Discovery 3 Module 4 Course Curriculum Picture Descriptions 4.0 ? Chapter Introduction 4.0.1 ? Introduction Diagram 1, Slideshow Slide 1 text ?Well-designed enterprise networks with many locations and users employ a logical addressing hierarchy.? Slide 2 text ?The use of classless addresses and Variable Length Subnet Masks(VLSM) facilitates network scalability. Slide 3 text ?Classless routing and Classless Inter-Domain routing or CIDR, address the problem of route summarization.? Slide 4 text ?Private addressing and Network Address Translation (NAT) preserve IPv4 addresses, providing flexibility and security in network design.? Slide 5 text ?After completion of this chapter, you should be able to: - Analyze the features and benefits of a hierarchical IP addressing structure - Plan and implement a VLSM IP addressing scheme - Plan a network using classless routing and CIDR - Configure and verify both static and dynamic NAT? 4.1 ? Using Hierarchical IP Network Address Scheme 4.1.1 ? Flat Hierarchical Networks 1 Diagram Diagram 1, Animation The animation depicts an image of three switches directly connected to each other with five computers directly connected to each switch. The topology alters to a hierarchical model with the switched network (Access layer) connecting to a router at the Distribution layer and the Distribution layer router connecting to the Core layer router which, in turn, connects to the network cloud. 4.1.2 ? Hierarchical Network Addressing 1 Diagram Diagram 1, Image The diagram depicts two scenarios, the first is the Non-hierarchical Addressing scenario and the second is the Hierarchical Addressing scenario. The scenario uses the same topology, only the addressing scheme changes. We have 3 switched LANS at the Access layer connected to the Distribution Layer router, which connects to the Core layer router which in turn connects to the network cloud. In the non-hierarchical addressing scheme, each network has non related IP addresses as follows. Connection to cloud: 192.168.100.0 Core to distribution layer: 192.168.5.0 LAN 1: 192.168.1.0 LAN 2: 10.22.5.0 LAN 3: 172.16.8.0 In the Hierarchical Addressing Scheme a logical grouping of networks exist. Connection to cloud: 10.1.0.0/16 Core to distribution layer: 10.1.1.0/24 LAN 1: 10.1.1.32/27 LAN 2: 10.1.1.64/27 LAN 3: 10.1.1.96/27 Note the subnet masks 4.1.3 ? Using Sub-netting to Structure the Network 2 Diagrams Diagram 1, Image The diagram depicts four boxed areas. 1: Broadcast Containment. A red line between the Distribution layer router and the Access layer switches marks the point of broadcast containment. 2: Security. A red line between the Access layer switch and the Distribution layer router marks the security demarcation point. The router offers security features. 3: Location. There are two sites labeled Site A and Site B depicted, with a serial link between them. 4: Logical Grouping. Two Access layer LANs (switched) are shown connected to another switch. The demarcation line is between this switch and the distribution layer router. The LANs are labeled Engineering and Accounting. Diagram 2, Activity Indicate whether a hierarchical addressing scheme using subnets should be used to structure the network. Match a YES or a NO to the statements below. 1. A mall business that has 10 employees uses a 12 port switch to connect to a router to access the ISP. 2. A large organization wants to break up some of their larger LANS to limit broadcasts and improve network performance. They do not want to purchase additional IP addresses from their ISP. 3. An organization has five locations in different states. All sites need to be connected. They have been assigned 5 IP addresses by their ISP but wish to separate users at each location based on the type of application they use at each location. 4. An organization has been assigned a single IP address and want s to break up the addresses into smaller chunks to be used by different departments within the organization. 5. A home user purchases a Linksys WRT300N with integrated router, 4 port switch and wireless access point. One computer will connect to the switch and other users will access the LInksys via wireless. Module 4.2 ? Using VLSM 4.2.1 ? Subnet Mask Two Diagrams Diagram 1, Tabular Subnet Mask Notation and Number of Possible Hosts NOTE: The number of hosts is calculated by taking the number 2 to the power of the number of host bits available and then subtracting 2. Dotted Decimal Subnet Mask - 255.0.0.0 Binary Subnet Mask - 11111111.00000000.00000000.00000000 Slash Notation - /8 Number of host Bits - 24 Hosts Possible 2^n-2 - 16777214 Dotted Decimal Subnet Mask - 255.128.0.0 Binary Subnet Mask - 