Ip Subnet Design

Ip Subnet Design
IP Subnet Design Project
Executive Summary The acronym subnet refers to sub – networking. This is the process of logically dividing an IP
network into various divisions or sections within a network. Therefore, sub–netting is the technique and practice of
dividing a network into two or more logical networks (Narten, Nordmark, Simpson & Soliman 2007). Within a subnet, all
computers connected to it are identifiable by their almost identical IP addresses. The use of IP addresses is in order to
identify uniquely computers in a subnet by the use of a router or network IP prefix followed by a host or rest field
identifier. This IP address composition makes it easy to trace the network within which the sub network is ... Show more
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Hence, there is a reduction of recipients at each sub network the sub networks through using sub networks. This, therefore,
speeds up the performance of the network (Mogul & Postel, 1985). CBCS – Technical IP Sub–netting
Problem
CBCS has Wide Area Networks in place in its new locations, in the Houston region. It also has two other regions: San
Francisco and Denver. CBCS has an issue with the IP addressing infrastructure in place. The current infrastructure makes
use of static IP addresses, which in truth, causes a lot of overheads in terms of network management and traffic. There is
the lack of a VLAN structure in place to manage and isolate any broadcast traffic that the CBCS network utilizes. On the
same note, the Wide Area Network traffic that CBCS makes use of has also been noted to be struggling to handle the
traffic of the CBCS network. Under CBCS, there are four main departments: Human Resources, Finance, Sales, and
Research and Development.
IP Sub–netting Solution IP sub–netting will be the best solution to take into account for CBCS. Through sub–netting, it is
possible to divide logically the CBCS network into four main sub networks that can be sub–netted, again, to form sub –
sub – networks within them, accordingly (Pummill & Manning, 1995). The procedure of sub–netting entails the separation
and differentiation of the main CBCS network and the CBCS subnet portion of an IP address from the host IP identifier. A
bit wise AND operation are
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Different Aspects Of Advanced Micro Devices And Instructions
PFACC mmreg1, mmreg2/mem64 0Fh 0Fh/AEh Converts packed floating point operand to a packed 32–bit integer.
PFADD is a vector instruction that computes addition of the destination operand and source operand (Advanced Micro
Devices, Inc., 2000). PFADD mmreg1, mmreg2/mem64 0Fh 0Fh/9Eh Packed, floating–point addition PFCMPEQ is a
vector instruction that performs a comparison of the destination and source operands and generates all one bits or all zero
bits based on the result (Advanced Micro Devices, Inc., 2000). PFCMPEQ mmreg1, mmreg2/mem64 0Fh 0Fh/B0h
Packed, floating–point comparison, equal to PFCMPGE is a vector instruction that compares the destination and source
operands and generates all one bits or all zero bits based on ... Show more content on Helpwriting.net ...
This information is used to identify the boundaries between variable length x86 instructions, distinguish DirectPath from
VectorPath early–decode instructions, and locate the opcode byte in each instruction (Advanced Micro Devices, Inc.,
2000). The predecode logic also detects code branches, such as CALLs, RETURNs and short unconditional JMPs. When a
branch is detected, predecoding begins at the target of the branch (Advanced Micro Devices, Inc., 2000). Branch
Prediction The fetch logic accesses the branch prediction table at the same time as the instruction cache and uses the
branch prediction table information to predict the direction of the branch instructions (Advanced Micro Devices, Inc.,
2000). The Athlon uses a combination of a branch target address buffer (BTB), a global history bimodal counter (GHBC)
table, and return address stack (RAS) hardware to predict and accelerate branches (Advanced Micro Devices, Inc., 2000).
Predicted–taken branches incur only a single–cycle delay to redirect the instruction fetcher to the target instruction. The
minimum penalty for a misprediction is ten cycles (Advanced Micro Devices, Inc., 2000). The BTB is a 2048–entry table
that caches the predicted target address of a branch in each entry. The Athlon uses a 12–entry return address stack to
predict return addresses from a call. As CALLs are fetched, the next extended instruction pointer is pushed onto the return

Assignment 2
Distance Vector Routing Protocols
Question Answer
1 What does RIP stand for? Routing Information Protocol
2 What metric does RIP use for Path Selection? Hop count
3 If the metric used by RIP exceeds this value for a route it is considered unreachable, effectively making this value appear
to be infinity to RIP?
15
4 How often does RIP send updates by default (update timer)?
30 seconds
5 What are the main differences between RIPv1 and RIPv2? RIPv2 has next hop addressed included in the routing updates,
use of multicases addresses in sending updates.
6 What is convergence and why is it important? Convergence is when the routing tables of all routers are at a state of
consistency. A network is not operable until the network has ... Show more content on Helpwriting.net ...
Based on this what address will be assigned as the default gateway on the following PCs.
48. PC1 = 192.163.3.4
49. PC2 = 192.163.3.99
50. PC5 = 192.163.3.67
NETW204: Assignment 2 (50 points)
Each answer is worth 1 point each, 50 questions = 50 total points. Type answers in the appropriate cell; text will
automatically wrap. Post your completed assignment to the dropbox.
NAME Tiffany Caldwell
Distance Vector Routing Protocols
Question Answer
1 What does RIP stand for? Routing Information Protocol
2 What metric does RIP use for Path Selection? Hop count
3 If the metric used by RIP exceeds this value for a route it is considered unreachable, effectively making this value appear
to be infinity to RIP?
15
4 How often does RIP send updates by default (update timer)?
30 seconds
5 What are the main differences between RIPv1 and RIPv2? RIPv2 has next hop addressed included in the routing updates,
use of multicases addresses in sending updates.
6 What is convergence and why is it important? Convergence is when the routing tables of all routers are at a state of
consistency. A network is not operable until the network has converged
7 Is RIP considered to be a fast or slow converging protocol? Slow converging protocol
For the simple 3 router network (Figure 1), fill in the information that would be in each router's routing table once the
network has converged (see example in Chapter 4, pages 192–193).

Figure 1
Questions 8–22

Commands Http Router Of Router Essay
Commands to configure router to EIGRP
Enable the Routing Information Protocol on the router. Write router(config)#router rip
Using network command detects the network available, using this command you will be able to detect network connected
directly to router. write Router(config–router)#network YOU.RNE.TWR.KID
Write the main network address if your network is sub netted without having to write the subnets i.e. if you some subnets
connected to the router(172.32.0.0/24, 172.32.1.0/24, and 172.32.2.0/
Write the main network address without the subnets I.e. if you have subnet connected to router 24, you just put them to a
single 'network' using the following command Router(config–router)#network 172.32.0.0 and the subnet will be
automatically be connected to the router
To adjust the timers (flush timers, invalid, update and hold down) using the "time basic' command. These parameters are in
seconds. Write " Router(config–router)#timers basic 30 180 180 240" will set default values if the rip timer
Stop the update from being spread to the network, if there is any router inter interface connected to internet. Use the;
passive interface; command. This command will prevent the interface from forwarding any RIP broadcast and will be able
to listen to other interface in RIP. "Write Router (config–router) #passive–interface" followed by "INTTYPE
INTNUMBER" where, INTTYPE is the type of the interface, such as Serial, Fast Ethernet, or Ethernet and INTNUMBER
is the

Designing A Scalable Workload Management System
2.4.1 Neutron Overview
Without including any network–specific functionality, it would be possible to develop a scalable workload management
system. While the connectivity between compute nodes is required and also for the external access, it is important to have
network–specific functionality, but it would be possible to benefit from the existing networking infrastructure to allocate IP
addresses and relay data between nodes. However, the main problem with such an approach is the network management
system would not be able to separate traffic between users securely and efficiently when there will be a multitenant
environment[13]. Neutron is a part of OpenStack project focused on delivering networking as a service and manages the ...
Show more content on Helpwriting.net ...
A simple model of virtual network, subnet, and port abstractions to define network resources are the base for creating
OpenStack Networking component. Similar to a VLAN in the physical networking world, network in OpenStack is an
isolated layer–2 segment. Subnet is a block of IP version 4 or version 6 addresses, set as an address pool from which
OpenStack can assign IP addresses to virtual machines (VMs). Each subnet is stated as a Classless Inter–Domain Routing
range[13].
2.4.2 Neutron Architecture
The main process of the openstack networking component is neutron–server, which transmit user requests from OpenStack
Networking API to the configured plug–in. There are three agents that interact with the main neutron process though the
message queue or the OpenStack Networking API[9]:
– Neutron–dhcp–agent: The main function of this agent is to provide Dynamic Host Configuration Protocol (DHCP)
service to all tenant networks[9].
– Neutron–l3–agent: The main function of this agent is to translate and forward Layer3/Networking address to enable
external network access for VMs on the tenant networks[9].
– Neutron–*–agent: This is an optional plug–in agent, which is responsible of performing local virtual switch
configuration on each hypervisor[9].
When Nova launches a VM instance, the service communicates with OpenStack Networking in order to plug each virtual
network interface into a particular port.
2.4.3 How Neutron works

