Networking, Statistic & Programming

Promo IKLAN MURAH di 2 Blog Ber-pagerank 2 dan dikunjungi lebih dari 6000 orang/bulan

2010-07-26T09:02:00.010+07:00

KABAR GEMBIRAAAAA!!!!!
Hanya dengan menyisihkan uang maksimal 50.000,- / bulan, iklan Anda akan ditayangkan di 2 blog sekaligus, yaitu Opan-Sopana.blogspot.com dan LOWONGANKERJA.bengkeldata.com

5 Kenapa Harus Beriklan disini :

Kenapa harus beriklan di blog Tahu Gezrot?
1. Iklan Anda ditayangkan di 2 blog sekaligus, yaitu Opan-Sopana.blogspot.com dan LOWONGANKERJA.bengkeldata.com

2. Pengunjung blog Opan-Sopana & LOWONGAN KERJA rata-rata dikunjungi > 100 orang per hari, jadi iklan Anda akan dilihat > 6.000 orang pengunjung per bulannya. Bayangkan jika Anda mendapatkan minimal 6 klien perharinya, berarti dapat 180 klien dalam sebulan. Coba Anda hitung berapa keuntungan yang akan Anda dapatkan :) Low bugdet high impact toh...

3. Masing-masing Blog ( Opan-Sopana & LOWONGAN KERJA) sudah memiliki Pagerank 2(maks. 10). Pagerank menunjukkan tingkat kepopuleran blog.

4. Pengunjung : Jumlah & Geografis
Jumlah pengunjung Blog Opan-Sopana.blogspot.com sudah menembus > 51.900 pengunjung dari 156 negara. Dimana jumlah pengunjung terbesar berasal dari Philipina, USA, Australia, Indonesia, India, Germany,United Kingdom & negara2 lainnya.

Jumlah pengunjung Blog LOWONGANKERJA.bengkeldata.com sudah menembus > 36.500 pengunjung dari 72 negara. Dimana jumlah pengunjung terbesar berasal dari Indonesia, USA, kemudian disusul Malaysia. Coba sekarang Anda Bayangkan berapa Negara yang akan melihat iklan Anda??? Info ini bisa dilihat di total visitor & Visitor location/peta pengunjung, kolom sebelah kanan. Segmented by negara pengunjung dapat Anda ketahui disini, sehingga iklan Anda bisa dilihat oleh berbagai negara.

5. Materi blog dinamis yang akan selalu akan di update.

Yaa.... kami menawarkan terbatas hanya untuk 6 orang pertama

Iklan Utama/Headline
Hanya Rp. 50.000,- / bulan, iklan ukuran 400 x 60 pixel

Iklan Kolom (lokasi kolom disamping kanan)
Hanya Rp. 30.000,- / bulan, iklan ukuran 125 x 125 pixel

Akan diprioritaskan untuk 6 orang pendaftar pertama.
setelah tanggal 8 Agustus 2010, investasi iklan berubah menjadi Rp. 100 ribu (Headline) & Rp. 60 ribu (iklan Kolom)

Mohon Maaf, NO iklan sex/sex toys & sara

Hubungi : Juragan Tahu Gezrot

Telp: 0819 4505 9000

Kirimkan materi iklan ke:
Email: opan@megamendung.com

Biaya iklan ditransfer ke :

Bank BCA KCP Jatibarang
No. Rekening: 3020274963
a.n. SOPANA

Analisis Data Statistik : Analisis Data Faktor-Faktor Konsumsi di Indonesia dengan Pendekatan ECM (Error Correction Model)

2010-07-26T02:10:00.001+07:00

Tujuan dari penelitian ini adalah mengkaji pengaruh variabel pendapatan nasional, inflasi, suku bunga dan jumlah uang beredar terhadap konsumsi masyarakat yang digambarkan oleh variabel pengeluaran konsumsi mayarakat. Peneilitan ini dilakukan di Indonesia pada kurun waktu tahun 1988 sampai 2005. Metode yang digunakan adalah dengan menggunakan Pendekatan ECM (Error Correction Model).

Besarnya tingkat pengaruh variabel pendapatan nasional, inflasi, suku bunga deposito riil dan jumlah uang beredar terhadap pengeluaran konsumsi di Indonesia dalam jangka pendek yaitu 69,98 % sedangkan sisanya dipengaruhi oleh variabel lain di luar model regresi yang tidak diteliti dalam penelitian ini.

Dalam jangka panjang variasi variabel independent mampu pengaruhi variasi dependent sebesar 0,984057 menunjukkan bahwa variabel independent lebih mampu menjelaskan variabel dependent sebesar 98,40% dan sisanya dijelaskan oleh variabel lain diluar model yang tidak diikutsertakan dalam penelitian ini.

Tanda koefisien koreksi kesalahan sebesar 0,69 menunjukkan bahwa 0,69 ketidaksesuaian antara pengeluaran konsumsi (Y) yang aktual dengan yang diinginkan akan dieliminasi atau dihilangkan dalam satu tahun. Hasil dari uji asumsi klasik ternyata ditemukan masalah asumsi klasik yaitu multikolinearitas, sedangkan heteroskedasitas dan autokorelasi tidak ada masalah.

Untuk menghilangkan masalah multikolinearitas, dilakukan dengan cara menghilangkan variabel jumlah uang beredar. Sehingga diperoleh spesifikasi model penelitian yang dipakai adalah tepat dan mampu menjelaskan hubungan jangka pendek dan jangka panjang. Dengan demikian persamaan tersebut sudah sahih dan tidak ada alasan untuk ditolak.

Berikut kesimpulan yang dapat di ambil dari penelitian ini:

Besarnya pengaruh variabel pendapatan nasional, inflasi, suku bunga deposito riil dan jumlah uang beredar terhadap pengeluaran konsumsi di Indonesia dalam jangka pendek yaitu 75,12 % sisanya dipengaruhi oleh variabel lain di luar model regresi yang tidak diteliti dalam penelitian ini.

Variabel pendapatan nasional pada jangka pendek dan jangka panjang secara statistik positif dan signifikan, berarti pendapatan nasional berpengaruh terhadap pengeluaran konsumsi masyarakat di Indonesia periode 1988-2005.

Variabel tingkat inflasi pada jangka pendek secara statistik tidak signifikan, berarti tingkat inflasi tidak berpengaruh terhadap pengeluaran konsumsi masyarakat di Indonesia periode 1988-2005

Variabel suku bunga deposito dalam jangka pendek tidak berpengaruh terhadap pengeluaran konsumsi. Dalam jangka panjang mempunyai hubungan yang tidak signifikan yang artinya tidak berpengaruh terhadap pengeluaran konsumsi

Variabel jumlah uang beredar dalam jangka pendek tidak berpengaruh terhadap pengeluaran konsumsi.

Berdasarkan pengujian secara serempak dengan menggunakan uji F menunjukkan bahwa variabel independent secara bersama-sama mempengaruhi variabel dependent, artinya pendapatan nasional, inflasi, suku bunga deposito dan jumlah uang beredar berpengaruh secara bersama-sama terhadap pengeluaran konsumsi masyarakat Indonesia.

Untuk konsultasi mengenai Olahdata lebih lanjut, silahkan menghubungi:

CS : 021-71088944 – 0819 4505 9000
YM : abays_khan
e-mail : info@bengkeldata.com

Beta Consulting

www.olah-data.com
www.bengkeldata.com

Keywords: Penelitian, pendapatan, inflasi, suku bunga, uang, pengeluaran, konsumsi, ECM, deposito, regresi, independent, dependent, uji asumsi klasik, multikolinearitas, heteroskedasitas, autokorelasi, uji F

Olahdata Skripsi: Analisis data statistik / Analisa data statistik apa yang cocok untuk skripsi, tesis, disertasi, penelitian Anda ? ( Part I )

2010-07-26T01:31:00.003+07:00

Apakah Anda sedang menyususn skripsi, tesis atau disertasi, atau penelitian? Dalam menyusun skripsi, tesis, disertasi, atau penelitian seringkali kita dihadapkan pada suatu permasalahan yang menjadi momok bagi mahasiswa/peneliti, yaitu analisis data / analisa data. Saya yakin bahwa Anda sebagai mahasiswa/peneliti memahami betul proses/tahapan dalam membuat skripsi, tesis atau disertasi, atau penelitian. Mulai dari tahapan mencari tema skripsi, tesis atau disertasi, atau penelitian, menetukan permasalahan, membuat kerangka pemikiran, membuat proposal, hingga menentukan lokasi penelitian. Nah ada satu lagi yang mungkin/bahkan ’lupa’ yaitu metode analisis data / analisa data yang digunakan. Apabila Anda belum memahami analisis data / analisa data –nya, maka Beta Consulting ( www.Bengkeldata.com / www.olah-data.com ) dengan senang hati akan membantu Anda untuk menentukan analisis data / analisa data yang cocok dengan skripsi, tesis atau disertasi, atau penelitian Anda.

Metode analisis data / analisa data apa yang cocok dengan skripsi, tesis, disertasi, atau penelitian Anda? Mari kita bahas contoh beberapa analisis data / analisa data yang cocok dengan skripsi, tesis, disertasi, atau penelitian. Yuukk Mariii :)

Pertama, misalkan skripsi, tesis, disertasi, atau penelitian Anda itu membahas bagaimana hubungan diantara dua variabel. Maka metode analisis data / analisa data yang digunakan adalah metode regresi sederhana. Apabila skripsi, tesis, disertasi, atau penelitian ada 3 variabel atau lebih, maka metode analisis data / analisa data yang digunakan adalah metode regresi berganda atau regresi multivariabel / regresi multivariat.

Tunggu dulu... Metode regresi sederhana / regresi berganda atau regresi multivariabel / regresi multivariat juga ada macamnya. Selain regresi sederhana, ada juga regresi logistik. Oke, sekali lagi Beta Consulting ( www.Bengkeldata.com / www.olah-data.com ) dengan senang hati akan membantu Anda untuk menentukan analisis data / analisa data yang cocok dengan skripsi, tesis atau disertasi, atau penelitian Anda.

Beta Consulting ( www.Bengkeldata.com / www.olah-data.com ) dengan senang hati dan didukung oleh tim yang berpengalaman, akan membantu Anda untuk menentukan analisis data / analisa data yang cocok dengan skripsi, tesis atau disertasi, atau penelitian Anda.

Hubungi kami di :

Beta Consulting

Telp: (021)71088944 – 0819 4505 9000
Email: info@bengkeldata.com

www.Bengkeldata.com
www.olah-data.com

follow us:
http://Twitter.com/olahdata

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CCNA Exploration 4.0 : Routing Protocols and Concepts - Chapter 7 Exam Answers

2008-11-27T14:53:00.002+07:00

Below is one of question CCNA Exam Protocol and Concepts chapter 7.

How are RIP v1 and RIP v2 similar to one another? (Choose three)
* They both use hop count as a metric.
* They both have the same metric value for infinite distance.
* They both broadcast their updates to their neighbors.
* They both send subnet mask information in their updates.
* They both provide for authentication of update sources.
* They both use split horizon to prevent routing loops.

If you want to know more,click here for DOWNLOAD

FREE POST BANNER & BLOG LINK

2008-10-22T14:00:00.005+07:00

NOW AVAILABLE!

In order to the feast of Idul Fitri (although it is late) and inspired by some bloggers I, like my game, I take it.

Namely to provide opportunities for bloggers who want to install his blog link and install the banner for free on this blog. Once again FREE !!!

With Motto: 'From FREE to give something that is also FREE'

Later, banners / blog that will be inserted into permanent alias I do not delete or in the rotation.

To Blog, I will make a link.

The banner for free, its position is in the bottom of the posting.
For example, such as Banner in this post.

I give this opportunity for 1 (one) month, since today (22 October 2008 to 22 November 2008).

This means that in the time range, each of my post, the banner has the right to participate selected to appear.

The criteria can be established blog that is:
1. Ads must be a blog.
2. Blog be active. At least 10 posting and 4 posts per month.
3. personal blog. (Permissible personal blog with the domain or use
blogspot,wordpress, etc.).
4. Unviolate applicable law.
5. Terms and conditions can change at any time and without notice.

Courtney was the only. Fortunately, there who want to backlink affection here. Or even make a review. Rev.'s about.

From the entrance application, I have the right to choose any of the show and that is not. Any decision could not be proceeded.

Please send your banner (width = "468" height = "60") that have been uploaded (for example, to imagecows.net), you simply send imagenya link and a link to your blog email: sopana888@gmail.com with the subject "Blog and Banner Free. "

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Wired vs Wireless Networking

2008-10-21T15:28:00.001+07:00

About Wired LANs
Wired LANs use Ethernet cables and network adapters. Although two computers can be directly wired to each other using an Ethernet crossover cable, wired LANs generally also require central devices like hubs, switches, or routers to accommodate more computers.
For dial-up connections to the Internet, the computer hosting the modem must run Internet Connection Sharing or similar software to share the connection with all other computers on the LAN. Broadband routers allow easier sharing of cable modem or DSL Internet connections, plus they often include built-in firewall support.

Installation
Ethernet cables must be run from each computer to another computer or to the central device. It can be time-consuming and difficult to run cables under the floor or through walls, especially when computers sit in different rooms. Some newer homes are pre-wired with CAT5 cable, greatly simplifying the cabling process and minimizing unsightly cable runs.
The correct cabling configuration for a wired LAN varies depending on the mix of devices, the type of Internet connection, and whether internal or external modems are used. However, none of these options pose any more difficulty than, for example, wiring a home theater system.
After hardware installation, the remaining steps in configuring either wired or wireless LANs do not differ much. Both rely on standard Internet Protocol and network operating system configuration options. Laptops and other portable devices often enjoy greater mobility in wireless home network installations (at least for as long as their batteries allow).

Cost
Ethernet cables, hubs and switches are very inexpensive. Some connection sharing software packages, like ICS, are free; some cost a nominal fee. Broadband routers cost more, but these are optional components of a wired LAN, and their higher cost is offset by the benefit of easier installation and built-in security features.

Reliability
Ethernet cables, hubs and switches are extremely reliable, mainly because manufacturers have been continually improving Ethernet technology over several decades. Loose cables likely remain the single most common and annoying source of failure in a wired network. When installing a wired LAN or moving any of the components later, be sure to carefully check the cable connections.
Broadband routers have also suffered from some reliability problems in the past. Unlike other Ethernet gear, these products are relatively new, multi-function devices. Broadband routers have matured over the past several years and their reliability has improved greatly.

Performance
Wired LANs offer superior performance. Traditional Ethernet connections offer only 10 Mbps bandwidth, but 100 Mbps Fast Ethernet technology costs little more and is readily available. Although 100 Mbps represents a theoretical maximum performance never really achieved in practice, Fast Ethernet should be sufficient for home file sharing, gaming, and high-speed Internet access for many years into the future.
Wired LANs utilizing hubs can suffer performance slowdown if computers heavily utilize the network simultaneously. Use Ethernet switches instead of hubs to avoid this problem; a switch costs little more than a hub.

Security
For any wired LAN connected to the Internet, firewalls are the primary security consideration. Wired Ethernet hubs and switches do not support firewalls. However, firewall software products like ZoneAlarm can be installed on the computers themselves. Broadband routers offer equivalent firewall capability built into the device, configurable through its own software.

About Wireless LANs
Popular WLAN technologies all follow one of the three main Wi-Fi communication standards. The benefits of wireless networking depend on the standard employed:
• 802.11b was the first standard to be widely used in WLANs.
• The 802.11a standard is faster but more expensive than 802.11b; 802.11a is more commonly found in business networks.
• The newest standard, 802.11g, attempts to combine the best of both 802.11a and 802.11b, though it too is more a more expensive home networking option.

Installation
Wi-Fi networks can be configured in two different ways:
• "Ad hoc" mode allows wireless devices to communicate in peer-to-peer mode with each other.
• "Infrastructure" mode allows wireless devices to communicate with a central node that in turn can communicate with wired nodes on that LAN.
Most LANs require infrastructure mode to access the Internet, a local printer, or other wired services, whereas ad hoc mode supports only basic file sharing between wireless devices.
Both Wi-Fi modes require wireless network adapters, sometimes called WLAN cards. Infrastructure mode WLANs additionally require a central device called the access point. The access point must be installed in a central location where wireless radio signals can reach it with minimal interference. Although Wi-Fi signals typically reach 100 feet (30 m) or more, obstructions like walls can greatly reduce their range.

Cost
Wireless gear costs somewhat more than the equivalent wired Ethernet products. At full retail prices, wireless adapters and access points may cost three or four times as much as Ethernet cable adapters and hubs/switches, respectively. 802.11b products have dropped in price considerably with the release of 802.11g, and obviously, bargain sales can be found if shoppers are persistent.

Reliability
Wireless LANs suffer a few more reliability problems than wired LANs, though perhaps not enough to be a significant concern. 802.11b and 802.11g wireless signals are subject to interference from other home applicances including microwave ovens, cordless telephones, and garage door openers. With careful installation, the likelihood of interference can be minimized.
Wireless networking products, particularly those that implement 802.11g, are comparatively new. As with any new technology, expect it will take time for these products to mature.

Performance
Wireless LANs using 802.11b support a maximum theoretical bandwidth of 11 Mbps, roughly the same as that of old, traditional Ethernet. 802.11a and 802.11g WLANs support 54 Mbps, that is approximately one-half the bandwidth of Fast Ethernet. Furthermore, Wi-Fi performance is distance sensitive, meaning that maximum performance will degrade on computers farther away from the access point or other communication endpoint. As more wireless devices utilize the WLAN more heavily, performance degrades even further.
Overall, the performance of 802.11a and 802.11g is sufficient for home Internet connection sharing and file sharing, but generally not sufficient for home LAN gaming.
The greater mobility of wireless LANs helps offset the performance disadvantage. Mobile computers do not need to be tied to an Ethernet cable and can roam freely within the WLAN range. However, many home computers are larger desktop models, and even mobile computers must sometimes be tied to an electrical cord and outlet for power. This undermines the mobility advantage of WLANs in many homes.

Security
In theory, wireless LANs are less secure than wired LANs, because wireless communication signals travel through the air and can easily be intercepted. To prove their point, some engineers have promoted the practice of wardriving, that involves traveling through a residential area with Wi-Fi equipment scanning the airwaves for unprotected WLANs. On balance, though, the weaknesses of wireless security are more theoretical than practical. WLANs protect their data through the Wired Equivalent Privacy (WEP) encryption standard, that makes wireless communications reasonably as safe as wired ones in homes.

source : http://compnetworking.about.com/cs/homenetworking/a/homewiredless_2.htm

How to Make Dynamips as Real Router

2008-10-13T16:11:00.004+07:00

Now Do ISCW Practicals on Real Internet Connections not on Simulated LAB

This post and the Contents inside it are for those people who are preparing for CCNP ISCW and in need of practicing on Real Devices on Real Connections. However Cisco Devices are not alway that easier to purchase.

I dont have any idea maybe some of you have already done this but since no one ever brought this topic before i decided to share this valuable Piece of work with you. i believe it will be worth for I.T related people

At CCNP level we must focus on Concepts and Concept is based on how much you implement them practically. Unless you do on real devices you cannot learn well. because the simulations or books will not produce results and you may not be able to tackle the problems in your work.
Lots of people go for Simulations. But i think Simulations are not good because of 2 reasons, they never are accurate, and they lack lots of IOS commands and features. this can affect our knowledge.

Lots of people asked before practicing PPPoE and DSL Labs along with IPSEC and GRE using Dynamips. Lot of time people practice such labs using LAN interfaces because they donot have Real Routers to connect to Real Internet Connections.

This small presentation will show you how you can connect Dynamips to internet and establish a Real Internet Connection using Dialer interface. Ofcouse you must have atleast running DSL connection and any small DSL Modem that is cheaper and easier to buy, that you connect to Dynamips.
Modem will be just a Layer 1 device only doing nothing other than Physically connecting Telephone line.

You will see how Dynamips will Negotiate and Initiate Internet connection with ISP
using PPP encapsulations,
Dynamips getting Real IP on its Dialer Interface.
Realtime Debugging outputs between Router interface and ISP
Leasing Ip Addreses to Physical Computers on LAN and Across Internet to WAN clients
Accepting Incoming Connections from Internet

all does Dynamips just like a Real Router. In this video the Dynamips acts as a ADSL ROUTER, Incoming VPN, PPPoE Internet Connetion, and a DHCP Server.

if you are interested in setup then consider these videos.

http://www.4shared.com/file/62580751/6f8892ff/dynamips_videos.html
refrence :http://www.sadikhov.com/forum/index.php?showtopic=137700

Risk Management

2008-07-28T15:00:00.000+07:00

Introduction
The traditional role of the risk manager as corporate steward is evolving as organizations face an increasingly complex and uncertain future. The mandate to clearly identify, measure, manage, and control risk has been expanded and integrated into best practice management of a bank. Today's risk manager is a key member of the senior executive team who helps define business opportunities from a risk-return perspective, presents unique ways of looking at them, has direct input into the configuration of products and services, and ensures the transparency of all the risks. Innovation necessitates new yardsticks for measuring and monitoring the resulting activities. The savvy corporate leader uses risk management as both a sword and a shield.
At the end of the last millennium, financial institutions and investors experienced increased volatility in the major financial and commodity markets, with many financial crises. At the start of the new millennium, we are in the midst of a technological revolution resulting in changes in the operation of markets, increased access to information, changes in the types of services available to investors, as well as major changes in the production and distribution of financial services.
If there is concern about an institution's ability to manage risk, then its share price will be penalized.
Risk is a cost of doing business for a financial institution and consequently best practice risk management is a benefit to our shareholders. To manage the risks facing an institution we must have a clearly defined set of risk policies and the ability to measure risk. But what do we measure? And how do we measure such risks? We must also have a best practice infrastructure. The starting point is that we need a framework.
This book provides such a framework. The content of the book is consistent with our own risk management strategy and experience. Our risk management strategy is designed to ensure that our senior management operates together in partnership to control risk while ensuring the independence of the risk management function. Improvements in analytic models and systems technology have greatly facilitated our ability to measure and manage risk. However, the new millennium brings new challenges. There are risks that we can identify and measure and there is the uncertainty of the unknown. The challenge facing risk managers is to minimize the consequences of the unknown. This book should help all risk and business managers address the issues arising from risk and uncertainty.

And you will get further information, please read this book:

Risk Management
By:
Michel Crouhy
Dan Galai
Robert Mark

Mc Graw-Hill Book

Let’s Study Risk Management. Enjoy it.

Database Jargon

2008-07-28T14:58:00.000+07:00

What is a Database?
A database is simply a collection of data, stored in an organized way.
For example, it may be:
_ an address list
_ employee details
_ details about items in stock.

A simple database could be used to store the names and addresses of those you send Christmas cards to, or details of your CD collection. A more complex database could be used to store the data that you need to run your company, e.g. supplier, customer, stock and order details.

Database Jargon
If you have never used a database before, you might not have encountered some of the database terminology that we will use in this book. Here are brief definitions of the terms you will come across.

Table: All the data on one topic is stored in a table. A table could contain details of the employees that work for your company. In a simple database, you might have only one table. More complex ones may consist of several tables.

Record: The data for a single item in a table is held in that item’s record. Using the employee table example, each employee would have their own record within the employee table.

Field: A field is a piece of information within a record e.g. in your employee table, things like surname, job title, date of birth, or salary grade.

Field name: The label, or name, given to a field.

Primary key: A special field that will contain a unique piece of information in each record e.g. StaffID.

Relationship: A relationship is used to link two tables through a common field.

Join: The process of linking two tables is often referred to as joining them.

Data definition: The process of defining, or describing, the data to be stored. This will involve specifying the data type (text, number, date), the field size, and how it is related to other tables.

Data manipulation: This term describes any work done on existing data within your tables. Sorting and extracting data and producing reports from it would all be examples of data manipulation.

Enjoy it.

Analisis Data Statistik : WHAT IS ECONOMETRICS?

2008-07-28T14:57:00.002+07:00

Literally interpreted, econometrics means “economic measurement.” Although measurement is an important part of econometrics, the scope of econometrics is much broader, as can be seen from the following quotations: Econometrics, the result of a certain outlook on the role of economics, consists of the application of mathematical statistics to economic data to lend empirical support to the models constructed by mathematical economics and to obtain numerical results.

Econometrics may be defined as the quantitative analysis of actual economic phenomena based on the concurrent development of theory and observation, related by appropriate methods of inference.