11111111.10000000.00000000.00000000 Slash Notation - /9 Number of host bits - 23 Hosts Possible 2^n-2 - 8388606 Dotted Decimal Subnet Mask - 255.192.0.0 Binary Subnet Mask - 11111111.11000000.00000000.00000000 Slash Notation - /10 Number of host bits - 22 Hosts Possible 2^n-2 - 4194302 Dotted Decimal Subnet Mask - 255.224.0.0 Binary Subnet Mask - 11111111.11100000.00000000.00000000 Slash Notation - /11 Number of host bits - 21 Hosts Possible 2^n-2 - 2097150 Dotted Decimal Subnet Mask - 255.240.0.0 Binary Subnet Mask - 11111111.11110000.00000000.00000000 Slash Notation - /12 Number of host bits - 20 Hosts Possible 2^n-2 - 1048574 Dotted Decimal Subnet Mask - 255.248.0.0 Binary Subnet Mask - 11111111.11111000.00000000.00000000 Slash Notation - /13 Number of host bits - 19 Hosts Possible 2^n-2 - 524286 Dotted Decimal Subnet Mask - 255.252.0.0 Binary Subnet Mask - 11111111.11111100.00000000.00000000 Slash Notation - /14 Number of host bits - 18 Hosts Possible 2^n-2 - 262142 Dotted Decimal Subnet Mask - 255.254.0.0 Binary Subnet Mask - 11111111.11111110.00000000.00000000 Slash Notation - /15 Number of host bits - 17 Hosts Possible 2^n-2 - 131070 Dotted Decimal Subnet Mask - 255.255.0.0 Binary Subnet Mask - 11111111.11111111.00000000.00000000 Slash Notation - /16 Number of host bits - 16 Hosts Possible 2^n-2 - 65534 Dotted Decimal Subnet Mask - 255.255.128.0 11111111.11111111.10000000.00000000 Slash Notation - /17 Number of host bits - 15 Hosts Possible 2^n-2 - 32766 Dotted Decimal Subnet Mask - 255.255.192.0 11111111.11111111.11000000.00000000 Slash Notation - /18 Number of host bits - 14 Hosts Possible 2^n-2 - 16382 Dotted Decimal Subnet Mask - 255.255.224.0 Binary Subnet Mask - 11111111.11111111.11100000.00000000 Slash Notation - /19 Number of host bits - 13 Hosts Possible 2^n-2 - 8190 Dotted Decimal Subnet Mask - 255.255.240.0 Binary Subnet Mask - 11111111.11111111.11110000.00000000 Slash Notation - /20 Number of host bits - 12 Hosts Possible 2^n-2 - 4094 Dotted Decimal Subnet Mask - 255.255.248.0 Binary Subnet Mask - 11111111.11111111.11111000.00000000 Slash Notation - /21 Number of host bits - 11 Hosts Possible 2^n-2 - 2046 Dotted Decimal Subnet Mask - 255.255.252.0 Binary Subnet Mask - 11111111.11111111.11111100.00000000 Slash Notation - /22 Number of host bits - 10 Hosts Possible 2^n-2 - 1022 Dotted Decimal Subnet Mask - 255.255.254.0 Binary Subnet Mask - 11111111.11111111.11111110.00000000 Slash Notation - /23 Number of host bits - 9 Hosts Possible 2^n-2 - 510 Dotted Decimal Subnet Mask - 255.255.255.0 Binary Subnet Mask - 11111111.11111111.11111111.00000000 Slash Notation - /24 Number of host bits - 8 Hosts Possible 2^n-2 - 254 Dotted Decimal Subnet Mask - 255.255.255.128 Binary Subnet Mask - 11111111.11111111.11111111.10000000< Slash Notation - /25 Number of host bits - 7 Hosts Possible 2^n-2 - 126 Dotted Decimal Subnet Mask - 255.255.255.192 Binary Subnet Mask - 11111111.11111111.11111111.11000000 Slash Notation - /26 Number of host bits - 6 Hosts Possible 2^n-2 - 62 Dotted Decimal Subnet Mask - 255.255.255.224 Binary Subnet Mask - 11111111.11111111.11111111.11100000< Slash Notation - /27 Number of host bits - 5 Hosts Possible 2^n-2 - 30 Dotted Decimal Subnet Mask - 255.255.255.240 Binary Subnet Mask - 11111111.11111111.11111111.11110000< Slash Notation - /28 Number of host bits - 4 Hosts Possible 2^n-2 - 14 Dotted Decimal Subnet Mask - 255.255.255.248 Binary Subnet Mask - 11111111.11111111.11111111.11111000< Slash Notation - /29 Number of host bits - 3 Hosts Possible 2^n-2 - 6 Dotted Decimal Subnet Mask - 255.255.255.252 Binary Subnet Mask - 11111111.11111111.11111111.11111100 Slash Notation - /30 Number of host bits - 2 Hosts Possible 2^n-2 ? 2 Diagram 2, Activity Subnet Mask Determine the Slash notation, and number of host bits for the following subnet masks. 255.255.255.224 255.255.255.248 255.255.255.252 255.255.128.0 255.255.255.128 4.2.2 ? Calculating Subnets Using Binary Representation Three Diagrams Diagram 1, Image The picture depicts process of ?anding? IP address and Subnet Mask to determine whether the destination Host is on the same network, or a different network. There are two hosts (HostA, HostB) connected to a switch. HostA sends HostB a message, the switch checks to see if HostB is on the same network as HostA. The network is determined by comparing the IP address to the Subnet Mask. HostB is on the same network so the Switch sends the message to HostB. The IP Address Subnet Mask Network Address. And their corresponding Binary equivalent are listed below. HostA Configuration IP Address - 192.168.1.44, 11000000.10101000.00000001.00101100 Subnet Mask - 255,255,255,0, 11111111.11111111.11111111.00000000 Network ? 