Converged Network Fundamentals V3 - RMAUG...Converged Network Fundamentals Workshop Student Guide...

V3 July 2011

Converged Network Fundamentals Workshop Student Guide

Developed by: Jim Reinhardt

Converged Network Fundamentals Workshop Student Guide Issue 2 April 2011

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Course Synopsis The intent of this course is to provide the students with basic information and practical experience necessary during the design, implementation, and maintenance converged voice and data networks. The content of this course may be helpful in the pursuit of certain basic certifications within the industry (ACA, CCNA, and Network+). Throughout this course, students may be expected to complete written exercises. Optimal number of students for this class ranges from 8 to 12. Lab equipment will include: 1) 3 to 4 each Avaya G250 2) 6 or more student laptops 3) Miscellaneous: a. Console cables b. Ethernet Cross-over cables c. Ethernet Straight cables Class Material will include: 1) Electronic copy of Courseware and Reference Material 2) Miscellaneous: a. Pens / pencils b. Notebooks / scratch-paper Classroom rules: 1) Ask Questions 2) Have Fun


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Course Prerequisites Students who have the ability will be expected to bring their laptops to class. Students will need the ability to install programs on their laptops. At a minimum they should have the following software loaded on their laptops prior to the start of class: Required

1. MS Office 2. Putty 3. Internet Browser

Optional

1. MS Visio or Visio Viewer 2. Wireshark


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Course Outline DAY 1

• Introductions / Logistics / Overview • Skill assessment • Skill Assessment Review

• Module 1: Basic Networking Discussion

o Ethernet and IP V4 Data Network Architecture

o OSI Model o Data Packet / Frame Transmission o ARP lab o Binary Calculations o Homework: Module 1 Review Exercise

DAY 2

• Homework Review

• Module 2: IP Address and Subnet Mask o IP Addressing Calculations o Network Address o Host Range o Broadcast Address o Homework: Module 2 Review Exercise

DAY 3

• Homework Review

• Module 3: VLSM (Variable Length Subnet Mask) o Network Design

• Module 4: Configuration Lab o Basic Network Configuration:

Interface configuration Static Route configuration

o Troubleshooting


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Basic Skills Assessment

Place the OSI Layer beside each description below

OSI Layer

Multiple Choice: An Ethernet switch distributes frames based on addresses at which layer

Layer 1 Layer 3

Layer 2

Layer 4

Transport Multiple Choice: Routers operate at which layer

Layer 1 Layer 3

Application Layer 2 Layer 4

Data Link True or False: FTP uses well known port 23

TRUE

Presentation FALSE

Physical Multiple Choice: Is 172.47.47.47 a public or private address

Public

Session Private

Network Multiple Choice: A 255.255.252.0 subnet mask can be represented with a

Slash 22 Slash 24 Multiple Choice: The _____ protocol automatically assigns IP Addresses to end points

DNS DHCP Slash 26 Slash 18

ARP FTP Slash 30 Slash 12


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Basic Skills Assessment Continued

True or False: Every port in a hub is a separate collision domain

TRUE Multiple Choice: Which class of Address is 224.11.14.11

Class A Class C

FALSE Class B Class D

Multiple Choice: An IP Address is a Layer _____ Address

1 7 True or False: Routers separate broadcast domains

TRUE

2 3 FALSE

Multiple Choice: A 255.255.255.192 subnet mask can be represented with a

Slash 22 Slash 24 Multiple Choice: The decimal representation of the binary value 11100000 is

127 224

Slash 26

Slash 18

128

192

True or False: HTTP uses wel known port 80

TRUE Multiple Choice: Which command will clear the arp cache on your PC

arp ‐d arp ‐c

FALSE arp ‐a arp ‐b

True or False: VLANs separate broadcast domains

TRUE True or False: Data is encapsulated in a frame at Layer 3

TRUE

FALSE FALSE


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Module 1 Basic Network Discussion


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Network Components


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Crossover Cables: The smallest Network consists of two endpoints. Two hosts can be connected without additional network equipment using an Ethernet crossover cable. Ethernet crossover cables swap the positions of pins 1, 2, 3, and 6 from one end of the cable to the other.