Econometrics may be defined as the social science in which the tools of economic theory, mathematics, and statistical inference are applied to the analysis of economic phenomena.
Econometrics is concerned with the empirical determination of economic laws.

The art of the econometrician consists in finding the set of assumptions that are both sufficiently specific and sufficiently realistic to allow him to take the best possible advantage of the data available to him.
Econometricians are a positive help in trying to dispel the poor public image of economics (quantitative or otherwise) as a subject in which empty boxes are opened by assuming the existence of can-openers to reveal contents which any ten economists will interpret in 11 ways.
The method of econometric research aims, essentially, at a conjunction of economic theory and actual measurements, using the theory and technique of statistical inference as a bridge pier.

METHODOLOGY OF ECONOMETRICS

How do econometricians proceed in their analysis of an economic problem?
That is, what is their methodology? Although there are several schools of thought on econometric methodology, we present here the traditional or classical methodology, which still dominates mpirical research in economics and other social and behavioral sciences.
Broadly speaking, traditional econometric methodology proceeds along the following lines:

1. Statement of theory or hypothesis.
2. Specification of the mathematical model of the theory
3. Specification of the statistical, or econometric, model
4. Obtaining the data
5. Estimation of the parameters of the econometric model
6. Hypothesis testing
7. Forecasting or prediction
8. Using the model for control or policy purposes.

For further information, please read a book: Basic Econometric 4th Ed. Gujarati.(2004)

Enjoy....

SQL Commands

2008-07-28T14:56:00.001+07:00

Below, given some Common SQL Command. Items bracketed [] are optional.

For a complete list of MySQL supported commands, visit the MySQL
website at http://www.mysql.com.

ALTER
ALTER TABLE table_name ADD [COLUMN] ...;

CREATE
CREATE DATABASE database_name;
CREATE TABLE table_name;

DELETE
DELETE FROM table_name [WHERE ...];

DROP
DROP DATABASE database_name;
DROP TABLE table_name;

GRANT
GRANT privilege ON table_name ►►
TO user [IDENTIFIED BY 'password'] [WITH GRANT OPTION];

INSERT
INSERT [INTO] table_name VALUES (...);

SELECT
SELECT ... [FROM table_name(s)] ►►
[WHERE ...] [GROUP BY ... ] [ORDER BY ...];

SET
SET PASSWORD FOR user@localhost = ►►
PASSWORD("password");
SET PASSWORD FOR user@"%.visibooks.com" = ►►
PASSWORD("password");
SET PASSWORD FOR user@"%" = PASSWORD("password");

SHOW
SHOW DATABASES;
SHOW TABLES;
UPDATE

UPDATE table_name SET column_name=value [WHERE ...];

USE
USE database_name;

Let’s study SQL ! Enjoy It !

CCNA Synopsis

2008-06-05T17:12:00.001+07:00

SYNOPSIS

The Cisco Certified Network Associate (CCNA) exam is an entry-level certification designed for network administrators and engineers. It can be taken in one of two forms. First, the CCNA can be taken in one, single exam - The CCNA Composite 640-802. Second, it can be taken as a two part exam designated Interconnecting Cisco Network Devices 1 and 2 (ICND1 and ICND2).

A majority of the professional networking world consider the CCNA to be the defacto Network Engineering exam that qualifies a user to operate a small to medium business network. Furthermore, the CCNA opens doors for new exams, such as the CCNP (Cisco Certified Network Professional) and CCDP (Cisco Certified Design Professional) certifications. Therefore, the exam is not in any way easy.

Overall, the CCNA has several subjects that it covers in detail, but the primary subjects of concern are Routing and Switching, including but not limited to the use of: IP, IGRP, Serial, Frame Relay, IP RIP, VLANs, RIP, Ethernet, and access lists. After attaining the CCNA, the exam is valid for three years before it must be recertified.

(Source : cramsession.com)

CCNA FAQ - Cisco Certified Network Associate Frequently Asked Question

2008-06-05T17:10:00.002+07:00

Q » What are the different types of Academies and their responsibilities in the Cisco Certified Networking Associate (CCNA) Program?
A »There are three types of Academies: Local Academies, Regional Academies and Cisco Academy Training Centers (CATCs).

Local Academies teach the curriculum to students. These are usually high schools/secondary schools and colleges, but can also be other agencies with education programs.

Regional Academies are usually colleges, but may also be high schools/secondary schools, education service centers and other agencies with education programs. Regional Academy instructors train and mentor Local Academy instructors in the operation of the program and the teaching of the curriculum.

CATCs are Cisco-selected Regional Academies that train instructors from other Regional Academies. CATC instructors get their training directly from Cisco.

CATCs and Regional Academies can also be Local Academies if they teach the curriculum to students.

Curriculum

Q » How many hours of instruction do the CCNA and CCNP online curricula include?
A » The online portion of the curriculum contains eight 70-hour blocks of study, which total - 560 hours. High schools/secondary schools typically teach the CCNA (courses 1- 4) in two academic years, whereas colleges and universities typically use one academic year to deliver the entire CCNA curriculum. The CCNP courses are typically taught over two academic years. The CCNP curriculum is 70-hours per course, lab-based with additional out-of-class lab time of 14 hours compulsory, for a total of 84 hours.

Q » How is the online curriculum provided to the students?
A » The online portion of the curriculum resides on a Web server at each Academy. Students access the curriculum from their workstations over the classroom LAN. Students also have access to view the curriculum online during non-classroom time.

Q » Does the curriculum teach students things that will be useful for maintaining non-Cisco networks?
A » The curriculum is a vendor-neutral. While Cisco equipment is used to practice and reinforce certain concepts, students gain broad-based knowledge of the operation of networks, the differences between various types of networking products, how to design and troubleshoot networks, and other general topics.

Q » What kinds of controls are there on course materials?
A » The curriculum is copyrighted. Access to the curriculum should be provided only to students enrolled in Networking Academy classes and the personnel within the institution offering the program.

Equipment

Q » What type of equipment is needed to teach the CCNA curriculum?
A » The labs include routers, switches, software, cables and first-year product support.

Q » Does Cisco donate lab equipment to schools in the program?
A » Cisco donates refurbished lab equipment to CCNA Regional Academies and to CCNA CATCs. Cisco also donates CCNA lab equipment to Academies in officially recognized Empowerment Zones and Least Developed Countries.

Q » Are multiple Academies allowed to share lab equipment?
A » Academies may share equipment, although Cisco believes that each institution is better able to provide a quality instructional experience if it has its own lab.

Instructor Training

Q » How are Academy instructors trained?
A » In the CCNA program, Local Academies are trained by Regional Academies, Regional Academies are trained by CATCs and CATCs are trained by Cisco.

Q » Where is the training for Academy instructors held?
A » Training is typically conducted at Regional Academies and CATC training sites.

Q » How long is CCNA instructor training?
A » Training for CCNA is at least 19 days.

Certification and Assessment

Q » When should a student take the Cisco Certified Network Associate exam?
A » Cisco recommends taking the CCNA exam after completing the fourth course of the CCNA curriculum.

Q » Where is the CCNA test given?
A » The CCNA test is given at Pearson VUE testing centers worldwide.

Q » What is Cisco Certified Network Professional (CCNP) Program?
A » The CCNP program follows CCNA. The curriculum consists of CCNP 1 - 4 and prepares students to obtain their CCNP certification. This is a more advanced curriculum covering Advanced Routing (CCNP 1), Remote Access (CCNP 2), Multi-Layer Switching (CCNP 3) and Network Troubleshooting (CCNP 4).

Q » What is Cisco Certified Internetworking Expert (CCIE) Program?
A » The CCIE program has no required curriculum, although many of Cisco's commercial classes taught by Cisco's Training Partners are suggested prerequisites to the CCIE exam. The exam consists of both written and practical exams. CCIEs have world-renowned internetworking expertise and are widely thought to be the most knowledgeable networking personnel available.

Q » Are CCNA students assessed as they progress through the CCNA curriculum?
A » Yes. Cisco has developed a state-of-the-art assessment system to provide students, instructors, administrators, and Cisco personnel with detailed data on student progress.

General Program Questions

Q » What sort of commitments do Academies make when they sign up for the program?
A » CATCs and Regional Academies sign contracts that list their commitments as well as Cisco’s commitments to them. For more information on the CATC and Regional Academy contract process, please fill out an online application. The relationship between Regional Academies and Local Academies is owned by the Regional Academy, which may or may not ask its Local Academies to sign a contract.

Q » What is the process of starting an Academy?
A » A school can begin the process of starting an Academy by filling out the Cisco Networking Academy Program online application.

Q » Where are the various Networking Academies located at?
A » For a complete list of Academies see the Academy Locator.

Q » Are for-profit training centers eligible for participation in the Program?
A » Yes, Cisco has created a special program to work with proprietary institutions known as the Accredited Career College Program. For details contact netacad-acc@cisco.com or the Accredited Career College FAQs.

(Source : cna.te.ugm.ac.id)

Interconnecting Cisco Network Devices, Part 2 (ICND2)

2008-06-05T12:00:00.003+07:00

Interconnecting Cisco Network Devices, Part 2 (ICND2), is a Cisco®-authorized, self-paced learning tool for CCNA® foundation learning. This book provides you with the knowledge needed to install, operate, and troubleshoot a small to medium-size branch office enterprise network, including configuring several switches and routers, connecting to a WAN, and implementing network security.
In Interconnecting Cisco Network Devices, Part 2 (ICND2), you will study actual router and switch output to aid your understanding of how to configure these devices. Many notes, tips, and cautions are also spread throughout the book. Specific topics include constructing medium-size routed and switched networks, OSPF and EIGRP implementation, access control lists (ACL), address space management, and LAN extensions into a WAN. Chapter-ending review questions illustrate and help solidify the concepts presented in the book.

Review the Cisco IOS® Software command structure for routers and switches
Build LANs and understand how to overcome problems associated with Layer 2 switching
Evaluate the differences between link-state and distance vector routing protocols
Configure and troubleshoot OSPF in a single area
Configure and troubleshoot EIGRP
Identify and filter traffic with ACLs
Use Network Address Translation (NAT) and Port Address Translation (PAT) to conserve IPv4 address space and implement IPv6
Connect different sites over WANs or the Internet using IPsec VPN, SSL VPN, leased line, and Frame Relay connections

http://rapidshare.com/files/116675573/ICND1_2ndEdition_jan2008.7z

(Reference: noc-rikasang.blogspot.com)

Interconnecting Cisco Network Devices, Part 1 (ICND1), 2nd Ed

2008-06-05T11:55:00.003+07:00

Interconnecting Cisco Network Devices, Part 1 (ICND1), Second Edition, is a Cisco®-authorized, self-paced learning tool for CCENT™ and CCNA® foundation learning. This book provides you with the knowledge needed to configure Cisco switches and routers to operate in corporate internetworks. By reading this book, you will gain a thorough understanding of concepts and configuration procedures required to build a multiswitch, multirouter, and multigroup internetwork that uses LAN and WAN interfaces for the most commonly used routing and routed protocols.
In Interconnecting Cisco Network Devices, Part 1 (ICND1), you will study installation and configuration information that network administrators need to install and configure Cisco products. Specific topics include building a simple network, Ethernet LANs, wireless LANs (WLANs), LAN and WAN connections, and network management. Chapter-ending review questions illustrate and help solidify the concepts presented in the book.

Understand the principles on which basic networks operate
Explore the operation and configuration of LANs
Extend the boundaries of the network by implementing and securing wireless connectivity
Configure routers to provide connectivity between different networks
Learn about IP addressing number conversion
Establish WAN interconnectivity using point-to-point links, DSL, and cable services
Configure Network Address Translation (NAT)
Use Cisco IOS® commands to determine the layout of a Cisco network topology
Manage the router startup and work with IOS configuration files and Cisco IOS images
This volume is in the Certification Self-Study Series offered by Cisco Press®. Books in this series provide officially developed self-study solutions to help networking professionals understand technology implementations and prepare for the Cisco Career Certifications examinations.

http://rapidshare.com/files/116675573/ICND1_2ndEdition_jan2008.7z

(Reference: noc-rikasang.blogspot.com)

Network Fundamentals - Chapter 11 Exam Answers

2008-06-05T11:53:00.001+07:00

1
Refer to the exhibit. What command will place the router into the correct mode to configure an appropriate interface to connect to a LAN?
UBAMA# configure terminal
UBAMA(config)# line vty 0 4
UBAMA(config)# line console 0
UBAMA(config)# interface Serial 0/0/0
* UBAMA(config)# interface FastEthernet 0/12
In a Cisco router, when do changes made to the running-configuration take effect?
after a system restart
as the commands are entered
when logging off the system
* when the configuration is saved to the startup-configuration3
Refer to the exhibit. A technician applies the configuration in the exhibit to a clean router. To verify the configuration, the technician issues the show running-config command in the CLI session with the router. What lines should the technician expect to see in the router output from the show running-config command?
enable password class
line console 0
password ccna
enable secret cisco
enable password class
line console 0
password ccna
enable secret 5 $1$v0/3$QyQWmJyT7zCa/yaBRasJm0
enable password class
line console 0
password ccna
enable secret cisco
enable password 7 14141E0A1F17
line console 0
password 7 020507550A
* enable secret 5 $1$v0/3$QyQWmJyT7zCa/yaBRasJm0
enable password 7 14141E0A1F17
line console 0
password 7 020507550A4
When network services fail, which port is most often used to access a router for management purposes?
AUX
Ethernet
* Console
Telnet
SSH5
Refer to the exhibit. A network administrator on HostA has problems accessing the FTP server. Layer three connectivity testing was successful from HostA to the S1 interface of RouterB. Which set of commands will allow the network administrator to telnet to RouterB and run debug commands?
RouterB(config)# enable secret class
RouterB(config)# line vty 0 4
RouterB(config-if)# login
RouterB(config)# enable secret class
RouterB(config)# line vty 0 2
RouterB(config-vty)# password cisco
RouterB(config-vty)# login
* RouterB(config)# enable secret class
RouterB(config)# line vty 0
RouterB(config-line)# password cisco
RouterB(config-line)# login
RouterB(config)# enable secret class
RouterB(config)# line aux 0
RouterB(config-line)# password cisco
RouterB(config-line)# login
RouterB(config)# enable secret class
RouterB(config)# line aux 0
RouterB(config-vty)# password cisco
RouterB(config-vty)# login6
Users in the network are experiencing slow response time when doing file transfers to a remote server. What command could be issued to determine if the router has experienced any input or output errors?
show running-config
show startup-config
* show interfaces
show ip route
show version
show memory7
Refer to the exhibit. Which names correctly identify the CLI mode represented by the prompt for Switch-East4#? (Choose two.)
line configuration mode
user executive mode
* global configuration mode
privileged executive mode
interface configuration mode
* enable mode8
What command is used to change the default router name to Fontana?
Router# name Fontana
Router# hostname Fontana
Router(config)# name Fontana
* Router(config)# hostname Fontana

9
The serial connection shown in the graphic needs to be configured. Which configuration commands must be made on the Sydney router to establish connectivity with the Melbourne site? (Choose three.)
* Sydney(config-if)# ip address 201.100.53.2 255.255.255.0
* Sydney(config-if)# no shutdown
Sydney(config-if)# ip address 201.100.53.1 255.255.255.224
* Sydney(config-if)# clock rate 56000
Sydney(config-if)# ip host Melbourne 201.100.53.2

10
Refer to the exhibit. The output is shown for the show ip route command executed on Router A. What does the IP address 192.168.2.2 represent?
Gateway for the 192.168.1.0 network
Gateway for the 192.168.3.0 network
IP assigned to the serial port on Router A
* IP assigned to the serial port on Router B

11
Refer to the exhibit. What additional command is required to allow remote access to this switch?
NA-SW1(config-if)# no shutdown
NA-SW1(config)# enable password password
* NA-SW1(config)# ip default-gateway address
NA-SW1(config-if)# description description

12
Immediately after a router completes its boot sequence, the network administrator wants to check the routers configuration. From privileged EXEC mode, which of the following commands can the administrator use for this purpose? (Choose two.)
show flash
show NVRAM
* show startup-config
* show running-config
show version

13
Which three terms correctly define the forms of help available within the Cisco IOS? (Choose three.)
hot keys
* context-check
context-sensitive
* structured check
command override
* command syntax check

14
Refer to the exhibit. A student is responsible for the IP addressing, configuration and connectivity testing of the network shown in the graphic. A ping from host B to host C results in a destination unreachable but a ping from host B to host A was successful. What two reasons could account for this failure based on the graphic and partial router output for the Dallas router? (Choose two.)
The host A is turned off.
The Fa0/0 interface on Dallas is shutdown.
The LAN cable is disconnected from host B.
* The S0/0/1 IP address of Dallas is improperly configured.
* The Fa0/0 interface on Dallas is in a different subnet than host B.
The clock rate is missing on the serial link between Dallas and NYC.

15
Which combination of keys would be used at the CLI prompt to interrupt a ping or traceroute process?
* Ctrl-C
Ctrl-P
Ctrl-R
Ctrl-Shift-6
Ctrl-Z

16
The connection between routers B and C has been successfully tested. However, after rebooting router C, the administrator noticed the response time between networks 10.10.3.0 and 10.10.4.0 is slower. Ping between the two routers is successful. A trace route indicates three hops from router B to router C. What else can be done to troubleshoot the problem?
Ping router B S0/1 connection from router C.
Trace the connection between router B to router C S0/1.
Issue a show ip route command in router B to verify routing is enabled.
* Issue a show interfaces command on router C.

17
In a Cisco device, where is the IOS file stored prior to system startup?
RAM
ROM
* Flash
NVRAM

18
On a Cisco router, which interface would be used to make the initial configuration?
toinen

19
A network administrator needs to keep the user ID, password, and session contents private when establishing remote CLI connectivity with a router to manage it. Which access method should be chosen?
Telnet
Console
AUX
* SSH
(Reference: noc-rikasang.blogspot.com)

Network Fundamentals - Chapter 10 Exam Answers

2008-06-05T11:50:00.002+07:00

1
Refer to the exhibit. A student working in the lab selects a cable that is wired as shown. Which connection types can successfully be made with this cable? (Choose two.)
*connecting a PC to a router's console port
connecting two routers together via their fast ethernet ports
connecting two switches together at gigabit speeds
*connecting a PC to a switch at Gigabit Ethernet speeds
connecting two devices with the same interface type at Fast Ethernet speeds2
Which three statements are true about the role of routers in the network? (Choose three.)
They propagate broadcasts.
They enlarge collision domains.
They segment broadcast domains.
* They interconnect different network technologies.
* Each router interface requires a separate network or subnet.
* They maintain a common frame format between LAN and WAN interfaces.3
Refer to the exhibit. A network administrator has decided to use packet capture software to evaluate all traffic from the student subnet on the way to the Internet. To ensure that all packets are captured, what network device should be used to connect the monitoring station to the network between R1 and R2?
* router
hub
switch
wireless access point4
A network administrator is required to use media in the network that can run up to 100 meters in cable length without using repeaters. The chosen media must be inexpensive and easily installed. The installation will be in a pre-existing building with limited cabling space. Which type of media would best meet these requirements?
STP
* UTP
coaxial
single-mode fiber
multimode fiber

5
Refer to the exhibit. Which three statements are true about the exhibited topology? (Choose three.)
Hosts B and C are in the same subnet.
* Five broadcast domains are present.
* Host B is using a crossover cable to connect to the router.
Four broadcast domains are present.
*Five networks are shown.
Host B is using a rollover cable to connect to the router.

6
Which three types of connectors are commonly associated with Cisco's V.35 serial cables? (Choose three.)
RJ 11
* DB 60
* Winchester 15 pin
DB 9
* smart serial
RJ 45

7
How many host addresses may be assigned on each subnetwork when using the 130.68.0.0 network address with a subnet mask of 255.255.248.0?
30
256
* 2046
2048
4094
4096

8
Refer to the exhibit. Based on the IP configuration shown, what would be the result of Host A and Host B attempting to communicate outside the network segment?
both host A and B would be successful
* host A would be successful, host B would fail
host B would be successful, host A would fail
both Host A and B would fail

9
Refer to the exhibit. What destination IP address will PC1 place in the header for a packet destined for PC2?
192.168.1.1
192.168.2.1
192.168.2.2
192.168.3.1
* 192.168.3.54

10
Refer to the exhibit. Which three statements identify the type of cabling that would be used in the segments that are shown? (Choose three.)
Segment 1 uses backbone cabling.
* Segment 1 uses a patch cable.
Segment 3 uses patch cabling.
Segment 3 uses vertical cabling.
* Segment 3 uses horizontal cabling.
* Segment 4 uses vertical cabling.

11
A company is planning to subnet its network for a maximum of 27 hosts. Which subnet mask would provide the needed hosts and leave the fewest unused addresses in each subnet?
255.255.255.0
255.255.255.192
* 255.255.255.224
255.255.255.240
255.255.255.248

12
Refer to the exhibit. A student is setting up a home network primarily used for extensive file transfers, streaming video, and gaming. Which network device is best suited to these types of activities in the topology shown?
wireless access point
router
hub
* switch

13
Refer to the exhibit. How many subnets are required to support the network that is shown?
2
* 3
4
5

14
What does the term "attenuation" mean in data communication?
* loss of signal strength as distance increases
time for a signal to reach its destination
leakage of signals from one cable pair to another
strengthening of a signal by a networking device

15
To establish a console connection from a computer to a Cisco router, which cable wiring option would be used?
* crossover cable
straight through cable
rollover cable
V.35 cable

16
What makes fiber preferable to copper cabling for interconnecting buildings? (Choose three.)
* greater distances per cable run
lower installation cost
*limited susceptibility to EMI/RFI
durable connections
*greater bandwidth potential
easily terminated

17
Refer to the exhibit. A network technician has been allocated the 192.168.1.0/24 private IP address range for use in the network that shown in the exhibit. Which subnet mask would be used to meet the host requirements for segment A of this internetwork?
* 255.255.255.128
255.255.255.192
255.255.255.224
255.255.255.240
255.255.255.248

18
Which subnet mask would be assigned to the network address of 192.168.32.0 to provide 254 useable host addresses per subnetwork?
255.255.0.0
* 255.255.255.0
255.255.254.0
255.255.248.0

19
What type of network cable is used between a terminal and a console port?
cross-over
straight-through
* roll-over
patch cable

20
Which option identifies the primary interface which would be used for initial configuration of a Cisco router?
AUX interface
Ethernet interface
serial interface
* console interface

21
What are three common methods for setting a UTP Ethernet port to MDI or MDIX operation? (Choose three.)
direct configuration of the device
* cable color code association
cable selection and configuration
use of cable testers to determine pinouts
* the enabling of the mechanism to electrically swap the transmit and receive pairs
* the automatic detection and negotiating of MDI/MDIX operation of the port

22
Which device is used to create or divide broadcast domains?
hub
switch
bridge
* router
repeater

23
Refer to the exhibit. Assume that all devices are using default settings. How many subnets are required to address the topology that is shown?
1
3
4
* 5
7