192.168.1.0, 11000000.10101000.00000001.00000000 HostB Configuration IP Address ? 192.168.1.66, 11000000.10101000.000000001.01000010 Subnet Mask ? 255.255.255.0, 11111111.11111111.11111111.00000000 Network 192.168.1.0, 11000000.10101000.00000001.00000000 Diagram 2, Image The picture illustrates the process of comparing IP address and Subnet Mask to determine subnets. There are two Hosts (HostA, HostB), The Subnet is determined by comparing the Last octet of both the IP Address and Subnet Mask. The IP Address Subnet Mask, Subnet. and their corresponding Binary equivalent are listed below. HostA Configuration IP Address 192.168.13.21/29, 11000000.10101000.00001101.00010101 Subnet Mask ? 255.255.255.248, 11111111.11111111.11111111.11111000 Subnet ? 192.168.13.16 Last Octet (00010000) HostB Configuration IP Address 192.168.13.25/29, 11000000.10101000.00001101.00011001 Subnet Mask ? 255.255.255.248, 11111111.11111111.11111111.11111000 Subnet ? 192.168.13.16 Last Octet (00010000) Diagram 3, Activity Identify whether the two hosts are on the same network Host - Host 1 IP Address - 172.16.5.72 Subnet Mask - 255.255.255.0 Slash Format - /24 Host - Host 2 IP Address - 172.16.5.79 Subnet Mask - 255.255.255.0 Slash Format - /24 Host - Host 1 IP Address - 192.168.19.35 Subnet Mask - 255.255.255.224 Slash Format - /27 Host - Host 2 IP Address - 192.168.19.48 Subnet Mask - 255.255.255.224 Slash Format - /27 Host - Host 1 IP Address - 10.128.14.14 Subnet Mask - 255.255.255.240 Slash Format - /28 Host - Host 2 IP Address - 10.128.14.19 Subnet Mask - 255.255.255.240 Slash Format - /28 Host - Host 1 IP Address - 192.168.3.68 Subnet Mask - 255.255.255.248 Slash Format - /29 Host - Host 2 IP Address - 192.168.3.74 Subnet Mask - 255.255.255.248 Slash Format - /29 4.2.3 ? Basic Subnetting Process Two Diagrams Diagram 1, Tabular Subnet ? 0 Network Address ? 192.168.1.0/26 Host range ? 192.168.1.1 ? 192.168.1.62 Broadcast address ? 192.168.1.63 Subnet ? 1 Network address ? 192.168.1.64/26 Host range ? 192.168.1.65 ? 192.168.1.126 Broadcast address ? 192.168.1.127 Subnet ? 2 Network address ? 192.168.1.128/26 Host range ? 192.168.1.129 ? 192.168.1.190 Broadcast address ? 192.168.1.191 Subnet ? 3 Network address ? 192.168.1.192/26 Host range ? 192.168.1.193 ? 192.168.1.254 Broadcast address ? 192.168.1.255 Diagram 2, Hands on Lab 4.2.4 ? Variable Length Subnet Masks (VLSM) Three Diagrams Diagram 1, Image The picture depicts the use of VLSM, to break up a subnet into an even smaller portion for use on Serial links. There are four Routers(R1, R2, R3, R4) R1 is connected to R2 via Serial link (Subnet Address: 192.168.20.192/30) R2 is connected to R3 via Serial link (Subnet Address: 192.168.20.196/30) R3 is connected to R4 via Serial link (Subnet Address: 192.168.20.200/30) LAN network addresses R1 network address: 192.168.20.0/27 R2 network address: 192.168.20.32/27 R3 network address: 192.168.20.64/27 R4 network address: 192.168.20.96/27 The following tables give a list of available subnets, and a list of Variable Length Subnets for the above network. 192.168.20.0 Subnets Subnet Number ? 0 Subnet Address ? 192.168.20.0/27 Subnet Number ? 1 Subnet Address - 192.168.20.32/27 Subnet Number ? 2 Subnet Address ? 192.168.20.64/27 Subnet Number ? 3 Subnet Address ? 192.168.20.96/27 Subnet Number ? 4 Subnet Address ? 192.168.20.128/27 Subnet Number ? 5 Subnet Address ? 192.168.20.160/27 Subnet Number ? 6 Subnet Address - 192.168.20.192/27 Subnet Number - 7 Subnet Address - 192.168.20.224/27 Subnets of 192.168.20.192 Subnet Number - 0 Subnet Address - 192.168.20.192/30 Subnet Number - 1 Subnet Address - 192.168.20.196/30 Subnet Number - 2 Subnet Address - 192.168.20.200/30 Subnet Number - 3 Subnet Address - 192.168.20.204/30 Subnet Number - 4 Subnet Address - 192.168.20.208/30 Subnet Number - 5 Subnet Address - 192.168.20.212/30 Subnet Number - 6 Subnet Address - 192.168.20.216/30 Subnet Number - 7 Subnet Address - 192.168.20.220/30 Diagram 2, Image The picture depicts three steps, which show how to apply VLSM to a network given the IP address 10.0.0.0/8. Step 1 10.0.0.0/8 has been subnetted using the subnet mask /16. There are five Routers (R1, R2, R3, R4, R5), which have been connected in a star. R1 is connected to R2, R3, R4, R5 R2 network address: 10.1.0.0/16 R3 network address: 10.2.0.0/16 R4 network address: 10.3.0.0/16 R5 network address: 10.4.0.0/16 Step 2 Any of the /16 subnets can be subnetted further. In this example 10.3.0.0/16 has been subnetted using the /24 mask. There are seven Routers (R1, R2, R3, R4, R5, R6, R7) R1-5 are as in the previous description. There are three Switches (S1, S2, S3) R4 is connected to S1 S1 is connected to R6 and R7 R6 is