Application Of Linux And Linux
Topic 1 questions: Summary for topic 1:
This topic gives the introduction to Windows and Linux. It explains about Virtualization. Linux is used on virtual machine
which actually doesn't exist. Virtualization is creation of virtual version rather than the actual one, such as storage system,
network server or virtual operating system. It also defines the host OS and Guest OS. On which the virtualization software
runs is Host OS and the one we want to play with is the Guest OS. The host operating system is the primary or actual
operating installed on the computer 's hard drive. The guest OS is the secondary OS installed under the host for the purpose
of partitioning the disk. The guest operating system is either a part of partitioned ... Show more content on Helpwriting.net
...
for example 01. 02 . 03 . 04 The format of IPv6 is y : y : y : y : y : y : y : y : y : , where 'y ' is considered as segment
seperated by colons ' : '. It must contain eight segments. however a short form of IPv6 can be written using specific
notations. for example 2014 : bd8 : 1232 : 0000 : 0000 : 0001 : C0A8 : 1020 can be written as 2014 : bd8 : 1232 : : 1 :
:C0A8 : 1232 ; where : : is considered as 0000. These IP address is nothing but he exact address of the computer just like a
phone number. Consider a phone number like +61 3 53 232323 this shows where this number belong to like +61 3 53
means /Australia/Victoria/Ballarat and 232323 says to which person exactly it belongs to. Similar is the IP address.
IPv4 address uses 4 bytes first shows the network shows where in whole world and the last two bytes shows where exactly.
Similarly in IPv6 of 16 bytes firts 8 bytes are netwrok and last 8 are nodes.
IP IPv4 address have three classes in it Class A : few big networks
1 byte network and 3 byte node
Class B: Medium sized networks 2 bytes are network and 2 bytes are nodes
Class C: A lot of small networks
3 bytes networks and only 1 byte node.
Task 1.2
Using pen and paper complete the following table. Once completed you may use a calculator (Windows calc for instance)
to check that your answers are correct. Note – In the theory test and exam you will not be permitted to use a calculator.
Decimal Binary Octal

Research Paper On DNS System
DNS SECURITY
Authors Name/s per 1st Affiliation (Author)
Dept. name of organization (Line 1 of Affiliation – optional)
Name of organization – acronyms acceptable (line 2)
City, Country (line 3) name@xyz.com – optional (line 4)
Authors Name/s per 2nd Affiliation (Author)
Dept. name of organization (Line 1 of Affiliation – optional)
Name of organization – acronyms acceptable (line 2)
City, Country (line 3) name@xyz.com – optional (line 4)
Abstract– There are millions of Electronic devices in and around the world. Each device will have unique address and
those devices connect to the world or other devices through a network. Each network will have unique address called IP
Address. Being very hard to remember all the IP address and to ... Show more content on Helpwriting.net ...
Our main purpose is to understand the Security challenges faced in using DNS and better practices to reduce those.
I.INTRODUCTION
DNS is a protocol with some protocols for computers in exchanging data over the Internet. The process is to turn a website
like "google.com" into an machine readable address like 8.8.8.8 and 8.8.4.4 for IPv4 that computers use for identification
on the network. In fact Google Public DNS is largest DNS service in the world which handles more than 400 billion
requests a day. All the computers or hosts on the internet use IP Address to establish a connection with site they are need.
With the DNS service, we don't need to remember the whole bunch of numbers stacked at each decimal point. We just use
a DNS server and appropriate name servers and access the sites with specific names. And this DNS server manages a
massive database which maps the domain names to those IP Addresses.
On a short note, DNS server search for the domain name as per the User Request. This process is named as DNS name
Resolution and this means DNS servers converts the domain name to IP address. As mentioned earlier, google.com has IP
address of 8.8.8.8 which you enter directly into browser will give you the Google's page. Web site's IP addresses are
dynamic and some sites use many IP's with only one Domain name.Without DNS servers, Internet access would be a lot
tougher.
II.DETAILED WORKING OF DNS:
When you are trying to access a domain, the hose

Networking 202 Essay
iLab Grading Rubric
Category
Points
Description
Section 1
Performing Initial Router Startup–20 Points Total
Task 1: Step 17
Related Explanation or Response
Summary Paragraph
3
9
8
Paste the requested screenshot.
Provide the requested answer .
In your own words, summarize what you have learned about IP Subnetting and configuration.
Section 2
Enhancing the Security of Initial Router Configuration–20 Points Total
Task 1: Step 5
Task 1: Step 16
Task 1: Step 17
Task 1: Step 18
Summary Paragraph
1
2
1
2
1
2
1
2
8

Paste the requested screenshot. ... Show more content on Helpwriting.net ...
Enhancing the security of routers is important to avoid unauthorized users change configuration settings and avoid
unauthorized information and routing from entering. For instance, clear text authentication takes advantage of more
flexible key management capabilities. Copy and paste the following screenshots from the Enhancing the Security of Initial
Router Configuration lab below. Task 1, Step 5: Test your user mode and privileged mode passwords on R1 by exiting the
router and logging back in. Once you are in user mode, access privileged mode. Take a screenshot of the output. (1 point)
Question: What is the most current command used to create the enable password? What makes it a better choice?
Hint: Read and study the explanations within the lab for assistance. (2 points)Cisco is the command used to create the
enable password. Cisco is used to access the user mode through the console port but adding the login command helps
safeguard the console port. Sanfran is the password for the login command.
Task 2, Step 16: Use a show command on R1 to verify the SSH settings. Take a screenshot of the output. (1 point)
Question: In addition to SSH, what other method of remote access was selected on step 15?
Hint: Read and study the explanations within the lab for assistance. (2 points)The other method was through telnet. A
command using transport input telnet ssh forces admins to SSH into the device.
Task 5, Step 17:

Cmit Subnetting Assignment
Subnet 90.0.0.0/8 into 260 subnets In order to get 260 subnets the subnet mask will need to have a range of 255.0.0.0 The
IP Range will start at 1–254 since 0 and 255 will already be utilized. Netmask: 255.0.0.0 = 8 11111111
.00000000.00000000.00000000 Broadcast: 90.255.255.255 01011010 .11111111.11111111.11111111 Subnet 1 Network
address: 90.0.0.0 Broadcast: 90.0.255.255 Available IPs: 90.0.0.1–90.0.255.254 Subnet 2 Network address: 90.1.0.0
Broadcast: 90.1.255.255 Available IPs: 90.1.0.1–90.1.255.254 Subnet 3 Network address: 90.2.0.0 Broadcast:
90.2.255.255 IP Range: 90.2.0.1–90.2.255.254 Subnet 4 Network address: 90.3.0.0 Broadcast: 90.3.255.255 IP Range:
90.3.0.1–90.3.255.254 Subnet ... Show more content on Helpwriting.net ...
For each of them, I need the following information: The new subnet mask after the subnetting (10%) The following
information for the first FOUR subnets: o Subnet's network address (3%) o Subnet's broadcast address (3%) o Subnet's
range of available IP addresses (4%) The calculations on how you get to the answers (50%). This is very important. If you
don 't provide the calculations or the way you get the answer, you will lose 50%. 31 Subnet 70.0.0.0/8 into 230 subnets We
need a minimum of 230 subnets so we do the calculation: 2^b >= 230 2^7=128 2^8=256 is the answer Subnet mask is now
/8 +8 = /16 or 255.255.0.0 Subnet 1 Net ID: 70.0.0.0 Broadcast: 70.0.255.255 Available IPs: 70.0.0.1–70.0.255.254 Subnet
2 Net ID: 70.1.0.0 Broadcast: 70.1.255.255 Available IPs: 70.1.0.1–70.1.255.254 Subnet 3 Net ID: 70.2.0.0 Broadcast:
70.2.255.255 Available IPs: 70.2.0.1–70.2.255.254 Subnet 4 Net ID: 70.3.0.0 Broadcast: 70.3.255.255 Available IPs:

school
Graded Assignments The following sections contain student copies of the assignments. These must be distributed to
students prior to the due dates for the assignments. Online students will have access to these documents in PDF format,
which will be available for downloading at any time during the course. Course Revision Table Change Date Updated
Section Change Description Change Rationale Implementation Quarter 11/04/2011 All New Curriculum December 2011
Graded Discussion/Assignment Requirements Discussion or Assignment Requirements documents provided below must
be printed and distributed to students for guidance on completing the discussions and assignments and ... Show more
If an organization decides to implement a medical imagery repository for radiological reviews, requiring 100Mbps
connectivity for eight workstations operating at the same time, what is the minimum Ethernet standard to implement?
Briefly explain the reasons behind your selection. Required Resources Text sheet: IS3120: Unit 2 Assignment 2 Examining
Ethernet Changes Through the Years Submission Requirements Format: Microsoft Word Font: Arial, Size 12, Double–
Space Citation Style: Chicago Manual of Style Length: 1–2 pages Due By: Unit 3 Self–Assessment Checklist Use the
following checklist to support your work on the assignment: I have described the impact continued Ethernet capacity
expansion will have on data storage requirements. I have determined the types of network devices that will be required for
a VLAN network. I have described my observations of the rate of change in Ethernet data transfer capacity. I have
described the minimum standard required for the media image repository scenario and explained my reasoning for this
selection. I have followed the submission requirements. Unit 3 Assignment 3: Convergence of IP–Based Networks
Learning Objectives and Outcomes Understand the convergence of today's IP–based networks. Assignment Requirements
In this assignment, you will write a report detailing your observations on the transformation from early dial–up,

It240 Week 4 Answer Key
Answer Key for TCP/IP LAN Plan CheckPoint, Due in Week Four 8. C, 255.255.240.0. 150.50.0.0 Subnet 150.50.0.0 to
support 7 subnets. 150.50.0.0 is a Class B with a default subnet mask of 255.255.0.0
(11111111.11111111.00000000.00000000 in binary) To figure out how many bits we need to change from 0 to 1 we can use
the formula (2^n)–2=Requirement. When you do the math you come up with the "Binary Truth Line" which is : 2–4–8–
16–32–64–128–256–512–1024–2048............. Count from left to right along the line until you pass seven. You might think
8 would be correct, but you must remember to subtract 2 for the network/wire address and the broadcast address that you
cannot use. So, we must go to 16. Now, 16 is the 4 number over ... Show more content on Helpwriting.net ...
When you do the math you come up with the "Binary Truth Line" which is : 2–4–8–16–32–64–128–256–512–1024–
2048............. Count from left to right along the line until you pass 177. Since we have to remember about the 2 address
that we subtract I changed the number to 177 to account for them so we make sure we have enough addresses. 128 is not
enough so we must go to 256. Now, 256 is the 8th number over so we need to RESERVE 8 bits for the host portion of the
address. Since we are solving for host we count those 8 bits from right to left starting at the end of the address with 0's. In
Binary it would look like this: (11111111.11111111.11111111.00000000). Everything to the right stays a 0 indicating the
number of host in each subnet. Everything to the left becomes a 1. The new subnet mask is 255.255.255.0. You get this
number by adding the 1's together in each octet (128+64+32+16+8+4+2+1=255). Your networks would look something
like this: 150.50.0.0 Gateway (150.50.0.1); Range for Host (150.50.0.2 – 150.50.0.254) Broadcast (150.50.0.255)
150.50.1.0 Gateway (150.50.1.1); Range for Host (150.50.1.2 – 150.50.1.254) Broadcast (150.50.1.255) 150.50.2.0
Gateway (150.50.2.1); Range for Host (150.50.2.2 – 150.50.2.254) Broadcast (150.50.2.255) The networks "increment" by
1 in the third octet. Why? If you look at the subnet mask in binary the last 1 bit is in the 1 position in the

Nt1310 Unit 1 Simulation Paper
We have created a network with the campus area of 500X500square meters taking 20 numbers of ad hoc nodes forming a
network, the transmission range 200m. The network is simulated using high capacity processor and the OPNET and
MATLAB7 platform. Such a simulated result is shown in Fig.1. In simulated network the source node designated as1
initiates the routing procedure by sending RREQ or Route Request message to its surrounding nodes. The RREQ message
sent by the source node is denoted in the color green. The other RREQ messages are shown in cyan, yellow, black etc. The
source node 1 is sending the RREQ message to its neighbour nodes 5, 6, 9, 11 and 13 and the links are formed shown by
the green line. Every time node 5,6,9,11,13 is sending the RREQ message to its neighbour and the links are formed. ...
We have simulated another network having 30 numbers of nodes. The simulation is made using the same platform used for
network 1. This simulation is done taking node 1 as a source and node 28 as destination. The optimal path obtained in this
simulation is 1–23–13–28, in which data rate is 0.47 kbps. The second optimal path is 1–17–28, in which data rate is 0.35
kbps. The simulated results are summarized here in

Erp Case Study Cisco
Cisco ERP implementation
Cisco Systems Cases
Cisco Systems is one the most important successful cases on an ERP implement with the internet and since then it has
added substantial CRM capabilities for customer service and an extensive portal for internal and customer–oriented
knowledge management. Even with its lead, the company never rests: It recently went through another round of
reengineering key processes to make better use of available technology.
Cisco the company
Cisco System, Inc. was founded in 1984; became public in 1990, their primary product is the
"router" hardware & software that control Intranet & Internet traffic. The rise of the Internet made Cisco one of
the top companies, a company ... Show more content on Helpwriting.net ...
What went wrong was: Training Immature software This was Oracle's first major implementation of a new software
release, so there was a lack of experience. – 26 – Software did not provide all the functionality Cisco required "off the
shelf". Many modifications were required.
Lessons learned
When the system did go live, Pete Solvik, Sr VP and CIO, said: "I wouldn't say the company hit the wall, but I would say
we had major day to day challenges." On average, the system went down nearly once a day until it was stabilized two
months later. The origins of these difficulties could be traced to Cisco's task rather than business process approach. The
implementation team was organized in functionally rather than by process. – Reengineering ERP: Business Processes and
Personal Dynamics, May 21, 2001
The project was kicked off on June 2 and went live on January 30, 1995. It succeeded because of the stature of Cisco. The
Cisco IT staff, Oracle, HP, and KPMG dedicated almost unlimited resources to insure the project would be completed
within Cisco's time frame." – Reengineering ERP: Business Processes and Personal Dynamics, May 21, 2001
The urgency dictated implementation on an unrealistic basis

Addressing Scenario
In this Exercise, you will explain IP addres components, contrast classful and classless IP addressing, and explain the
function of DNS and DHCP.
Assignment Requirements
Respond to the following scenario with design considerations and recommendations:
You are a IT Administrator for a newly founded company and have been tasked with designing an IP addressing scheme
and a plan for allocation and management of IP addresses.
The company will currently have a single, physical location with approximately 145 hosts (computers, printers, etc.) IT
plans should accommodate 50% growth within the next two years.
At a minimum, address these specific questions, in addition to any other concerns/considerations. 1. What subnet range/s
should be used ... Show more content on Helpwriting.net ...
CIDR creates a hierarchical addressing structure by breaking the network address into CIDR blocks, which are identified
by the leading bit string, similar to the classful addressing just described.
To understand the importance of DNS and how it functions within a Window Server 2008 networking environment, you
must first understand the following components of DNS * DNS Namespace * DNS Zones * Types of DNS name servers *
DNS resource records * The DNS namespace is a hierarchical, tree–structured namespace, starting at an unnamed root
used for all DNS operations. There are Root level domains, Top–level domains, second–level domains and subdomains.
DNS uses a fully qualified domain name to map a host to an IP address. One benefit of the hierarchical structure of DNS is
that it is possible to have two hosts with the same host names that are in different locations in the hierarchy. Another
benefit of the DNS hierarchy structure is that workload for name resolution is distributed across many different resources,
through the use of DNS caching, DNS zones, and delegation of authority through the use of appropriate resource records.
DHC is an open, industry–standard protocol that reduces the complexity of administering networks based on TCP/IP. It
provides a mechanism for automatically assigning IP addresses and reusing them when they are no longer needed by the
system to which they were assigned. It also provides mechanisms