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Hubs: Hubs are normally the lowest cost networking device that supports more than two endpoints. Hubs operate at layer 1 or the physical layer. Hubs can operate at 10 to 100 Meg. Some disadvantages of hubs include: 1) Common Collision Domain 2) All Frames are transmitted through all ports 3) Transmission is half duplex Note: Because all frames are transmitted through all ports, hubs can be useful for packet captures


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Ethernet Switches: While Ethernet Switches are more expensive than hubs, they do provide additional features that may be necessary depending on the requirements of the network. While most home networking switches are “unmanaged” most enterprise switches are “managed” switches with VLAN capabilities. Unlike hubs, every port on a switch is a separate collision domain, but a common broadcast domain, unless they are in separate VLANs. Switches can operate at speeds ranging from 10 Meg to 1 Gig with half or full duplex modes on access ports. Switches maintain a “CAM” table that records MAC addresses per port. Switches are normally a layer 2 device; however, many switches have the ability to route between subnets can operate at Layer 3.


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Routers: Where hubs and switches are used to connect endpoints to other endpoints, thus, creating a network; Routers are used to connect networks. Routers separate broadcast domains, and operate at layer 3. Routers can have various types of interfaces including, but not limited to: 1) Ethernet (10Meg) 2) Fast Ethernet (100Meg) 3) Gigabit Ethernet ( 1Gig) 4) T1 (1.5 Meg) These interfaces are connected to each other through the processor of the router


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What to Remember:

• Crossover Cables o Can be used to connect 2 endpoints without a hub or switch o Swaps position of pins 1,2,3, and 6

• Hubs o Low cost method to connect two or more endpoints o Layer 1 device o All ports are a common collision domain & common broadcast domain o Can operate at speeds 10-100Meg o Half Duplex o All traffic is transmitted through all ports (good for packet captures)

• Switches o Normally a Layer 2 Device; however, there are Layer 3 switches o Each port is a separate collision domain, but common broadcast domain unless VLANs

are used o Can operate at speeds 10Meg-1Gig (Some faster speeds can be extended across the

“fabric” in stackable systems (switching “fabric” is sometimes used to refer to the proprietary components used to connect multiple modules of common manufacturer and model)

• Routers o Layer 3 device o Used to connect multiple networks o Can have multiple interfaces


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OSI Model and Data Encapsulation


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Packet and Frame Transmission


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ARP Lab 1) Pair up in teams of 2-3 students with PCs 2) All students join the network

a. If no production network is available set up lab switch b. Use static IP addressing if using lab switch

3) All students bring up a “cmd” prompt on their PC and type and enter “ipconfig” command and note IP Address of:

a. Student’s PC b. Default Gateway c. Teammate’s PC

4) At the command line issue “arp –a” command and note all entries 5) At the command line issue “ping aaa.bbb.ccc.ddd” where aaa.bbb.ccc.ddd is

the IP Address of your teammate / teammates 6) At the command line issue “arp –a” command and note all entries 7) At the command line issue “arp –d” 8) At the command line issue “arp –a” command and note all entries


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Binary Calculations: Using the base 10 number system, we increment the tens column by a value of one and start over at zero in the ones column when we exceed 9 in the ones column. Computer logic uses binary which is a base 2 number system. Binary representation adds a “1” to the left of the current column every time the count exceeds 1. In the examples below, we’ll count from 0 to 31 and show the binary equivalent: 0=0 1=1 2=10 3=11 4=100 5=101 6=110 7=111 8=1000 9=1001 10=1010 11=1011

12=1100 13=1101 14=1110 15=1111 16=10000 17=10001 18=10010 19=10011 20=10100 21=10101 22=10110 23=10111

24=11000 25=11001 26=11010 27=11011 28=11100 29=11101 30=11110 31=11111

Bit Weight Line: Each zero or one is considered a bit. There are 8 bits in a byte. In a byte, we can achieve any value from 0 to 255. As you might have noticed in the examples on the previous page the value the 1 doubles each time it is moved to the left. A “bit weight line” is similar to an Abacus for calculating binary values. Every time a 1 is placed under a value on the line, that value is “turned on”, a zero means that value is “turned off”. Think of the ones and zeroes as on and off switches.

