

Class	Format	Default Subnet Mask
A	network.host.host.host	255.0.0.0
B	network.network.host.host	255.255.0.0
C	network.network.network.host	255.255.255.0

For example:

Class	First Host	Network Address	Default Subnet Mask
A	10.0.0.1	10.0.0.0	255.0.0.0
B	176.17.0.1	176.17.0.0	255.255.0.0
C	192168..1	192.168.1.0	255.255.255.0

Example of Class A

Example of Class B

Example for Class C

Lets see focus on Class C IP address subnetting calculation. Let's find the subnet of network address for 190.160.1.0 using the subnet mask for

255.255.255.192 (/26)

255.255.255.224 (/27)

255.255.255.240 (/28)

255.255.255.240 (/29)

255.255.255.240 (/30)

Firstly you have to find following answer before to decide the subnetting for each subnet must.

How many subnets?

How many hosts per subnet?

What the valid subnets?

What the broadcast address for each subnet?

What are the valid hosts?

For 255.255.255.192 (/26)

How many subnets?

Since 192 is 2 bits on ( 11000000 ) , so the number of subnets would be 2²= 4

How many hosts per subnet?
- Since 192 is 6's bits off (11000000), so the number of host per subnets would be 2⁶ – 2 = 62 hosts

What the valid subnets?

256 – 192 = 64 , so our subnets are 0,64,128 and 192
- What the broadcast address for each subnet?

Because the next subnet is 64 so the broadcast address is 63

What are the valid hosts?

This valid hosts can be find by selecting the IP address between subnet and broadcast of 0,64,128,192

Subnet	0	64	128	192
First Host	1	65	129	193
Last Host	62	126	190	254
Broadcast	63	127	191	255

The possible network as below

For 255.255.255.192 (/27)

1. How many subnets?

Since 224 is 2 bits on ( 11100000 ) , so the number of subnets would be 2³= 8

2. How many hosts per subnet?

Since 224 is 5's bits off (11100000), so the number of host per subnets would be 2⁵ – 2 = 30 hosts

What the valid subnets?

256 – 224 = 32 , so our subnets are 0,32,64,96,128,160,192 and 224

4. What the broadcast address for each subnet?

Because the next subnet is 32 so the broadcast address is 31

5. What are the valid hosts?

This valid hosts can be find by selecting the IP address between subnet and broadcast of 0,32,64,96,128,160,192 and 224

Subnet	0	32	64	96	128	160	192	224
First Host	1	33	65	97	129	161	193	225
Last Host	30	62	94	126	150	190	222	254
Broadcast	31	63	95	127	159	191	223	255

For 255.255.255.192 (/28)

1. How many subnets?

Since 240 is 2 bits on ( 11110000 ) , so the number of subnets would be 2⁴= 16

2. How many hosts per subnet?

Since 240 is 4's bits off (11110000), so the number of host per subnets would be 2⁴ – 2 = 14 hosts

What the valid subnets?

256 – 240 = 16 , so our subnets are 0,16,32,48.64,80,96,112,128,144,160,176,192.208,224 and 240

4. What the broadcast address for each subnet?

Because the next subnet is 16 so the broadcast address is 15

5. What are the valid hosts?

This valid hosts can be find by selecting the IP address between subnet and broadcast of 0,16,32,48.64,80,96,112,128,144,160,176,192.208,224 and 240

Subnet	0	32	64	96	128	160	192	224
First Host	1	33	65	97	129	161	193	225
Last Host	30	62	94	126	150	190	222	254
Broadcast	31	63	95	127	159	191	223	255

For summary of class C, to make easy for you , you may remember this simple table of subnetting

Subnet Mask	Mask	Bit ON	Bit OFF	Block size	Subnet	Hosts
/25	128	1	7	128	2	126
/26	192	2	6	64	4	62
/27	224	3	5	32	8	30
/28	240	4	4	16	16	14
/29	248	5	3	8	32	6
/30	252	6	2	4	64	2

Subnetting Class B Addresses

Let's now focus on Class B IP address subnetting calculation. Let's find the subnet of network address for 170.160.0.0 using the subnet mask for

a. 255.255.128.0 (/17)

b. 255.255.192.0 (/18)

c. 255.255.240.0 (/20)

The method to find the subnetting for Class B is quite same as it is for Class C. But you must remember that you need start your finding in the third octet.