connected to S2 R7 is connected to S3 R2 network address: 10.1.0.0/16 R3 network address: 10.2.0.0/16 R4 network address: 10.3.0.0/16 R6 network address: 10.3.1.0/24 R7 network address: 10.3.2.0/24 R5 network address: 10.4.0.0/16 Step 3 In this example 10.3.2.0/24 has been subnetted using the /28 mask. There are eight Routers (R1, R2, R3, R4, R5, R6, R7, R8) There are six Switches (S1, S2, S3, S4, S5, S6) R1 is connected to R2, R3, R4, and R5 R4 is connected to S1 S1 is connected to R6 and R7 R6 is connected to S2 R7 is connected to S3 and S4 S4 is connected to R8 R8 is connected to S5 and S6 R2 network address: 10.1.0.0/16 R3 network address: 10.2.0.0/16 R4 network address: 10.3.0.0/16 R6 network address: 10.3.1.0/24 R7 network address: 10.3.2.0/24 S3 network address: 10.3.2.16/28 S4 network address: 10.3.2.32/28 S5 network address: 10.3.2.48/28 S6 network address: 10.3.2.64/28 R5 network address: 10.4.0.0/16 Diagram 3, Activity Determine which /Slash Format would be applied to allow for the given number of hosts listed below. 25 100 1000 5 45 400 12 2 4.2.5 ? Implementing VLSM Addressing Five Diagrams Diagram 1, Image The picture depicts a network, which has not had a VLSM addressing scheme applied. The picture identifies the amount of required and wasted addresses on each segment. Four Routers (Atlanta HQ, Perth HQ, Sydney HQ, Corpus HQ) Four Switches (S1, S2, S3, S4) each with three hosts attached Atlanta HQ is connected to S1 Atlanta HQ is connected to Sydney HQ via Serial link Sydney HQ is attached to S2 Sydney HQ is attached to Corpus HQ via Serial link Sydney HQ is attached to Perth HQ via Serial link Perth HQ is attached to S3 Corpus HQ is attached to S4 Headquarters ? Atlanta HQ Actual Requirements ? 58 host addresses Total Wasted Addresses ? 4 addresses Headquarters ? Perth HQ Actual Requirements ? 26 host addresses Total Wasted Addresses ? 36 addresses Headquarters ? Sydney HQ Actual Requirements ? 10 host addresses Total Wasted Addresses ? 52 addresses Headquarters ? Corpus HQ Actual Requirements ? 10 host addresses Total Wasted Addresses ? 52 addresses Headquarters ? WAN Links Actual Requirements ? 2 host addresses (each) Total Wasted Addresses ? 60 addresses Diagram 2, Tabular The picture depicts 5 steps, which have been used to calculate and apply VLSM addressing. Step 1 List the network requirements from largest to smallest Three point-to-point WAN links require two addresses each. Name?required addresses ? AtlantaHQ-58 Name?required addresses ? PerthHQ-28 Name?required addresses ? SydneyHQ-10 Name?required addresses ? CorpusHQ-10 Name?required addresses ? WAN1-2 Name?required addresses ? WAN2-2 Name?required addresses ? WAN3-2 Step 2 Use the next available address 192.168.15.128/28 Name?required addresses ? AtlantaHQ-58 Subnet address ? 192.168.15.0 Address range ? .1 ? .62 Broadcast Address ? .63 Network/prefix ? 192.168.15.0/26 Name?required addresses ? PerthHQ-28 Name?required addresses ? SydneyHQ-10 Name?required addresses ? CorpusHQ-10 Name?required addresses ? WAN1-2 Name?required addresses ? WAN2-2 Name?required addresses ? WAN3-2 Step 3 Borrow 2 more bits with a /30 mask. Name?required addresses ? AtlantaHQ-58 Subnet address ? 192.168.15.0 Address range ? .1 ? .62 Broadcast Address ? .63 Network/prefix ? 192.168.15.0/26 Name?required addresses ? PerthHQ-28 Subnet address ? 192.168.15.64 Address range - .65 - .94 Broadcast Address - .95 Network/prefix ? 192.168.15.64/27 Name?required addresses ? SydneyHQ-10 Name?required addresses ? CorpusHQ-10 Name?required addresses ? WAN1-2 Name?required addresses ? WAN2-2 Name?required addresses ? WAN3-2 Step 4 This creates subnets: 192.168.15.128, 192.168.15.32, 192.168.15.136. Name?required addresses ? AtlantaHQ-58 Subnet address ? 192.168.15.0 Address range ? .1 ? .62 Broadcast Address ? .63 Network/prefix ? 192.168.15.0/26 Name?required addresses ? PerthHQ-28 Subnet address ? 192.168.15.64 Address range - .65 - .94 Broadcast Address - .95 Network/prefix ? 192.168.15.64/27 Name?required addresses ? SydneyHQ-10 Subnet address ? 192.168.15.96 Address range - .97 - .110 Broadcast Address - .111 Network/prefix ? 192.168.15.96/28 Name?required addresses ? CorpusHQ-10 Subnet address ? 192.168.15.112 Address range - .113 - .126 Broadcast Address - .127 Network/prefix ? 192.168.15.112/28 Name?required addresses ? WAN1-2 Name?required addresses ? WAN2-2 Name?required addresses ? WAN3-2 Step 5 Use all three subnets, one for each WAN. Name?required addresses ? AtlantaHQ-58 Subnet address ? 192.168.15.0 Address range ? .1 ? .62 Broadcast Address ? .63 Network/prefix ? 192.168.15.0/26 Name?required addresses ? PerthHQ-28 Subnet address ? 