In this example, we turned on the values for 128, 32, 8, 2, and 1. The sum of these values is 171. 128+32+8+2+1=171


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Your Turn: Turn on the necessary values to achieve the indicated sum, by placing a 1 or a 0 in the appropriate cells. 149

128 64 32 16 8 4 2 1

192 128 64 32 16 8 4 2 1

127 128 64 32 16 8 4 2 1

240 128 64 32 16 8 4 2 1

129 128 64 32 16 8 4 2 1

10 128 64 32 16 8 4 2 1

224 128 64 32 16 8 4 2 1


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Module 1 Homework: Take a few minutes and use the Internet to find the answers to the homework questions below. If you already know the information from previous training or job experience, then answer the questions without using the Internet. 1) identify the well-known ports for the following protocols:

Protocol Well-Known Port Protocol Well-Known Port Protocol Well-Known Port FTP Telnet HTTP SFTP SSH HTTPS TFTP SMTP SNMP 2) What are the two sub-layers of the Data Link layer of the OSI Model? 3) How many bits are in an IP Address 4) What protocol below allows for devices throughout the enterprise network to

have a common time reference a. NTP b. RTP

c. UDP d. RTCP


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Module 1 Review

Fill in the Well Known Port for Each Protocol

Well Known Port Multiple Choice: An Ethernet switch operates at which Layer

Layer 1 Layer 3

Layer 2 Layer 4

TFTP Multiple Choice: Routers operate at which layer

Layer 1 Layer 3

HTTPS Layer 2 Layer 4

SNMP True or False: Each port on a switch is a separate collision domain

TRUE

SMTP FALSE

Telnet Multiple Choice: A Crossover Cable swaps the position of which pins

1,2 & 4,5 4,5 & 7,8

FTP 1,2 & 3,6 5,6 & 7,8

HTTP Multiple Choice: the binary value 10001011 has a decimal

value of:

172 139 Multiple Choice: The _____ protocol resolves IP Addresses to MAC Addresses

DNS DHCP 130 128

ARP FTP 201 192


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Module 2 IP Addressing / Subnetting


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Before we begin we have to understand the design rules:

• First rule: We must know how big our network is going to be • Second rule: We need to know if our network needs to talk to other networks

The bit weight line: This will be used to help us calculate our values The subnet mask: This designates the size of our network (we will be discussing this in length)


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A Sample IP Address And Subnet Mask:

• The IP address

o 192.168.10.97 (seen by us as dotted decimal) o 11000000101010000000101001100001(seen by the computer as 32 bit

binary) • The subnet mask

o 255.255.252.0 (what does this do?) o /22 (how do we get this from 255.255.252.0)

• The Network / Subnet ID • The host range

o What addresses are on this subnet • The Broadcast address


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The IP Address: Converting from binary to dotted decimal and from dotted decimal to binary

Convert from Dotted Decimal to Binary Dotted Decimal Binary

192.11.13.6 220.9.14.0 10.10.78.12 160.9.0.255 127.0.0.1 172.196.197.1

Convert from Binary to Dotted Decimal Binary Dotted Decimal

00001010000010100100111000001100 11000000000010110000110100000110 10101100110001001100010100000001 11011100000010010000111000000000 01111111000000000000000000000001 10100000000010010000000011111111


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The Subnet Mask: The rules and converting from dotted decimal to slash or CIDR notation Subnet Masking Rules: • Subnet masks are built in binary using consecutive ones from left to right

o Example 11111111000000000000000000000000 • The shortest subnet mask is eight bits, in dotted decimal 255.0.0.0

o This creates the largest subnet • The longest subnet mask is thirty bits, in dotted decimal 255.255.255.252

o This creates the smallest subnet • Since the subnet mask is built from left to right using consecutive ones, only

certain values will be valid and cannot be preceded by a lesser value o Valid values are: 0, 128, 192, 224, 240, 248, 252, 254, and 255


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Converting Dotted Decimal To Slash Notation: • Dotted decimal or traditional notation of the subnet mask is much like the

dotted decimal notation of the IP address w/ the exception of the rules we just discussed

• Each octet of 255 in the subnet mask represents eight bits (example 255.255.0.0 is the equivalent of /16 or in binary 11111111111111110000000000000000)

• Our Example o /22 would appear in binary as: 11111111111111111111110000000000 o Remember each set of eight bits is an octet

11111111111111111111110000000000 /22 would appear in dotted decimal as: 255.255.252.0

Let’s ensure we have this down!!