Network Fundamentals - Chapter 9 Exam Answers

2008-06-05T11:42:00.007+07:00

1

In the graphic, Host A has reached 50% completion in sending a 1 KB Ethernet frame to Host D when Host B wishes to transmit its own frame to Host C. What must Host B do?
Host B can transmit immediately since it is connected on its own cable segment.
Host B must wait to receive a CSMA transmission from the hub, to signal its turn.
Host B must send a request signal to Host A by transmitting an interframe gap.
* Host B must wait until it is certain that Host A has completed sending its frame.
2 Ethernet operates at which layers of the OSI model? (Choose two.)
Network layer
Transport layer
* Physical layer
Application layer
Session layer
* Data-link layer
3 Which of the following describe interframe spacing? (Choose two.)
* the minimum interval, measured in bit-times, that any station must wait before sending another frame
the maximum interval, measured in bit-times, that any station must wait before sending another frame
the 96-bit payload padding inserted into a frame to achieve a legal frame size
the 96-bit frame padding transmitted between frames to achieve proper synchronization
* the time allowed for slow stations to process a frame and prepare for the next frame
the maximum interval within which a station must send another frame to avoid being considered unreachable
4 What three primary functions does data link layer encapsulation provide? (Choose three.)
* addressing
* error detection
* frame delimiting
port identification
path determination
IP address resolution
5 When a collision occurs in a network using CSMA/CD, how do hosts with data to transmit respond after the backoff period has expired?
* The hosts return to a listen-before-transmit mode.
The hosts creating the collision have priority to send data.
The hosts creating the collision retransmit the last 16 frames.
The hosts extend their delay period to allow for rapid transmission.
6 What are three functions of the upper data link sublayer in the OSI model? (Choose three.)
recognizes streams of bits
* identifies the network layer protocol
* makes the connection with the upper layers identifies the source and destination applications
* insulates network layer protocols from changes in physical equipment
determines the source of a transmission when multiple devices are transmitting
7 What does the IEEE 802.2 standard represent in Ethernet technologies?
MAC sublayer
Physical layer
* Logical Link
Control sublayer
Network layer
8 Why do hosts on an Ethernet segment that experience a collision use a random delay before attempting to transmit a frame?
A random delay is used to ensure a collision-free link.
A random delay value for each device is assigned by the manufacturer.
A standard delay value could not be agreed upon among networking device vendors.
* A random delay helps prevent the stations from experiencing another collision during the transmission.
9 Refer to the exhibit. Which option correctly matches the frame field type with the contents that frame field includes?
* header field - preamble and stop frame
data field - network layer packet
data field - physical addressing
trailer field - FCS and SoF
10 Host A has an IP address of 172.16.225.93 and a mask of 255.255.248.0. Host A needs to communicate with a new host whose IP is 172.16.231.78. Host A performs the ANDing operation on the destination address. What two things will occur? (Choose two.)
Host A will change the destination IP to the IP of the nearest router and forward the packet.
Host A will broadcast an ARP request for the MAC of its default gateway.
A result of 172.16.225.0 will be obtained.
* Host A will broadcast an ARP request for the MAC of the destination host.
A result of 172.16.224.0 will be obtained.
* A result of 172.16.225.255 will be obtained.
11 Which of the following is a drawback of the CSMA/CD access method?
* Collisions can decrease network performance.
It is more complex than non-deterministic protocols.
Deterministic media access protocols slow network performance.
CSMA/CD LAN technologies are only available at slower speeds than other LAN technologies.
12 Ethernet operates at which layer of the TCP/IP network model?
application
physical
transport
internet
data link
* network access
13 What is the primary purpose of ARP?
translate URLs to IP addresses
* resolve IPv4 addresses to MAC addresses
provide dynamic IP configuration to network devices
convert internal private addresses to external public addresses
14

Refer to the exhibit. The switch and workstation are administratively configured for full-duplex operation. Which statement accurately reflects the operation of this link?
* No collisions will occur on this link.
Only one of the devices can transmit at a time.
The switch will have priority for transmitting data.
The devices will default back to half duplex if excessive collisions occur.
15

Refer to the exhibit. Host_A is attempting to contact Server_B. Which statements correctly describe the addressing Host_A will generate in the process? (Choose two.)
A packet with the destination IP of Router_B.
A frame with the destination MAC address of Switch_A.
A packet with the destination IP of Router_A.
* A frame with the destination MAC address of Router_A.
* A packet with the destination IP of Server_B.
A frame with the destination MAC address of Server_B.
16 Which statements correctly describe MAC addresses? (Choose three.)
dynamically assigned
* copied into RAM during system startup
layer 3 address
* contains a 3 byte OUI
* 6 bytes long
32 bits long
17 Which two features make switches preferable to hubs in Ethernet-based networks? (Choose two.)
reduction in cross-talk
* minimizing of collisions
support for UTP cabling
* division into broadcast domains
increase in the throughput of communications
18 What are the two most commonly used media types in Ethernet networks today? (Choose two.)
* coaxial thicknet
copper UTP
coaxial thinnet
optical fiber
* shielded twisted pair
19 Convert the binary number 10111010 into its hexadecimal equivalent. Select the correct answer from the list below.
85
90
* BA
A1
B3
1C
20 After an Ethernet collision, when the backoff algorithm is invoked, which device has priority to transmit data?
the device involved in the collision with the lowest MAC address
the device involved in the collision with the lowest IP address
* any device in the collision domain whose backoff timer expires first
those that began transmitting at the same time

Refer to the exhibit. What command was executed on a host computer to produce the results shown?
route PRINT
* arp -a
arp -d
netstat
telnet
(Reference: noc-rikasang.blogspot.com)

Network Fundamentals - Chapter 8 Exam Answers

2008-06-05T11:29:00.006+07:00

1 Which type of cable run is most often associated with fiber-optic cable?
* backbone cable
horizontal cable
patch cable
work area cable2
In most business LANs, which connector is used with twisted-pair networking cable?
BNC
RJ-11
* RJ-45
Type F3
When is a straight-through cable used in a network?
when connecting a router through the console port
when connecting one switch to another switch
* when connecting a host to a switch
when connecting a router to another router4
With the use of unshielded twisted-pair copper wire in a network, what causes crosstalk within the cable pairs?
* the magnetic field around the adjacent pairs of wire
the use of braided wire to shield the adjacent wire pairs
the reflection of the electrical wave back from the far end of the cable
the collision caused by two nodes trying to use the media simultaneously
5
Which OSI layer is responsible for binary transmission, cable specification, and physical aspects of network communication?
Presentation
Transport
Data Link
* Physical
6
An installed fiber run can be checked for faults, integrity, and the performance of the media by using what device?
light injector
* OTDR
TDR
multimeter
7
XYZ Company is installing new cable runs on their data network. What two types of cable would most commonly be used for new runs? (Choose two.)
* coax
Cat4 UTP
Cat5 UTP
* Cat6 UTP
STP
8
What is a possible effect of improperly applying a connector to a network cable?
Data will be forwarded to the wrong node.
* Data transmitted through that cable may experience signal loss.
An improper signaling method will be implemented for data transmitted on that cable.
The encoding method for data sent on that cable will change to compensate for the improper connection.
9
What are three measures of data transfer? (Choose three.)
goodput
* frequency
amplitude
* throughput
crosstalk
**bandwidth
10
Refer to the exhibit. Which type of Category 5 cable is used to make an Ethernet connection between Host A and Host B?
coax cable
rollover cable
* crossover cable
straight-through cable

11
In LAN installations where potential electrical hazards or electromagnetic interference may be present, what type of media is recommended for backbone cabling?
coax
* fiber
Cat5e UTP
Cat6 UTP
STP
12
Which fiber connector supports full duplex Ethernet?

*

13
Which method of signal transmission uses radio waves to carry signals?
electrical
optical
* wireless
acoustic
14
Which characteristics describe fiber optic cable? (Choose two.)
* It is not affected by EMI or RFI.
Each pair of cables is wrapped in metallic foil.
It combines the technique of cancellation, shielding and twisting to protect data.
It has a maximum speed of 100 Mbps.
* It is the most expensive type of LAN cabling.
15
Which of the following is a characteristic of single-mode fiber-optic cable?
* generally uses LEDs as the light source
relatively larger core with multiple light paths
less expensive than multimode
generally uses lasers as the light source
16
What is considered a benefit of wireless as a media choice?
* more host mobility
lower security risks
reduced susceptibility to interference
less impact of the surroundings on the effective coverage area
17
What characteristic of UTP cable helps reduce the effects of interference?
the metal braiding in the shielding
the reflective cladding around core
* the twisting of the wires in the cable
the insulating material in the outer jacket
18
What is a primary role of the Physical layer in transmitting data on the network?
* create the signals that represent the bits in each frame on to the media
provide physical addressing to the devices
determine the path packets take through the network
control data access to the media

(Reference: noc-rikasang.blogspot.com)

Network Fundamentals - Chapter 7 Exam Answers

2008-06-05T11:27:00.001+07:00

1
Which options are properties of contention-based media access for a shared media? (Choose three.)
* non-deterministic
* less overhead
one station transmits at a time
* collisions exist
devices must wait their turn
token passing2
What is a primary purpose of encapsulating packets into frames?
provide routes across the internetwork
format the data for presentation to the user
* facilitate the entry and exit of data on media
identify the services to which transported data is associated3
Refer to the exhibit. How many unique CRC calculations will take place as traffic routes from the PC to the laptop?
1
* 2
4
84
What is true concerning physical and logical topologies?
The logical topology is always the same as the physical topology.
Physical topologies are concerned with how a network transfers frames.
Physical signal paths are defined by Data Link layer protocols.
* Logical topologies consist of virtual connections between nodes.5

Refer to the exhibit. Assuming that the network in the exhibit is converged meaning the routing tables and ARP tables are complete, which MAC address will Host A place in the destination address field of Ethernet frames destined for www.server?
00-1c-41-ab-c0-00
* 00-0c-85-cf-65-c0
00-0c-85-cf-65-c1
00-12-3f-32-05-af6
What are three characteristics of valid Ethernet Layer 2 addresses? (Choose three.)
* They are 48 binary bits in length.
They are considered physical addresses.
* They are generally represented in hexadecimal format.
They consist of four eight-bit octets of binary numbers.
They are used to determine the data path through the network.
* They must be changed when an Ethernet device is added or moved within the network.

7

Refer to the exhibit. A frame is being sent from the PC to the laptop. Which source MAC and IP addresses will be included in the frame as it leaves RouterB? (Choose two.)
source MAC - PC
source MAC - S0/0 on RouterA
* source MAC - Fa0/1 on RouterB
* source IP - PC
source IP - S0/0 on RouterA
source IP - Fa0/1 of RouterB

8
What determines the method of media access control? (Choose two.)
network layer addressing
* media sharing
application processes
* logical topology
intermediary device function

9
What is the purpose of the preamble in an Ethernet frame?
is used as a pad for data
identifies the source address
identifies the destination address
marks the end of timing information
* is used for timing synchronization with alternating patterns of ones and zeros

10
What statements are true regarding addresses found at each layer of the OSI model? (Choose two.)
* Layer 2 may identify devices by a physical address burned into the network card
Layer 2 identifies the applications that are communicating
* Layer 3 represents a hierarchical addressing scheme
Layer 4 directs communication to the proper destination network
Layer 4 addresses are used by intermediary devices to forward data

11
Which statements describe the logical token-passing topology? (Choose two.)
Network usage is on a first come, first serve basis.
* Computers are allowed to transmit data only when they possess a token.
Data from a host is received by all other hosts.
* Electronic tokens are passed sequentially to each other.
Token passing networks have problems with high collision rates.

12
Which sublayer of the data link layer prepares a signal to be transmitted at the physical layer?
LLC
* MAC
HDLC
NIC

13
What is true regarding media access control? (Choose three.)
* Ethernet utilizes CSMA/CD
* defined as placement of data frames on the media
contention-based access is also known as deterministic
802.11 utilizes CSMA/CD
* Data Link layer protocols define the rules for access to different media
controlled access contains data collisions

14
What is a characteristic of a logical point-to-point topology?
The nodes are physically connected.
The physical arrangement of the nodes is restricted.
* The media access control protocol can be very simple.
The data link layer protocol used over the link requires a large frame header.

15
What two facts are true when a device is moved from one network or subnet to another? (Choose two.)
The Layer 2 address must be reassigned.
The default gateway address should not be changed.
* The device will still operate at the same Layer 2 address.
Applications and services will need additional port numbers assigned.
* The Layer 3 address must be reassigned to allow communications to the new network.

16
What is a function of the data link layer?
provides the formatting of data
provides end-to-end delivery of data between hosts
provides delivery of data between two applications
* provides for the exchange data over a common local media

17
Which three factors should be considered when implementing a Layer 2 protocol in a network? (Choose three.)
the Layer 3 protocol selected
* the geographic scope of the network
the PDU defined by the transport layer
* the physical layer implementation
* the number of hosts to be interconnected

18
What is the primary purpose of the trailer in a data link layer frame?
define the logical topology
provide media access control
* support frame error detection
carry routing information for the frame

19
A network administrator has been asked to provide a graphic representation of exactly where the company network wiring and equipment are located in the building. What is this type of drawing?
logical topology
* physical topology
cable path
wiring grid
access topology

20

Refer to the exhibit. Which statement describes the media access control methods that are used by the networks in the exhibit?
All three networks use CSMA/CA
None of the networks require media access control.
* Network 1 uses CSMA/CD and Network 3 uses CSMA/CA.
Network 1 uses CSMA/CA and Network 2 uses CSMA/CD.
Network 2 uses CSMA/CA and Network 3 uses CSMA/CD.

(Reference: noc-rikasang.blogspot.com)

Network Fundamentals - Chapter 6 Exam Answers

2008-06-05T11:25:00.001+07:00

1
How many bits make up an IPv4 address?128
64
48
* 32
2
Refer to the exhibit. A network administrator has to develop an IP addressing scheme that uses the 192.168.1.0 /24 address space. The network that contains the serial link has already been addressed out of a separate range. Each network will be allocated the same number of host addresses. Which network mask will be appropriate to address the remaining networks?255.255.255.248
255.255.255.224
* 255.255.255.192
255.255.255.240
255.255.255.128
255.255.255.2523
Refer to the exhibit. The network administrator has assigned the internetwork of LBMISS an address range of 192.168.10.0. This address range has been subnetted using a /29 mask. In order to accommodate a new building, the technician has decided to use the fifth subnet for configuring the new network (subnet zero is the first subnet). By company policies, the router interface is always assigned the first usable host address and the workgroup server is given the last usable host address. Which configuration should be entered into the workgroup server's properties to allow connectivity to the network?

IP address: 192.168.10.38 subnet mask: 255.255.255.240 default gateway: 192.168.10.39
IP address: 192.168.10.38 subnet mask: 255.255.255.240 default gateway: 192.168.10.33
* IP address: 192.168.10.38 subnet mask: 255.255.255.248 default gateway: 192.168.10.33
IP address: 192.168.10.39 subnet mask: 255.255.255.248 default gateway: 192.168.10.31
IP address: 192.168.10.254 subnet mask: 255.255.255.0 default gateway: 192.168.10.1

4
Refer to the exhibit. Which network prefix will work with the IP addressing scheme shown in the graphic.

/24
/16
/20
* /27
/25
/28

5
Refer to the exhibit. A network administrator discovers that host A is having trouble with Internet connectivity, but the server farm has full connectivity. In addition, host A has full connectivity to the server farm. What is a possible cause of this problem?

The router has an incorrect gateway.
Host A has an overlapping network address.
Host A has an incorrect default gateway configured.
Host A has an incorrect subnet mask.
* NAT is required for the host A network.

6
What subnet mask would a network administrator assign to a network address of 172.30.1.0 if it were possible to have up to 254 hosts?

255.255.0.0
* 255.255.255.0
255.255.254.0
255.255.248.0

7
Given the IP address and subnet mask of 172.16.134.64 255.255.255.224, which of the following would describe this address?

This is a useable host address.
This is a broadcast address.
* This is a network address.
This is not a valid address.

8
What three facts are true about the network portion of an IPv4 address? (Choose three.)

identifies an individual device
is identical for all hosts in a broadcast domain
* is altered as packet is forwarded
* varies in length
* is used to forward packets
uses flat addressing

9
What is the network address of the host 172.25.67.99 /23 in binary?

10101100. 00011001.01000011.00000000
10101100. 00011001.01000011.11111111
* 10101100. 00011001.01000010.00000000
10101100. 00011001.01000010.01100011
10101100. 00010001.01000011. 01100010
10101100. 00011001.00000000.00000000

10
A router interface has been assigned an IP address of 172.16.192.166 with a mask of 255.255.255.248. To which subnet does the IP address belong?

172.16.0.0
172.16.192.0
172.16.192.128
* 172.16.192.160
172.16.192.168
172.16.192.176

11
Refer to the exhibit. Why would the response shown be displayed after issuing the command ping 127.0.0.1 on a PC?

* The IP settings are not properly configured on the host.
Internet Protocol is not properly installed on the host.
There is a problem at the physical or data link layer.
The default gateway device is not operating.
A router on the path to the destination host has gone down.

12
What two things will happen if a router receives an ICMP packet which has a TTL value of 1 and the destination host is several hops away? (Choose two.)

The router will discard the packet.
The router will decrement the TTL value and forward the packet to the next router on the path to the destination host.
* The router will send a time exceeded message to the source host.
The router will increment the TTL value and forward the packet to the next router on the path to the destination host.
* The router will send an ICMP Redirect Message to the source host.

13
Refer to the exhibit. An administrator is designing security rules for the 192.168.22.0 /24 network. Each security rule requires an address and mask pair to represent the machines that will be affected by the rule. Which two address and mask pairs will be the most precise way to reference the servers and hosts separately. (Choose two.)

192.168.22.40 255.255.255.248
192.168.22.40 255.255.255.252
192.168.22.100 255.255.255.128
* 192.168.22.100 255.255.255.254
* 192.168.22.1 255.255.255.192
192.168.22.101 255.255.255.224

14
Which of the following are features of IPv6? (Choose three.)

* larger address space
faster routing protocols
* data types and classes of service
* authentication and encryption
improved host naming conventions
same addressing scheme as IPv4

15
Which process do routers use to determine the subnet network address based upon a given IP address and subnet mask?

binary adding
hexadecimal anding
binary division
binary multiplication
* binary ANDing

16
What is the primary reason for development of IPv6?

security
header format simplification
* expanded addressing capabilities
addressing simplification

17
Which three IP addresses are private? (Choose three.)

172.168.33.1
* 10.35.66.70
* 192.168.99.5
* 172.18.88.90
192.33.55.89
172.35.16.5

18
Which statements are true regarding IP addressing? (Choose two.)

NAT translates public addresses to private addresses destined for the Internet.
Only one company is allowed to use a specific private network address space.
* Private addresses are blocked from public Internet by router.
Network 172.32.0.0 is part of the private address space.
* IP address 127.0.0.1 can be used for a host to direct traffic to itself.

19
Which IPv4 subnetted addresses represent valid host addresses? (Choose three.)

172.16.4.127 /26
* 172.16.4.155 /26
* 172.16.4.193 /26
172.16.4.95 /27
172.16.4.159 /27
* 172.16.4.207 /27

20
What is a group of hosts called that have identical bit patterns in the high order bits of their addresses?

an internet
* a network
an octet
* a radi

21
Refer to the exhibit. Host A is connected to the LAN, but it cannot get access to any resources on the Internet. The configuration of the host is shown in the exhibit. What could be the cause of the problem?

The host subnet mask is incorrect.
The default gateway is a network address.
The default gateway is a broadcast address.
* The default gateway is on a different subnet from the host.

22
Which of the following network devices are recommended to be assigned static IP addresses? (Choose three.)

LAN workstations
* servers
* network printers
* routers
remote workstations
laptops

(Reference: noc-rikasang.blogspot.com)

Network Fundamentals - Chapter 5 Exam Answers

2008-06-05T11:24:00.001+07:00

1
In an IPv4 environment, what information is used by the router to forward data packets from one interface of a router to another?
* destination network address
source network address
source MAC address
well known port destination address

2
What information is added during encapsulation at OSI Layer 3?
source and destination MAC
source and destination application protocol
source and destination port number
*source and destination IP address

3
In a connectionless system, which of the following is correct?
The destination is contacted before a packet is sent.
* The destination is not contacted before a packet is sent.
The destination sends an acknowledgement to the source that indicates the packet was received.
The destination sends an acknowledgement to the source that requests the next packet to be sent.

4
Which IP packet field will prevent endless loops?
type-of-service
identification
flags
* time-to-live
header checksum

5
Which portion of the network layer address does a router use to forward packets?
host portion
broadcast address
* network portion
gateway address

6

Refer to the exhibit. Using the network in the exhibit, what would be the default gateway address for host A in the 192.133.219.0 network?
192.135.250.1
192.31.7.1
192.133.219.0
* 192.133.219.1

7
If the default gateway is configured incorrectly on the host, what is the impact on communications?
The host is unable to communicate on the local network.
* The host can communicate with other hosts on the local network, but is unable to communicate with hosts on remote networks.
The host can communicate with other hosts on remote networks, but is unable to communicate with hosts on the local network.
There is no impact on communications.

8
What is the purpose of a default gateway?
physically connects a computer to a network
provides a permanent address to a computer
identifies the network to which a computer is connected
identifies the logical address of a networked computer and uniquely identifies it to the rest of the network
* identifies the device that allows local network computers to communicate with devices on other networks

9
What type of routing uses information that is manually entered into the routing table?
dynamic
interior
* static
standard

10
When the destination network is not listed in the routing table of a Cisco router, what are two possible actions that the router might take? (Choose two.)
* The router sends an ARP request to determine the required next hop address.
The router discards the packet.
The router forwards the packet toward the next hop indicated in the ARP table.
*The router forwards the packet to the interface indicated by the source address.
The router forwards the packet out the interface indicated by the default route entry.

11 What are the key factors to consider when grouping hosts into a common network? (Choose three.)
* gateways
purpose
physical addressing
software version
* geographic location
* ownership

12
What is a component of a routing table entry?
the MAC address of the interface of the router
the destination Layer 4 port number
the destination host address
*the next-hop address

13
Which intermediary devices could be used to implement security between networks? (Choose two.)
* router
hub
switch
* firewall
access point
bridge

14
What are three common problems with a large network? (Choose three.)
too few broadcasts
*performance degradation
* security issues
* limited management responsibility
host identification
protocol compatibility

15
Refer to the exhibit. All devices shown in the exhibit have factory default settings. How many broadcast domains are represented in the topology that is shown?
3
* 4
5
7
8
11

16
Which three statements are true about routes and their use? (Choose three.)
If no route to the destination network is found, the packet is returned to the previous router.
If the destination network is directly connected, the router forwards the packet to the destination host.
* If multiple network entries exist for the destination network, the most general route is used to forward the packet.
* If no route exists for the destination network and a default route is present, the packet is forwarded to the next-hop router.
* If the originating host has a default gateway configured, the packet for a remote network can be forwarded using that route.
If a host does not have a route manually configured for the destination network, the host will drop the packet.

17
Refer to the exhibit. A network administrator is troubleshooting a connectivity problem and needs to determine the address that is used to forward network packets out the network. Using the netstat -r command, the administrator would identify which address as the address to which all hosts send packets that are destined for an outside network?
10.10.10.26
127.0.0.1
* 10.10.10.6
10.10.10.1
224.0.0.0

18
Refer to the exhibit. A network administrator notices that there are too many broadcasts on the network. What two steps can the network administrator take to resolve this problem? (Choose two.)
* Replace S2 with a router.
Place all servers on S1.
Disable TCP/IP broadcasts.
* Subnet the 192.168.0.0 /24 network.
Disable all unused interfaces on the switches.

19
Refer to the exhibit. The network in the exhibit is fully operational. What two statements correctly describe the routing for the topology that is shown? (Choose two.)
* 192.168.0.2 is the next-hop address that is used by R3 to route a packet from the 10.0.0.0 network to the 172.16.0.0 network.
10.0.0.1 is the next-hop address that is used by R1 to route a packet from the 192.168.12.0 network to the 10.0.0.0 network.
192.168.0.1 is the next-hop address that is used by R1 to route a packet from the 192.168.12.0 network to the 172.16.0.0 network.
172.16.0.1 is the next-hop address that is used by R3 to route a packet from the 10.0.0.0 to the 172.16.0.0 network.
* 192.168.0.1 is the next-hop address that is used by R2 to route a packet from the 172.16.0.0 network to the 192.168.12.0 network.
192.168.0.2 is the next-hop address that is used by R2 to route a packet from the 172.16.0.0 network to the 192.168.12.0 network.

20
What two characteristics are commonly associated with dynamic routing protocols? (Choose two.)
require no device configuration
* provide routers with up-to-date routing tables
require less processing power than static routes require
consume bandwidth to exchange route information
* prevent manual configuration and maintenance of the routing table

21
What statement describes the purpose of a default route?
A host uses a default route to transfer data to another host on the same network segment.
A host uses a default route to forward data to the local switch as the next hop to all destinations.
A host uses a default route to identify the Layer 2 address of an end device on the local network.
* A host uses a default route to transfer data to a host outside the local network when no other route to the destination exists.

(Reference: noc-rikasang.blogspot.com)

Network Fundamentals - Chapter 4 Exam Answers

2008-06-05T11:20:00.002+07:00

1 Refer to the exhibit. What two pieces of information can be determined from the output that is shown? (Choose two.)
The local host is using well-known port numbers to identify the source ports.
* A termination request has been sent to 192.135.250.10.
Communication with 64.100.173.42 is using HTTP Secure.
* The local computer is accepting HTTP requests.
192.168.1.101:1042 is performing the three-way handshake with 128.107.229.50:80.