192.168.15.64 Address range - .65 - .94 Broadcast Address - .95 Network/prefix ? 192.168.15.64/27 Name?required addresses ? SydneyHQ-10 Subnet address ? 192.168.15.96 Address range - .97 - .110 Broadcast Address - .111 Network/prefix ? 192.168.15.96/28 Name?required addresses ? CorpusHQ-10 Subnet address ? 192.168.15.112 Address range - .113 - .126 Broadcast Address - .127 Network/prefix ? 192.168.15.112/28 Name?required addresses ? WAN1-2 Subnet address ? 192.168.15.128 Address range - .129 - .130 Broadcast address - .131 Network/prefix ? 192.168.15.128/30 Name?required addresses ? WAN2-2 Subnet address ? 192.168.15.132 Address range - .133 - .134 Broadcast Address - .135 Network/prefix ? 192.168.15.132/30 Name?required addresses ? WAN3-2 Subnet address ? 192.168.15.136 Address range - .137 - .138 Broadcast Address - .139 Network/prefix ? 192.168.15.136/30 Diagram 3, The picture depicts a pie with six different colored pieces cut out (P1, P2, P3, P4, P5, P6), which represents the network address 192.168.1.0/24 broken up into six Variable Length subnets. Piece - P1 Network Address ? 192.168.1.0/25 Hosts - 126 Range - .1 - .127 Piece - P2 Network Address ? 192.168.1.128/26 Hosts ? 62 Range - .129 - .191 Piece ? P3 Network Address - 192.168.1.192/28 Hosts ? 14 Range - .193 - .207 Piece - P4 Network Address ? 192.168.1.208/28 Hosts ? 14 Range ? .209 - .223 Piece ? P5 Network Address ? 192.168.1.224/30 Hosts ? 2 Range ? .225 - .227 Piece ? P6 Network Address ? Unused Diagram 4, Activity Create an addressing scheme for the given requirements IP Address: 172.16.66.0/24 The fist subnet is given Subnet 2 requires 25 Hosts Subnet 3 requires 25 Hosts Subnet 4 requires 12 Hosts Subnet 5 requires 6 Hosts Subnet 6 requires 2 Hosts Subnet 1 Host Requirements ? 25 /Slash ? /27 # of hosts ? 30 Subnet ? 172.16.6.0 Host Range - .1 - .30 Broadcast - .31 Diagram 5, Hands on Lab 4.3 - Using Classless Routing and CIDR 4.3.1 - Classful and Classless Routing Four Diagrams Diagram 1, Tabular Image shows two separate tables. Table 1 Class A 1st Octet: Network 2nd Octet: Host 3rd Octet: Host 4th Octet: Host Subnet Mask: 255.0.0.0 or /8 Class B 1st Octet: Network 2nd Octet: Network 3rd Octet: Host 4th Octet: Host Subnet Mask: 255.255.0.0 or /16 Class C 1st Octet: Network 2nd Octet: Network 3rd Octet: Network 4th Octet: Host Subnet Mask: 255.255.255.0 or /24 Table 2 Number of Networks and Hosts per Network for Each Class Address Class: Class A First Octet Range: 0 to 127 Number of Possible Networks: 128 (2 are reserved) Number of Host per Network: 16,777,214 Address Class: Class B First Octet Range: 128 to 191 Number of Possible Networks: 16, 348 Number of Host per Network: 65,534 Address Class: Class C First Octet Range: 192 to 223 Number of Possible Networks: 2,097,152 Number of Host per Network: 254 Diagram 2, Image Router (R1) is connected from its S0/0/0 port via serial connection to S0/0/0 port of another router (R2). The network address of the serial connection 172.16.2.0/24. R2 is connected from its S0/0/1 port via serial connection to S0/0/1 port of router (R3). The network address of the serial connection 192.168.1.0/24. R1 is connected via Fa0/0 to a switch on the network 172.16.1.0/24. R2 is connected via Fa0/0 to a switch on the network 172.16.3.0/24. R3 is connected via Fa0/0 to a switch on the network 10.1.0.0/16. When R1 sends an update to R2, R2 applies its serial 0/0/0 /24 mask to the 172.16.1.0 routing updates from R1. When sends updated to R3, R3 applies the classfull/16 mask to the 172.16.0.0 routing update from R2. Diagram 3, Image Company 1 has 1000 employees, its network address is 172.16.0.0/22 (1022 Hosts). It is connected via router (Company 1) which is connected to an ISP. The ISP network address is 172.16.0.0/16 (65,534 Hosts). Company 2 has 500 employees, its network address is 172.16.20.0/23 (510 Hosts). It is connected via router (Company 2) which is connected to the same ISP as Company 1. Diagram 4, Animation Router (R1) is connected from its S0/0/0 port via serial connection to S0/0/0 port of router (R2). The network address of the serial connection 172.16.3.0/24. R2 is connected from its S0/0/1 port via serial connection to S0/0/1 port of router (R3). The network address of the serial connection 192.168.1.0/24. R1 is connected via Fa0/0 to a switch on the network 172.16.1.0/24. R1 connected via Fa0/1 to another switch with the network address 172.16.2.0/24. R2 is connected via Fa0/0 to a switch on the network 172.16.0.0/24. R3 is connected via Fa0/0 to a switch on the network 10.1.0.0/16. R1 says ?I must advertise out my route information.? R1 sends an update packet to all networks which it is directly connected to 172.16.1.0/24, 172.16.2.0/24 and 172.16.3.0/24. When R2 receives the update it says ?I will summarize all routes from R1, and my 172.16.0.0 route and will send it to R3.? R2 sends a summary route to R3 with the 172.16.0.0/24 information. 