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Example /22: We know the first and second octets in dotted decimal are both 255 because all the “binary switches” are “on” or 1’s 1111111111111111 = 255.255 (this is only 16 ones, we still need 6 more for a /22) Let’s examine how we came up with the value of 252 in the third octet (remember the bit weight line)

Now simply add the bit values: 128+64=192 192+32=224 224+16=240 240+8=248 248+4=252


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The Subnet Mask: Converting from CIDR to dotted decimal and from dotted decimal to CIDR

Convert from Dotted Decimal to CIDR or Slash, use NL to indicate “not legal” Dotted Decimal CIDR or Slash 255.255.192.0

255.255.255.128 255.0.192.0

255.255.255.254 254.0.0.0

255.255.255.0 Now from CIDR to Dotted Decimal, use NL to indicate “not legal”

CIDR Dotted Decimal /9

/27 /22 /31 /12 /30

The Significance of The Last “On Switch”:

The value above the last “on switch” indicates the legal values by which the Network Ids will increment by beginning with Zero in their respective octet Our Sample IP Address and Subnet Mask was: IP Address: 192.168.10.97

Subnet Mask : /22 which placed the last “on switch” under the 4 in the third octet as indicated above. With this subnet mask, our network addresses (NID) will increment by 4 beginning with 0 in the third Octet. The first and second Octets will remain constant (i.e 192.168.x.0) The very first NID would be 192.168.0.0 The second would be 192.168.4.0 The third would be 192.168.8.0 The fourth would be 192.168.12.0 Since the third octet of our IP Address is between 8 and 12 (192.168.10.97) our IP Address is in the 192.168.8.0 /22 network


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Mathematical Shortcut • Determine the significant octet and the number by which the network IDs will increment • Divide the number from the IP Address by the number the network IDs will increment by (Network ID-NID) • Add the number the network IDs will increment by to the product of the multiplier and divisor from the previous step (Next NID) • Back off 1 host address from the next NID (Broadcast Address-Bcast) • Identify all addresses between the NID and Bcast (Host Range)


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Determine the significant octet and the number by which the network IDs will increment:

192.168.10.97 192 (first octet) 168 (second octet) 10 (third octet) 97 (fourth octet)


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Divide the number from the IP Address by the number the network IDs will increment by(Network ID-NID)


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Add the number the number the network IDs will increment by to the product of the multiplier and divisor from the previous step(Next NID)


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Back off 1 host address from the next NID (Broadcast Address-Bcast)


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Identify all addresses between the NID and Bcast (Host Range)


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IP Subnetting Exercise

address subnet mask nid hr bcast class public / private assignable

14.111.46.90 27 bit 177.89.40.40 22 bit 211.229.250.1 17 bit 224.28.40.99 23 bit 49.49.255.255 12 bit 78.99.99.100 18 bit 7.11.7.11 30 bit 192.11.13.6 28 bit

190.190.14.255 22 bit 46.9.8.7 29 bit

127.114.96.17 28 bit 11.11.254.255 14 bit 65.64.63.62 26 bit

176.176.176.176 25 bit 172.16.144.9 23 bit 172.15.156.10 24 bit 119.119.17.86 22 bit 128.128.255.255 9 bit


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IP Subnetting Homework

address subnet mask nid hr bcast class public / private assignable

14.65.63.255 /18

223.14.11.255 255.255.252.0

192.11.13.14 /30

127.65.135.8 255.255.224.0

135.148.143.90 255.255.255.192

172.168.4.44 /21

28.128.255.255 255.0.0.0

28.128.255.255 /9

241.127.10.10 255.255.255.0

145.148.143.34 /26

235.221.1.1 255.255.192.0

212.67.14.127 /26

43.46.127.255 255.192.0.0

101.102.103.104 /29


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Bit Value 128 64 32 16 8 4 2 1subnet mask na na na na na na na 255.0.0.0CIDR notation na na na na na na na /8

