Ccna Lec3 Subnetting Superneting Basic

Slide 1

ROUTER IS A DEVICE WHICH ENSURES COMMUNICATION BETWEEN DIFFERENT NETWORKS

TWO INTERFACES OF A ROUTER CANNOT HAVE SAME NETWORK ID

Break large networks into smaller subnets

Break large networks into smaller subnets

S0 ---200.100.100.0/24S1 ---200.100.100.0/241000 Packets destined to 200.100.100.2

S0 ---200.100.150.0/24S1 ---200.100.100.0/241000 Packets destined to 200.100.100.2

192.1.1.1/25192.1.1.1/27192.1.1.1/30168.1.1.1/2410.1.1.1/247.1.1.1/20

#

1816NetworkHost172001010110011111111101011000001000011111111000100000000000000000000101000000000000000000000Subnets not in usethe default00000010Subnet Mask Without Subnets172.16.2.160 255.255.0.0NetworkNumber18Purpose: This graphic explains how routers use addresses that have no subnet mask. Emphasize: Explain how masking works at the bit level. Zero bits mask host information.Note: This is an easy place to lose students. At this point, they need to learn several abstract mathematical concepts before we can show them how to lay out an IP-addressed network. To the novice, these techniques may seem unrelated, making the presentation confusing. To a more experienced audience, these techniques will be familiar.

19Network number extended by eight bitsSubnet Mask with Subnets16NetworkHost172.16.2.160 255.255.255.017220101011001111111110101100000100001111111100010000111111110000001010100000000000000000000000000010SubnetNetworkNumber12819222424024825225425519Slide 1 of 2Purpose: This figure shows how the router determines an address when subnetting is used.Emphasize: This example makes a Class B address space look like a collection of Class C address spaces.Now the logical AND allows us to extract the subnet number as well as the assigned network number.An exercise follows that tests the students understanding of subnet masks.

20Subnet Mask with Subnets (cont.)NetworkHost172.16.2.160 255.255.255.192101011001111111110101100000100001111111100010000111111110000001010100000110000001000000000000010SubnetNetwork number extended by ten bits161722128NetworkNumber12819222424024825225425512819222424024825225425520Slide 2 of 2Purpose: This figure shows how the router determines an address when subnetting is used.Emphasize: This example is different from the previous example in that the the subnet and host are divided within an octet.Transition: An exercise follows that tests the students understanding of subnet masks.

21Subnet Mask ExerciseAddressSubnet MaskClassSubnet172.16.2.1010.6.24.2010.30.36.12255.255.255.0255.255.240.0255.255.255.021Layer 1 of 2Purpose: This exercise is for the students to take the given IP addresses and associated subnet masks and perform a logical AND to extract the subnet number. Provide time in class and review the answers after most of the students have finished.The answers are given in the following figure.

22Subnet Mask Exercise AnswersAddressSubnet MaskClassSubnet172.16.2.1010.6.24.2010.30.36.12255.255.255.0255.255.240.0255.255.255.0BAA172.16.2.010.6.16.010.30.36.022Layer 2 of 2Purpose: The answers to the exercise are given in the figure.Note: Students can also find the answers to this exercise in the Appendix D, Answers.23Broadcast Addresses172.16.1.0172.16.2.0172.16.3.0172.16.4.0

172.16.3.255(Directed Broadcast)255.255.255.255(Local Network Broadcast)X172.16.255.255(All Subnets Broadcast)

23Purpose: This figure explains how broadcast addresses work.Emphasize: A range of addresses is needed to allocate address space. A valid range of addresses is between subnet zero and the directed broadcast.The following RFCs provide more information about broadcasts:RFC 919, Broadcasting Internet DatagramsRFC 922, Broadcasting IP Datagrams in the Presence of SubnetsCiscos support for broadcasts generally complies with these two RFCs. It does not support multisubnet broadcasts that are defined in RFC 922.

24Addressing Summary Example101011001111111110101100000100001111111100010000111111110000001010100000110000001000000000000010101011000001000000000010101111111010110000010000000000101000000110101100000100000000001010111110HostMaskSubnetBroadcastLastFirst172.16.2.160255.255.255.192172.16.2.128172.16.2.191172.16.2.129172.16.2.19012345678916172216024Layer 9 of 9Emphasize: In layer 9, convert binary back to dotted decimal.

25IP Host Address:172.16.2.121Subnet Mask: 255.255.255.0Subnet Address = 172.16.2.0Host Addresses = 172.16.2.1172.16.2.254Broadcast Address = 172.16.2.255Eight Bits of SubnettingNetworkSubnetHost10101100000100000000001011111111172.16.2.121:255.255.255.0:1010110011111111Subnet:1010110000010000000100001111111100000010000000101111111101111001 0000000000000000Class B Subnet ExampleBroadcast:Network25Purpose: This figure shows an example of a Class B network with a subnet.

26

Subnet Planning

Other Subnets192.168.5.16192.168.5.32192.168.5.48

20 Subnets5 Hosts per SubnetClass C Address: 192.168.5.026Purpose: This figure explains how to plan subnet numbers.Emphasize: What if this were a Class B address? How many bits would we have for subnetting then? Where do you want to draw the line now?Alternatives to review: Creating the subnet at the octet boundary is easier to work withmore host bits and more subnet bits.Explain that the decision is really a guess on how you think your network will growwill it have more subnets or more hosts?RFC 1219: Mirroring hedges the subnetting decision by buying time. Do not use mirroring if you intend to use route summarization or variable-length subnet masking (VLSM); they are incompatible with mirroring.