4.3.2 - CIDR and Route Summarization Three Diagrams Diagram 1, Image Route Summarisation Example Router (R1) is associated with the networks192.168.48.0/24 192.168.49.0/24 192.168.50.0/24 192.168.51.0/24. Router (R2) is associated with the networks 192.168.52.0/24 192.168.53.0/24 192.168.54.0/24 192.168.55.0/24. Router (R3) is associated with the network 192.168.56.0/24 192.168.57.0/24 to 192.168.63.0/24 R1 is connected to router (R4) via a serial link with the network address 192.168.48.0/22. R2 connected to R4 via a serial link with the network address 192.168.52.0/22. R3 connected to R4 via a serial link with the network address 192.168.56.0/21. R4 is connected to an ISP via serial connection with the network address of 192.168.48.0/20. Diagram 2, Image Two routers are connected to a third router, which connects to a fourth router within an ISP cloud. The networks connected to the first two routers are 192.168.4.0/24 and 192.168.6.0/24 respectively. The addresses of their links to the third router are 192.168.5.0/24 and 192.168.7.0/24 respectively. The connection between the third and fourth routers is the summary route 192.168.4.0/22. Summary Route All of these four networks have the first 22 bits in common: 192.168.4.0 = 11000000 10101000 000001 00 00000000 192.168.5.0 = 11000000 10101000 000001 01 00000000 192.168.6.0 = 11000000 10101000 000001 10 00000000 192.168.7.0 = 11000000 10101000 000001 11 00000000 These four networks are advertised as 192.168.4.0/22 or 192.168.4.0 255.255.252.0. Diagram 3, Activity Determine if the IP address with the CIDR information is a subnet or a route summary. Choose either Subnet or Route Summary based on the IP address provided. 172.24.0.0/14 (Subnet or Route Summary) 192.168.17.192/26 (Subnet or Route Summary) 10.24.0.0/16 (Subnet or Route Summary) 172.17.4.0/24 (Subnet or Route Summary) 10.0.100.0/24 (Subnet or Route Summary) 172.128.0.0/12 (Subnet or Route Summary) 192.168.0.0/23 (Subnet or Route Summary) 4.3.3 - Calculating Route Summarization Three Diagrams Diagram 1, Image Summarization Process Step 1 List the IP addresses you want to summarise 172.20.0.0 10101100.00010100.00000000.00000000 172.21.0.0 10101100.00010101.00000000.00000000 172.22.0.0 10101100.00010110.00000000.00000000 172.23.0.0 10101100.00010111.00000000.00000000 Step 2 The number of matching bits equals 14 e.g., the first 14 bits of each of these addresses if the same. Step 3 Copy the matching bits and add zero bits to determine the network address. 172.20.0.0 10101100.00010100.00000000.00000000 Diagram 2, Activity Select the best summary route for the contiguous address groups shown. Select the answer that represents a summarization of each group of networks. Group 1 192.168.0.0 /24 192.168.1.0 /24 192.168.2.0 /24 192.168.3.0 /24 Group 2 172.16.0.0 /16 172.17.0.0 /16 Group 3 10.3.5.0/27 10.3.5.32/27 10.3.5.64 /27 10.3.5.96 /27 10.3.5.128 /27 10.3.5.160 /27 10.3.5.192 /27 10.3.5.224 /27 Group 1 summarise to: 192.168.0.0/22 192.168.4.0/22 192.168.1.0/26 192.168.4.0/26 Group 2 summarise to: 172.16.0.0/15 172.16.0.0/17 172.17.0.0/15 172.17.0.0/17 Group 3 summarise to: 10.3.5.64/24 10.3.5.64/25 10.3.5.64/26 10.3.5.64/28 Diagram 3, Hands on Lab 4.3.4 - Discontiguous Subnets Three Diagrams Diagram 1, Image Router (R1) is connected from its S0/0/0 port to S0/0/0 port of router (R2). The network address of the serial connection 172.16.100.4/30. R2 is connected from its S0/0/1 port to S0/0/1 port of router (R3). The network address of the serial connection 172.16.100.8/30. R1 is connected via Fa0/1 to a switch (S1) on the network 192.168.3.0/26. R2 is connected via Fa0/0 to a switch (S2) on the network 192.168.2.0/24. R3 is connected via Fa0/0 to a switch (S4) on the network 192.168.3.64/26. R3 is connected via Fa0/1 to a switch (S3) on the network 192.168.3.128/26. Diagram 2, Image Router (R1) is connected from its S0/0 port which is addressed 172.16.100.5 via serial connection to S0/0 port of router (R2). The network address of the serial connection 172.16.100.4/30. R2 is connected from its S0/1 port via serial connection to S0/1 port of router (R3) which is addressed 172.16.100.10. The network address of the serial connection 172.16.100.8/30. R1 is connected via Fa0/1 to a switch on the network 192.168.3.0/26. R2 is connected via Fa0/0 to a switch on the network 192.168.2.0/24. R3 is connected via Fa0/0 to a switch on the network 192.168.3.64/26. R3 is connected via Fa0/1 to a switch on the network 192.168.3.128/26. There is