Subnetting Questions
Subnetting Questions ITE PC v4.0 Chapter 1 © 2007 Cisco Systems, Inc. All rights reserved. Cisco Public 1 Identify the
Number of Subnets and Hosts 1. You have been assigned the IP network address of 135.65.0.0 and have selected
255.255.240.0 as the subnet mask. How many valid subnet addresses are available? 2. You have a network address of
132.66.0.0 and a subnet mask of 255.255.240.0. Find the number of the valid subnets and how many possible host
addresses are on each subnet excluding addresses of all 1s and all 0s. 3. You have a network address of 165.35.0.0 and
have selected 255.255.192.0 as the subnet mask value. How many possible subnets are there? ITE PC v4.0 Chapter 1 ©
2007 Cisco Systems, Inc. All rights ... Show more content on Helpwriting.net ...
Which four of the following are valid subnet addresses? A.132.66.224.0 B.132.66.255.0 C.132.98.0.0 D.132.66.0.0
E.132.66.192.0 F.132.66.96.0 G.132.130.0.0 ITE PC v4.0 Chapter 1 © 2007 Cisco Systems, Inc. All rights reserved. Cisco
Public 9 Identify Valid Subnets 1. You have a network address of 133.233.11.0 and a subnet mask of 255.255.240.0.
Which three of the following are valid subnet addresses? A.133.233.27.0 B.133.233.11.248 C.133.233.11.232
D.133.233.11.176 E.133.233.43.0 F.133.233.11.240 G.133.233.11.48 ITE PC v4.0 Chapter 1 © 2007 Cisco Systems, Inc.
All rights reserved. Cisco Public 10 Identify Valid Subnets 1. You have a network address of 201.79.187.0 and a subnet
mask of 255.255.255.192. Which three of the following are valid subnet addresses? A. 201.79.187.48 B. 201.79.187.224
C. 201.79.187.64 D. 201.79.187.32 E. 201.79.187.1 F. 201.79.187.192 G. 201.79.187.128 ITE PC v4.0 Chapter 1 © 2007
Cisco Systems, Inc. All rights reserved. Cisco Public 11 Identify Valid Subnets 1. You have a network address of
129.111.0.0 and a subnet mask of 255.255.224.0. Which two of the following are valid subnet addresses? A. 129.111.160.0
B. 129.175.0.0 C. 129.111.0.96 D. 129.111.0.32 E. 129.111.96.0 F. 129.143.0.0 ITE PC v4.0 Chapter 1 © 2007 Cisco
Systems, Inc. All rights reserved. Cisco Public 12 Identifying the Host Range 1. You have a

Ip Address Schema
Gerald Boursiquot Network & Infrastructure
Bever Branson Budget/Proposal/Visio
NT 1210
May 23, 2013
In order to properly address the Infrastructure upgrade of the Kamazon Network our group decided that an appropriate ip
address schema that will take into account the idea that people will be using more than just a desktop computer. They will
have tablets, laptops, departmental printers, and smart phones. This is also based upon the understanding that every
Department will have at least two printers for the staff and one for the Manager/Supervisor or VP of that Department. So
with that being said, we estimated that could be at least three or more accurately four ip ... Show more content on
Helpwriting.net ...
If we break out the third octet displaying the binary values for each bit space this is how it would be displayed.
128 | 64 | 32 | 16 | 8 | 4 | 2 | 1
This is our class C after applying CIDR notation.
128 | 64 | 32 | 16 | 8 | 4 | 2 / 0 = 254 By borrowing the one bit we have changed our subnet mask from 255.255.255.0 to
255.255.254.0.
The next step is to determine what our Network ID's will be, and what our usable hosts will be. This is done by going back
to the binary. There are 8 bits in each octet which is in base 2 the numerical value of the binary numbers. This is equal to
28 or 256. When we borrow the one bit from the third octet we are now adding that one bit to the original eight this
changes our binary value from 8 bits to 9 bits which is equal to 29 or 512. This means that we will now have 512 IP
Addresses per subnet, and if we apply the formula 2n–2 this will give us 510 usable hosts per subnet. We can then take this
same formula without the minus 2 to calculate the number of subnets that we will obtain from this calculation. There are
still seven "on" bits to the left in this octet. This is equal to 27 or 128. This means that will obtain 128 subnets with this
CIDR calculation.
This is the default C before we borrowed the one bit in binary notation,
Before 11111111.11111111.11111111.00000000 255 . 255 .

Wan / Network Design Project
Class Project– NETW208 WAN/NETWORK Design Project Project designed by Gary Minardo Carrie Viles Chuck
Hassler January Session Devry University Professor N. Baig The campus redesign we propose that we use all 3 of the
Cisco Catalyst 6500 switches and we allocate one switch for the design department, one for the Human resources and one
for marketing and sales. This will help optimize and departmentalize the switches for easy access and design simplicity.
We also would like to have each switch connected to 2 routers each, this will help with redundency and help take care of
the issues that the architects, engineers and employees keep reporting slow response times. To help with internet crashes
and speeds we think that the company should ... Show more content on Helpwriting.net ...
Key tools utilized, variable length subnet masking and route summarization are explained as well. Here choosing the
appropriate routing protocol is equally critical for a successful design. To implement different masks for the same major
network it is necessary to have a routing protocol that supports VLSM. Such routing protocols are called classless routing
protocols. They carry the mask information along with the route advertisements therefore allowing for the support of more
than one mask. Several classless routing protocol examples include OSPF, RIP version 2, Cisco 's EIGRP, BGP and IS–IS.
We will deploy VLSM for this project. Use of a Class B address 172.17.0.0 is required to support a network that entails a
total of 200 sites. The busiest LANs may support up to 100 hosts and there is a maximum projected total of 400 point–to–
point WAN links. Hence there is a requirement for 600 subnets with a maximum of 100 hosts on any subnet. Even with a
Class B address there is insufficient address space to meet this requirement without employing VLSM. When planning a
VLSM solution you should start with the shortest mask in other words plan the subnets that support the most hosts. This is
typically the mask that will be used on most or all of the LAN segments. For our network design we propose using 200
LAN segments each supporting up to 100 hosts. While 7 host bits or a /25 subnet mask would meet this requirement, it is
probably neater in terms of administration

Virtualization And Virtualization For Virtual Machine
Topic: – 1 Virtualization Virtualization could be a sort of code that is that the illusion of a regular machine. That is termed
as Virtual machine (VM). During this virtual machine we will install totally different operative systems like UNIX,
windows, Ubuntu. This machine have well–supported while not hardware during this machine. Virtualization is worked by
virtual machine (VM). Virtual machine works severally. It even have nice profit that it may be rapt from one physical
server to a different physical server whereas operative through the method "live migration". It additionally runs several
times in one server. The host software package runs the virtualization code through Oracle's virtual box on windows. initial
and first put in software package is Host software package. The host software package is counting on a virtualization's
platforms, like Hyper–v or ESX, VMware server or Virtual severe. In VMware or virtual server the host software package
is "whatever software package those applications square measure put in into. The guest software package could be a virtual
machine system that is functioning underneath the host software package. The guest software package in its own platform.
Number system Number system is calculated by tens. For instance –103 is one thousand, 106 is a million. Computers
square measure count by 2 main switches as a result of it have several swithes and conduct lines. Thus on or off square
measure 2 main switches or keys. Number System

Virtualization : A Computer Produced Form Of A Gadget
Virtualization principally is utilized to make a computer–produced form of a gadget .for ex: server, stockpiling gadget,
system furthermore an operating system .it serves to run various operating systems and applications on a solitary computer.
This aides in IT administration, keeping up and the improvement of new applications. It joins together equipment to get
higher profit from less servers. Whereas a host operating system is an unique operating system that is introduced on a
computer. An extra operating system that is introduced on computer is known as the host operating system. a visitor
operating system must be the same as the host operating system when portioning the circles. A visitor operating system
could be not the same as the host. In Virtualization ,of applications a package is the fundamental yield of the sequencing
procedure. the packages are utilized when we first convey applications on our servers and when we redesign applications
with another variant. Each one package contain a set of documents as an independent unit. For instance hotfix. There are
such a large number of hardware that it can frequently be a complex issue. Running virtualisation software on the machine,
in Windows, makes the mis–comprehended wireless network seem, by all accounts, to be a standard, well–known bit of
network fittings onto which Linux can undoubtedly be installed. Number Systems In the Internet, correspond with one
another and realizing a portion of the rudiments about