2711111000IP Host Address:192.168.5.121Subnet Mask: 255.255.255.248NetworkSubnetHost192.168.5.121:1100000011111111Subnet:1100000010101000101010001111111100000101000001011111111101111001 01111000255.255.255.248:Class C Subnet Planning ExampleSubnet Address = 192.168.5.120Host Addresses = 192.168.5.121192.168.5.126Broadcast Address = 192.168.5.127Five Bits of SubnettingBroadcast:NetworkNetwork1100000010101000000001010111111127Purpose: This figure shows an example of a Class C network with a subnet. Emphasize: Contrast the Class C network subnet mask with the previous Class B example.

28Exercise192.168.10.0/27

? SNM? Block Size?- Subnets

29Exercise/27

? SNM 224? Block Size = 256-224 = 32?- Subnets

Subnets10.010.3210.64FHID10.110.33LHID10.3010.62Broadcast10.3110.6330Exercise192.168.10.0/30

? SNM? Block Size?- Subnets

31Exercise/30

? SNM 252? Block Size = 256-252 = 4?- Subnets

Subnets10.010.410.8FHID10.110.5LHID10.210.6Broadcast10.310.732ExerciseMaskSubnetsHost/26???/27???/28???/29???/30???33ExerciseMaskSubnetsHost/26192462/27224830/282401614/29248326/3025264234Exam QuestionFind Subnet and Broadcast address192.168.0.100/2734Bs = 32Subnet 192.168.0.96B/c 192.168.0.12735Exercise192.168.10.54 /29Mask ?Subnet ?Broadcast ?35Mask = 248BS = 8Nw= 48B/c - 5536Exercise192.168.10.130 /28Mask ?Subnet ?Broadcast ?36Mask = 240BS=16NW-128B/c-14337Exercise192.168.10.193 /30Mask ?Subnet ?Broadcast ?37Mask = 252BS=4NW-192B/c-19538Exercise192.168.1.100 /26Mask ?Subnet ?Broadcast ?38Mask = 192BS=64NW-64B/c-12739Exercise192.168.20.158 /27Mask ?Subnet ?Broadcast ?39Mask = 224BS=32NW-128B/c-15940Class B172.16.0.0 /19Subnets ?Hosts ?Block Size ?

41Class B172.16.0.0 /19Subnets 23 -2 = 6Hosts 213 -2 = 8190Block Size 256-224 = 32

Subnets0.032.064.096.0FHID0.132.164.196.1LHID31.25463.25495.254127.254Broadcast31.25563.25595.255127.25542Class B172.16.0.0 /27Subnets ?Hosts ?Block Size ?








Subnets0.00.1281.01.1282.02.128FHID0.10.1291.11.1292.12.129LHID0.1260.2541.1261.2542.1262.254Broadcast0.1270.2551.1271.2552.1272.25550Find out Subnet and Broadcast Address172.16.85.30/2050Mask = 240BS = 256-240= 16Network - 172.16.80.0B/c 172.16.95.255

51Find out Subnet and Broadcast Address172.16.85.30/2951Mask = 248BS = 256-248= 8Network - 172.16.85.24B/c 172.16. 85.31

52Find out Subnet and Broadcast Address172.30.101.62/2352Mask = 254BS = 256-254= 2Network - 172.30.100.0B/c 172.30.101.255

53Find out Subnet and Broadcast Address172.20.210.80/2453Mask = 255BS = 256-255= 1Network - 172.20.210.0B/c 172.20.210.255

54ExerciseFind out the mask which gives 100 subnets for class B542x = 12827=128Mask

255.255.254.0

55ExerciseFind out the Mask which gives 100 hosts for Class B557 bits to be left from host part255.255.255.12856Class A10.0.0.0 /10Subnets ?Hosts ?Block Size ?

57Class A10.0.0.0 /10Subnets 22 -2 = 2Hosts 222 -2 = 4194302Block Size 256-192 = 64

Subnets10.010.6410.12810.192FHID10.0.0.110.64.0.110.128.0.110.192.0.1LHID10.63.255.25410.127.255.25410.191.255.25410.254.255.254Broadcast10.63.255.25510.127.255.25510.191.255.25510.254.255.25558Class A10.0.0.0 /18Subnets ?Hosts ?Block Size ?

59Class A10.0.0.0 /18Subnets 210 -2 = 1022Hosts 214 -2 = 16382Block Size 256-192 = 64

Subnets10.0.0.010.0.64.010.0.128.010.0.192.0FHID10.0.0.110.0.64.110.0.128.110.0.192.1LHID10.0.63.25410.0.127.25410.0.191.25410.0.254.254Broadcast10.0.63.25510.0.127.25510.0.191.25510.0.254.25560Broadcast Addresses ExerciseAddressClassSubnetBroadcast201.222.10.60 255.255.255.248Subnet Mask15.16.193.6 255.255.248.0128.16.32.13 255.255.255.252153.50.6.27 255.255.255.12860Layer 1 of 2Emphasize: Have the students calculate the subnet numbers and the broadcast address for each subnet from the given IP addresses and subnet masks.

61Broadcast Addresses Exercise Answers153.50.6.127AddressClassSubnetBroadcast201.222.10.60 255.255.255.248C201.222.10.63201.222.10.56Subnet Mask15.16.193.6 255.255.248.0A15.16.199.25515.16.192.0128.16.32.13 255.255.255.252B128.16.32.15128.16.32.12153.50.6.27 255.255.255.128B153.50.6.061Layer 2 of 2Purpose: The answers to the exercise are given in the figure.Note: Students can also find the answers to this exercise in the Appendix D, Answers.

Calculate Summary Route for :192.168.1.1/24192.168.2.1/24192.168.3.1/24192.168.0.0/22

Summary Route ??

Ccna Lec3 Subnetting Superneting Basic

Documents

Transcript of Ccna Lec3 Subnetting Superneting Basic