# subnets (A or classless) na na na na na na na 1# of subnets (B) na na na na na na na na# of subnets (C) na na na na na na na na

# of hosts per subnet/net na na na na na na na 16777214

Bit Value 128 64 32 16 8 4 2 1subnet mask 255.128.0.0 255.192.0.0 255.224.0.0 255.240.0.0 255.248.0.0 255.252.0.0 255.254.0.0 255.255.0.0CIDR notation /9 /10 /11 /12 /13 /14 /15 /16

# subnets (A or classless) 2 4 8 16 32 64 128 256# of subnets (B) na na na na na na na 1# of subnets (C) na na na na na na na na

# of hosts per subnet/net 8388606 4194302 2097150 1048574 524286 262142 131070 65534

Bit Value 128 64 32 16 8 4 2 1subnet mask 255.255.128.0 255.255.192.0 255.255.224.0 255.255.240.0 255.255.248.0 255.255.252.0 255.255.254.0 255.255.255.0CIDR notation /17 /18 /19 /20 /21 /22 /23 /24

# subnets (A or classless) 512 1024 2048 4096 8192 16384 32768 65536# of subnets (B) 2 4 8 16 32 64 128 256# of subnets (C) na na na na na na na 1

# of hosts per subnet/net 32766 16382 8190 4094 2046 1022 510 254

Bit Value 128 64 32 16 8 4 2 1subnet mask 255.255.255.128 255.255.255.192 255.255.255.224 255.255.255.240 255.255.255.248 255.255.255.252 na naCIDR notation /25 /26 /27 /28 /29 /30 na na

# subnets (A or classless) 131072 262144 524288 1048576 2097152 4194304 na na# of subnets (B) 512 1024 2048 4096 8192 16384 na na# of subnets (C) 2 4 8 16 32 64 na na

# of hosts per subnet/net 126 62 30 14 6 2 na na

Class A (1-126) 0 x x x x x x xClass B (128-191) 1 0 x x x x x xClass C (192-223) 1 1 0 x x x x x

Class D (224-239 multicast) 1 1 1 0 x x x xClass E (240 -255 experimental) 1 1 1 1 x x x xPrivate Addressing Range

Class A:10.0.0.0-10.255.255.255

Class B:172.16.0.0-172.31.255.255

Class C:192.168.0.0-192.168.255.255

3rd Octet

4th Octet

1st Octet

2nd Octet


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Module 3 VLSM

Variable Length Subnet Mask


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VLSM Design:

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Slice up the Network:


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Document the Network:

VLSM Network Layout Network Subnet Mask NID Host Range Bcast Location 1 Data Location 1 Voice Location 2 Data Location 2 Voice Location 3 Data Location 3 Voice WAN Link FastEth Link


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Module 4 Configuration Lab


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G250 Basic Setup:

Task: Command: G250 Setup

1) Telnet or Console into the G250 and clear any previous configs

G250: G250-???(super)# nvram init Note: if you were connected via telnet, you will need to establish a console connection

2) Set the system name G250: G250-???(super)# hostname Team_X (where X is your team number)

3) Create the interface to be used as the default gateway

G250: Team_X-???(super)# Interface Vlan 1

4) Set the IP Address for this interface

G250: Team_X-???(super-if:Vlan 1)# ip address Y.Y.Y.Y 255.255.Z.Z (where Y and Z are from design) Note: The following message will appear / Follow the instructions from the system The Primary management interface has changed. Please copy the running configuration to the start-up configuration file, and reset the device.