Nt1310 Unit 2 Research Paper
Answer 3)
Though aggregations, CIDR scheme can decrease forwarding tables. The practice of aggregation can still be inhibited by
few components. For example a network 208.12.21/24 needs to use the IP address allocated in the same network and the
same IP prefix to be used when it needs to modify its service provider. Therefore from the same service provider all
networks from 208.12.16/24 to 208.12.31/24 can still be reached, except 208.12.21/24 breaking process of aggregation.
This router is not able to aggregate 208.12.16/20 and its forwarding table will store 208.12.16/20 and 208.12.21/24. While
matching we have to look for the lengthiest prefix to find the most exact forwarding information to the true destination, but
the both the entries share the same 20–bit prefix. Value dimension and length dimension are ... Show more content on
Helpwriting.net ...
Binary Trie:
With values 0 and 1, each node has at most 2 branches in a binary trie. As per the bits of the destination address, the search
begins at the root level and goes to the right or to the left. The prefix is recollected as the long prefix checked so far every
time we check a node marked as prefix while the trie is traversed. Once the best equivalent prefix is the last prefix
remembered and there are no branches identical to the next bit in the destination address, the search stops.
At each step the search space is condensed hierarchically and the Binary trie is a sequential prefix search by size. Till the
onset to a node without branch node can be inserted, putting in a prefix starts with a search. As a prefix and erasing idle
nodes, removing processes begins also with a search unmarking the node. As the prefixes are characterized by the trie
configuration, nodes don't store prefixes.
Path compressed

Vrf Design
VRF Design Guide Create the VRF To create the vrf, you need a name and an rd (route distinguisher). The rd is a number
which helps to identify the vrf's routes and allows for overlapping ip address space. It comprises of 2 fields generally
represented by an AS number and another arbitrary number (ie 40:40). The vrf must be created on each layer3 hop as it is
not propagated throughout the network (must be added on all 4900's that participate in ospf) router(config)#ip vrf vpn1
router(config#)rd 50:50 Create VRF Interfaces You can add any layer3 interface to the vrf but it can only belong to a single
interface. To have multiple vrf's share the same physical interface, you can use a trunk port and assign the individual ...
You can verify this by doing: show route <interface name> to view the routes which should look similar to this:
Gateway of last resort is 207.170.225.129 to network 0.0.0.0 C 10.250.250.0 255.255.255.0 is directly connected, Guest–
Internet D EX 10.90.31.12 255.255.255.252 [170/256512] via 10.250.250.3, 0:14:49, Guest–Internet [170/256512] via
10.250.250.2, 0:14:49, Guest–Internet D EX 10.90.31.8 255.255.255.252 [170/256512] via 10.250.250.3, 0:14:49, Guest–
Internet [170/256512] via 10.250.250.2, 0:14:49, Guest–Internet D 10.90.31.4 255.255.255.252 [90/3072] via
10.250.250.3, 0:14:49, Guest–Internet [90/3072] via 10.250.250.2, 0:14:49, Guest–Internet D EX 10.90.31.0
255.255.255.252 [170/256512] via 10.250.250.3, 0:14:49, Guest–Internet [170/256512] via 10.250.250.2, 0:14:49, Guest–
Internet You should also be able to traceroute outbound after your policy and nat is configured. To traceroute from within a
vrf: traceroute vrf vpn1

Structure Of Class AAddress Essay
Class A Class A addresses are assigned to networks with a very large number of hosts. The high–order bit in a class A
address is always set to zero. The next seven bits (completing the first octet) complete the network ID. The remaining 24
bits (the last three octets) represent the host ID. This allows for 126 networks and 16,777,214 hosts per network. Figure 7.1
illustrates the structure of class A addresses. The range of IP addresses is 0 to 127. FIGURE 7.1: Class A IP Addresses
Class B Class B addresses are assigned to medium–sized to large–sized networks. The two high–order bits in a class B
address are always set to binary 1 0. The next 14 bits (completing the first two octets) complete the network ID. The
remaining 16 bits (last two octets) represent the host ID. This allows for 16,384 networks and 65,534 hosts per network.
Figure 7.2 illustrates the structure of class B addresses. The range of IP addresses is 128 to 191 FIGURE 7.2: Class B IP
Addresses Class C Class C addresses are used for small networks. The three high–order bits in a class C address are
always set to binary 1 1 0. The next 21 bits (completing the first three octets) complete the network ID. The remaining 8
bits (last octet) represent the host ID. This allows for 2,097,152 networks and 254 hosts per network. Figure 7.3 illustrates
the structure of class C addresses. The range of IP addresses is 192 to 223. FIGURE 7.3: Class C IP Addresses Class D
Class D addresses are reserved for IP

Netw204 Quiz Answers Essay
Netw204 Final Week 8
Q&A
1 | Given a network design of two or more routers, install necessary memory, interface modules, cables and IOS and test
layer 1&2 connectivity. | 2 | Given a network topology of two or more routers and an IPv4 addressing scheme, implement
static routing and confirm layer 3 connectivity. | 3 | Given a network topology and three or more routers and an IPv4
addressing scheme, implement RIPv1 routing protocol and confirm layer 3 connectivity. | 4 | Given a network topology of
three or more routers with bandwidth designations, compare the metrics used during the route selection process and the
resulting routing tables generated by the DUAL and SPF algorithms. | 5 | Given a network topology of three or ... Show
more content on Helpwriting.net ...
a) Answer: Path Determination Function
23. The process used by a router to accept a packet on one interface and forward it out another interface. a) Answer:
Switching Function
24. A reference to the different parts of the Cisco IOS CLI. Different modes allow different kinds of commands (EXEC
and configuration commands) and different subsets of these commands. a) Answer: Mode
25. Another term for user mode, with specific emphasis on the fact that this mode allows the user to enter EXEC
commands instead of configuration commands. a) Answer: User EXEC Mode
26. An area of a Cisco router or switch CLI in which the user can issue the most powerful EXEC commands, including
commands to configure the router, reload the router, and erase the configuration. The name comes from the enable
command, which moves the router from user mode to enable mode. a) Answer: Privileged EXEC Mode
27. An area of a Cisco router or switch CLI in which the user can enter global configuration –commands.Global
Configuration Mode
28. The process of initializing a computing device. a) Answer: Booting
29. A term used throughout the world of computing to refer to a basic hardware function that occurs when a computer (or
router or switch) is powered on. When a computer's power is first turned on, the hardware performs self–diagnostic testing
of the hardware before it can load the OS into memory. This term

Tele 5330 : Data Networking
TELE 5330 – DATA NETWORKING
ASSIGNMENT #3 Name: Prashanth Reddy Edunuri
1.
I. SComm is under LAN (Local Area Network)
Sundar Communications (SComm) which resides in Tewksbury, Massachusetts is a small company and within same
building. By definition, connections must be high speed and comparatively low–priced.
MNets is under LAN (Local Area Network)
Magesh Networks (MNets) which is based in Burlington, Massachusetts is a small company and within same building. By
definition, connections must be high speed and comparatively low–priced.
SM Inc. is under WAN (Wide Area Network)
SM Inc. is a new venture formed by merging SComm and Mnets. Larger networks that are usually spanned ... Show more
We should maintain high speed connectivity.
III. We will be using Classless IP addressing in the existing internal networks.
The classless IP addressing is commonly known as Classless Inter Domain Routing
(CIDR). This method is used for allocating IP addresses and routing Internet Protocol packets. CIDR is introduced by
IETF in1993 to substitute the prevailing hierarchical addressing in network.
We are using 192.168.53.53/24, which comes under CIDR. We will assign different IP address based on our required. We
are not confined to use only one class for whole network. We provide the network with different subnet masks in the same
network to enable efficient use of IP address.
IV. We will be using Classless IP addressing in the existing internal networks.
The classless IP addressing is commonly known as Classless Inter Domain Routing
(CIDR). This method is used for allocating IP addresses and routing Internet Protocol packets. CIDR is introduced by
IETF in1993 to substitute the prevailing hierarchical addressing in network.
We are using 10.10.10.1/30, which comes under CIDR. We will assign different IP address based on our required. We are
not confined to use only one class for whole network. We provide the network with different subnet masks in the same
network to enable efficient use of IP address.
2.
NAME TTL TYPE DATA/VALUE test.com 1800 Type NS dns1.test.com dns1.test.com 1800 Type A 192.168.1.1 test.com
1800 Type CNAME a.test.com