For . 255.255.128.0 (/17)

1. How many subnets?

Since 128 is 1 bits on ( 10000000 ) , so the number of subnets would be 2¹= 2

2. How many hosts per subnet?

Since 0 is 15's bits off (10000000.00000000), so the number of host per subnets would be 2¹⁵ – 2 = 32766 hosts

What the valid subnets?

256 – 128 = 128 , so our subnets are 0,128

4. What the broadcast address for each subnet?

Because the next subnet is 128.1 so the broadcast address is 127.255

5. What are the valid hosts?

This valid hosts can be find by selecting the IP address between subnet and broadcast of 0.0

Subnet	0.0	128.0
First Host	0.1	128.1
Last Host	127.254	255.254
Broadcast	127.255	255.255

For . 255.255.192.0 (/18)

1. How many subnets?

Since 192 is 2 bits on ( 11000000 ) , so the number of subnets would be 2²= 4

2. How many hosts per subnet?

Since 0 is 14's bits off (11000000.00000000), so the number of host per subnets would be 2¹⁴ – 2 = 16382 hosts

What the valid subnets?

256 – 192 = 164 , so our subnets are 0,64,128,192

4. What the broadcast address for each subnet?

Because the next subnet is 64.1 so the broadcast address is 63.255

5. What are the valid hosts?

This valid hosts can be find by selecting the IP address between subnet and broadcast of 0.0 to 192.255

Subnet	0.0	64.0	128.0	192.0
First Host	0.1	64.1	128.1	192,1
Last Host	63.254	127.254	191.254	255.254
Broadcast	63.255	127.255	191.255	255.255

For . 255.255.240.0 (/20)

1. How many subnets?

Since 249 is 4 bits on ( 11110000 ) , so the number of subnets would be 2⁴= 16

2. How many hosts per subnet?

Since 0 is 12's bits off (11110000.00000000), so the number of host per subnets would be 2¹² – 2 = 4094 hosts

What the valid subnets?

256 – 240 = 16 , so our subnets are 0,16,32,48,64,80,96,112,128,144,160,176,192,208,224,240

4. What the broadcast address for each subnet?

Because the next subnet is 240.1 so the broadcast address is .255

5. What are the valid hosts?

This valid hosts can be find by selecting the IP address between subnet and broadcast of 0.0

First nine of the subnet

Subnet	0.0	16.0	32.0	48.0	64.0	80.0	96.0	112.0	128.0
First Host	0.1	16.1	32.1	48.1	64.1	80.1	96.1	112.1	128.1
Last Host	15.254	31.254	47.254	63.254	79.254	95.254	111.254	127.254	143.254
Broadcast	15.255	31.255	47.255	63.255	79.255	95.255	111.255	127.255	143.255

Subnetting Class A Address

Now lets us see the Class A subnetting and the Class A as followed

255.0.0.0 (/8)

255.128.0.0 (/9)

255.1920 (/10)

255.224.0 (/11)

255.240.0 (/12)

255.248.0.0 (/13)

255.252.0.0 (/14)

255.254.0.0 (/15)

255.255.0.0 (/16)

For example , lets us discuss one of the subnet (/8) or 255.0.0.0 or

11111111.00000000.00000000.00000000.

1. How many subnets?

Since 255 is 8 bits on ( 11111111 ) , so the number of subnets would be 2⁸= 256

2. How many hosts per subnet?

Since 0 is 16's bits off (00000000.00000000.00000000), so the number of host per subnets would be 2¹⁶ – 2 = 65534 hosts

What the valid subnets?

