Monday, February 14, 2011

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;


  1. Fundamental idea of IP address and Subnet Mask

  2. Understand the Classfull and Classless Concept.

  3. Understand the Classes of IP address concepts

  4. Understand the Variable Length Subnet Masks (VLSM) concept

  5. 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




  1. Classfull

  2. 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.




  1. How many subnets?

  2. How many hosts per subnet?

  3. What the valid subnets?

  4. What the broadcast address for each subnet?

  5. 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 , 2x – 2 = 55 hosts ,

    2x - 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 , 2x – 2 = 25 hosts ,

    2x - 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

    192.160.10.96/27
    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

    192.160.10.160/27
    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

    192.160.10.244/27
    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 , 2x – 2 = >12 hosts ,

    2x - 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



    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

    192.160.10.96/27
    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

    192.160.10.160/27
    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

    192.160.10.244/27
    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 :




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

  2. 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.

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

  4. 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

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






























































































ST

EP

3
192

.
20

.
0

.
0

1
1
0
0
0
0
0
0
.
0
0
0
1
0
1
0
0
.
0
0
0
0
0
0
0
0
.
0
0
0
0
0
0
0
0

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





  1. The maximum IP address for 255.255.255.240?


    1. 14

    2. 15

    3. 16

    4. 30





  1. The maximum subnet for 255.255.255.240?


    1. 14

    2. 15

    3. 16

    4. 30





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


    1. 2

    2. 3

    3. 4

    4. 5









  1. 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





  1. 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





  1. 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





  1. 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





  1. How many host can you aspect with the IP address of 192.168.100.10/28?


    1. 4

    2. 8

    3. 16

    4. 32





  1. What the block size of the subnet 192.168.10.10/29?


    1. 8

    2. 16

    3. 32

    4. 64





  1. 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





  1. 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




  2. 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





  1. 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
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