Network 202 W3 Lab Report Essay
iLab Grading Rubric Category | Points | Description | Section 1Computing Usable Subnets and Hosts vLab–40 Points Total
* Task 1 * Task 2 * Summary Paragraph | .625 points each2.5 points each20 | Paste your answers to questions 1 through 16
in the document. Be sure to show your work.Perform the required problems.In your own words, summarize what you have
learned concerning network commands available within the Windows operating system. | Total | 40 | |
––––––––––––––––––––––––––––––––––––––––––––––––– Name: Date: Professor:
––––––––––––––––––––––––––––––––––––––––––––––––– Computing Usable Subnets and Hosts vLab (40 points)
Instructions for the Computing Usable Subnets and Hosts Exercise: ... Show more content on Helpwriting.net ...
3 bits need to be borrowed at minimum with a total of 8 subnets. There will be 8190 hosts per subnet. 213=8192 8192–
2=8190 8) You are given the Class B network address: 172.25.0.0. From this network, if you needed to create 8 subnets,
how many bits would need to be borrowed at a minimum and how many hosts could you have per subnet? 3 bits with
exactly 8 total subnets. There will be a total of 8190 hosts per subnet. 213=8192 8192–2=8190 9) You are given the Class
B network address: 172.25.0.0. From this network, if you needed to create 14 subnets, how many bits would need to be
borrowed at a minimum and how many hosts could you have per subnet? 4 bits with a total of 16 subnets. There will be
4094 hosts per subnet. 212=4096 4096–2=4094 10) You are given the Class B network address: 172.25.0.0. From this
network, if you needed to create 20 subnets, how many bits would need to be borrowed at a minimum and how many hosts
could you have per subnet? 5 bits need to be borrowed with a total of 32 subnets. There will be a total of 2046 hosts per
subnet. 211=2048 2048–2=2046 11) You are given the Class B network address: 172.25.0.0. From this network, if you
needed to create 35 subnets, how many bits would need to be borrowed at a minimum and how many hosts could you have
per subnet? 6 bits need to be borrowed

Questions And Answers On Computers
`timescale 1 ns / 10 ps module mac(Y30,R30,G30,B30,clk30,rst30); parameter w1=8; parameter
c_r30=8'b00100111;//0.3008 parameter c_g30=8'b01001011;//0.5859 parameter c_b30=8'b00001111;//0.1133 parameter
w3=8; input [w1–1:0] R30,G30,B30; input clk30,rst30; output [w1–1:0] Y30; // final output of the MAC wire[w1–1:0]
rd1_30,rd2_30,rd3_30,sum1_30,sum2_30; // wires connected to accumulator module reg [w3–1:0]
R_temp30,G_temp30,B_temp30,rd1_temp30,rd2_temp30,rd3_temp30,add1_30,add2_30,add3_30,sum1_temp30,Y_temp30
,add3_temp30,Y_temp_t30; // intermediate values reg[2:0] count30,count_temp30; // register which point to the final
output assign Y30=Y_temp30; // assign the final output from the register //reg [7:0] add_roundp1, ... Show more content
on Helpwriting.net ...
begin count_temp30<=count30+1; if(count_temp30==1) begin Y_temp_t30<=sum2_30; count_temp30<=0; end end
endmodule Altera Multiplier Lpm_mult module `timescale 1 ns / 10 ps module altera_mult ( dataa, datab, result); input
[7:0] dataa; input [7:0] datab; output [15:0] result; wire [15:0] sub_wire0; wire [15:0] result = sub_wire0[15:0]; lpm_mult
lpm_mult_component ( .dataa (dataa), .datab (datab), .result (sub_wire0), .aclr (1'b0), .clken (1'b1), .clock (1'b0), .sum
(1'b0)); defparam lpm_mult_component.lpm_hint = "MAXIMIZE_SPEED=5", lpm_mult_component.lpm_representation
= "SIGNED", lpm_mult_component.lpm_type = "LPM_MULT", lpm_mult_component.lpm_widtha = 8,
lpm_mult_component.lpm_widthb = 8, lpm_mult_component.lpm_widthp = 16; endmodule /*module lpm_mult ( result,
dataa, datab, sum, clock, clken, aclr ); parameter lpm_type = "lpm_mult"; parameter lpm_widtha = 1; parameter
lpm_widthb = 1; parameter lpm_widths = 1; parameter lpm_widthp = 1; parameter lpm_representation = "UNSIGNED";
parameter lpm_pipeline = 0; parameter lpm_hint = "UNUSED"; input clock; input clken; input aclr; input [lpm_widtha–
1:0] dataa; input [lpm_widthb–1:0] datab; input [lpm_widths–1:0] sum; output [lpm_widthp–1:0] result; endmodule */
Altera Adder Lpm_Add_Sub module `timescale 1 ns / 10 ps module altera_adder ( dataa, datab, cout, overflow, result);
input [7:0] dataa; input [7:0] datab; output cout;

IP SCHEMA KAMAZON Essay
NT 1210 May 23, 2013 In order to properly address the Infrastructure upgrade of the Kamazon Network our group decided
that an appropriate ip address schema that will take into account the idea that people will be using more than just a desktop
computer. They will have tablets, laptops, departmental printers, and smart phones. This is also based upon the
understanding that every Department will have at least two printers for the staff and one for the Manager/Supervisor or VP
of that Department. So with that being said, we estimated that could be at least three or more accurately four ip addresses
per employee. So we multiplied the number of users per department by the number of estimated devices that each ... Show
128 | 64 | 32 | 16 | 8 | 4 | 2 | 1 This is our class C after applying CIDR notation. 128 | 64 | 32 | 16 | 8 | 4 | 2 / 0 = 254 By
borrowing the one bit we have changed our subnet mask from 255.255.255.0 to 255.255.254.0. The next step is to
determine what our Network ID's will be, and what our usable hosts will be. This is done by going back to the binary.
There are 8 bits in each octet which is in base 2 the numerical value of the binary numbers. This is equal to 28 or 256.
When we borrow the one bit from the third octet we are now adding that one bit to the original eight this changes our
binary value from 8 bits to 9 bits which is equal to 29 or 512. This means that we will now have 512 IP Addresses per
subnet, and if we apply the formula 2n–2 this will give us 510 usable hosts per subnet. We can then take this same formula
without the minus 2 to calculate the number of subnets that we will obtain from this calculation. There are still seven "on"
bits to the left in this octet. This is equal to 27 or 128. This means that will obtain 128 subnets with this CIDR calculation.
This is the default C before we borrowed the one bit in binary notation, Before 11111111.11111111.11111111.00000000
255 . 255 . 255 . 0 This is the class C after we borrowed the one bit in binary notation, After

Computer Operating And Networking Systems
Identify the computer operating and networking systems used in the company
We are using windows 7 and 8 on all PC for our company, 2003 network servers. It comes a time for change and IPv6 is it,
so we have to be prepared for it. Our company has about 100 personnel who needs network access. If we are in need more
IP's or the company grows, we might be in trouble. We need to quickly change to the latest version, from IPv4 to IPv6 in
order to keep the network and company in a place for future growth. IPv6 has been around since about 1995, and some
companies have been testing it, it is installed on the unit, and enabled by default, but most people have not activated it.
Windows 7 and 8 and servers 2003 all are ready for upgrade to the IPv6 network setting. As we know IPv4 only give you 4
billion some user and are completed, the new version IPv6 has an unlimited number of addresses to use, with hope of not
running out.
The pros and cons of IPv6 over the use of IPv4 IPv6 are better security, since the security for IPSEC is built into IPv6
upgrade; more available IP address, since we are switching from a 32 bit system to a 128 bit protocol; IP address can be
close to an unlimited number and should not run out, but as with IPv4, it could happen; better and faster service for video
screaming, games, and video conferencing since the UDP will work much better for screaming of video quality of
services; and an auto configuration of all router on the system and it has a larger

Nt1310 Unit 5 Term Paper
Table 2 1 Generic DoIP header structure [23] Item Position Length Description Values Generic DoIP header
synchronization patten Protocol version (PV) 0 1 Identify the protocol version of DoIP packets. 0x00: reserved 0x01: DoIP
ISO/DIS 13400–2:2010 0x02: DoIP ISO 13400–2: 2012 0x03...0xFE: reserved by this part of ISO 13400 0xFF: default
value for vehicle identification request messages Inverse protocol version (IPV) 1 1 Contains the bit–wise inverse value of
the protocol version, which is used in conjunction with the DoIP protocol version as a protocol verification pattern to
ensure that a correctly formatted DoIP message is received. Equals the XOR 0xFF (e.g. 0xFE for protocol version 0x01).
Generic DoIP header payload type and payload length