256 – 255 = 1 , so our subnets are 0,1,2,3,4,5 up to 256 subnets. The subnet would be 10.0.0.0, 10.1.0.0 up to 10.255.0.0

4. What the broadcast address for each subnet?

Because the next subnet is 10.1.0.0 so the broadcast address is 10.0.255.255

5. What are the valid hosts?

These valid hosts can be find by selecting the IP address between subnet and broadcast.

Subnet	10.0.0.0	10.1.0.0	…….	10.254.0.0	10.255.0.0
First Host	10.0.0.1	10.1.0.1	…….	10.254.0.1	10.255.0.1
Last Host	10.0.255.254	10.1.255.254	…….	10.254.255.254	10.255.255.254
Broadcast	10.0.255.255	10.1.255.255	…….	10.254.255.255	10.255.255.255

Variable Length Subnet Masks (VLSM)

As you working more with subnet problems, you will come across with a network too big or large and not practical to be use or implement subnetting. So, VLSM concept is creating subnet in a subnet itself. By applying these VLSM methods the network become more convenient and more practical as the need of a certain network. VLSM also help you to utilize and design a better network without wasting the IP addresses usage.

As we mention above, VLSM is another important and useful network configuration that need to be discuss by example.

Example 1:

You need to design a network with different location and host, lets say

Kuala Lumpur : 55 host

Putra Jaya : 25 host

Serdang : 12 host

STEP 1.

Firstly you need calculate the biggest number of hosts and followed with the smaller number of hosts.

For Kuala Lumpur, we need 55 host

So , 2^x – 2 = 55 hosts ,

2^x- 2 => 55 , from this we can fin the of X , X represent the number of bits

2⁶– 2 => 55 , the value of 6 is selected because the nearest number to number 55.

So, 2⁶– 2 => 55 host and the host bit is 6. As 6 host bit will the host addresses and the other 2 bit will be the network addresses.

How many subnets?

Since 192 is 2 bits on ( 11000000 ) , so the number of subnets would be 2²= 4.

What the valid subnets?

So, the valid subnet is 256 – 192 = 64 , so our subnets are 0,64,128 and 192

Now you can aspect you will have 4 subnet starting with 0, 64, 128 and 192. So VLSM will help you to reduce wasting IP addresses. We will start to create valid subnet under this 4 main subnet.

For, Kuala Lumpur with 55 hosts, you have known that the host bit is 6, there for you will adding 2 bit as the network bit , therefore now the network address will become :

192.168.10.0/24 to 192.168.10.0/26

Subnet Mask	Bit Network	Number of Host	Host required
255.255.255.192	/26	62	55
255.255.255.128	/25	126
255.255.255.0	/24	254
Octet 1	. Octet 2	.Octet 3	.Octet 4
11111111	.11111111	.11111111	.00000000	/24
11111111	.11111111	.11111111	.11000000	/26

You can see an additional 2 bit added to Octet 4

	Valid Subnet	Number of Subnet	Network ID	Broadcast
192.160.10.0/24	192.160.10.0/26	0	192.160.10.0/26	192.160.10.63/26
	192.160.10.64/26	1	192.160.10.64/26	192.160.10.127/26
	192.160.10.128/26	2	192.160.10.128/26	192.160.10.191/26
	192.160.10.192/26	3	192.160.10.192/26	192.160.10.255/26

From this you can used the subnet number 55 for Kuala Lumpur.

Valid Subnet

Valid Range

Subnet Mask

Network ID

Broadcast

192.160.10.0/26

192.168.10.1 – 192.160.10.62

255.255.255.192

192.160.10.0/26

192.160.10.63/26

For Putra Jaya with 25 hosts,

So , 2^x – 2 = 25 hosts ,

2^x- 2 = 25 , from this we can fin the of X , X represent the number of bits

2⁵– 2 => 25 , the value of 5 is selected because the nearest number to number 25.

So, 2⁵– 2 = 32 host and the host bit is 5. As 5 host bit will the host addresses and the other 3 bit will be the network addresses.

How many subnets?