2 After a web browser makes a request to a web server that is listening to the standard port, what will be the source port number in the TCP header of the response from the server?
13
53
* 80
1024
1728

3 Which information is found in both the TCP and UDP header information?
sequencing
flow control
acknowledgments
*source and destination

4 Which is an important characteristic of UDP?
acknowledgement of data delivery
* minimal delays in data delivery
high reliability of data delivery
same order data delivery

5 Refer to the exhibit. Host A is using FTP to download a large file from Server 1. During the download process, Server 1 does not receive an acknowledgment from Host A for several bytes of transferred data. What action will Server 1 take as a result?
create a Layer 1 jam signal
* reach a timeout and resend the data that needs to be acknowledged
send a RESET bit to the host
change the window size in the Layer 4 header

6 Refer to the exhibit. In line 7 of this Wireshark capture, what TCP operation is being performed?
* session establishment
segment retransmit
data transfer
session disconnect

7 Refer to the exhibit. The initial TCP exchange of data between two hosts is shown in the exhibit. Assuming an initial sequence number of 0, what sequence number will be included in Acknowledgment 2 if Segment 6 is lost?
2
3
6
* 1850
3431
3475

8 Why is flow control used for TCP data transfer?
to synchronize equipment speed for sent data
to synchronize and order sequence numbers so data is sent in complete numerical order
* to prevent the receiver from being overwhelmed by incoming data
to synchronize window size on the server
to simplify data transfer to multiple hosts

9 During a TCP communication session, if the packets arrive to the destination out of order, what will happen to the original message?
The packets will not be delivered.
The packets will be retransmitted from the source.
* The packets will be delivered and reassembled at the destination.
The packets will be delivered and not reassembled at the destination.

10 With TCP/IP data encapsulation, which range of port numbers identifies all well-known applications?
0 to 255
256 to 1022
* 0 to 1023
1024 to 2047
49153 to 65535

11 Which transport layer protocol provides low overhead and would be used for applications which do not require reliable data delivery?
TCP
IP
* UDP
HTTP
DNS

12 What are two features of the User Datagram Protocol (UDP)? (Choose two.)
flow control
* low overhead
* connectionless
connection-oriented
sequence and acknowledgements

13 What mechanism is used by TCP to provide flow control as segments travel from source to destination?
sequence numbers
session establishment
* window size
acknowledgments

14 What is dynamically selected by the source host when forwarding data?
destination logical address
source physical address
default gateway address
* source port

15 Why are port numbers included in the TCP header of a segment?
to indicate the correct router interface that should be used to forward a segment
to identify which switch ports should receive or forward the segment
to determine which Layer 3 protocol should be used to encapsulate the data
* to enable a receiving host to forward the data to the appropriate application
to allow the receiving host to assemble the packet in the proper order

16 Which two options represent Layer 4 addressing? (Choose two.)
identifies the destination network
identifies source and destination hosts
* identifies the communicating applications
* identifies multiple conversations between the hosts
identifies the devices communicating over the local media

17 Which three features allow TCP to reliably and accurately track the transmission of data from source to destination?
encapsulation
* flow control
connectionless services
* session establishment
* numbering and sequencing
best effort delivery

18 Which OSI model layer is responsible for regulating the flow of information from source to destination, reliably and accurately?
application
presentation
session
* transport
network

19 Based on the transport layer header shown in the diagram, which of the following statements describe the established session? (Choose two.)
This is a UDP header.
* This contains a Telnet request.
This contains a TFTP data transfer.
The return packet from this remote host will have an Acknowledgement Number of 43693.
* This is a TCP header.

20 Which event occurs during the transport layer three-way handshake?
The two applications exchange data.
* TCP initializes the sequence numbers for the sessions.
UDP establishes the maximum number of bytes to be sent.
The server acknowledges the bytes of data received from the client.

(Reference: noc-rikasang.blogspot.com)

Network Fundamentals - Chapter 3 Exam Answers

2008-06-05T11:18:00.001+07:00

1 A network administrator is designing a network for a new branch office of twenty-five users. What are the advantages of using a client-server model? (Choose two.)
* centralized administration
does not require specialized software
* security is easier to enforce
lower cost implementation
provides a single point of failure

2 Which of the following are benefits of peer-to-peer networks? (Choose three.)
centralized security
* easy to create
very scalable
* no centralized equipment required
* centralized administrator not required
centralized control of assets

3 Which two protocols are used to control the transfer of web resources from a web server to a client browser? (Choose two.)
ASP
FTP
HTML
* HTTP
* HTTPS
IP

4 Which statements are correct concerning the role of the MTA in handling email? (Choose three.)
routes email to the MDA on other servers
* receives email from the client's MUA
receives email via the POP3 protocol
* passes email to the MDA for final delivery
* uses SMTP to route email between servers
delivers email to clients via the POP3 protocol

5 What is the automated service that matches resource names with the required IP address?
HTTP
SSH
FQDN
* DNS
Telnet
SMTP

6 What three protocols operate at the Application layer of the OSI model? (Choose three.)
ARP
* DNS
PPP
* SMTP
* POP
ICMP

7 What are the two forms of application layer software? (Choose two.)
* applications
processes
* services
dialogs
syntax
daemons

8 What are three properties of peer-to-peer applications? (Choose three.)
* acts as both a client and server within the same communication
requires centralized account administration
* hybrid mode includes a centralized directory of files
* can be used in client-server networks
does not require specialized software
centralized authentication is required

9 Which application layer protocols correctly match a corresponding function? (Choose two.)
DNS dynamically allocates IP addresses to hosts
* HTTP transfers data from a web server to a client
POP delivers email from the client to the server email server
SMTP supports file sharing
Telnet provides a virtual connection for remote access

10 Which are valid DNS resource record types? (Choose three.)
* A
* NS
MTA
* MX
FQDN
MAU

11 Which email components are used to forward mail between servers? (Choose two.)
MDA
IMAP
* MTA
POP
* SMTP
MUA

12 What is the primary disadvantage of telnet?
does not support authentication
consumes network bandwidth
utilizes client resources
not widely available
* does not support encryption
requires remote access

13 What are three common HTTP message types? (Choose three.)
* GET
PUSH
* POST
HTML
UPLOAD
* PUT

14 Which of the following are examples of TCP/IP application layer protocols? (Choose three.)
* a terminal emulation protocol that supports remote console connections with various network devices
a protocol created by IBM that makes it easier for mainframes to connect to remote offices
* a protocol responsible for transporting electronic mail on TCP/IP networks and the Internet
a protocol that controls the rate at which data is sent to another computer
* a protocol that exchanges network management information between a network device and a management console

(Reference: noc-rikasang.blogspot.com)

Outing Protocols and Concepts - Cisco CCNA Chapter 1 Exam Answers

2008-06-05T11:14:00.001+07:00

1. If a router cannot find a valid configuration file during the startup sequence, what will occur?

The startup sequence will reset.
*The router will prompt the user for a response to enter setup mode.
The startup sequence will halt until a valid configuration file is acquired.
The router will generate a default configuration file based on the last valid configuration.
The router will monitor local traffic to determine routing protocol configuration requirements.

2.

Refer to the exhibit. The network administrator has configured the router with the interface IP addresses shown for the directly connected networks. Pings from the router to hosts on the connected networks or pings between router interfaces are not working. What is the most likely problem?

The destination networks do not exist.
The IP addresses on the router interfaces must be configured as network addresses and not host addresses.
* The interfaces must be enabled with the no shutdown command.
Each interface must be configured with the clock rate command.

3.

Refer to the exhibit. What can be concluded from the routing table output in the exhibit? (Choose two.)

This router only has two interfaces.
The router interfaces are not operational yet.
This router is configured to forward packets to remote networks.
* The FastEthernet0/0 and Serial0/0/0 interfaces of this router were configured with an IP address and the no shutdown command.
* An IP packet received by this router with a destination address of 198.18.9.1 will be forwarded out of the Serial0/0/0 interface.

4.

Refer to the exhibit. The frame shown in the exhibit was received by the router. The router interfaces are operational. How will the router process this frame? (Choose two.)

The router will change the source and destination IP address in the packet before forwarding the frame.
* The router will change the frame type to one supported by the WAN link before forwarding the frame.
The router will use the destination MAC address to determine which interface to forward the packet.
The router will look up the MAC address of the S0/0/0 interface in the ARP table and add it to the frame before forwarding.
* The frame was received on the Fa0/0 interface of the router and will be switched to the S0/0/0 interface.
The frame was received on the S0/0/0 interface of the router and will be switched to the Fa0/0 interface.

5. Passwords can be used to restrict access to all or parts of the Cisco IOS. Select the modes and interfaces that can be protected with passwords. (Choose three.)

* VTY interface
* console interface
Ethernet interface
secret EXEC mode
* privileged EXEC mode
router configuration mode

6. Which two statements correctly describe the components of a router? (Choose two.)

RAM permanently stores the configuration file used during the boot sequence.
* ROM contains diagnostics executed on hardware modules.
NVRAM stores a backup copy of the IOS used during the boot sequence.
* Flash memory does not lose its contents during a reboot.
ROM contains the most current and most complete version of the IOS.
Flash contains boot system commands to identify the location of the IOS

7.

Refer to the exhibit. After host 2 is connected to the switch on the LAN, host 2 is unable to communicate with host 1. What is the cause of this problem?

The subnet mask of host 2 is incorrect.
* Host 1 and host 2 are on different networks.
The switch needs an IP address that is not configured.
The router LAN interface and host 1 are on different networks.
The IP address of host 1 is on a different network than is the LAN interface of the router.

8. Which are functions of a router? (Choose three.)

* packet switching
extension of network segments
* segmentation of broadcast domains
* selection of best path based on logical addressing
election of best path based on physical addressing

9.

Refer to the exhibit. All routers have a route in its routing table to each network that is shown in the exhibit. Default routes have not been issued on these routers. What can be concluded about how packets are forwarded in this network? (Choose two.)

If RouterC receives a packet that is destined for 10.5.1.1, it will be forwarded out interface Fa0/0.
* If RouterA receives a packet that is destined for 192.168.3.146, it will be forwarded out interface S0/0/1.
* If RouterB receives a packet that is destined for 10.5.27.15, it will be forwarded out interface S0/0/1.
If RouterB receives a packet that is destined for 172.20.255.1, it will be forwarded out interface S0/0/0.
If RouterC receives a packet that is destined for 192.16.5.101, it will be forwarded out interface S0/0/1.

10.

The serial connection shown in the graphic needs to be configured. Which configuration commands must be made on the Sydney router to establish connectivity with the Melbourne site? (Choose three.)
* Sydney(config-if)# ip address 201.100.53.2 255.255.255.0
* Sydney(config-if)# no shutdown
Sydney(config-if)# ip address 201.100.53.1 255.255.255.224
* Sydney(config-if)# clock rate 56000
Sydney(config-if)# ip host Melbourne 201.100.53.2

11.

Refer to the exhibit. What can be concluded from the output of the running-configuration of a router?

The passwords are encrypted.
The current configuration was saved to NVRAM.
The configuration that is shown will be the one used on the next reboot.
* The commands that are displayed determine the current operation of the router.

12.

Refer to the exhibit. Host A pings host B. When R4 accepts the ping into the Ethernet interface, what two pieces of header information are included? (Choose two.)

source IP address: 192.168.10.129
source IP address: BBBB.3333.5677
destination IP address: 192.168.10.33
* destination IP address: 192.168.10.134
* destination MAC address: 9999.DADC.1234

13. What is the outcome of entering these commands?

R1(config)# line vty 0 4
R1(config-line)# password check123
R1(config-line)# login

ensures that a password is entered before entering user EXEC mode
* sets the password to be used for connecting to this router via Telnet
requires check123 to be entered before the configuration can be saved
creates a local user account for logging in to a router or switch

14. Which of the following is the correct flow of routines for a router startup?

* load bootstrap, load IOS, apply configuration
load bootstrap, apply configuration, load IOS
load IOS, load bootstrap, apply configuration, check hardware
check hardware, apply configuration, load bootstrap, load IOS

15. What three processes does a router execute when it receives a packet from one network that is destined for another network? (Choose three.)

* decapsulates the Layer 3 packet by stripping off the Layer 2 frame header
uses the destination MAC Address in the IP Header to look up the next-hop address in the routing table
leaves the Layer 2 frame header intact when decapsulating the Layer 3 packet
* uses the destination IP Address in the IP header to look up the next-hop address in the routing table
* encapsulates the Layer 3 packet into the new Layer 2 frame and forwards it out the exit interface
encapsulates the Layer 3 packet into a special Layer 1 frame and forwards it to the exit interface

16. The network administrator needs to connect two routers directly via their FastEthernet ports. What cable should the network administrator use?

straight-through
rollover
* cross-over
serial

17. Which two statements describe characteristics of load balancing? (Choose two.)

Load balancing occurs when a router sends the same packet to different destination networks. Load balancing occurs when a router sends the same packet to different destination networks.
* Load balancing allows a router to forward packets over multiple paths to the same destination network.
* Unequal cost load balancing is supported by EIGRP.
If multiple paths with different metrics to a destinations exist, the router cannot support load balancing.

18. What information about the router and its startup process can be gathered from the output of the show version command? (Choose three.)

* the last restart method
the command buffer contents
the amount of NVRAM and FLASH used
* the configuration register settings
* the location from where the IOS loaded

19.

Which interfaces in the exhibit could be used for a leased line WAN connection? (Choose two.)

* 1
2
3
* 4
5
6

20. From what location can a router load the Cisco IOS during the boot process? (Choose two.)

RAM
* TFTP server
NVRAM
setup routine
* Flash memory
terminal

21. A network administrator has just entered new configurations into Router1. Which command should be executed to save configuration changes to NVRAM?

Router1# copy running-config flash
Router1(config)# copy running-config flash
Router1# copy running-config startup-config
* Router1(config)# copy running-config startup-config
Router1# copy startup-config running-config
Router1(config)# copy startup-config running-config

22. What is the default sequence for loading the configuration file?

NVRAM, FLASH, ROM
FLASH, TFTP,CONSOLE
* NVRAM, TFTP, CONSOLE
FLASH, TFTP, ROM

23. What header address information does a router change in the information it receives from an attached Ethernet interface before information is transmitted out another interface?

only the Layer 2 source address
only the Layer 2 destination address
only the Layer 3 source address
only the Layer 3 destination address
* the Layer 2 source and destination address
the Layer 3 source and destination address

Routing Protocols and Concepts - Cisco CCNA Chapter 2 Exam Answers

2008-06-05T11:05:00.001+07:00

1 A static route that points to the next hop IP will have what administrative distance and metric in the routing table?

administrative distance of 0 and metric of 0
administrative distance of 0 and metric of 1
* administrative distance of 1 and metric of 0
administrative distance of 1 and metric of 1

2 What address can be used to summarize networks 172.16.1.0/24, 172.16.2.0/24, 172.16.3.0/24, and 172.16.4.0/24?

* 172.16.0.0/21
172.16.1.0/22
172.16.0.0 255.255.255.248
172.16.0.0 255.255.252.0

3

The routers in the diagram use the subnet assignments shown. What is the most efficient route summary that can be configured on Router3 to advertise the internal networks to the cloud?

192.1.1.0/26 and 192.1.1.64/27
192.1.1.128/25
192.1.1.0/23 and 192.1.1.64/23
192.1.1.0/24
* 192.1.1.0/25
192.1.1.0/24 and 192.1.1.64/24

4

Refer to the exhibit. What is the significance of the /8 in the route to the 10.0.0.0 network?

It indicates that there are 8 hops between this router and the 10.0.0.0 network.
It represents the time, in milliseconds, it takes for a ping to reply when sent to the 10.0.0.0 network.
It indicates that there are 8 subnets in the destination network to which the router can forward packets.
* It indicates the number of consecutive bits, from the left, in the destination IP address of a packet that must match 10.0.0.0 to use that route.

5

Refer to the exhibit. How will packets destined to the 172.16.0.0 network be forwarded?

* Router1 will perform recursive lookup and packet will exit S0/0.
Router1 will perform recursive lookup and packet will exit S0/1.
There is no matching interface associated with network 172.16.0.0 so packets will be dropped.
There is no matching interface associated with network 172.16.0.0 so packets will take gateway of last resort and exit out S0/2.

6 A network administrator enters the following command into Router1: ip route 192.168.0.0 255.255.255.0 S0/1/0. Router1 then receives a packet that is destined for 192.168.0.22/24. After finding the recently configured static route in the routing table, what does Router1 do next to process the packet?

drops the packet because the destination host is not listed in the routing table
looks up the MAC address of the S0/1/0 interface to determine the destination MAC address of the new frame
performs a recursive lookup for the IP address of the S0/1/0 interface before forwarding the packet
* encapsulates the packet into a frame for the WAN link and forwards the packet out the S0/1/0 interface

7
Refer to the exhibit. Given the output in the exhibit, how would a clock rate be determined for this link?

The rate would be negotiated by both routers.
A rate would not be selected due to the DCE/DTE connection mismatch.
The rate configured on the DTE determines the clock rate.
* The rate configured on the DCE determines the clock rate.

8
Refer to the exhibit. Which set of commands will configure static routes that will allow the WinterPark and the Altamonte routers to deliver packets from each LAN and direct all other traffic to the Internet?

* WinterPark(config)# ip route 0.0.0.0 0.0.0.0 192.168.146.1
Altamonte(config)# ip route 10.0.234.0 255.255.255.0 192.168.146.2
Altamonte(config)# ip route 0.0.0.0 0.0.0.0 s0/1
WinterPark(config)# ip route 0.0.0.0 0.0.0.0 192.168.146.1
Altamonte(config)# ip route 10.0.234.0 255.255.255.0 192.168.146.2
Altamonte(config)# ip route 198.18.222.0 255.255.255.255 s0/1
WinterPark(config)# ip route 172.191.67.0 255.255.255.0 192.168.146.1
WinterPark(config)# ip route 0.0.0.0 0.0.0.0 192.168.146.1
Altamonte(config)# ip route 10.0.234.0 255.255.255.0 192.168.146.2
WinterPark(config)# ip route 172.191.67.0 255.255.255.0 192.168.146.1
Altamonte(config)# ip route 10.0.234.0 255.255.255.0 192.168.146.2
Altamonte(config)# ip route 0.0.0.0 0.0.0.0 s0/0

9 Hosts on two separate subnets cannot communicate. The network administrator suspects a missing route in one of the routing tables. Which three commands can be used to help troubleshoot Layer 3 connectivity issues? (Choose three.)

* ping
show arp
* traceroute
* show ip route
show interface
show cdp neighbor detail

10
Refer to the exhibit. A company network engineer is assigned to establish connectivity between the two Ethernet networks so that hosts on the 10.1.1.0/24 subnet can contact hosts on the 10.1.2.0/24 subnet. The engineer has been told to use only static routing for these company routers. Which set of commands will establish connectivity between the two Ethernet networks?

R1(config)# ip route 10.1.2.0 255.255.255.0 192.168.0.1
R2(config)# ip route 10.1.1.0 255.255.255.0 192.168.0.2
R1(config)# ip route 10.1.2.0 255.255.255.0 192.168.0.2
R2(config)# ip route 10.1.1.0 255.255.255.0 192.168.0.1
* R1(config)# ip route 10.1.1.0 255.255.255.0 192.168.0.2
R2(config)# ip route 10.1.2.0 255.255.255.0 192.168.0.1
R1(config)# ip route 10.1.1.0 255.255.255.0 192.168.0.1
R2(config)# ip route 10.1.2.0 255.255.255.0 192.168.0.2
R1(config)# ip route 0.0.0.0 0.0.0.0 10.1.2.1
R2(config)# ip route 0.0.0.0 0.0.0.0 10.1.1.1

11 Which of the following are displayed by the Router# show cdp neighbors command? (Choose three.)

load
* platform
reliability
* holdtime
* local interface

12
Refer to the exhibit. What two commands are required to provide connectivity between the 192.168.1.0 and 10.0.0.0 networks without requiring recursive lookup? (Choose two.)

* A(config)# ip route 10.0.0.0 255.0.0.0 s 0/1/0
A(config)# ip route 10.0.0.0 255.0.0.0 172.16.40.2
A (config)# ip route 10.0.0.0 255.0.0.0 s 0/0/0
* B(config)# ip route 192.168.1.0 255.255.255.0 s 0/0/0
B (config)# ip route 192.168.1.0 255.255.255.0 172.16.40.1
B(config)# ip route 192.168.1.0 255.255.255.0 s 0/1/0

13
Refer to the exhibit. What two commands will change the next-hop address for the 10.0.0.0/8 network from 172.16.40.2 to 192.168.1.2? (Choose two.)

A(config)# no network 10.0.0.0 255.0.0.0 172.16.40.2
A(config)# no ip address 10.0.0.1 255.0.0.0 172.16.40.2
* A(config)# no ip route 10.0.0.0 255.0.0.0 172.16.40.2
A(config)# ip route 10.0.0.0 255.0.0.0 s0/0/0
* A(config)# ip route 10.0.0.0 255.0.0.0 192.168.1.2

14 What happens to a static route entry in a routing table when the outgoing interface is not available?

* The route is removed from the table.
The router polls neighbors for a replacement route.
The route remains in the table because it was defined as static.
The router redirects the static route to compensate for the loss of the next hop device.

15 A router has one static route to each destination network. Which two scenarios would require an administrator to alter the static routes that are configured on that router? (Choose two.)

* The destination network no longer exists.
The destination network is moved to a different interface on the same router.
The path between the source and destination is upgraded with a higher bandwidth link.
* A topology change occurs where the existing next-hop address or exit interface is not accessible.
The remote destination network interface has to be down for 15 minutes of maintenance.

16 Why is it advisable to enter a next-hop IP address when creating a static route whose exit interface is an Ethernet network?

Adding the next-hop address eliminates the need for the router to do any lookups in the routing table before forwarding a packet.
* In a multi-access network, the router cannot determine the next-hop MAC address for the Ethernet frame without a next-hop address.
Using a next-hop address in a static route provides a route with a lower metric.
In multi-access networks, using a next-hop address in a static route makes that route a candidate default route.

17 The output of the Router# show interfaces serial 0/1 command displays the following:

Serial0/1 is up, line protocol is down.

What is the most likely cause for the line protocol being down?

Serial0/1 is shutdown.
There is no cable connecting the routers.
The remote router is using serial 0/0.
* No clock rate has been set.

18
Refer to the exhibit. Which static route should be configured on Router1 so that host A will be able to reach host B on the 172.16.0.0 network?

ip route 192.168.0.0 172.16.0.0 255.255.0.0
ip route 172.16.0.0 255.255.0.0 192.168.0.1
ip route 172.16.0.0 255.255.0.0 S0/0/1
* ip route 172.16.0.0 255.255.0.0 S0/0/0

19 What two devices are responsible for converting the data from the WAN service provider into a form acceptable by the router? (Choose two).

the serial port of the router
* a modem
a switch
the ethernet port of the router
* a CSU/DSU device
a DTE device

20
Which of the following is true regarding CDP and the graphic shown?

CDP running on Router D will gather information about routers A, B, C, and E.
* By default, Router A will receive CDP advertisements from routers B and C.
If routers D and E are running different routing protocols, they will not exchange CDP information.
Router E can use CDP to identify the IOS running on Router B.

21 Which two statements describe functions or characteristics of CDP? (Choose two.)

* It starts up automatically and allows the device to detect directly connected neighbor devices that use CDP.
It operates at the network layer and allows two systems to learn about each other.
It creates a topology map of the entire network.
* It allows systems to learn about each other even if different network layer protocols are configured.
It forwards advertisements about routes for faster convergence.

22 Which piece of information is available from examining the output of the command show ip interface brief?

Interface speed and duplex
Interface MTU
Errors
Interface MAC address
* Interface IP address

Download Cisco Certification Exams Questions - Pass4Sure

2008-06-05T10:53:00.002+07:00

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Pass4Sure : 640-802 Exam preparation

2008-06-05T10:28:00.002+07:00

With the complete collection of and answers pass4sure has assembled to take you through Q&A your 640-802 Exam preparation, you will cover every field and category helping to ready you for your successful Cisco Certification. Each Q & A set will test your existing knowledge of Cisco fundamentals, and is complemented by our interactive Testing Engine.