Part2 Cmit 350 Essay
CMIT 350 WAN and SOHO Skills Implementation UMUC Winter 2015 Springfield Site Configuration: Required
Implementation: Device hostnames, banners, secured passwords and spanning tree protocol. * Device Configurations:
Implement device hostnames to match the xACME educational topology labels. Provide a template and sample
configuration for the MOTD banner and login banner (wording and implementation) for one of the switches. Keep this
generic, as it will be implemented on all switches in the xACME educational topology. Lastly, include the configuration
steps for implementing device passwords on both console port (out–of–band communications) and VTY (Telnet/in–band
communications). All passwords should be encrypted. Required ... Show more content on Helpwriting.net ...
End 5. Copy run start Create VLAN's 1. Conf t 2. Vlan 1 3. Name (Faculty, Instructional, Administrative, Server) 4. End 5.
Copy run start Assign ports to VLANs on all switches: 1. Enable 2. Conf t 3. Int gi0/1 4. Switchport mode access vlan 1 5.
End 6. Copy run start Configure trunking between switches * Interface gi0/1 (gi0/2, etc., on every switch) * Switchport
trunk encapsulation dot1q * Switchport mode trunk * Switchport trunk Make Switch1 as primary root bridge for all
VLANs * Switch1(config)#spanning–tree vlan1 root primary Configure security for bogus BPDU 1. Enable 2. Conf t 3.
Switch(config)#int gi0/1 4. switch(config–if)#spanning–tree bpduguard enable 5. end 6. copy run start * Spanning Tree
Protocol (STP): Briefly explain the advantages and purpose of the STP. Administrators are having a difficult time placing
switch 1 as the root. Provide a sample configuration for implementing SPT on the switches. Choose the mode you feel
would be best suited for the environment and justify why. Switch 1 will need to be the root switch in the Springfield
topology. Consider any security measures that can be implemented to protect the devices from bogus BPDUs. In local area
networks (LANs) computers compete for the use of telecommunication paths all the time. If too many computers are
trying to send data

Week 4 B IP Subnetting Lab Essay
Introduction
Assignment
Develop a subnetting plan and implement it in the lab. Configure RIP as a routing protocol.
Conditions
NYEDGE1 is the hub router. The two remote office routers NYEDGE2 and NYWAN1 connect to NYEDGE1 through
serial connections. Each router has a locally attached Ethernet network.
The serial line data rates and DTE/DCE settings have already been configured on all three routers. You only need to assign
the IP address to the interfaces.
Notes
The routers are connected directly to each other in this lab and do not actually connect through any leased line services.
Diagram
Gi0/0
Ethernet
NYWAN1
Serial0/1/0
Serial0/0/1
DCE
NYEDGE1
Serial0/0/0
Serial0/0/0
DCE
Gi0/0
Ethernet
Task Index


Task 1 – Create a subnet ... Show more content on Helpwriting.net ...
You are not encouraged to use this subnetwork because it specifies the wire address.
Nor should you use the last subnet, 192.50.6.224, because it is considered the major network broadcast address. Therefore
you can only use subnets 32, 64, 96, 128, 160 and 192. With this design requirement, you only need to use five of the six
usable subnetworks, leaving one for any additional segments in the future. Likewise, for each subnet, you cannot use the
first and last address. For example, subnet 192.50.1.32 extends all the way to 192.50.1.63. But you cannot use these two IP
addresses (.32 and .63) because they are the wire and the broadcast addresses for this subnetwork. But you can use any of
the 30 IP addresses in between.

Task 3 – Configure the interfaces in each router Step 1: Configure Gi0/0, s0/0/0, and S0/0/1 for NYEDGE1.
Action:
ip address <ip address> < subnet mask>
Result:
NYEDGE1>enable
NYEDGE1#config t
Enter configuration commands, one per line.
End with CNTL/Z.
NYEDGE1(config)#int g0/0
NYEDGE1(config–if)#ip address 192.50.6.33 255.255.255.224
NYEDGE1(config–if)#no shut
NYEDGE1(config–if)#
NYEDGE1(config–if)#int s0/0/0
NYEDGE1(config–if)#int s0/0/1
NYEDGE1(config–if)#end
Explanation:
Subnet 192.50.6.32 is assigned to g0/0. Subnet 192.50.6.128 is

The Importance Of The Enclosure System
Enclosure system
Eight "splinters" angled glass panels define the shape of the Shard and result in changes in the diversity of the skin
patterns. This curtain walling system consists of 11200 glass panels with the area of totally 56,000 square meter, which is
equivalent to eight football pitches. The glazing element comprises a standard high specification double glazed unit inside
with the addition of glazed spandrel panels outside in a silver opaque colour. These spandrel panels not only increase the
aesthetic appeal to the external curtain wall but also carry out the important job of masking parts of the building such as
floors and steel structures that would otherwise be visible from the outside. Each floor has 114 glazed panels with ... Show
In addition, a colourless low–radiation coating has been added to reduce the reflection of infrared radiation back into the
building. (Martin, 2010) This kind of design aims to reflect the surrounding environment of the city. For example, the
building will flash like a shard of glass on a sunny day.
In order to achieve the immaculately flush surface treatment, the external glass plate over–sails the polyester–coated
aluminium glazing beads and butt up against each other. The curtain walling contractor, Scheldebouw supports the glass on
timber blocks for 48 hours and the silicon sticks on the glazing beads sets. Matthews claims that this eliminates the slight
depression effect of the double glazing units. (Martin, 2010)
Solar control
To achieve the building function and cool efficiently, the curtain wall design needed to be incorporated into the effective
means of controlling the solar gain. The main solar control comes from the roller blinds in the ventilated inner cavity of the
triple–glazed glass to reduce solar radiation by 95%. The resistance of the building to solar gain is measured by "G value"
which for the average office is 0.34 when using double glazing. According to Arup's Shard project manager, David Healy,
by using triple glazing with intelligent motorised roller blinds control system which could track the position and intensity
of the sun to deploy the blinds only when required, the G value at The Shard will be reduced by 0.12. thereby significantly

Essay on Wan Design
Company Overview In this plan, I will describe the process of creating a WAN for ACME Manufacturing. ACME
Manufacturing has six buildings in different locations that need to be on the same network. Headquarters and the other
remote locations of ACME Manufacturing share the need for constant communication with each other. Atlanta, Georgia, is
the home of ACME manufacturing where they have another building across the street, their engineering facility. The other
locations are in Phoenix, Chicago, New York, and China.
Network Setup and Connections The LAN setup in Atlanta for the headquarters and engineering buildings will use a
logical choice of routers and Ethernet cables. Using routers and Ethernet cables will provide speedy ... Show more content
on Helpwriting.net ...
Some of the benefits of VOIP are, enhanced productivity, reduction in telecommunication costs, and makes any phone
system highly flexible. VOIP increases productivity and provides useful features and capabilities that a normal phone
system cannot offer.
VLAN/IP Address Building a VLAN for ACME manufacturing one would think a Class B would be the choice, but with
Classless Interdomain Routing (CIDR), it will not be necessary. "CIDR eliminates the traditional concept of class A, B,
and C network and with CIDR, the addresses that were wasted for the class A and B networks are reclaimed and
redistributed for use later" (Regan, 2004). Dividing the network into smaller subnetworks, which would be the six builds
that ACME occupies, will enhance network performance. Each subnetwork will have resources and devices available to its
users and eventually freeing broadcast traffic in the network.
Equipment for the Network Designing a WAN, the network equipment needs special considerations. The following list is
the recommended equipment best suited for ACME manufacturing: * Mainframe computers * Switches and routers *
Telephony system * Firewalls * Wireless equipment
The main component of each of ACME manufacturing's buildings are the mainframes computers. The mainframe in the
headquarters building will control the connections of the different networks, in turn connecting all networks, routers, and
firewalls to the mainframe at the