a host connected to the switch which is connected to R2 named H1. There is also a host connected to the switch that comes from the Fa0/1 port of R3, this hosts name in H2. Its address is 192.168.3.130. R2 Routing Table Gateway of last resort is not set 172.16.0.0/30 is subnetted, 2 subnets C 172.16.100.8 is directly connected, Serial0/1 C 172.16.100.4 is directly connected, Serial0/0 C 192.168.2.0/24 is directly connected, FastEthernet0/0 R 192.168.3.0/24 [120/1] via 172.16.100.10, 00:00:05, Serial0/1 [120/1] via 172.16.100.5, 00:00:18, Serial0/0 H1 says ?I?m sending a message to 192.168.3.130.? H1 sends out its packets which are propagated through the network and dropped at R1 but forwarded by R3 to H2 on the 192.168.3.128/26 network. Diagram 3, Hands on Lab 4.3.5 - Subnetting and Addressing Best Practices Single Diagram Diagram 1, Image The image shows an example of a hierarchical addressing scheme created using best practice. The core router is connected to four routers with /16 networks. These then connect to complex networks using the best practice hierarchical addressing schemes. 4.4.1 ? Private IP Address Space 3 Diagrams Diagram 1, Image The diagram depicts three network clouds each with a dedicated router. Each edge router is connected to one of three other routers that form the Internet cloud. The first cloud has a 192.168.1.0 Class C Private Network address. The second cloud has a 10.0.0.0 Class A Private Network address and the third cloud has a 172.16.0.0 Class B Private Network address. The first cloud has a switch and three computers connected. The second cloud has three switches and 10 computers connected. The third cloud has two switches and 7 computers connected. Diagram 2, Image The diagram depicts four routers connected in a star topology configuration. Router 1 has connection to R2 network address 10.1.0.0/16. R1 is also connected to R3 on network address 10.3.0.0/16. R1 is also connected to R4 on network address 10.2.0.0/16. R1 is directly connected to a network cloud with the network address 10.0.0.0.8. Diagram 3, Activity Determine if the IP address is public or private. 172.16.35.2 209.165.200.226 192.168.3.5 10.168.21.3 209.165.202.130 209.165.201.30 192.168.11.5 4.4.2 ? NAT at the Enterprise Edge 1 Diagram Diagram 1, Image The diagram depicts a router (R1) and a switch (S1) directly connected. Connected to the switch are two computers named H1 and H3. H1 is the source and has the IP address 192.168.1.106 and this forms part of the Inside Global Addressing scheme. Router R1 is the border router for this network. Connected to the router R1, by serial link, is the ISP router (cloud). On the other side of the cloud is the destination H2 with an IP address 209.165.200.226. This network forms part of the Outside Global Addressing scheme. H1?s address before the translation is 192.168.1.106 and after the translation it is 209.165.202.129. 4.4.3 ? Static and Dynamic NAT 4 Diagrams Diagram 1, Image The diagram depicts two scenarios, Static NAT and Dynamic NAT. The topology for this scenario is as follows: H3 (within a cloud) has the IP address 200.165.202.130 configured and there is a serial link from the cloud to router R1. R1 has a switch directly connected and two computers H1 and H2 are connected to the switch. H1 and H2 have been configured with the IP addresses 192.168.2.18 and 192.168.2.19 respectively. Also connected to router R1 is the web server with the IP address 192.168.1.200 configured. Static NAT: H3 sends a message out to R1 and the source IP address is 209.165.202.130 and the destination IP address is 209.165.200.225 ? the inside global address (R1). The destination IP is changed to 192.168.1.200 (web server) the web server responds with a destination address of H3 (the requester) and a source address of the inside local address (web server), R1 maps the Inside local address to the inside global address which becomes the new source address. Dynamic NAT: As with Static NAT. Differences explained in text body Diagram 2, Image The diagram depicts two scenarios, the static NAT and the Dynamic NAT. The static Nat has a man sitting at his desk in front of a computer. The speech bubble appears above his head, ?Static NAT maps a single private address to a specific public address.? The dynamic NAT scenario depicts a man sitting at his desk in front of a computer, the speech bubble appears above his head, ?Dynamic NAT maps multiple private addresses to multiple public address.? Static NAT output: R1# show running-config Ip nat inside source static 172.31.232.14 209.165.202.130 Interface fastethernet 0/0 Ip address 172.31.252.182 255.255.225.0 Ip nat inside Interface serial 0/0/0 Ip address 209.165.202.1 