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5 CD-ROMs - Live instructor-led classroom sessions

2008-06-05T10:22:00.000+07:00

Download all to a folder somewhere, make sure all those files are in the same folder. We do not need to open any of this .r1 etc etc files at all. The main files is the last file which is in a format or rar (iso) unzip that to the same folder again. This will unzip many files but the one we want is says START and hurray is running.......

After opening all video's have a cup of coffee...may be a little boring covers allot of everything....watch with a friend...may become more interesting.

5 CD-ROMs - Live instructor-led classroom sessions with full Audio, Video and demonstration components
Detailed Description
Schedule our instructor led classroom trainings at your convenience and never miss another lecture or fall behind. You are in complete control. We invited the Best ComTIA A+ Trainers in the industry to help us develop the ultimate training & certification program which includes everything you will need to fully prepare ou for the New CCNA 640-801 Certification exam.
The CCNA certification (Cisco Certified Network Associate) indicates a foundation in and apprentice knowledge of networking. CCNA certified professionals can install, configure, and operate LAN, WAN, and dial access services for small networks (100 nodes or fewer), including but not limited to use of these protocols: IP, IGRP, Serial, Frame Relay, IP RIP, VLANs, RIP, Ethernet, Access Lists.

Level 1: Internetworking & TCP/IP
Internetworking
Intro to Internetworking
The "OSI" Model
Ethernet
"Wireless": SOS
TCP/IP
The DoD Model
The IP Address
IP Classes
SOS
Level 2: IP Subnetting & Internetwork Operation System
IP Subnetting
What is Subnetting?
Classless Inter-Domain Routing (CIDR)
Subnetting
Variable Length Subnet Masks
Troubleshooting TCP/IP
Internetwork Operating System
Introduction to IOS
Passwords
Banners, Hostnames & Descriptions
Configuring Interfaces
Saving & Verifying Configuration
Level 3: IP Routing Protocols
IP Routing
Bringing it Together
Building Route Tables
Distance -- Vector Routing Protocols
Interior Gateway Routing Potocol
Verifying
More Routing Protocols
EIGRP
Open Shortest Path First
Level 4: Layer 2 Switching, Virtual LANs, & IOS
Layer 2 Switching
Spanning Tree Protocol
Switch Types
Configuring the Switch
Virtual LANs
VLAN Concepts
VLAN Truking Protocol
Configuring
Advanced IOS
Router Components
Backup and Restore
Cisco Discovery Protocol
Using Telnet
Host Names
Level 5: Access Lists & Wide Area Networks
Access Lists
Types of Access Lists
Creating an Access List
Using the Access List
Wide Area Networks
WAN Basics
HDLC
PPC
Frame Relay
ISDN
SS/CC

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VALUE AT RISK (VAR) (PART 1)

2008-06-02T14:40:00.000+07:00

Value-at-Risk (VaR)

by Simon Benninga and Zvi Wiener

VALUE-AT-RISK
Value-at-Risk (VaR) measures the worst expected loss under normal market conditions over a specific time interval at a given confidence level. As one of our references states: “VaR answers the question: how much can I lose with x% probability over a pre-set horizon” (J.P. Morgan, RiskMetrics–Technical Document). Another way of expressing this is that VaR is the lowest quantile of the potential losses that can occur within a given portfolio during a specified time period. The basic time period T and the confidence level (the quantile) q are the two major parameters that should be chosen in a way appropriate to the overall goal of risk measurement. The time horizon can differ from a few hours for an active trading desk to a year for a pension fund. When the primary goal is to satisfy external regulatory requirements, such as bank capital requirements, the quantile is typically very small (for example, 1% of worst outcomes). However for an
internal risk management model used by a company to control the risk exposure the typical number is around 5% (visit the internet sites in references for more details). A general introduction to VaR can be found in Linsmeier, [Pearson 1996] and [Jorion 1997]. In the jargon of VaR, suppose that a portfolio manager has a daily VaR equal to $1 million at 1%. This means that there is only one chance in 100 that a daily loss bigger than $1 million occurs under normal market conditions.

A REALLY SIMPLE EXAMPLE
Suppose portfolio manager manages a portfolio which consists of a single asset. The return of the asset is normally distributed with annual mean return 10% and annual standard deviation 30%. The value of the portfolio today is $100 million. We want to answer various simple questions about the end-of-year distribution of portfolio value:
1. What is the distribution of the end-of-year portfolio value?
2. What is the probability of a loss of more than $20 million dollars by year end (i.e., what is the probability
that the end-of-year value is less than $80 million)?
3. With 1% probability what is the maximum loss at the end of the year? This is the VaR at 1%.
We start by loading Mathematica ’s statistical package:
Needs["Statistics‘Master‘"]
Needs["Statistics‘MultiDescriptiveStatistics‘"]
We first want to know the distribution of the end-ofyear portfolio value:
Plot[PDF[NormalDistribution[110,30],x],{x,0,200}];
50 100 150 200
0.002
0.004
0.006
0.008
0.01
0.012
The probability that the end-of-year portfolio value is less than $80 is about 15.9%.
CDF[NormalDistribution[110.,30],80]0.158655
With a probability of 1% the end-of-year portfolio value will be less than 40.2096; this means that the VaR of the distribution is 100 - 40.2096 = 59.7904.
Quantile[NormalDistribution[110.,30],0.01]40.2096
We can formalize this by defining a VaR function which takes as its parameters the mean mu and standard deviation sigma of the distribution as well as the VaR level x.
ClearAll[VaR];
VaR[mu_,sigma_,x_]:=
100-Quantile[NormalDistribution[mu,sigma],x]
VaR[110,30,0.01]59.7904
Vol. 7 No. 4 1998 MathematicainEducationandResearch 1

VALUE AT RISK (VAR) (PART 2)

2008-06-02T14:38:00.000+07:00

Value-at-Risk (VaR)

by Simon Benninga and Zvi Wiener

VALUE-AT-RISK (VAR)
LOGNORMAL DISTRIBUTIONS
As explained in our previous articles, the lognormal distribution is a more reasonable distribution for many asset prices (which can not become negative) than the normal distribution. This is not a problem: Suppose that the natural logarithm of the portfolio value is normally distributed with annual mean m and annual standard deviation s. Denoting the value of the portfolio by v it follows that the logarithm of the portfolio value at time T , vt , is normally distributed:

Log[vT ] ~ Normal CLog[v] + Km - s22 OT ,sT G The term s22 T appears due to Ito’s Lemma (see [Hull1997]). In our case, this means that v = 100, m = 10%,s = 30%. Thus the end-of-year log of the portfolio value is distributed NormalDistribution[ Log[ 100 ] + (0.1-0.3ˆ2/2), 0.3] = NormalDistribution[
4.666017,0.3]. This means that the probability that the end-of-year value of the portfolio is less than 80 is
given by: CDF[NormalDistribution[Log[100]+ (0.1-0.3ˆ2/2),0.3],Log[80]] 0.176926
Similarly the VaR function has to be redefined:
ClearAll[lognormalVaR];
lognormalVaR[mu_,sigma_,x_]:=100-Exp[Quantile[NormalDistribution[
Log[100]+(mu-sigmaˆ2/2),sigma],x]];
lognormalVaR[0.10,0.30,0.01]47.4237
Thus a portfolio whose initial value is $100 million and whose annual returns are lognormally distributed with parameters mu = 10% and sigma = 30%, has an annual VaR equal to $47.42 million at 1%.
Most VaR calculations are not concerned with annual value at risk. The main regulatory and management concern is with loss of portfolio value over a much shorter time period (typically several days or perhaps weeks).
It is clear that the distribution formula Log[vT ] ~ Normal[Log[v ] + (m - s22 )T ,sT ] can be used to calculate the VaR over any horizon. Recall that T is measured in annual terms; if there are 250 business days in a year, then the daily VaR corresponds to T = 1/250 (for many fixed income instruments one should use 1/360, 1/365, or 1/365.25 depending on the market convention):
ClearAll[lognormalVaR];
lognormalVaR[mu_,sigma_,x_,T_]:=100-Exp[Quantile[NormalDistribution[Log[100]+(musigmaˆ2/2)*T,sigma*T],x]];
lognormalVaR[0.10,0.30,0.01,1/250]
lognormalVaR[0.10,0.30,0.01,5/250]
lognormalVaR[0.10,0.30,0.01,21/250]
0.256831
1.27758
5.25717
The daily VaR of the portfolio at 1% is $256,831. The probability that the firm will lose more than this amount on its portfolio over the course of a single day is less than 1%. Similarly, the weekly and the monthly VaRs at 1% are $1.28 and $5.26 million.

VALUE AT RISK (VAR) (PART 3)

2008-06-02T14:37:00.000+07:00

Value-at-Risk (VaR)

by Simon Benninga and Zvi Wiener

A THREE-ASSET PROBLEM: THE IMPORTANCE OF THE VARIANCE-COVARIANCEMATRIX
As can be seen from the above examples, VaR is not–in principle, at least–a very complicated concept. In the implementation of VaR, however, there are two big practical problems (both problems are discussed in much greater detail in the material available on the J.P. Morgan Web site):
1. The first problem is the estimation of the parameters of asset return distributions. In “real world” applications of VaR, it is necessary to estimate means, variances, and correlations of returns. This is a not-inconsiderable problem! In this section we illustrate the importance of the correlations between asset returns. In the following section we give a highly-simplified example of the estimation of return distributions from market data.
For example you can imagine that a long position in Deutschmarks and a short position in Dutch guldens
is less risky than one leg only, because of a high probability that profits of one position will be mainly offset by losses of another.
2. The second problem is the actual calculation of position sizes. A large financial institution may have thousands of loans outstanding. The data base of these loans may not classify them by their riskiness, nor
even by their term to maturity. Or–to give a second example–a bank may have offsetting positions in foreign currencies at different branches in different locations.
A long position in Deutschmarks in New York may be offset by a short position in Deutschmarks in
Geneva; the bank’s risk–which we intend to measure by VaR–is based on the net position.
We start with the problem of correlations between asset returns. We continue the previous example, but assume that there are three risky assets. As before the parameters of the distributions of the asset returns are known: all the means: m1, m2, m3, as well as the variance covariance matrix of the returns:
S =ÊÁÁÁÁÁÁË
s11
s21
s31
s12
s22
s32
s13
s23
s33
ˆ˜˜˜˜˜˜¯
The matrix S is of course symmetric; sij is the covariance of the returns of assets i and j (if i = j , s]ii is the variance of asset i’s return).
Suppose that the total portfolio value today is $100 million, with $30 million invested in asset 1, $25 Million 2 MathematicainEducationandResearch Vol. 7 No. 4 1998 VALUE-AT-RISK (VAR) in asset 2, and $45 million in asset 3. Then the return distribution of the portfolio is given by:
mean return = x.m = x1m1 + x2m2 + x3m3
variance of return = x.S.xT ,where x = {x1, x2, x3} = {0.3, 0.25, 0.45} is the vector of proportions invested in each of the three assets. Assuming that the returns are normally distributed (meaning that prices are lognormally distributed), we may calculate the VaR by:
Clear[x,S,means,portfolioMean,portfolioSigma];
X=Table[x[i],{i,3}];
S=Table[sigma[i,j],{i,3},{j,3}];
means=Table[mu[i],{i,3}];
portfolioMean[X_,means_,initial_]:=X.means*initial;
portfolioSigma[X_,s_,initial_]:=Sqrt[X.s.X]*initial;
To implement this:
Clear[VaR];
X={0.3,0.25,0.45};
means={0.1,0.12,0.13};
initial=100;
VaRlevel=0.01;
S={{0.1,0.04,0.03},{0.04,0.2,-0.04},
{0.03,-0.04,0.6}};
Print["The portfolio mean = ",
portfolioMean[X,1+means,initial] ]
Print["The portfolio sigma = ",
portfolioSigma[X,S,initial]]
VaR[X_,s_,initial_,level_]:=initial-Quantile[NormalDistribution[portfolioMean[X,means,initial],
portfolioSigma[X,s,initial]],VaRlevel]
Print["The portfolio VaR at the ",
VaRlevel*100, "% level is ",
VaR[X,S,initial,VaRlevel] ]
The portfolio mean = 111.85
The portfolio sigma = 38.4838
The portfolio VaR at the 1. % level is 177.677
USING EMPIRICAL DATA TO CALCULATE THE VAR
In this section we use market data to do some VaR calculations. Our data consists of 40 business days of Data for a market index, foreign interest rates and foreign exchange rates. We use this data to do three kinds of VaR calculations: Historic simulations, variance-covariance calculations, and Monte Carlo simulations.
Data Description and Preliminary Calculations We consider the VaR of a portfolio manager who has invested in only two assets: A domestic stock index and a foreign bond. For simplicity we assume that the value of this portfolio is wholly determined by only several parameters:
Í The current price of the stock index
Í The foreign market interest rate (the bond is assumed to have a zero coupon, so that only the interest rate
until the bond’s maturity determines its price in foreign currency).
Í The time until the bond’s maturity; along with the foreign market interest rate, this will determine the bond’s foreign currency price.
Í The exchange rate between the foreign and the domestic currency.
We use historic price and return data to do our VaR calculations. It is convenient to translate all dates to a numerical format. We set January 1, 1997 to be day number 0 and then any day is given by a number of days since the initial date (we assume that all days are business days for simplicity).
Needs["Miscellaneous‘Calendar‘"];
Needs["Statistics‘Master‘"]
Needs["Statistics‘MultiDescriptiveStatistics‘"]
Day0 ={1997, 1, 1};
dayN[ day_]:=DaysBetween[ Day0, day];
Thus January 13, 1997 is the day number 12 and so on.
dayN[{1997, 1,13}]12
Consider February 10, 1997. Suppose that on February 10, 1997, the stock index value is 293., the foreign interest rate is 5.3%, and the exchange rate is 3.4; February 10, 1997 is the 40th business day since January 1, 1997. Thus we write the data for this day as a list: {40, 293., 5.3,3.4}. Obviously any real market data must contain much more information, however will use this simple example to illustrate the VaR approach.
The whole data set with which we will be working is the following (in a real-world situation, we would obviously use much more data about many more assets):
dataVAR ={
{1, 282., 5.28, 3.5},{2, 283., 5.26, 3.47},
{3, 285., 5.23, 3.46},{4, 280., 5.24, 3.45},
{5, 282., 5.25, 3.45},{6, 281., 5.24, 3.46},
{7, 282., 5.24, 3.45},{8, 286., 5.25, 3.43},
{9, 285., 5.25, 3.47},{10, 286., 5.26, 3.443},
{11, 288., 5.27, 3.42},{12, 289., 5.28, 3.42},
{13, 290., 5.28, 3.41},{14, 289., 5.28, 3.42},
{15, 291., 5.29, 3.46},{16, 293., 5.31, 3.41},
{17, 294., 5.32, 3.40},{18, 290., 5.34, 3.49},
{19, 287., 5.35, 3.47},{20, 288., 5.34, 3.48},
{21, 289., 5.35, 3.46},{22, 281., 5.36, 3.44},
{23, 283., 5.23, 3.45},{24, 285., 5.24, 3.42},
{25, 288., 5.25, 3.41},{26, 289., 5.26, 3.41},
{27, 287., 5.26, 3.43},{28, 285., 5.28, 3.42},
{29, 290., 5.27, 3.44},{30, 291., 5.27, 3.42},
{31, 289., 5.27, 3.37},{32, 288., 5.29, 3.39},
{33, 290., 5.28, 3.41},{34, 293., 5.31, 3.44},
{35, 292., 5.32, 3.41},{36, 293., 5.28, 3.42},
{37, 293., 5.30, 3.42},{38, 293., 5.31, 3.44},
{39, 292., 5.32, 3.41},{40, 293., 5.30, 3.4}};
Suppose that the financial institution holds two shares of the stock index portfolio and a short position in a zerocoupon foreign bond having face value 100 rubles with maturity May 8, 2000. We represent this portfolio as a list:
Vol. 7 No. 4 1998 MathematicainEducationandResearch 3
VALUE-AT-RISK (VAR)
portfolio = { {STOCK,{}, 2},
{FBOND,{dayN[{2000, 5, 8}]}, -1}}
Note that the empty paramters list in the stock part means that there are no parameters and we always keep
the same portfolio of stocks. The list of parameters of the bond part consists of one number - day of maturity. In general both lists can have many parameters describing a specific portfolio of stocks or bonds.

VALUE AT RISK (VAR) (PART 4)

2008-06-02T14:36:00.001+07:00

Value-at-Risk (VaR)

by Simon Benninga and Zvi Wiener

Risk Mapping
“Risk mapping” is a nice wording for a pricing function. In a general case we might decompose the pricing function for each type of financial instrument into simple risk factors (perhaps using the option pricing techniques explained in our previous articles). However, in our simple example the current value of each instrument is a simple one-dimensional function of current market data. For example, the price of the stock is just its market value and the price of the bond isac its discounted future payoff translated to a local currency according to the current exchange rate. Next, we define the pricing functions for the stock index
and the foreign bond. Suppose we are given a list of market data, market={dayN[{1997,3,6}], 283.,5.26, 3.47}. The stock market price is the second item in this list, whereas the bond price in the local currency is
given by 100*3.47*exp[-5.26%*(time to maturity in years)]. We use Mathematica ’s calendar functions to define two dollar pricing functions:
Clear[stock,fbond]
stock[ param_, market_]:= market[[2]];
fbond[ param_,market_]:=market[[4]]*100*Exp[-market[[3]]/100*(param[[1]]-market[[1]])/365.25];
junk={dayN[{1997,8,9}],355.,5.77,6.};
stock[{},junk]
fbond[{1223},junk]
355.
512.08
Next we define
pricingFun[instr_]:=Switch[instr, STOCK,stock, FBOND,fbond]
valueP[ portf_, mrkt_]:=Module[{valueList, i},
valueList =Table[Apply[ pricingFun[ portf[[i,1]] ],
{portf[[i,2]], mrkt}]*portf[[i,3]],
{i,Length[ portf]}];
Apply[ Plus,valueList]
];
Check that it works
mrkt9Feb97 = {40, 293., 5.30, 3.4};
currentPrice = valueP[ portfolio, mrkt9Feb97]
299.63
If everything is correct we should get 299.63 since the current value of our stock position is 586 = 2 * 293, and the bond position has a value of -286.37 = -100 * 3.4 *e-0.053*1183/365.25.
For a VaR calculation we have to fix two parameters: the time horizon and the confidence level (1 - quantile). We choose a one day time horizon and an 80This time horizon is a typical one for financial institutions, however the confidence level 80% is very low, but it will allow us to use a short data file. For a higher precision much more data is necessary.
Historical Simulation
The first method we are presenting here is the historical simulation. We pretend that the change in the market conditions from today to tomorrow are the same as the to the changes that took place some time in the past. There is an important question on what type of current changes are the same as historic changes. The main difficulty is to distinguish between multiplicative and additive types of market variables. We provide here a simplistic scheme assuming that all changes are additive, for a detailed explanation
of this problem see [Grundy, Wiener 1996].
This procedure pretends that the change in market parameters from today to tomorrow will be the same as it was some time ago. The routine below returns a list of the historical prices. Note that this procedure is completely different from just pricing the current portfolio according to old data, in which case the global change in the market level is lost. Parameters of the historical simulation function are: portf - the current portfolio, mrkt - the current market data, histData - the historical data.
histSimul[ portf_, mrkt_, histData_]:=Module[{histPrice={}, i},
For[ i=1, i
AppendTo[ histPrice,
valueP[portf,
mrkt+histData[[i+1]]-histData[[i]]]
];
];
histPrice
];
To use this simulation for VaR measurement we use the following function, which requires in addition to the parameters described above the quantile.
HistApproach[ portf_, mrkt_, hData_, quant_]:=
Module[{currPrice, histPr, changes},
currPrice = valueP[ portf, mrkt];
histPr = histSimul[ portf, mrkt, hData];
changes = histPr - currPrice;
{Quantile[ changes, quant], changes, histPr}
];
The outcome of this function is a list of three objects.
The first one is the VaR, the second one is a list of pretended market changes in the value of our portfolio. The third is a list of resulting portfolio values.
4 MathematicainEducationandResearch Vol. 7 No. 4 1998
VALUE-AT-RISK (VAR)
HistApproach[portfolio, mrkt9Feb97,
dataVAR, 0.2]
{-3.0144, {4.30181, 4.52345, -9.10677, 4.05111,
-2.97692, 2.80083, 9.73534, -5.41111, 4.32482,
5.98798, 2.05111, 2.80083, -2.88394, 0.682656,
8.35297, 2.89323, -15.4328, -4.26466, 1.02308,
3.73534, -14.2647, 1.90504, 6.57746, 6.89323,
2.05111, -5.72633, -3.0144, 8.18042, 3.64322,
0.170379, -3.53993, 2.18042, 3.71166, 0.577459,
0.743977, 0.143757, -1.63311, 0.577459,
2.61594}, {303.932, 304.153, 290.523, 303.681,
296.653, 302.431, 309.365, 294.219, 303.955,
305.618, 301.681, 302.431, 296.746, 300.313,
307.983, 302.523, 284.197, 295.365, 300.653,
303.365, 285.365, 301.535, 306.207, 306.523,
301.681, 293.904, 296.615, 307.81, 303.273,
299.8, 296.09, 301.81, 303.342, 300.207,
300.374, 299.774, 297.997, 300.207, 302.246}}
Thus the simulated daily VaR at 20% level is a loss of $3.0144.