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5) Exit the Interface G250: Team_X-???(super-if:Vlan 1)# exit

6) Copy the running configuration to the start-up configuration

G250: Team_X-???(super)# copy running-config startup-config

7) Reset the Gateway G250: Team_X-???(super)# reset

Note: Follow the prompts from the system to complete the reset

8) Console back into the Gateway and log in

G250: Login: root G250: Password: ****

9) Enter the CSU Config mode for the T1 Circuit

G250: Team_X-???(super)# controller t1 2/1

10) Set the linecoding G250: Team_X-???(super-controller:2/1)#

linecode b8zs 11) Set the framing G250: Team_X-???(super-controller:2/1)#

framing esf 12) Set up the channel groups

G250: Team_X-???(super-controller:2/1)# channel-group 1 timeslots 1-24 speed 64

13) Set the clock source

G250: Team_X-???(super-controller:2/1)# clock source line

14) Exit the CSU config G250: Team_X-???(super-controller:2/1)# exit 15) Enter the Serial G250: Team_X-???(super)# interface serial 2/1:1


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Interface of the T1 16) Set the IP Address of this interface

G250: Team_X-???(super-if:Serial 2/1:1)# ip address X.X.X.X 255.255.255.252 (where X is from design)

17) Exit the interface G250: Team_X-???(super-if:Serial 2/1:1)# exit 18) Create the route statement to send all traffic out through the Serial interface

G250: Team_X-???(super)# ip route 0.0.0.0 0.0.0.0 serial 2/1:1

19) Identify the Media Gateway Controllers

G250: Team_X-???(super)# set mgc list 192.168.0.130

20) Copy the running configuration to the start-up configuration

G250: Team_X-???(super)# copy running-config startup-config


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TU VQRSP XYZW

PHONE/EXIT PAGELEFT

PAGERIGHT

OPTIONS

HOLD

TRANSFER

CONFERENCE

DROP

REDIAL

MUTE

HEADSET

SPEAKER

0SW IP 1 2 3

64 5

97 8

#* 0

A BC D EF

J K L MN OGH I

TU VQRSP XYZW

ETR

ALM

T ST

ACT

T RUNK LINE

1 2 3

MDM

ALM

CPU

PW R

SYST

EM

CONSOLE USB

RST ASB

V2V1

V4

G250-DCP

V3


4

ETRCCA

15 6

V10

11

2 12DCPETH LAN

ETH WAN ALM

T EST

ACT

1 2 3 4

3

4

ALMTSTAC T

AVAYA

MODUL E

SIG

E1/T 1

ALMTST

ACT

AVAYA

MODUL E

SIG

E1/T 1

W AN

1 LAN 10/3SERVICESCCA 10/12

USB

1 0/2L AN1 0/4

MDMALMCPUPWR

RST ASB

V1

SYST

EM

COMPACT FLASH

CAR DIN USE

V2

V3

AUDIOG43

0


50

Six Gateway VLSM Network Layout (Appendix B) Network Subnet Mask NID Host Range Bcast Location 1 Data Location 1 Voice FastEth Link to 2 Location 2 Data Location 2 Voice FastEth Link to 3 Location 3 Data Location 3 Voice WAN Link to 4 Location 4 Data Location 4 Voice FastEth Link to 5 Location 5 Data Location 5 Voice FastEth Link to 6 Location 6 Data Location 6 Voice


51

Router Configuration

Summary Checklist

(Appendix C)

1. Design 1.1. Ensure the design covers the following details

1.1.1. Subnet and VLAN assignments for 1.1.1.1. Data 1.1.1.2. Voice 1.1.1.3. Servers if necessary

1.1.2. Uplink Subnets 1.1.2.1. Interface types 1.1.2.2. Interface IP Address assignments

1.2. Design should include a graphic topology diagram 2. Connectivity

2.1. Connect cabling per design on the LANs 2.2. Connect cabling per design on the WANs

3. Configuration 3.1. Configure Hostname 3.2. Configure Interfaces

3.2.1. Logical 3.2.1.1. Data VLANs 3.2.1.2. Voice VLANs 3.2.1.3. Other VLANs as necessary

3.2.2. Physical 3.2.2.1. Ethernet / Fast Ethernet / Gigabit Ethernet 3.2.2.2. Serial / T1

3.3. Configure Routes 3.3.1. Static: remember to include all other networks 3.3.2. Automatic Using routing protocols

3.3.2.1. RIP V2 3.3.2.2. OSPF


52

Gateway Password Recovery

(Appendix D)

login: root password: ggdaseuaimhrke


53

Thank You!!

Email Questions or Comments to: [email protected]

Date post:	10-Mar-2020
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