255.255.255.0 Ip nat outside Dynamic NAT R1# show running-config Access-list 1 permit 172.31.232.0 0.0.0.255 Ip nat pool pub-addr 209.165.202.131 209.165.202.140 netmask 255.225.255.0 Ip nat inside source list 1 pool pub-addr Interface fastethernet 0/0 Ip address 172.31.202.182 255.255.255.0 Ip nat inside Interface serial 0/0/0 Ip address 209.165.202.1 255.255.255.0 Ip nat outside Diagram 3, Hands On Lab Diagram 4, Hands on Lab 4.4.4 ? Using PAT 3 Diagrams Diagram 1, Image The diagram depicts a web server within a network cloud. The network cloud is connected by serial link to router R1. Router R1 has a switch connected and two computers named H1 and H2. H2 sends a HTTP message with the source IP port address of 192.168.2.19 and source port number of 3012. The destination IP address is 209.165.202.130 and the destination port 80. The inside local address is 192.168.2.19:3012 and the inside global address is 209.165.202.2. The outside global address is 209.165.201.130:80 and the outside local address is 209.165.201.130:80. The unique public address consists of the serial interface IP address plus a port number. H1 sends a http message and the source IP port address is 192.168.2.18 and the source port number is 4177. The destination IP address is 209.165.202.130 and the destination port number is 80. The inside local address is 192.168.2.18 and the inside global address is 209.165.202.2:4177. The outside global address is 209.165.201.130:80 and the inside local address is 209.165.201.130:80. Diagram 2, Image The diagram depicts two aspects of using PAT configuring dynamic PAT and the verifying PAT. Dynamic PAT: The output from the show running-config command is listed below: R1# show running-config Access-list 1 permit 172.31.232.0 0.0.0.255 Ip nat inside source list 1 interface serial 0/0/0 overload Interface fastethernet 0/0 Ip address 172.31.252.182 255.255.255.0 Ip nat inside A man sitting at a desk in front of a computer has a speech bubble above his head, the bubble says, ? I have to configure PAT since we are converting of our private addresses into one public address.? When the Verifying PAT button is selected, the output on the switch when the show running-config command is issued is as follows: R# show ip nat translation Pro Inside Global Inside Local --- 209.165.202.130 172.31.252.14 Icmp 209.165.202.131:512 172.31.232.1:512 Udp 209.165.202.131:1067 172.31.232.2:1067 Tcp 209.165.202.131:1028 172.31.232.3:1028 Outside Global Outside Global ------ 209.465.202.1:512 209.65.202.1:51 209.165.202.2:53 209.165.202.2:53 209.165.202.3:80 209.165.202.3:80 Diagram 3, Hands On Lab 4.5 - Chapter Summary 4.5.1 - Summary Single Diagram Diagram 1, Slideshow Slide 1 text ?Image shows hierarchical network where three separate LANs are connected to routers which converge to a single router before connecting to the network cloud. A single broadcast domain is a non-hierarchical or flat network. A hierarchical addressing structure logically groups networks into smaller sub-networks. A hierarchical network design simplifies network management and improves scalability and performance. ? Slide 2 text ?With basic or standard subnetting, each subnet is the same size and has the same number of hosts. Variable Length Subnet Masking (VLSM) enables routers to use route summarization to reduce the size of routing tables. Variable Length Subnet Masking (VLSM) enables different masks for each subnet. A subnet can be further subnetted, creating sub-subnets. VLSM requires classless routing protocols. When implementing VLSM, ensure room for growth in the number of subnets and hosts available.? Slide 3 Text ?Classful IP addressing determines the subnet mask of a network address by the value of the first octet. With CIDR the network address is not determined by the class of the address, instead it is determined by the prefix length. Route summarization groups contiguous subnets using a single address and shorter mask to reduce the number of routes advertised. Route summarization, route aggregation, or supernetting are done at network boundaries on a boundary router. The use of classful routing protocols can create the issue of discontiguous networks.? Slide 4 text ?Image shows how routers use NAT translation to forward packets. Private addresses are used and routed internally but are not routed on the Internet. NAT translates private addresses into public addresses that route into the Internet. Static NAT maps a single inside local address to a single inside global (public) address. Dynamic NAT uses an available pool of public addresses and assigns them to inside local addresses. PAT translates multiple local addresses to a single global IP address.?