VALUE AT RISK (VAR) (PART 5)

2008-06-02T14:35:00.001+07:00

Value-at-Risk (VaR)

by Simon Benninga and Zvi Wiener

Variance Covariance Approach
This method is based on the assumption that the short term changes in the market parameters and in the value of the portfolio are normal. This method also reflects the fact that the market parameters are not independent, however it is restricted to the first degree of dependence - correlation. First based on historical data we build a vector of average daily changes in each parameter, and a historical variance-covariance matrix of these changes. Second we calculate a linear (so-called delta) approximation of our portfolio for small changes in the market (first term of the Taylor series). Third, we measure the variance of the portfolio assuming normally distributed changes in the market with this variance covariance matrix. Fourth, we
measure the lowest quantile of P&L (profit and loss) for the required level of confidence.
To calculate the additive covariance matrix we use
AddChangesMx = Drop[dataVAR,1] - Drop[dataVAR,-1];
AddMeansVec = Mean[ AddChangesMx ];
AddCovMx = CovarianceMatrix[ AddChangesMx ];
sqrtAddCovar = MatrixPower[ AddCovMx, 1/2 ];
This completes the first step. We also recommend to verify that the covariance matrix is positive definite (and well-defined), since otherwise you may experience some problems with stability.
The mean vector and covariance matrix of the market data are
AddMeansVec
{1, 0.282051, 0.000512821, - 0.0025641}
AddCovMx // MatrixForm
ÊÁÁÁÁÁÁÁÁÁË
0 0 0 0
0 6.10256 -0.0122537 -0.00328408
0 -0.0122537 0.000662888 -0.0000302294
0 -0.00328408 -0.0000302294 0.0007132
ˆ˜˜˜˜˜˜˜˜˜¯
The first line and column correspond to the time change, which is deterministic and thus uncorrelated with the rest of the data.
Since our portfolio is not linear in the market parameters we should measure its sensitivity to a small hanges in each parameter. There are different ways to do this. We demonstrate below a way when we price the portfolio under the current market data, then we increase one parameter by 1% of its value, and price the portfolio again. Then take the finite difference and estimate the derivative.
Denote the price of a portfolio by P, this is a deterministic function of the market data P(x). To measure the
sensitivity we consider P(x + Dix) - P(x) where Dix = (0, 0, ..., 0, 0.01xi , 0, ..., 0). To estimate the partial derivative of P with respect to the i-th argument we calculate
¶P(x)
¶xi
ª
P(x + Dix) - P(x)
0.01xi
.
sensitivity[ portf_, mrkt_]:=
Module[{currentPrice, unitVec,
eps=0.01, deriv, i},
currentPrice = valueP[ portf,mrkt];
unitVec = Table[ 1, {4}];
deriv = Table[(valueP[ portf,
mrkt*ReplacePart[ unitVec, 1+eps, i]]-
currentPrice)/(mrkt[[i]]*eps), {i, 4}]
];
Applying this function to our portfolio we get:
sensitivity[ portfolio, mrkt9Feb97 ]
{-0.0415553, 2., 9.26722, - 84.2265}
We will also need the matrix of second derivatives of the pricing function. Using finite differences it can be calculated as:
ClearAll[secondDeriv];
secondDeriv[portf_, mrkt_]:=
Module[{zerVec= Table[ 0, {4}],
unitVec=Table[ 1, {4}], eps=0.01, i,j},
Table[
(valueP[ portf,
mrkt*(unitVec+ReplacePart[ zerVec, eps, i]+
ReplacePart[ zerVec, eps, j])]-
valueP[ portf,
mrkt*(unitVec+ReplacePart[ zerVec, eps, i]+
ReplacePart[ zerVec, -eps, j])]-
valueP[ portf,
mrkt*(unitVec+ReplacePart[zerVec, -eps, i]+
ReplacePart[ zerVec, eps, j])]+
valueP[ portf,
mrkt*(unitVec+ReplacePart[zerVec, -eps, i]+
ReplacePart[ zerVec, -eps, j])])/4/
(mrkt[[i]]*eps)/(mrkt[[j]]*eps),{i,4},{j,4}]
];
Vol. 7 No. 4 1998 MathematicainEducationandResearch 5
VALUE-AT-RISK (VAR)
secondDeriv[ portfolio,mrkt9Feb97]//MatrixForm
ÊÁÁÁÁÁÁÁÁÁË
-6.02975 ¥ 10-6 0. -0.00649451 -0.0122218
0. 0. 0. 0.
-0.00649451 0. -0.300412 2.72799
-0.0122218 0. 2.72799 0.
ˆ˜˜˜˜˜˜˜˜˜¯
Zeros indicate that the part of our portfolio that depends on the stock market is not influenced directly by the interest and exchange rates. It still is influenced by correlations between differnt market factors, but indirectly only.
Price of a portfolio is a function of the market data, say P(x), where x is the vector of the market data. The
current parameters of the market are known x0, however tomorrow the market will move to a new vector x1. The important simplifying assumption of the variance covariance approach is that the changes of the parameter vector are assumed to be normally distributed. Then we can write
P(x1) = P(x0) + P¢(x0)(x1 - x0)+
1
2
P¢¢(x0)(x1 - x0)2 + o(x1 - x0)
,
We keep the second order term due to Ito’s lemma. Assume that the market parameter x follows a simple arithmetic Brownian motion. dx=mdt + sdB, then by Ito’s lemma we can keep only terms up to the order dt by choosing
P(x1) ª P(x0) + P¢(x0)(mDt +sDB)+
1
2
P¢¢(x0)s2 Dt +o(Dt )
We assume that all components of the vector Dx = x1 - x0 are normally distributed, and the covariance matrix is known, s2= S(x,t). Then the right hand side is a linear combination of normally distributed random variables. As soon as all means are very small (note that daily expected changes are close to zero), the change in value can be approximated by a normally distributed random variable with mean and variance defined below:
mP = E(P(x1) - P(x0)) = Jm.P¢(x0) +
1
2
tr(P¢¢(x0).S)NDt
s2
P = var(P(x1) - P(x0)) = P¢(x0).S.(P¢(x0))T Dt
Here the superscript T denotes transposition and we use the standard vector product of lists. Then the lower qquantile can be approximated by a q-quantile of the normal distribution N(mP ,sP )
VaR(q) = mP + sP × Quantile(N(0, 1), q)
The sign plus here corresponds to the lowest quantile, since, the Quantile(N(0,1), q) is negative for q < sens =" sensitivity[" secderiv="secondDeriv[">

VALUE-AT-RISK (VAR) (PART 6)

2008-06-02T14:30:00.000+07:00

Value-at-Risk (VaR)

by Simon Benninga and Zvi Wiener

Monte Carlo Approach
This method is based on the assumption that we have some information about the joint distribution of market changes. Then using this distribution we can draw randomly a large number of scenarios and price the portfolio for each scenario. A rich set of scenarios will give a good approximation for the distribution of final value of the portfolio. The lowest q-quantile of this distribution can be used as an approximation to VaR. Moreover this method allows a dynamic improvement. One can run a small set of simulations, get a preliminary result and then improve it by running additional simulations if necessary. In the example below we choose the simplest form of the joint distribution - all market parameters are distributed jointly normal with the same mean and covariance as we have measured above. In a general case one can provide any reasonable distribution and use the same method.
nor[mu_,sig_]:=
Random[ NormalDistribution[mu,sig] ];
MCvarAdd[portf_, mrkt_, n_:10]:=
Module[
{tbl, simulatedParam, values},
tbl = Table[ sqrtAddCovar.Table[ nor[0,1],
{Length[AddCovMx]}] + AddMeansVec, {n} ];
simulatedParam = Table[mrkt,{n}] + tbl;
values =
Map[valueP[ portf,#]&, simulatedParam ];
{Mean[values], StandardDeviation[values],
StandardErrorOfSampleMean[values],
values, simulatedParam }
];
To generate a vector of correlated normally distributed random variables we generate first a vector of independent random variables (normally distributed) and then multiply it by the square root of the covariance matrix. It is left as an exercise to verify that this leads to the required correlation. The tbl table is the resulting table of market changes.
The following function incorporates all the necessary steps for Monte Carlo approach to VaR. The parameters of this function are: portf portfolio, mrkt current market 6 MathematicainEducationandResearch Vol. 7 No. 4 1998
VALUE-AT-RISK (VAR)
data, quant quantile, n number of generated scenarios.
This function returns the VaR estimate and the list of all changes in the value of the portfolio that were observed in this simulation.
MCApproachAdd[ portf_, mrkt_, quant_, n_:10 ]:=
Module[{currentValue, changes},
currentValue = valueP[ portf,mrkt];
changes = MCvarAdd[portf, mrkt, n][[4]] -
currentValue;
{Quantile[ changes, quant], changes}
];
To get consistent results this function can be combined with the random seed generator. For 1000 simulations (about 1 minute of computational time) we get:
SeedRandom[1];
MCApproachAdd[ portfolio, mrkt9Feb97, 0.2,
1000][[1]]
-3.71068
Running this simulation again we get a similar result:
SeedRandom[7];
MCApproachAdd[ portfolio, mrkt9Feb97,
0.2, 1000][[1]]
-3.76359
Discussion
The historical simulation method is useful when the amount of data is not very large and we do not have enough information about the profit and loss distribution. It is usually very time consuming, but its main advantage is that it catches all recent market crashes. This feature is very important for risk measurement.
The variance covariance method is the fastest. However it relies heavily on several assumptions about the
distribution of market data and linear approximation of the portfolio. It is probably the best method for quick estimates of VaR. However one should be very careful when using this method for a non-linear portfolio, especially in the case of high convexity in options or bonds. The Monte Carlo simulation method is very slow, but it is probably the most powerful method. It is flexible enough to incorporate private information together with historical observations. There are many methods of speeding calculations, so-called variance reduction techniques. The results of all three methods are similar and our goal was to demonstrate a very basic approach to risk measurement techniques using Mathematica.
REFERENCES
BEDER, TANYA (1996), VAR: Seductive But Dangerous, Financial Analyst Journal, September-October, pp. 12–24.
GRUNDY, BRUCE D. and ZVI WIENER. 1996. The Analysis of VAR, Deltas and State Prices: A New Approach. mimeo The Wharton School, University of Pennsylvania.
HULL, JOHN C. Options Futures, and Other Derivatives, Third Edition. Prentice-Hall, 1997.
JORION, PHILIPPE, Value at Risk, the New Benchmark for Controlling Market Risk, McGraw-Hill, 1997.
LINSMEIER, THOMAS and NEIL PEARSON. Risk Measurement: An Introduction to Value at Risk. mimeo, University of Illinois, 1996.
Several Web sites offer information on value-at-risk:
http://pw2.netcom.com/ bschacht/varbiblio.html—an excellent source of VaR related materials.
http://www.riskmetrics.reuters.com/—Reuters materials on riskmetrics.
http://www.jpmorgan.com/RiskManagement/RiskMetrics/RiskMetrics.html
ABOUT THE AUTHORS
The authors acknowledge grants from Wolfram Research, the Krueger and Eshkol Centers at the Hebrew University, and the Israeli Academy of Science. Wiener’s research has benefited from a grant from the Israel Foundations Trustees, the Alon Fellowship and the Eshkol grant.
Simon Benninga is professor of finance at Tel-Aviv University (Israel) and the Wharton School of the University of Pennsylvania. He is the author of Financial Modeling (MIT Press, 1997) and of Corporate Finance: A Valuation Approach (with Oded Sarig, McGraw-Hill, 1997); he is also the editor of the European Finance Review.
Simon Benninga
Faculty of Management
Tel-Aviv University, Tel-Aviv, Israel
benninga@post.tau.ac.il
http://finance.wharton.upenn.edu/ benninga Zvi Wiener is assistant professor of finance at the business school of the Hebrew University of Jerusalem. His finance research concentrates on the pricing of derivative securities, Value-at-Risk, computational finance and stochastic dominance. He wrote this article while visiting at the Olin School of Business at Washington University in St. Louis.
Zvi Wiener
Finance Department, Business School
Hebrew University, Jerusalem, Israel
mswiener@mscc.huji.ac.il
http://pluto.mscc.huji.ac.il/ mswiener/zvi.html
ELECTRONIC SUBSCRIPTIONS
Included in the distribution for each electronic subscription is the file varisk.nb, containing Mathematica code for the material described in this article.
Vol. 7 No. 4 1998 MathematicainEducationandResearch 7
VALUE-AT-RISK (VAR)
8 MathematicainEducationandResearch Vol. 7 No. 4 1998

Better Performance - LEFT JOIN or NOT IN?

2008-05-27T17:53:00.000+07:00

Which method of T-SQL is better for performance LEFT JOIN or NOT IN when writing query? Answer is : It depends! It all depends on what kind of data is and what kind query it is etc. In that case just for fun guess one option LEFT JOIN or NOT IN. If you need to refer the query which demonstrates the mentioned clauses, review following two queries.

USE AdventureWorks;
GO
SELECT ProductID
FROM Production.Product
WHERE ProductID
NOT IN (
SELECT ProductID
FROM Production.WorkOrder);
GO
SELECT p.ProductID
FROM Production.Product p
LEFT JOIN Production.WorkOrder w ON p.ProductID = w.ProductID
WHERE w.ProductID IS NULL;
GO

Reference : Pinal Dave (http://www.SQLAuthority.com)

SQL SERVER Database Coding Standards and Guidelines Complete List Download

2008-05-27T16:38:00.000+07:00

Just like my previous series of SQL Server Interview Questions and Answers Complete List Download, I have received many comments and emails regarding this series. Once I go through all the emails and comments, I will make summary of them and integrate them with my series. I have also received emails asking me to create PDF for download. I have created that as well. Please feel free to download it and use it.

Please ask me any questions you might have. Contact me if you are interested in writing mini series with me.

Download SQL SERVER Database Coding Standards and Guidelines Complete List

Complete Series of Database Coding Standards and Guidelines
SQL SERVER Database Coding Standards and Guidelines - Introduction
SQL SERVER - Database Coding Standards and Guidelines - Part 1
SQL SERVER - Database Coding Standards and Guidelines - Part 2
SQL SERVER Database Coding Standards and Guidelines Complete List Download

Other popular Series
SQL Server Interview Questions and Answers Complete List Download
SQL SERVER - Data Warehousing Interview Questions and Answers Complete List Download

Taken from : http://blog.sqlauthority.com

Delete Duplicate Records - Rows - Readers Contribution

2008-05-27T16:32:00.000+07:00

This works in 2000. WARNING: According to MS, SET ROWCOUNT will not work with INSERT, DELETE, and UPDATE in later versions.

– Create table with a number of values between zero and nine
select a+b+c as val
into dbo.rmtemp
from (select 0 a union all select 1 union all select 2 union all select 3) a
, (select 0 b union all select 1 union all select 2 union all select 3) b
, (select 0 c union all select 1 union all select 2 union all select 3) c

– Show what you’ve got
select val,count(*) row_count from dbo.rmtemp group by val

– Limit processing to a single row
set rowcount 1
– While you’ve got duplicates, delete a row
while (select top 1 val from dbo.rmtemp group by val having count(*) > 1) is not null
delete from dbo.rmtemp where val in (select top 1 val from dbo.rmtemp group by val having count(*) > 1);
– Remove single row processing limit
set rowcount 0

– Confirm that only uniques remain
select val,count(*) row_count from dbo.rmtemp group by val

– Clean up
drop table dbo.rmtemp

Reference : Pinal Dave (http://www.SQLAuthority.com)

Four Basic SQL Statements - SQL Operations

2008-05-27T16:29:00.000+07:00

There are four basic SQL Operations or SQL Statements.

SELECT - This statement selects data from database tables.

UPDATE - This statement updates existing data into database tables.

INSERT - This statement inserts new data into database tables.

DELETE - This statement deletes existing data from database tables.

If you want complete syntax for this four basic statement, please download FAQ (PDF) from SQL SERVER - Download FAQ Sheet - SQL Server in One Page

Reference : Pinal Dave (http://www.SQLAuthority.com

COMMON SQL PROBLEMS

2008-05-27T16:25:00.000+07:00

COMMON SQL PROBLEMS - 1

NOT USING AN INDEX:

 In most cases, an index should be used for each table in the query. Generally, when an index isn’t used the entire table is scanned. This is bad :>(
 Know what tables are indexed and how. Understand how/when indexes are used. [More slides on this.]
 Understand how certain predicate constructions prevent use of an index. [More slides on this.]
 Use ‘showplan’ to confirm expectations about index usage.
 If an apparently obvious index was not used…understand why.

COMMON SQL PROBLEMS - 2
 Avoid joining too many tables.
 Much depends on the indexes used and the efficiency of those indexes.
 Max is about 6-7 tables on IRF2, but only if joined properly on narrow clustered index keys.
 Avoid joining more than 3 really big tables (> 10^6 rows)
 Avoid excessively complex predicate
 It is easy to write predicates that prevent proper index usage. Avoid this. [More slides on this.]
 Avoid combining two or more special predicate statements like ‘GROUP BY’ with an aggregate and/or ‘SORT’ and/or ‘COMPUTE BY’ and/or ‘HAVING’ et cetera.
 Avoid multiple OR operators in predicates.
 Avoid more than one subquery in a predicate.

COMMON SQL PROBLEMS - 3
 Some properly formed queries legitimately ask the RDBMS server to do a lot of work and may take time to execute.
 Know the size of the objects in the query and try to understand how much work is being requested.
 Start queries off more simply with fewer tables and/or more simple or more restrictive predicates to develop a performance baseline.

COMMON SQL PROBLEMS - 4
 When using cursors, acquire a large stock of garlic, crucifixes and wooden stakes.
 Use cursors ONLY when absolutely necessary. There are always unpredictable performance consequences to the use of cursors.
 Never use cursors when ‘set SQL’ will suffice.
 If cursors must be used, be attentive to transaction blocking issues.
 Some cursors operations require specific types of indexes to support them (like unique or clustered).
 Keep it simple…never use cursors.

By
Don Madden & Jerome Roa
NSIT/CSI

Ordering in SQL

2008-05-27T16:07:00.000+07:00

Ordering Guarantees in SQL Server 2005

SQL is a declarative language that returns multi-sets (sets allowing duplicates) of rows. The operations exposed in SQL, such as join, filter, group by, and project, do not inherently have any ordering guarantees. ANSI SQL does expose an ORDER BY clause for the top-most SELECT scope in a query, and this can be used to return rows in a presentation order to the user (or to a cursor so that you can iterate over the results of query). This is the only operation in ANSI SQL that actually guarantees row order.

Microsoft SQL Server provides additional ordering guarantees beyond ANSI, mostly for backwards-compatibility with previous releases of the product. For example, variable assignment in the top-most SELECT list when an ORDER BY is specified is done in the presentation order of the query.

Example:

SELECT @a = @a + col FROM Table ORDER BY col2;

SQL Server also contains a non-ANSI operator called TOP. TOP allows you to limit a result set to a certain number or percentage of the result set. If an ORDER BY is used in the same scope, it qualifies rows based on the ORDER BY. So, TOP 1 … ORDER BY col will return the “first” row from that result set based on the order by list. However, SQL Server does not guarantee that the rows will be returned in that order from the intermediate result set. It only guarantees which rows actually qualify. You’d need to put an ORDER BY at the top of the query to guarantee the output order returned to the client. (In a previous blog entry, I noted how SQL 2005 actually doesn’t bother processing TOP 100 PERCENT … ORDER BY since it is “meaningless” under this definition).

Other operations in SQL Server also have this “which rows qualify” semantic. ROW_NUMBER, RANK, DENSERANK, and NTILE contain an OVER clause in which an ORDER BY can be specified. This order by guarantees the output of the operation, but not the order in which the rows are output. So, the following is valid output for the row_number query – the function outputs row values as if it had been evaluated in a specific order.

SELECT col2, ROW_NUMBER() OVER (ORDER BY col1) FROM T

Col2 Col1 row_num

1 1 1

3 3 3

2 2 2

Now, in practice, we don’t currently generate a plan that would return rows in this order for this specific query. However, different query plans for the same query can return rows in different order, so it is important to understand what is guaranteed and what is not. When you build an assumption in your SQL application during development about one query plan, then deploy it and a customer with their own data in the database gets a different plan, you can very quickly learn about this dependency the hard way. (A more plausible possibility for this example is that we could assign the row numbers backwards if we knew the exact rowcount from a previously completed operation in the SQL query plan).

Essentially, the SQL Server Query Optimizer is guaranteeing that the internal operator in the query tree will process its input in a particular order. There is no corresponding guarantee that the output of that operator will imply that the next operator in the query tree is performed in that order. The reordering rules can and will violate this assumption (and do so when it is inconvenient to you, the developer ;). Please understand that when we reorder operations to find a more efficient plan, we can cause the ordering behavior to change for intermediate nodes in the tree. If you’ve put an operation in the tree that assumes a particular intermediate ordering, it can break.

by Conor
(Taken from http://blogs.msdn.com/queryoptteam/)

How to Read Statistics Profile in SQL

2008-05-27T15:51:00.001+07:00

In SQL Server, “Statistics Profile” is a mode in which a query is run where you can see the number of invocations of each physical plan operator in the tree. Instead of running a query and just printing the output rows, this mode also collects and returns per-operator row counts. Statistics Profile is used by the SQL Query Optimizer Team to identify issues with a plan which can cause the plan to perform poorly. For example, it can help identify a poor index choice or poor join order in a plan. Oftentimes, it can help identify the needed solution, such as updating statistics (as in the histograms and other statistical information used during plan generation) or perhaps adding a plan hint. This document describes how to read the statistics profile information from a query plan so that you can also debug plan issues.

A simple example query demonstrates how to retrieve the statistics profile output from a query:

use nwind
set statistics profile on
select * from customers c inner join orders o on c.customerid = o.customerid;

The profile output has a number of columns and is a bit tricky to print in a regular document. The key pieces of information that it prints are the plan, which looks like this:

StmtText
------------------------------------------------------------------------------------------------------
select * from customers c inner join orders o on c.customerid = o.customerid
|--Hash Match(Inner Join, HASH:([c].[CustomerID])=([o].[CustomerID]),
|--Clustered Index Scan(OBJECT:([nwind].[dbo].[Customers].[aaaaa_PrimaryKey] AS [c]))
|--Clustered Index Scan(OBJECT:([nwind].[dbo].[Orders].[aaaaa_PrimaryKey] AS [o]))

Other pieces of useful information are the estimated row count and the actual row count for each operator and the estimated and actual number of invocations of this operator. Note that the actual rows and # of executions are physically listed as early columns, while the other columns are listed later in the output column list (so you typically have to scroll over to see them).

Rows Executes
-------------------- --------------------
1078 1
1078 1
91 1
1078 1

EstimateRows
------------------------
1051.5834
1051.5834
91.0
1078.0

EstimateExecutions
------------------------
NULL
1.0
1.0
1.0

Other fields, such as the estimated cost of the subtree, the output columns, the average row size, also exist in the output (but are omitted for space in this document).

Note: The output from statistics profile is typically easiest to read if you set the output from your client (Query Analyzer or SQL Server Management Studio) to output to text, using a fixed-width font. You can then see the columns pretty easily and you can even move the results into a tool like Excel if you want to cluster the estimates and actual results near each other by rearranging the columns (SSMS also can let you do this).

There are a few key pieces of information needed to understand the output from Statistics profile. First, query plans are generally represented as trees. In the output, children are printed below their parents and are indented:
StmtText
------------------------------------------------------------------------------------------------------
select * from customers c inner join orders o on c.customerid = o.customerid
|--Hash Match(Inner Join, HASH:([c].[CustomerID])=([o].[CustomerID]),
|--Clustered Index Scan(OBJECT:([nwind].[dbo].[Customers].[aaaaa_PrimaryKey] AS [c]))
|--Clustered Index Scan(OBJECT:([nwind].[dbo].[Orders].[aaaaa_PrimaryKey] AS [o]))

In this example, the scan of Customers and Orders are both below a hash join operator. Next, the “first” child of the operator is listed first. So, the Customers Scan is the first child of the hash join. Subsequent operators follow, in order. Finally, query execution plans are executed in this “first” to “last” order. So, for this plan, the very first row is returned from the Customers table. (A more detailed discussion of operators will happen later in the article). Notice that the output from the graphical showplan is similar but slightly different:

(see attachment tree.jpg).

In this tree representation, both Scans are printed to the right on the screen, and the first child is above the other operators. In most Computer Science classes, trees are printed in a manner transposed from this:

Hash Join
/ \
Customers Orders

The transposition makes printing trees in text easier. In this classical view, the tree is evaluated left to right with rows flowing bottom to top.

With an understanding of the nuances of the query plan display, it is possible to understand what happens during the execution of this query plan. The query returns 1078 rows. Not coincidentally, there are also 1078 orders in this database. Since there’s a Foreign Key relationship between Orders and Customers, it requires that a match exist for each order to each customer. So, the 91 rows in Customers match the 1078 rows in Orders to return the result.

The query estimates that the join will return 1051.5834 rows. First, this is a bit less than the actual (1078) but is not a substantial difference. Given that the Query Optimizer is making educated guesses based on sampled statistical information that may itself be out-of-date, this estimate is actually pretty good. Second, the number is not an integer because we use floating point for our estimates to improve accuracy on estimates we make. For this query, the number of executions is 1 for both the estimate and actual. This won’t always be the case, but it happens to be true for this query because of the way hash joins work. In a hash join, the first child is scanned and a hash table is built. Once the hash join is built, the second child is then scanned and each row probes the hash table to see if there is a matching row.

Loops join does not work this way, as we’ll see in a slightly modified example.

select * from customers c with (index=1) inner loop join orders o with (index=1) on c.customerid = o.customerid

In this example, I’ve forced a loop join and the use of clustered indexes for each table. The plan now looks like this:

StmtText
-----------------------------------------------------------------
select * from customers c with (index=1) inner loop join orders o
|--Nested Loops(Inner Join, WHERE:([nwind].[dbo].[Orders].[Cust
|--Clustered Index Scan(OBJECT:([nwind].[dbo].[Customers].
|--Table Spool
|--Clustered Index Scan(OBJECT:([nwind].[dbo].[Orders

Beyond the different join algorithm, you’ll notice that there is now a table spool added to the plan. The spool is on the second child (also called the “inner” child for loops join because it is usually invoked multiple times). The spool scans rows from its child and can store them for future invocations of the inner child. The actual row count and execution count from the statistics profile is a bit different from the previous plan:

Rows Executes
-------------------- --------------------
1078 1
1078 1 <--- Loop Join
91 1 <--- Scan of Customers
98098 91 <--- Spool
1078 1 <--- Scan of Orders

In this plan, the second child of the loop join is scanned 91 times returning a total number of 98098 rows. For the actual executions, the total number of rows is to sum of all invocations of that operator, so it is 91*1078=98098. This means that the inner side of this tree is scanned 91 times. Nested Loops joins require rescans of the inner subtree (Hash Joins do not, as you saw in the first example). Note that the spool causes only one scan of the Orders table, and it only has one execution as a result. It isn’t hard to see that there are far more rows touched in this plan compared to the hash join, and thus it shouldn’t be a huge surprise that this plan runs more slowly.

Note: When comparing the estimated vs. actual number of rows, it is important to remember that the actual counts need to be divded by the actual number of executions to get a value that is comparable to the estimated number of rows returned. The estimate is the per-invocation estimate.