Since 224 is 3 bits on ( 11100000 ) , so the number of sub-subnets would be 2¹= 2.

What the valid subnets?

So, the valid subnet is 256 – 224 = 32 , so our subnets are 0,32,64,96,128,160, and 192

Subnet Mask	Bit Network	Number of Host	Host required
255.255.255.224	/27	32	25
255.255.255.192	/26	62	55
255.255.255.128	/25	126
255.255.255.0	/24	254
Octet 1	. Octet 2	.Octet 3	.Octet 4
11111111	.11111111	.11111111	.00000000	/24
11111111	.11111111	.11111111	.11000000	/26
11111111	.11111111	.11111111	.11100000	/27

You can see an additional 1 bit added to Octet 4, and new subnet is 192.168.10.0/26 to 192.168.10.0/27, subnet mask is 255.255.255.224. When you added a bit and now number of sub-subnet possible is 2¹= 2. You can see each /26 can assign 2 subnet or sub-subnet.

Valid Subnet	Sub-subnet	Number of Subnet	Network ID	Broadcast
192.160.10.64/26	192.160.10.64/27	1	192.160.10.64/27	192.160.10.95/27
192.160.10.64/26	192.160.10.96/27	0	192.160.10.96/27	192.160.10.127/27
192.160.10.128/26	192.160.10.128/27	1	192.160.10.128/27	192.160.10.159/27
192.160.10.128/26	192.160.10.160/27	0	192.160.10.160/27	192.160.10.191/27
192.160.10.192/26	192.160.10.192/27	1	192.160.10.192/27	192.160.10.223/27
192.160.10.192/26	192.160.10.244/27	0	192.160.10.224/27	192.160.10.255/27

Finally the number of network addresses for Putra Jaya can be summarized as:

Valid Subnet

Valid Range

Subnet Mask

Network ID

Broadcast

192.160.10.64/27

192.168.10.65 – 192.160.10.95

255.255.255.224

192.160.10.64/27

192.160.10.95/27

For Serdang with 12 host

So , 2^x – 2 = >12 hosts ,

2^x- 2 => 12 , from this we can fin the of X , X represent the number of bits

2⁴– 2 =>12, the value of 4 is selected because the nearest number to number 12.

So, 2⁴– 2 = 14 host and the host bit is 4. As 4 hosts bit will be additional to host addresses and the other 4 bit will be added to network addresses.

How many subnets?

Since 240 is 4 bits on (11110000) , so the number of sub-subnets (add one bit to subnet) would be 2¹= 2.

What the valid subnets?

So, the valid subnet is 256 – 240 = 16, so our subnets are 0,16,32,48,64,80, and 192

Subnet Mask	Bit Network	Number of Host	Host required
255.255.255.240	/28	14	12
255.255.255.224	/27	30	25
255.255.255.192	/26	62	55
255.255.255.128	/25	126
255.255.255.0	/24	254
Octet 1	. Octet 2	.Octet 3	.Octet 4
11111111	.11111111	.11111111	.00000000	/24
11111111	.11111111	.11111111	.11000000	/26
11111111	.11111111	.11111111	.11100000	/27

Valid Subnet	Sub-subnet	Number of Subnet	Network ID	Broadcast
192.160.10.64/26	192.160.10.64/27	1	192.160.10.64/27	192.160.10.95/27
192.160.10.64/26	192.160.10.96/27	0	192.160.10.96/27	192.160.10.127/27
192.160.10.128/26	192.160.10.128/27	1	192.160.10.128/27	192.160.10.159/27
192.160.10.128/26	192.160.10.160/27	0	192.160.10.160/27	192.160.10.191/27
192.160.10.192/26	192.160.10.192/27	1	192.160.10.192/27	192.160.10.223/27
192.160.10.192/26	192.160.10.244/27	0	192.160.10.224/27	192.160.10.255/27

Finally the number of network addresses for Serdang can be summarized as

Valid Subnet

Valid Range

Subnet Mask

Network ID

Broadcast

192.160.10.96/27

192.168.10.113 – 192.160.10.126

255.255.255.224

192.160.10.112/27

192.160.10.127/27

We selected this range of subnet because we need only 12 host so we can used the unused subnet at /27

Summary :

The valid IP address 192.168.10.0/26 is used for Kuala Lumpur

By using VLSM methods we optimize one of the subnet 192.168.10.64/26 to sub-subnet of 192.168.10.64/27 and 192.168.10.96/27.