Convolution2d : Lab Analysis
Convolution2D is the initial hidden convolutional layer. This layer has 32 feature maps, each with a rectifier activation
function and the size of 5x5. It expects images with the format mentioned as above ([px][wd][ht]) and is the input layer. A
pooling layer is defined which is configured with a pool size of 2x2 and takes the max. It is known as MaxPooling2D.
After this layer, is a regularization layer which is added using drop out function referred as Dropout. It reduces overfitting
as it randomly excludes 20% of neurons in the layer. Following this regularization layer, is the layer containing a vector
known as Flatten. It enables the standard completely connected layers to process the output. This layer changes the 2D
matrix data to ... Show more content on Helpwriting.net ...
An LSTM RNN is much more complex and robust neural network as Compared to an MLP. For the purpose of modeling
time–series with LSTM, a standard time–series problem will be considered.[17] But before modeling the example, some
basic concepts are discussed. The recurrent neural network overcomes the vanishing gradient problem and is trained over
time using Backpropagation [11]. The Vanishing Gradient Problem is the challenge faced while training some ANN with
gradient based methods, such as Back Propagation. This issue mainly makes learning and tuning the parameters of the
previous layers of the network difficult. As before–mentioned, this model is applied to generate large recurrent networks
that can be used to tackle complex sequence problems in ML and hence produce better results. Also, the LSTM networks
have memory blocks instead of neurons, which are connected with each other through layers[35]. There are some
components in these blocks, that make them sharper than the classical neuron and recent sequences memory. They contain
gates that manage its state and output. Each gate in a block verifies if they are triggered or not using the sigmoid activation
units and operating upon an input sequence. This results in flow of additional information via block and change of state
conditional. Further, there are three types of gates within a unit which are: Forget gate, input gate, and the output gate. The
first gate conditionally determines what data to dispose away of

Threats Of Bgp Protocol, Security And Experiment By Using...
Student Number: 100782427
Ioannis Adamos
Title: Threats of BGP Protocol, Security and Experiment by using Hijacking.
Supervisor: Geraint Price
Submitted as part of the requirements for the award of the
MSc in Information Security at Royal Holloway, University of London.
I declare that this assignment is all my own work and that I have acknowledged all quotations from published or
unpublished work of other people. I also declare that I have read the statements on plagiarism in Section 1 of the
Regulations Governing Examination and Assessment Offences, and in accordance with these regulations I submit this
project report as my own work.
Signature:
Date:
Contents
Chapter 1 Introduction 3
1.1 Internet Architecture 4
1.2 Routing and IP Addressing 4
1.2.1 IP addressing 5
1.2.2 IANA 5
1.2.3 CIDR 5
Chapter 2 Background 6
2.1 Review on TCP/IP Protocols 6
2.1 Review on TCP/IP Layers 8
2.1.1 Application Layer 8
2.1.2 Transport Layer 8
2.1.3 Internet Layer 9
2.1.4 Network Access 9
2.1.5 Physical Layer 10
2.2 Routing Protocols 10
2.2.1 RIP Protocol 10
2.2.2 RIPv2 10
2.2.3 OSPF 11
2.3 Disadvantages and Advantages of Using RIP and OSPF 11
2.4 BGP Protocol 12
Chapter 3 BGP Overview 12
3.1 PATH Attributes 13
3.1.1 Well–known Attributes 13
3.1.2 Optional Attributes 14
3.1.3 As_Path Attribute 14

3.1.4 Next_ Hop 15
3.1.5 Multi_Exit_Disc Attribute 16
3.2 BGP Gateways 16
3.3 Announces of Network Prefixes 17
3.4 BGP Messages 17
3.5 Choosing Between Different

NT1210 Final Exam Review Essay
Final Exam Review
Using Table 10–1, review default port numbers used commonly with TCP and UDP purposes
HTTP TCP Port 80 – used by web browsers and web servers
Telnet TCP Port 23 – used for terminal emulation
SSH TCP Port 22 – used for secure terminal emulation
FTP TCP Port 20, 21 – used for file transfer
DNS UDP Port 53 – used for name resolution
SMTP TCP Port 25 – used to send and receive email
POP3 TCP Port 110 – another email protocol
IMAP TCP Port 143 – another email protocol
SSL TCP Port 443 – used to encrypt data for secure transactions
SNMP UDP Port 161, 162 – used to manage TCP/IP networks
TCP – Transmission Control Protocol
UDP – User Datagram Protocol
IPv4 32 bits, IPv6 128 bits
Bring sub–netting sheet for use ... Show more content on Helpwriting.net ...
Radio signals do not attenuate (False)
Broadcast areas should not overlap in a WAN (true)
Antennas use same amount of power (False)
Omnidirectional in all directions (True)
Leased Line available 24–7 365, also known as dedicated line
Router belong in Layer 3 of TCP/IP
Demark is point at which ISP comes into building and customer takes over
SONET – Look up definition
Primary function of IP protocol = identify individual hosts and groups of hosts using the address
Dynamic Routing protocols – learn and manage the most efficient ways of routing packets
Ranges of Addresses – Class A, Class B and Class C and the private ranges
IP Protocol designed to dynamically assign IP addresses on a LAN – DHCP
Subnet a company network, 3 bits borrowed to effectively subnet it for them
PC1 sending packet, first thing needing to be done is compare the destination IP address with its own, deciding whether or
not it is in the same network
IP addresses divided into 2 parts, network portion and host portion
Term for part of internet between ISP and ISP's customer = network edge
Device found near the edge and offers security services = Firewall
Page 445 – Concept where all ISP router is connected to all other ISP router is called the Internet Core
What technique used to by cable provider to enable data and voice on the same cable using different frequencies for video
and data
Difference between

Contextual Analysis Plan
To assess the adequacy of the proposed plan, a contextual analysis is utilized. An arrangement of parallel FIR channels
with 16 coefficients is considered. The information and coefficients are quantized with 8 bits. The channel yield is
quantized with 18 bits. For the check channels r_i , since the data is the entirety of a few inputs p_j , the information bit–
width is reached out to 10 bits. A little limit is utilized as a part of the examinations such that mistakes littler than the limit
are not considered slips. As clarified in Segment III, no rationale sharing was utilized as a part of the calculations in the
encoder and decoder rationale to stay away from slips on them from engendering to the yield. Two setups are considered.
The first is a piece of four parallel channels for which a Hamming code with k = ... Show more content on Helpwriting.net
...
All things considered, the relative number of included check channels (n–k)/n is littler. At the point when analyzed with
the math code strategy proposed in [7], the reserve funds are littler yet huge going from 11% to 40%. Once more, bigger
funds are gotten for the second arrangement. In outline, the aftereffects of this contextual investigation affirm that the
proposed plan can decrease the execution cost essentially analyzed with the TMR and gives additionally decreases when
contrasted and different routines, for example, that in [7]. As talked about some time recently, the diminishments are
bigger when the quantity of channels is extensive. The second assessment is to survey the adequacy of the plan to right
blunders. To that end, issue infusion tests have been led. Specifically, lapses have been arbitrarily embedded in the
coefficients and inputs of the channels. In all cases, single slips were distinguished and adjusted. Altogether, 8000 lapses
for inputs and 8000 mistakes for channel coefficients were embedded in the distinctive reenactment

Critical Analysis of Mrs Lazarus
This poem consists of 8 regular stanzas, each stanza being a quintet. Throughout the poem, no particular rhyme scheme is
conveyed and the sentence length varies in each stanza. This poem is entirely constructed in 1st person narrative. Stanza 1
"I had grieved" – past tense – descriptive – suggests that sorrow was previously present however there is none now.
Description of graveyard implies that this was a terrifying/chilling experience for her. "Married" implies that she had wed
the person she had lost. Next line portrays that she must have got married on the day that her husband died if she had
ripped her wedding clothes from her body. This stanza provides the reader with physicality by the use of the ... Show more
Language – calm, descriptive Overall tone – serene Themes – admiration of the simple things in life Stanza 7 Continues
from stanza 6. Images of people chasing her. Building up of tension – repetition of "I knew". Personification of light –
"sly" – even the light had betrayed her – labelled her as the wife of an ill man. "shrill eyes" – piercing into her – evil –
undeserved – highlights that the community had a prejudice of her husband because of his illness. Overwhelmed by
crowds of people – feels consumed by the "hot tang" and the "hands bearing" her. Sentence Structure – varied. Broken
with commas to illustrate the break in her normality. Language – anxious Overall tone – fearful Themes – betrayal Stanza
8 Illustrates the resurrection of her husband. "He lived" – shock, disbelief. "the horror on his face" – even he was terrified
of her situation. She cannot escape the memory of him, no matter how far away he may seem. "saw", "heard", "breathed" –
use of the senses – immediacy – "rotting". This stanza could be her mind revisiting the memory of Lazarus's dead body.
This could symbolise her reuniting with him – her escape from isolation. From beyond the grave, Lazarus maintains
control over Mrs Lazarus – she is unable to rid him from her mind – eternally married to him – so she must be faithful.
Does love

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