As a more complicated example, we can try something with a few more operators and see how things work on one of the TPC-H benchmark queries (Query 8, for those who are interested, on a small-scale 100MB database):

SELECT O_YEAR,
SUM(CASE WHEN NATION = 'MOROCCO'
THEN VOLUME
ELSE 0
END) / SUM(VOLUME) AS MKT_SHARE
FROM ( SELECT datepart(yy,O_ORDERDATE) AS O_YEAR,
L_EXTENDEDPRICE * (1-L_DISCOUNT) AS VOLUME,
N2.N_NAME AS NATION
FROM PART,
SUPPLIER,
LINEITEM,
ORDERS,
CUSTOMER,
NATION N1,
NATION N2,
REGION
WHERE P_PARTKEY = L_PARTKEY AND
S_SUPPKEY = L_SUPPKEY AND
L_ORDERKEY = O_ORDERKEY AND
O_CUSTKEY = C_CUSTKEY AND
C_NATIONKEY = N1.N_NATIONKEY AND
N1.N_REGIONKEY = R_REGIONKEY AND
R_NAME = 'AFRICA' AND
S_NATIONKEY = N2.N_NATIONKEY AND
O_ORDERDATE BETWEEN '1995-01-01' AND '1996-12-31' AND
P_TYPE = 'PROMO BURNISHED NICKEL' AND
L_SHIPDATE >= CONVERT(datetime,(1156)*(30),121) AND L_SHIPDATE < CONVERT(datetime,((1185)+(1))*(30),121)
) AS ALL_NATIONS
GROUP BY O_YEAR
ORDER BY O_YEAR

As the queries get more complex, it gets harder to print them in a standard page of text. So, I’ve truncated the plan somewhat in this example. Notice that the same tree format still exists, and the main operators in this query are Scans, Seeks, Hash Joins, Stream Aggregates, a Sort, and a Loop Join. I’ve also included the actual number of rows and actual number of executions columns as well.

Rows Executes Plan
0 0 Compute Scalar(DEFINE:([Expr1028]=[Expr1026]/[Expr1027]))
2 1 |--Stream Aggregate(GROUP BY:([Expr1024]) DEFINE:([Expr1026]=SUM([par
2 1 |--Nested Loops(Inner Join, OUTER REFERENCES:([N1].[N_REGIONKEY]
10 1 |--Stream Aggregate(GROUP BY:([Expr1024], [N1].[N_REGIONKEY
1160 1 | |--Sort(ORDER BY:([Expr1024] ASC, [N1].[N_REGIONKEY] A
1160 1 | |--Hash Match(Inner Join, HASH:([N2].[N_NATIONKEY
25 1 | |--Clustered Index Scan(OBJECT:([tpch100M].[
1160 1 | |--Hash Match(Inner Join, HASH:([N1].[N_NATI
25 1 | |--Index Scan(OBJECT:([tpch100M].[dbo].
1160 1 | |--Hash Match(Inner Join, HASH:([tpch10
1000 1 | |--Index Scan(OBJECT:([tpch100M].[
1160 1 | |--Hash Match(Inner Join, HASH:([t
1160 1 | |--Hash Match(Inner Join, HAS
1432 1 | | |--Hash Match(Inner Join
126 1 | | | |--Clustered Index
0 0 | | | |--Compute Scalar(D
224618 1 | | | |--Clustered I
0 0 | | |--Compute Scalar(DEFINE
45624 1 | | |--Clustered Index
15000 1 | |--Index Scan(OBJECT:([tpch10
2 10 |--Clustered Index Seek(OBJECT:([tpch100M].[dbo].[REGION].[

I’ll point out a few details about the statistics profile output. Notice that the Compute Scalars (also called Projects) return zero for both columns. Since Compute Scalar always returns exactly as many rows as it is given from its child, there isn’t any logic to count rows again in this operator simply for performance reasons. The zeros can be safely ignored, and the values for its child can be used instead. Another interesting detail can be seen in the last operator in this printout (the Seek into the Region table). In this operator, there are 10 executions but only 2 rows returned. This means that even though there were 10 attempts to find rows in this index, only two rows were ever found. The parent operator (the Nested Loops near the top) has 10 rows coming from its first (left) child and only 2 rows output by the operator, which matches what you see in the seek. Another interesting tidbit can be found if you look at the estimates for the Seek operator:

Est.# rows Est. #executes
1.0 20.106487

The SQL Server Query Optimizer will estimate a minimum of one row coming out of a seek operator. This is done to avoid the case when a very expensive subtree is picked due to an cardinality underestimation. If the subtree is estimated to return zero rows, many plans cost about the same and there can be errors in plan selection as a result. So, you’ll notice that the estimation is “high” for this case, and some errors could result. You also might notice that we estimate 20 executions of this branch instead of the actual 10. However, given the number of joins that have been evaluated before this operator, being off by a factor of 2 (10 rows) isn’t considered to be too bad. (Errors can increase exponentially with the number of joins).

SQL Server supports executing query plans in parallel. Parallelism can add complexity to the statistics profile output as there are different kinds of parallelism that have different impacts on the counters for each operator. Parallel Scans exist at the leaves of the tree, and these will count all rows from the table into each thread even through each thread only returns a fraction of the rows. The number of executions (the second column in the output) will also have 1 execution for each thread. So, it is typical to just divide the number of threads into the total number of rows to see how many rows were actually returned by the table. Parallel zones higher in the tree usually work the same way. These will have N (where N is the degree of parallelism) more executions than the equivalent non-parallel query. There are a few cases where we will broadcast one row to multiple threads. If you examine the type of the parallelism exchange operation, you can identify these cases and notice that one row becomes multiple rows through the counts in the statistics profile results.

The most common use of the statistics profile output is to identify areas where the Optimizer may be seeing and using incomplete or incorrect information. This is often the root cause of many performance problems in queries. If you can identify areas where the estimated and actual cardinality values are far apart, then you likely have found a reason why the Optimizer is not returning the “best” plan. The reasons for the estimate being incorrect can vary, but it can include missing or out-of-date statistics, too low of a sample rate on those statistics, correlations between data columns, or use of operators outside of the optimizer’s statistical model, to name a few common cases.

by QueryOptTeam
Taken From : http://blogs.msdn.com/queryoptteam/

TIPS : an Intro to CHAID Analysis

2008-05-27T15:42:00.000+07:00

CHAID Analysis : Cereal Case*

Below, step by step we give tips & tricks how to make CHAID Analysis with SPSS Software.

Firstly, from SPSS menu, click :
Analyse  Classify  Tree

Then, input ‘Preferred Breakfast’ into Dependent Variabel coloumn, and else variables to Independent Variables.

Click Ok.

Then Click Define Variables Properties at Classification Tree window
Click All Target Category and input to Exclude Coloumn. Click Continue.

At Classification Tree : Output, give checklist at ‘Tree sub menu’, ‘Topdown’ (in Orientation ), ‘Tables & Chart’ (in Node Contents), ‘Automatics’ (in Scale) and ‘Independent variables statistics’, ‘Node Definition’, then click Continue.

At Statistics submenu, give checklist at all item at Model & Node Performancesubmenu, then click continue.

We have result as below:

Model Summary

Specifications Growing Method CHAID
Dependent Variable Preferred breakfast
Independent Variables Age category, Gender, Marital status, Lifestyle
Validation NONE
Maximum Tree Depth 3
Minimum Cases in Parent Node 100
Minimum Cases in Child Node 50
Results Independent Variables Included Age category, Marital status, Lifestyle
Number of Nodes 13
Number of Terminal Nodes 8
Depth 2

Misclassification Costs

Observed Predicted
Breakfast Bar Oatmeal Cereal
Breakfast Bar .000 1.000 1.000
Oatmeal 1.000 .000 1.000
Cereal 1.000 1.000 .000
Dependent Variable: Preferred breakfast

For Cereal. Look at big Chart i.e node 5,8,9 with N=3, 12, 31, with Sum=139. look at table Classification below.

Classification

Observed Predicted
Breakfast Bar Oatmeal Cereal Percent Correct
Breakfast Bar 112 34 85 48.5%
Oatmeal 13 251 46 81.0%
Cereal 84 116 139 41.0%
Overall Percentage 23.8% 45.6% 30.7% 57.0%
Growing Method: CHAID
Dependent Variable: Preferred breakfast

(*) Data Cereal taken from SPSS

Tips: Minitab Tutorials

2008-05-27T15:01:00.001+07:00

All versions of Minitab Statistical Software include step-by-step tutorials accessible through the Help menu.
Minitab also provides the following additional free tutorials:
Meet Minitab
Meet Minitab is a concise guide to help you quickly get started using Minitab. It shows you how to:
• Manage and manipulate data and files
• Produce graphs
• Analyze data and assess quality
• Design an experiment
• Generate reports
• Customize your Minitab software
• Access the Help files provided with Minitab
Meet Minitab also includes a thorough and easy-to-use reference section.
You can download a free electronic copy of Meet Minitab (PDF) or order a hard copy (book version) of the guide. Meet Minitab is also available to download in other languages.
Accessing the Power of Minitab
Tips and tricks to help you quickly harness the power of Minitab Statistical Software and save time.
Answers Knowledgebase
Search our Answers Knowledgebase for detailed information on performing statistical operations, as well as answers to hundreds of the most frequently asked technical support questions.
For articles about using Minitab for basic and applied statistics, visit our Help With Statistics page.

Taken from : http://www.minitab.com/resources/tutorials/

C++ Programming (Part 1)

2008-05-27T14:25:00.001+07:00

C++ is a general-purpose, platform-neutral programming language that supports object-oriented programming and other useful programming paradigms, including procedural programming, object-based programming, generic programming, and functional programming.
C++ is viewed as a superset of C, and thus offers backward compatibility with this language. This reliance on C provides important benefits:
• Reuse of legacy C code in new C++ programs
• Efficiency
• Platform neutrality
• Relatively quick migration from C to C++
Yet it also incurs certain complexities and ailments such as manual memory management, pointers, unchecked array bounds, and a cryptic declarator syntax, as described in the following sections.
As opposed to many other programming languages, C++ doesn't have versions. Rather, it has an International ANSI/ISO Standard, ratified in 1998, that defines the core language, its standard libraries, and implementation requirements. The C++ Standard is treated as a skeleton on which vendors might add their own platform-specific extensions, mostly by means of code libraries. However, it's possible to develop large-scale applications using pure standard C++, thereby ensuring code portability and easier maintenance.
The primary reason for selecting C++ is its support of object-oriented programming. Yet even as a procedural programming language, C++ is considered an improvement over ANSI C in several aspects. C programmers who prefer for various reasons not to switch to object-oriented programming can still benefit from the migration to C++ because of its tighter type-safety, strongly typed pointers, improved memory management, and many other features that make a programmer's life easier. Let's look at some of these improvements more closely:
• Improved memory management. In C, you have to call library functions to allocate storage dynamically and release it afterwards. But C++ treats dynamically allocated objects as first-class citizens: it uses the keywords new and delete to allocate and deallocate objects dynamically.
• User-defined types are treated as built-in types. For example, a struct or a union's name can be used directly in declarations and definitions just as a built-in type:
• struct Date
• {
• int day;
• int month;
• int year;
• };
•
• Date d; //In C, 'struct' is required before Date
void func(Date *pdate); //ditto
• Pass-by-reference. C has two types of argument passing: by address and by value. C++ defines a third argument-passing mechanism: passing by reference. When you pass an argument by reference, the callee gets an alias of the original object and can modify it. In fact, references are rather similar to pointers in their semantics; they're efficient because the callee doesn't get a copy of the original variable, but rather a handle that's bound to the original object. Syntactically, however, references look like variables that are passed by value. Here's an example:
• Date date;
• void func(Date &date_ref); //func takes Date by reference
func(date); // date is passed by reference, not by value
• Default argument values. C++ allows you to declare functions that take default argument values. When the function call doesn't provide such an argument, the compiler automatically inserts its respective default value into the function call. For example:
• void authorize(const string & username, bool log=true);
authorize(user); // equivalent to: authorize(user, true);
In the function call above, the programmer didn't provide the second argument. Because this argument has a default value, the compiler silently inserted true as a second argument.
• Mandatory function prototypes. In classic C, functions could be called without being previously declared. In C++, you must either declare or define a function before calling it. This way, the compiler can check the type and number of each argument. Without mandatory prototypes, passing arguments by reference or using default argument values wouldn't be possible because the compiler must replace the arguments with their references or add the default values as specified in the prototype.
A well-formed C++ program must contain a main() function and a pair of matching braces:
int main()
{}
Though perfectly valid, this program doesn't really do anything. To get a taste of C++, let's look at a more famous example:
#include
int main()
{
std::cout<<"hello world!"<}
If you're a C programmer with little or no prior experience in C++, the code snippet above might shock you. It's entirely different from the equivalent C program:
#include
int main()
{
printf("hello world!\n");
}
Let's take a closer look at the C++ example. The first line is a preprocessor directive that #includes the standard header in the program's source file:
#include
contains the declarations and definitions of the standard C++ I/O routines and classes. Taking after C, the creators of C++ decided to implement I/O support by means of a code library rather than a built-in keyword. The following line contains the main() function. The program consists of a single line:
std::cout<<"hello world!"<Let's parse it:
• std::cout is the qualified name of the standard output stream object, cout. This object is automatically created whenever you #include .
• The overloaded insertion operator << comes next. It takes an argument and passes it on to cout. In this case, the argument is a literal string that we want to print on the screen.
• The second argument, std::endl, is a manipulator that appends the newline character after the string and forces buffer flushing.
As trivial as this program seems, it exposes some of the most important features of C++. For starters, C++ is an object-oriented language. Therefore, it uses objects rather than functions to perform I/O. Secondly, the standard libraries of C++, including , are declared in the namespace std (std stands for standard. An exhaustive discussion about namespaces is available in the "Namespaces" section.). Thus, instead of declaring standard functions, classes, and objects globally, C++ declares them in a dedicated namespace—thereby reducing the chances of clashing with user code and third-party libraries that happen to use the same names.

Path Analysis

2008-05-27T13:15:00.000+07:00

Overview
Path analysis is an extension of the regression model, used to test the fit of the correlation matrix against two or more causal models which are being compared by the researcher. The model is usually depicted in a circle-and-arrow figure in which single arrows indicate causation. A regression is done for each variable in the model as a dependent on others which the model indicates are causes. The regression weights predicted by the model are compared with the observed correlation matrix for the variables, and a goodness-of-fit statistic is calculated. The best-fitting of two or more models is selected by the researcher as the best model for advancement of theory.
Path analysis requires the usual assumptions of regression. It is particularly sensitive to model specification because failure to include relevant causal variables or inclusion of extraneous variables often substantially affects the path coefficients, which are used to assess the relative importance of various direct and indirect causal paths to the dependent variable. Such interpretations should be undertaken in the context of comparing alternative models, after assessing their goodness of fit discussed in the section on structural equation modeling (SEM packages are commonly used today for path analysis in lieu of stand-alone path analysis programs). When the variables in the model are latent variables measured by multiple observed indicators, path analysis is termed structural equation modeling, treated separately. We follow the conventional terminology by which path analysis refers to single-indicator variables.
Key Concepts and Terms
Note that path estimates may be calculated by OLS regression or by MLE maximum likelihood estimation, depending on the computer package. Two-Stage Least Squares (2SLS), discussed separately, is another path estimation procedure designed to extend the OLS regression model to situations where non-recursivity is introduced because the researcher must assume the covariances of some disturbance terms are not 0 (this assumption is discussed below). Click here for a separate discussion.
• Path model. A path model is a diagram relating independent, intermediary, and dependent variables. Single arrows indicate causation between exogenous or intermediary variables and the dependent(s). Arrows also connect the error terms with their respective endogenous variables. Double arrows indicate correlation between pairs of exogenous variables. Sometimes the width of the arrows in the path model are drawn in a width which is proportional to the absolute magnitude of the corresponding path coefficients (see below).
• Causal paths to a given variable include (1) the direct paths from arrows leading to it, and (2) correlated paths from endogenous variables correlated with others which have arrows leading to the given variable. Consider this model:

This model has correlated exogenous variables A, B, and C, and endogenous variables D and E. Error terms are not shown. The causal paths relevant to variable D are the paths from A to D, from B to D, and the paths reflecting common anteceding causes -- the paths from B to A to D, from C to A to D, and from C to B to D. Paths involving two correlations (C to B to A to D) are not relevant. Likewise, paths that go backward (E to B to D, or E to B to A to D) reflect common effects and are not relevant.
• Exogenous and endogenous variables. Exogenous variables in a path model are those with no explicit causes (no arrows going to them, other than the measurement error term). If exogenous variables are correlated, this is indicated by a double-headed arrow connecting them. Endogenous variables, then, are those which do have incoming arrows. Endogenous variables include intervening causal variables and dependents. Intervening endogenous variables have both incoming and outgoing causal arrows in the path diagram. The dependent variable(s) have only incoming arrows.
• Path coefficient/path weight. A path coefficient is a standardized regression coefficient (beta) showing the direct effect of an independent variable on a dependent variable in the path model. Thus when the model has two or more causal variables, path coefficients are partial regression coefficients which measure the extent of effect of one variable on another in the path model controlling for other prior variables, using standardized data or a correlation matrix as input. Recall that for bivariate regression, the beta weight (the b coefficient for standardized data) is the same as the correlation coefficient, so for the case of a path model with a variable as a dependent of a single exogenous variable (and an error residual term), the path coefficient in this special case is a zero-order correlation coefficient.
Consider this model, based on Bryman, A. and D. Cramer (1990). Quantitative data analysis for social scientists, pp. 246-251.

This model is specified by the following path equations:
Equation 1. satisfaction = b11age + b12autonomy + b13 income + e1
Equation 2. income = b21age + b22autonomy + e2
Equation 3. autonomy = b31age + e3
where the b's are the regression coefficients and their subscripts are the equation number and variable number (thus b21 is the coefficient in Equation 2 for variable 1, which is age.
Note: In each equation, only (and all of) the direct priors of the endogenous variable being used as the dependent are considered. The path coefficients, which are the betas in these equations, are thus the standardized partial regression coefficients of each endogenous variable on its priors. That is, the beta for any path (that is, the path coefficient) is a partial weight controlling for other priors for the given dependent variable.
Formerly called p coefficients, now path coefficients are called simply beta weights, based on usage in multiple regression models. Bryman and Cramer computed the path coefficients = standardized regression coefficients = beta weights, to be:

Correlated Exogenous Variables. If exogenous variables are correlated, it is common to label the corresponding double-headed arrow between them with its correlation coefficient.
Disturbance terms.The residual error terms, also called disturbance terms, reflect unexplained variance (the effect of unmeasured variables) plus measurement error. Note that the dependent in each equation is an endogenous variable (in this case, all variables except age, which is exogenous). Note also that the independents in each equation are all the variables with arrows to the dependent.
The effect size of the disturbance term for a given endogenous variable, which reflects unmeasured variables, is (1 - R2), and its variance is (1 - R2) times the variance of that endogenous variable, where R2 is based on the regression in which it is the dependent and those variables with arrows to it are independents. The path coefficient is SQRT(1 - R2).
The correlation between two disturbance terms is the partial correlation of the two endogenous variables, using as controls all their common causes (all variables with arrows to both). The covariance estimate is the partial covariance: the partial correlation times the product of the standard deviations of the two endogenous variables.
• Path multiplication rule: The value of any compound path is the product of its path coefficients. Imagine a simple three-variable compound path where education causes income causes conservatism. Let the regression coefficient of income on education be 1000: for each year of education, income goes up $1,000. Let the regression coefficient of conservatism on income be .0002: for every dollar income goes up, conservativism goes up .0002 points on a 5-point scale. Thus if education goes up 1 year, income goes up $1,000, which means conservatism goes up .2 points. This is the same as multiplying the coefficients: 1000*.0002 = .2. The same principle would apply if there were more links in the path. If standardized path coefficients (beta weights) were used, the path multiplication rule would still apply, but the the interpretation is in standardized terms. Either way, the product of the coefficients along the path reflects the weight of that path.
• Effect decomposition. Path coefficients may be used to decompose correlations in the model into direct and indirect effects, corresponding, of course, to direct and indirect paths reflected in the arrows in the model. This is based on the rule that in a linear system, the total causal effect of variable i on variable j is the sum of the values of all the paths from i to j. Considering "satisfaction" as the dependent in the model above, and considering "age" as the independent, the indirect effects are calculated by multiplying the path coefficients for each path from age to satisfaction:
age -> income -> satisfaction is .57*.47 = .26
age -> autonomy -> satisfaction is .28*.58 = .16
age -> autonomy -> income -> satisfaction is .28*.22 x .47 = .03
total indirect effect = .45
That is, the total indirect effect of age on satisfaction is plus .45. In comparison, the direct effect is only minus .08. The total causal effect of age on satisfaction is (-.08 + .45) = .37.
Effect decomposition is equivalent to effects analysis in regression with one dependent variable. Path analysis, however, can also handle effect decomposition for the case of two or more dependent variables.
In general, any bivariate correlation may be decomposed into spurious and total causal effects, and the total causal effect can be decomposed into a direct and an indirect effect. The total causal effect is the coefficient in a regression with all of the model's prior but not intervening variables for x and y controlled (the beta coefficient for the usual standardized solution, the partial b coefficient for the unstandardized or raw solution). The spurious effect is the total effect minus the total causal effect. The direct effect is the partial coefficient (beta for standardized, b for unstandardized) for y on x controlling for all prior variables and all intervening variables in the model. The indirect effect is the total causal effect minus the direct effect, and measures the effect of the intervening variables. Where effects analysis in regression may use a variety of coefficients (partial correlation or regression, for instance), effect decomposition in path analysis is restricted to use of regression.
For instance, imagine a five-variable model in which the exogenous variable Education is correlated with the exogenous variable Skill Level, and both Education and Skill Level are correlated with the exogenous variable Job Status. Further imagine that Education and each of the other two exogenous variables are modeled to be direct causes of Income and also of Median House Value, which are the two dependent variables. We might then decompose the correlation of Education and Income:
1. Direct effect of Education on Income, indicated by the path coefficient of the single-headed arrow from Education to Income.
2. Indirect effect due to Education's correlation with Skill Level, and Skill Level's direct effect on Income, indicated by multiplying the correlation of Education and Skill Level by the path coefficient from Skill Level to Income.
3. Indirect effect due to Education's correlation with Job Status, and Job Status's direct effect on Income, indicated by multiplying the correlation of Education and Job Status by the path coefficient from Job Status to Income.
As a second example decomposition for the same five-variable model is a bit more complex if we wish to break down the correlation of the two dependent variables, Income and Median House Value. Since here somewhat implausibly the two dependents are modeled not to have a direct effect from Income to House Value, the true correlation is hypothesized to be zero and all correlations are spurious.
4. The spurious direct effect of Education as a common anteceding variable directly causing both dependents, indicated by multiplying the path coefficient from Education to Income by the path coefficient of Education to House Value.
5. The spurious direct effect of Skill Level as a common anteceding variable directly causing both dependents, indicated by multiplying the path coefficient from Skill Level to Income by the path coefficient of Skill Level to House Value.
6. The spurious direct effect of Job Status as a common anteceding variable directly causing both dependents, indicated by multiplying the path coefficient from Job Status to Income by the path coefficient of Job Status to House Value.
7. The spurious indirect effect of Education and Skill Level as a common antecedings variable directly causing both dependents, indicated by multiplying the path coefficient from Education to Income by the correlation of Education and Skill Level by the path from Skill Level to House Value and adding the product of the path from Skill Level to Income by the correlation of Education and Skill Level by the path from Education to Median House Value.
8. The spurious indirect effect of Education and Job Status as a common anteceding variables directly causing both dependents, indicated by multiplying the path coefficient from Education to Income by the correlation of Education and Job Status by the path from Job Status to House Value and adding the product of the path from Job Status to Income by the correlation of Education and Job Status by the path from Education to Median House Value..
9. The spurious indirect effect of Skill Level and Job Status as a common anteceding variables directly causing both dependents, indicated by multiplying the path coefficient from Skill Level to Income by the correlation of Skill Level and Job Status by the path from Job Status to House Value and adding the product of the path from Job Status to Income by the correlation of Skill Level and Job Status by the path from Skill Level to Median House Value..
10. The residual effect is the difference between the correlation of Income and Median House Value and the sum of the spurious direct and indirect effects.
Correlated exogenous variables. The path weights connecting correlated exogenous variables are equal to the Pearson correlations. When calculating indirect paths, not only direct arrows but also the double-headed arrows connecting correlated exogenous variables, are used in tracing possible indirect paths, except:
Tracing rule: An indirect path cannot enter and exit on an arrowhead. This means that you cannot have a direct path composed of the paths of two correlated exogenous variables.
• Significance and Goodness of Fit in Path Models
o To test individual path coefficients one uses the standard t or F test from regression output.
o To test the model with all its paths one uses a goodness of fit test from a structural equation modeling program. If a model is correctly specified, including all relevant and excluding all irrelevant variables, with arrows correctly indicated, then the sum of path values from i to j will equal the regression coefficient for j predicted on the basis of i. That is, for standardized data, where the bivariate regression coefficient equals the correlation coefficient, the sum of path coefficients (standardized) will equal the correlation coefficient. This means one can compare the path-estimated correlation matrix with the observed correlation matrix to assess the goodness-of-fit of path models. As a practical matter, goodness-of-fit is calculated by entering the model and its data into a structural equation modeling program such as LISREL or AMOS, which compute a variety of alternative goodness-of-fit coefficients, discussed separately.
o To modify the path model on uses modification indexes (MI) to add arrows and uses nonsignificance of path coefficients to drop arrows, in a model-building and model-trimming process discussed in the section on structural equation modeling.