The valid IP address 192.168.10.64/27 is used for Putra Jaya.

The valid IP address 192.168.10.96/27 is used for Serdang.

Classless interdomian routing (CIDR)

Classless interdomian routing is a method to customize or modified way to use prefix based standard in analyzing an IP address. This CIDR technique makes capable to simplest a complex and multiple networks, a mass of addresses into a single route. CIDR will provide us the information to the IP address is noted as four octets and subnet mask. For example:

178.163.10.0/24

192.163.10.0/24

202.163.10.0/24

222.163.10.0/24

Route Summarization

What you are going to learn from this topic is to route summarize or aggregation of many networks addresses to one or single address with shorter subnet mask. The purpose of this technique is to reduce space the size of the routing table. As we that noted CIDR also a same method of router summarization and supernetting.

There some difference about CIDR, CIDR do not take into account the restriction of classful limitations and permits any summarization with masks lesser compare to the default classful mask.

This route summarization helps the route to reduce the resource and bandwidths utilization for routing updates. As the consequent the time to updating the routing table becomes faster and more efficient.

Noted that the only routing protocol applied this supernetting is the classless IP addressing. Because of classless routing protocol utilities the network address and mask when updating routing

By using the summarization technique Router A will allowed all IP with /24 to be forward to Router B. How to use this summarization technique you need to find out the block size.

Calculation Route Summarization

How about when you have only 4 IP address! For example, what the best IP address summarization for Router B ?

The most important find out the block first. So you can summarize 192.160.0.0/20 using a mask of 255.255.240.0, so block size is 256-240 = 16.

	Octet1	.Octet2	.Octet3	.Octet4	Mask
192.160.4.0/24	11000000	.10100000	.00000100	.00000000	255.255.252.0
192.160.5.0/24	11000000	.10100000	.00000101	.00000000	255.255.253.0
192.160.6.0/24	11000000	.10100000	.00000110	.00000000	255.255.254.0
192.160.7.128/25	11000000	.10100000	.00000111	.10000000	255.255.255.128
192.160.8.192/26	11000000	.10100000	.00001000	.11000000	255.255.255.192
192.160.0.0/20	11000000	.10100000	.00000000	.00000000	255.255.240.0

Another example:

There few step to be followed to makes the summarization, Step 1 list the networks in binary arrangement. Step 2, find the leftmost matching to decide the mask or /14 or 255.252.0.0

		Octet1	.Octet2	.Octet3	.Octet4	Mask
S T E P 1	192.20.0.0/16	11000000	.00010100	.00000000	.00000000	255.255.252.0
	192.21.0.0/16	11000000	.00010101	.00000000	.00000000	255.255.253.0
	192.22.0.0/16	11000000	.00010110	.00000000	.00000000	255.255.254.0
	192.23.0.0/16	11000000	.00010111	.00000000	.00000000	255.255.255.128
S T E P 2			Select the same bits in this case bit 6's are same
S T E P 2	192.20.0.0/14	11000000	.00010100	.00000000	.00000000	255.252.0.0

For the result, we can show the summarization of 192.20.0.0/14 as

192

Copy the leftmost matching bit up to 14 bits

Add zero bits

Routing Table Structure

The understanding of router table structure is to ensure that we know what is happening when we punch the key board. In this session we will learn how the router works will each IP address and interfaces. Three major things will cover are the Level 1 Router, Level 2 router and Next Hop IP address or Exit interface. As we known that Cisco router is design in hierarchical methods, the purpose of this is to makes the router works faster and accurately.

Let we looks a simple topology for easy and better understanding about routing table structure.

Router	Ethernet	Serial 1	Serial 0
R1	E0 (192.162.1.1)		S0 ( 192.162.2.1)
R2	E0(192.162.3.1)	S1 (172.162.1.1)	S0 ( 192.162.2.2)
R3	E0 (192.162.4.2)	S1(172.162.1.2)

Form the topology , we can see the connection between router 1,2 and 3 where the interface that involved can be see clearly: the serial 0 and 1 and Ethernet. By provide the diagram we can easily the configuration but it applied for simple networks only. Basically this diagram is suitable for starting point for new learner. Simply we can describe the network easily and clearly with any confusing.

The Router R1 and R2 share same network ( 192.162.0.0/16) with 24 subnet. For Route R2 and R3 linked by 172.162.1.0/24. Router R3 in addition has 192.162.4.0/24 and known as disconnected or discontiguous network of 192.162.0.0. From the diagram we can also clearly the connection of serial and Ethernet with their IP address. E.g. R1 : EO – 192.162.1.1, S0 – 192.162.2.1 ; R2 : S0 – 192.162.2.2, S1 – 172.162.1.1, E0 – 192.162.3.1 ; R3 : S1 – 172.162.1.2 , E0 – 192.162.4.2. Hoping by using this table we will not makes any mistake we configuration the router.

Routing Table lookup Process

Routing Behavior

Summary

The two type of IP address are :
- Classfull
- Classless

Classfull is usually use and always interchange to Classless

Class A IP addresses with valid dotted decimal : A.X.X.X

Class B IP addresses with valid dotted decimal : B.B.X.X

Class C IP addresses with valid dotted decimal : C.C.C.X

IP address divided into network address (A,B,C) and host address ( X ) .

TEST ONE

The maximum IP address for 255.255.255.240?
1. 14
2. 15
3. 16
4. 30

The maximum subnet for 255.255.255.240?
1. 14
2. 15
3. 16
4. 30

If you have 16 subnets, how many bits must you borrow from the host field?
1. 2
2. 3
3. 4
4. 5

The mask for /28 for Class C is?
1. 255.255.255.255
2. 255.255.255.254
3. 255.255.255.240
4. 255.255.255.224

A network address of 172.17.0.0/18 provides how may subnets and hosts?
1. 2 subnets and 30 hosts
2. 2 subnets and 32 hosts
3. 4 subnets and 30 hosts
4. 4 subnets and 32 hosts

The IP address 172.17.45.5/30 belongs to?
1. 172.17.40.0
2. 172.17.40.4
3. 172.17.40.8
4. 172.17.40.12

The subnetwork for IP address 172.16.50.0/21 is?
1. 172.17.36.0
2. 172.17.48.0
3. 172.17.64.0
4. 172.17.72.0

How many host can you aspect with the IP address of 192.168.100.10/28?
1. 4
2. 8
3. 16
4. 32

What the block size of the subnet 192.168.10.10/29?
1. 8
2. 16
3. 32
4. 64

What is the broadcast IP address that available with in subnet 192.168.10.10/29?
1. 192.168.10.0 255.255.255.255
2. 192.168.10.15 255.255.255.248
3. 192.168.10.27 255.255.255.248
4. 192.168.10.31 255.255.255.240

What is the broadcast IP address that available with in subnet 192.168.10.10/28?
1. 192.168.10.15 255.255.255.240
2. 192.168.10.15 255.255.255.248
3. 192.168.10.27 255.255.255.248
4. 192.168.10.31 255.255.255.240

Find the valid subnet for 172.110.122.1/25?
1. 172.110.122.0
2. 172.110.16.0
3. 172.110.32.0
4. 172.110.64.0

What is the valid host address for 172.17.16.0/22.?
1. 172.17.18.255 255.255.252.0
2. 172.17.17.0 255.255.255.255
3. 172.17.100.0 255.255.240.0
4. 172.17.200.0 255.255.255.0

CCNA NOTE BOOK

Monday, February 14, 2011

Management in IP addressing

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