Assumptions
• Linearity: relationships among variables are linear (though, of course, variables may be nonlinear transforms).
• Additivity: there are no interaction effects (though, of course, variables may be interaction crossproduct terms)
• Interval level data for all variables, if regression is being used to estimate path parameters. As in other forms of regression modeling, it is common to use dichotomies and ordinal data in practice. If dummy variables are used to code a categorical variable, one must be careful that they are represented as a block in the path diagram (ex., if an arrow is drawn to one dummy it must be drawn to all others in the set). If an arrow were to be drawn from one dummy variable to another dummy variable in the same set, this would violate the recursivity assumption discussed below.
• Residual (unmeasured) variables are uncorrelated with any of the variables in the model other than the one they cause.
• Disturbance terms are uncorrelated with endogenous variables. As a corollary of the previous assumption, path analysis assumes that for any endogenous variable, its distubance term is uncorrelated with any other endogenous variable in the model. This is a critical assumption, violation of which may make regression inappropriate as a method of estimating path parameters. This assumption may be violated due to measurement error in measuring an endogenous variable; when an endogenous variable is actually a direct or indirect cause of a variable which the model states is the cause of that endogenous variable (reverse causation); or when a variable not in the model is a cause of an endogenous variable and a variable the model specifies as a cause of that endogenous variable (spurious causation).
• Low multicollinearity (otherwise one will have large standard errors of the b coefficients used in removing the common variance in partial correlation analysis).
• No underidentification or underdetermination of the model is required. For underidentified models there are too few structural equations to solve for the unknowns. Overidentification usually provides better estimates of the underlying true values than does just identification.
o Recursivity: all arrows flow one way, with no feedback looping. Also, it is assumed that disturbance (residual error) terms for the endogenous variables are uncorrelated. Recursive models are never underidentified.
• Proper specification of the model is required for interpretation of path coefficients. Specification error occurs when a significant causal variable is left out of the model. The path coefficients will reflect the shared covariance with such unmeasured variables and will not be accurately interpretable in terms of direct and indirect effects. In particular, if a variable specified as prior to a given variable is really consequent to it, "we can do ourselves considerable damage" (Davis, 1985: 64) because if a variable is consequent it would be estimated to have no path effect, whereas when it is included as a prior variable in the model, this erroneously changes the coefficients for other variables in the model. Note, however, that while interpretation of path coefficients is inaccurate under specification error, it is still possible to compare the relative fit of two models, perhaps both with specification error.
• Appropriate correlation input. When using a correlation matrix as input, it is appropriate to use Pearsonian correlation for two interval variables, polychoric correlation for two ordinals, tetrachoric for two dichotomies, polyserial for an interval and an ordinal, and biserial for an interval and a dichotomy.
• Adequate sample size is needed to assess significance. Kline (1998) recommends 10 times as many cases as parameters (or ideally 20 times). He states that 5 times or less is insufficient for significance testing of model effects.
• The same sample is required for all regressions used to calculate the path model. This may require reducing the data set down so that there are no missing values for any of the variables included in the model. This might be achieved by listwise dropping of cases or by data imputation.

Taken from : http://www2.chass.ncsu.edu/garson/pa765/path.htm

Central Limit Theorem

2008-05-26T11:51:00.000+07:00

The central limit theorem (CLT) states that the sum of a large number of independent and identically-distributed random variables will be approximately normally distributed (i.e., following a Gaussian distribution, or bell-shaped curve) if the random variables have a finite variance. Formally, a central limit theorem is any of a set of weak-convergence results in probability theory. They all express the fact that any sum of many independent identically distributed random variables will tend to be distributed according to a particular "attractor distribution".

Since many real populations yield distributions with finite variance, this explains the prevalence of the normal probability distribution. For other generalizations for finite variance which do not require identical distribution, see Lindeberg condition, Lyapunov condition, Gnedenko and Kolmogorov states.

History

Tijms (2004, p.169) writes:

“The central limit theorem has an interesting history. The first version of this theorem was postulated by the French-born mathematician Abraham de Moivre, who, in a remarkable article published in 1733, used the normal distribution to approximate the distribution of the number of heads resulting from many tosses of a fair coin. This finding was far ahead of its time, and was nearly forgotten until the famous French mathematician Pierre-Simon Laplace rescued it from obscurity in his monumental work Théorie Analytique des Probabilités, which was published in 1812. Laplace expanded De Moivre's finding by approximating the binomial distribution with the normal distribution. But as with De Moivre, Laplace's finding received little attention in his own time. It was not until the nineteenth century was at an end that the importance of the central limit theorem was discerned, when, in 1901, Russian mathematician Aleksandr Lyapunov defined it in general terms and proved precisely how it worked mathematically. Nowadays, the central limit theorem is considered to be the unofficial sovereign of probability theory.”

A thorough account of the theorem's history, detailing Laplace's foundational work, as well as Cauchy's, Bessel's and Poisson's contributions, is provided by Hald.^[1] Two historic accounts, one covering the development from Laplace to Cauchy, the second the contributions by von Mises, Pólya, Lindeberg, Lévy, and Cramér during the 1920s, are given by Hans Fischer.^[2] See Bernstein (1945) for a historical discussion focusing on the work of Pafnuty Chebyshev and his students Andrey Markov and Aleksandr Lyapunov that led to the first proofs of the C.L.T. in a general setting.

Classical central limit theorem

The central limit theorem is also known as the second fundamental theorem of probability. (The Law of large numbers is the first.) Let X₁, X₂, X₃, ... be a set of n independent and identically distributed random variables having finite values of mean µ and variance σ² > 0. The central limit theorem states that as the sample size n increases, the distribution of the sample average approaches the normal distribution with a mean µ and variance σ²/n irrespective of the shape of the original distribution.

Let the sum of the random variables be S_n, given by

S_n = X₁ + ... + X_n. Then, defining

the distribution of Z_n converges towards the standard normal distribution N(0,1) as n approaches ∞ (this is convergence in distribution).^[3] This means: if Φ(z) is the cumulative distribution function of N(0,1), then for every real number z, we have

or,

where

is the sample mean. ^[3] ^[4]

Proof of the central limit theorem

For a theorem of such fundamental importance to statistics and applied probability, the central limit theorem has a remarkably simple proof using characteristic functions. It is similar to the proof of a (weak) law of large numbers. For any random variable, Y, with zero mean and unit variance (var(Y) = 1), the characteristic function of Y is, by Taylor's theorem,

where o (t² ) is "little o notation" for some function of t that goes to zero more rapidly than t². Letting Y_i be (X_i − μ)/σ, the standardized value of X_i, it is easy to see that the standardized mean of the observations X₁, X₂, ..., X_n is

By simple properties of characteristic functions, the characteristic function of Z_n is

But, this limit is just the characteristic function of a standard normal distribution, N(0,1), and the central limit theorem follows from the Lévy continuity theorem, which confirms that the convergence of characteristic functions implies convergence in distribution.

Convergence to the limit

If the third central moment E((X₁ − μ)³) exists and is finite, then the above convergence is uniform and the speed of convergence is at least on the order of 1/n^½ (see Berry-Esséen theorem).

The convergence normal is monotonic, in the sense that the entropy of Z_n increases monotonically to that of the normal distribution, as proven by Artstein, Ball, Barthe and Naor^{[citation needed]}.

Pictures of a distribution being "smoothed out" by summation (showing original density of distribution and three subsequent summations, obtained by convolution of density functions):

(See Illustration of the central limit theorem for further details on these images.)

A graphical representation of the central limit theorem can be formed by plotting random means of a population. Consider A_n. A_n will represent the mean of a random sample and X_n represents a single random variable from the sample:

A_n = (X₁ + ... + X_n) / n. Derive A_n from 1 to whichever sample size.

A₁ = (X₁) / 1

A₂ = (X₁ + X₂)/ 2

A₃ = (X₁ + X₂ + X₃)/3

For the CLT, it is recommended to plot the means upwards to 30 points (sample size 30).If we standardize A_n by setting Z_n = (A_n − μ) / (σ / n^½), we obtain the same variable Z_n as above, and it approaches a standard normal distribution.

The Central Limit Theorem, as an approximation for a finite number of observations, provides a reasonable approximation only when close to the peak of the normal distribution; it requires a very large number of observations to stretch into the tails.

The Central Limit theorem applies in particular to sums of independent and identically distributed discrete random variables. A sum of discrete random variables is still a discrete random variable, so that we are confronted to a sequence of discrete random variables whose cumulative probability distribution function converges towards a cumulative probability distribution function corresponding to a continuous variable (namely that of the normal distribution). This means that if we build a histogram of the realisations of the sum of n independent identical discrete variables, the curve that joins the centers of the upper faces of the rectangles forming the histogram converges toward a gaussian curve as n approaches . The binomial distribution article details such an application of the central limit theorem in the simple case of a discrete variable taking only two possible values.

Relation to the law of large numbers

The law of large numbers as well as The Central Limit Theorem are partial solutions to a general problem: "What is the limiting behavior of S_n as n approaches infinity?" In mathematical analysis, asymptotic series is one of the most popular tools employed to approach such questions.

Suppose we have an asymptotic expansion of f(n):

dividing both parts by and taking the limit will produce a₁ - the coefficient at the highest-order term in the expansion representing the rate at which f(n) changes in its leading term.

Informally, one can say: "f(n) grows approximately as ". Taking the difference between f(n) and its approximation and then dividing by the next term in the expansion we arrive to a more refined statement about f(n):

here one can say that: "the difference between the function and its approximation grows approximately as " The idea is that dividing the function by appropriate normalizing functions and looking at the limiting behavior of the result can tell us much about the limiting behavior of the original function itself.

Informally, something along these lines is happening when S_n is being studied in classical probability theory. Under certain regularity conditions, by The Law of Large Numbers, and by The Central Limit Theorem, where ξ is distributed as N(0,σ²) which provide values of first two constants in informal expansion:

It could be shown^{[citation needed]} that if X₁, X₂, X₃, ... are i.i.d. and for some then hence is the largest power of n which if serves as a normalizing function would provide a non-trivial (non-zero) limiting behavior. Interestingly enough, The Law of the Iterated Logarithm tells us what is happening "in between" The Law of Large Numbers and The Central Limit Theorem. Specifically it says that the normalizing function intermediate in size between n of The Law of Large Numbers and of The Central Limit Theorem provides a non-trivial limiting behavior.

Alternative statements of the theorem

Density functions

The density of the sum of two or more independent variables is the convolution of their densities (if these densities exist). Thus the central limit theorem can be interpreted as a statement about the properties of density functions under convolution: the convolution of a number of density functions tends to the normal density as the number of density functions increases without bound, under the conditions stated above.

Since the characteristic function of a convolution is the product of the characteristic functions of the densities involved, the central limit theorem has yet another restatement: the product of the characteristic functions of a number of density functions tends to the characteristic function of the normal density as the number of density functions increases without bound, under the conditions stated above.

An equivalent statement can be made about Fourier transforms, since the characteristic function is essentially a Fourier transform.

Products of positive random variables

The central limit theorem tells us what to expect about the sum of independent random variables, but what about the product? Well, the logarithm of a product is simply the sum of the logs of the factors, so the log of a product of random variables that take only positive values tends to have a normal distribution, which makes the product itself have a log-normal distribution. Many physical quantities (especially mass or length, which are a matter of scale and cannot be negative) are the product of different random factors, so they follow a log-normal distribution.

Whereas the central limit theorem for sums of random variables requires the condition of finite variance, the corresponding theorem for products requires the corresponding condition that the density function be square-integrable (see Rempala 2002).

Lyapunov condition

Let X_n be a sequence of independent random variables defined on the same probability space. Assume that X_n has finite expected value μ_n and finite standard deviation σ_n. We define

Assume that the third central moments

are finite for every n, and that

(This is the Lyapunov condition). We again consider the sum , its expected value is and its standard deviation is s_n, if we standardize it by setting

then the distribution of Z_n converges to the standard normal distribution N(0,1).

Lindeberg condition

In the same setting and with the same notation as above, we can replace the Lyapunov condition with the following weaker one (from Lindeberg in 1920). For every ε > 0

\varepsilon s_n \right) = 0" class="tex" v:shapes="_x0000_i1058" border="0" height="50" width="381">

where E( U : V > c) is E( U 1{V > c}), i.e., the expectation of the random variable U 1{V > c} whose value is U if V > c and zero otherwise. Then the distribution of the standardized sum Z_n converges towards the standard normal distribution N(0,1).

Non-independent case

There are some theorems which treat the case of sums of non-independent variables, for instance:

Applications and examples

There are a number of useful and interesting examples arising from the central limit theorem. Below are brief outlines of two such examples and here are a large number of CLT applications, presented as part of the SOCR CLT Activity.

The probability distribution for total distance covered in a random walk (biased or unbiased) will tend toward a normal distribution.
Flipping a large number of coins will result in a normal distribution for the total number of heads (or equivalently total number of tails).

Signal processing

Signals can be smoothed by applying a Gaussian filter, which is just the convolution of a signal with an appropriately scaled Gaussian function. Due to the central limit theorem this smoothing can be approximated by several filter steps that can be computed much faster, like the simple moving average.

The central limit theorem implies that to achieve a Gaussian of variance σ² n filters with windows of variances with must be applied.

Notes

^ Andreas Hald, History of Mathematical Statistics from 1750 to 1930, Ch.17.
^ Hans Fischer: (1) "The Central Limit Theorem from Laplace to Cauchy: Changes in Stochastic Objectives and in Analytical Methods"; (2) "The Central Limit Theorem in the Twenties".
^ ^a ^b For decades, large sample size was set as n > 29; however, research since 1990, has indicated larger samples, such as 100 or 250, might be needed if the population is skewed far from normal: the more skew, the larger the sample needed. The conditions might be rare, but critical when they occur: computer animations are used to illustrate the cases. The cutoff with n > 29 has allowed Student-t tables to format in limited pages; however, that sample size might be too small. See below "Using graphics and simulation.." by Marasinghe et al, and see "Identification of Misconceptions in the Central Limit Theorem and Related Concepts and Evaluation of Computer Media as a Remedial Tool" by Yu, Chong Ho and Dr. John T. Behrens, Arizona State University & Spencer Anthony, Univ. of Oklahoma, Annual Meeting of the American Educational Research Association, presented April 19, 1995, paper revised in Feb 12, 1997, webpage (accessed 2007-10-25): CWisdom-rtf.
^ Marasinghe, M., Meeker, W., Cook, D. & Shin, T.S.(1994 August), "Using graphics and simulation to teach statistical concepts", Paper presented at the Annual meeting of the American Statistician Association, Toronto, Canada.

References

Henk Tijms, Understanding Probability: Chance Rules in Everyday Life, Cambridge: Cambridge University Press, 2004.
S. Artstein, K. Ball, F. Barthe and A. Naor, "Solution of Shannon's Problem on the Monotonicity of Entropy", Journal of the American Mathematical Society 17, 975-982 (2004).
S.N.Bernstein, On the work of P.L.Chebyshev in Probability Theory, Nauchnoe Nasledie P.L.Chebysheva. Vypusk Pervyi: Matematika. (Russian) [The Scientific Legacy of P. L. Chebyshev. First Part: Mathematics] Edited by S. N. Bernstein.] Academiya Nauk SSSR, Moscow-Leningrad, 1945. 174 pp.
G. Rempala and J. Wesolowski, "Asymptotics of products of sums and U-statistics", Electronic Communications in Probability, vol. 7, pp. 47-54, 2002.

Source : http://en.wikipedia.org/wiki/Central_limit_theorem

Introduction to Probability

2008-05-26T11:45:00.000+07:00

Charles M. Grinstead

Swarthmore College

J. Laurie Snell

Dartmouth College

Probability theory began in seventeenth century France when the two great French mathematicians, Blaise Pascal and Pierre de Fermat, corresponded over two problems from games of chance. Problems like those Pascal and Fermat solved continued to influence such early researchers as Huygens, Bernoulli, and DeMoivre in establishing a mathematical theory of probability. Today, probability theory is a wellestablished branch of mathematics that finds applications in every area of scholarly activity from music to physics, and in daily experience from weather prediction to predicting the risks of new medical treatments.

This text is designed for an introductory probability course taken by sophomores, juniors, and seniors in mathematics, the physical and social sciences, engineering, and computer science. It presents a thorough treatment of probability ideas and techniques necessary for a firm understanding of the subject. The text can be used in a variety of course lengths, levels, and areas of emphasis.

For use in a standard one-term course, in which both discrete and continuous probability is covered, students should have taken as a prerequisite two terms of calculus, including an introduction to multiple integrals. In order to cover Chapter 11, which contains material on Markov chains, some knowledge of matrix theory is necessary.

The text can also be used in a discrete probability course. The material has been organized in such a way that the discrete and continuous probability discussions are presented in a separate, but parallel, manner. This organization dispels an overly rigorous or formal view of probability and offers some strong pedagogical value in that the discrete discussions can sometimes serve to motivate the more abstract continuous probability discussions. For use in a discrete probability course, students should have taken one term of calculus as a prerequisite.

Very little computing background is assumed or necessary in order to obtain full benefits from the use of the computing material and examples in the text. All of the programs that are used in the text have been written in each of the languages TrueBASIC, Maple, and Mathematica.

This book is on the Web at http://www.dartmouth.edu/˜chance, and is part of the Chance project, which is devoted to providing materials for beginning courses in probability and statistics. The computer programs, solutions to the odd-numbered exercises, and current errata are also available at this site. Instructors may obtain all of the solutions by writing to either of the authors, at jlsnell@dartmouth.edu and cgrinst1@swarthmore.edu

For detail, please look at : http://www.dartmouth.edu/~chance/teaching_aids/books_articles/probability_book/amsbook.mac.pdf

Some Example in Probability Theory

2008-05-26T11:33:00.000+07:00

Problem:	A spinner has 4 equal sectors colored yellow, blue, green and red. What are the chances of landing on blue after spinning the spinner? What are the chances of landing on red?	Top of Form Bottom of Form
Solution:	The chances of landing on blue are 1 in 4, or one fourth.
	The chances of landing on red are 1 in 4, or one fourth.

This problem asked us to find some probabilities involving a spinner. Let's look at some definitions and examples from the problem above.

Definition	Example
An experiment is a situation involving chance or probability that leads to results called outcomes.	In the problem above, the experiment is spinning the spinner.
An outcome is the result of a single trial of an experiment.	The possible outcomes are landing on yellow, blue, green or red.
An event is one or more outcomes of an experiment.	One event of this experiment is landing on blue.
Probability is the measure of how likely an event is.	The probability of landing on blue is one fourth.

In order to measure probabilities, mathematicians have devised the following formula for finding the probability of an event.

Probability Of An Event

P(A) =	The Number Of Ways Event A Can Occur
	The Total Number Of Possible Outcomes

The probability of event A is the number of ways event A can occur divided by the total number of possible outcomes. Let's take a look at a slight modification of the problem from the top of the page.

Experiment 1:

A spinner has 4 equal sectors colored yellow, blue, green and red. After spinning the spinner, what is the probability of landing on each color?

Outcomes:

The possible outcomes of this experiment are yellow, blue, green, and red.

Probabilities:

P(yellow)	=	number of ways to land on yellow	=	1
		total number of colors		4

P(blue)	=	number of ways to land on blue	=	1
		total number of colors		4

P(green)	=	number of ways to land on green	=	1
		total number of colors		4

P(red)	=	number of ways to land on red	=	1
		total number of colors		4

Experiment 2:

A single 6-sided die is rolled. What is the probability of each outcome? What is the probability of rolling an even number? of rolling an odd number?

Top of Form

Bottom of Form

Outcomes:

The possible outcomes of this experiment are 1, 2, 3, 4, 5 and 6.

Probabilities:

P(1)	=	number of ways to roll a 1	=	1
		total number of sides		6

P(2)	=	number of ways to roll a 2	=	1
		total number of sides		6

P(3)	=	number of ways to roll a 3	=	1
		total number of sides		6

P(4)	=	number of ways to roll a 4	=	1
		total number of sides		6

P(5)	=	number of ways to roll a 5	=	1
		total number of sides		6

P(6)	=	number of ways to roll a 6	=	1
		total number of sides		6

P(even)	=	# ways to roll an even number	=	3	=	1
		total number of sides		6		2

P(odd)	=	# ways to roll an odd number	=	3	=	1
		total number of sides		6		2

Experiment 2 illustrates the difference between an outcome and an event. A single outcome of this experiment is rolling a 1, or rolling a 2, or rolling a 3, etc. Rolling an even number (2, 4 or 6) is an event, and rolling an odd number (1, 3 or 5) is also an event.

In Experiment 1 the probability of each outcome is always the same. The probability of landing on each color of the spinner is always one fourth. In Experiment 2, the probability of rolling each number on the die is always one sixth. In both of these experiments, the outcomes are equally likely to occur. Let's look at an experiment in which the outcomes are not equally likely.

Experiment 3:

A glass jar contains 6 red, 5 green, 8 blue and 3 yellow marbles. If a single marble is chosen at random from the jar, what is the probability of choosing a red marble? a green marble? a blue marble? a yellow marble?

Outcomes:

The possible outcomes of this experiment are red, green, blue and yellow.

Probabilities:

P(red)	=	number of ways to choose red	=	6	=	3
		total number of marbles		22		11

P(green)	=	number of ways to choose green	=	5
		total number of marbles		22

P(blue)	=	number of ways to choose blue	=	8	=	4
		total number of marbles		22		11

P(yellow)	=	number of ways to choose yellow	=	3
		total number of marbles		22

The outcomes in this experiment are not equally likely to occur. You are more likely to choose a blue marble than any other color. You are least likely to choose a yellow marble.

Experiment 4:

Choose a number at random from 1 to 5. What is the probability of each outcome? What is the probability that the number chosen is even? What is the probability that the number chosen is odd?

Outcomes:

The possible outcomes of this experiment are 1, 2, 3, 4 and 5.

Probabilities:

P(1)	=	number of ways to choose a 1	=	1
		total number of numbers		5

P(2)	=	number of ways to choose a 2	=	1
		total number of numbers		5

P(3)	=	number of ways to choose a 3	=	1
		total number of numbers		5

P(4)	=	number of ways to choose a 4	=	1
		total number of numbers		5

P(5)	=	number of ways to choose a 5	=	1
		total number of numbers		5

P(even)	=	number of ways to choose an even number	=	2
		total number of numbers		5

P(odd)	=	number of ways to choose an odd number	=	3
		total number of numbers		5

The outcomes 1, 2, 3, 4 and 5 are equally likely to occur as a result of this experiment. However, the events even and odd are not equally likely to occur, since there are 3 odd numbers and only 2 even numbers from 1 to 5.

Summary:

The probability of an event is the measure of the chance that the event will occur as a result of an experiment. The probability of an event A is the number of ways event A can occur divided by the total number of possible outcomes. The probability of an event A, symbolized by P(A), is a number between 0 and 1, inclusive, that measures the likelihood of an event in the following way:

If P(A) > P(B) then event A is more likely to occur than event B.
If P(A) = P(B) then events A and B are equally likely to occur.

Source: http://www.mathgoodies.com/lessons/vol6/intro_probability.html