INSTRUCTIONS FOR THE 2011 DIVERSITY IMMIGRANT VISA PROGRAM (DV-2011)
The congressionally mandated Diversity Immigrant Visa Program is administered on an annual basis by the Department of State and conducted under the terms of Section 203(c) of the Immigration and Nationality Act (INA). Section 131 of the Immigration Act of 1990 (Pub. L. 101-649) amended INA 203 and provides for a class of immigrants known as ―diversity immigrants.‖ Section 203(c) of the INA provides a maximum of 55,000 Diversity Visas (DVs) each fiscal year to be made available to persons from countries with low rates of immigration to the United States. The annual DV program makes visas available to persons meeting simple, but strict, eligibility requirements. A computer-generated, random lottery drawing chooses selectees for DVs. The visas are distributed among six geographic regions, with a greater number of visas going to regions with lower rates of immigration, and with no visas going to nationals of countries sending more than 50,000 immigrants to the United States over the period of the past five years. Within each region, no single country may receive more than seven percent of the available DVs in any one year.
For DV-2011, natives of the following countries1 are not eligible to apply because the countries sent a total of more than 50,000 immigrants to the United States in the previous five years: BRAZIL, CANADA, CHINA (mainland-born), COLOMBIA, DOMINICAN REPUBLIC, ECUADOR, EL SALVADOR, GUATEMALA, HAITI, INDIA, JAMAICA, MEXICO, PAKISTAN, PERU, PHILIPPINES, POLAND, SOUTH KOREA, UNITED KINGDOM (except Northern Ireland) and its dependent territories, and VIETNAM. Persons born in Hong Kong SAR, Macau SAR, and Taiwan are eligible. For DV-2011, no countries have been added or removed from the previous year’s list of eligible countries. The Department of State implemented the electronic registration system beginning with DV-2005 in order to make the DV process more efficient and secure. The Department utilizes special technology and other means to identify those who commit fraud for the purposes of illegal immigration or those who submit multiple entries. DV REGISTRATION PERIOD Entries for the DV-2011 DV lottery must be submitted electronically between noon, Eastern Daylight Time (EDT) (GMT-4), Friday, October 2, 2009, and noon, Eastern Standard Time (EST) (GMT-5), Monday, November 30, 2009. Applicants may access the electronic DV Entry Form (E-DV) at during the registration period. Paper entries will not be accepted. Applicants are strongly encouraged not to wait until the last week of the registration period to enter. Heavy demand may result in website delays. No entries will be accepted after noon, EST, on November 30, 2009.
1 The term "country" in this notice includes countries, economies, and other jurisdictions explicitly listed at the end of these instructions.
2
REQUIREMENTS FOR ENTRY
To enter the DV lottery, you must be a native of one of the listed countries. See List Of Countries By Region Whose Natives Qualify.
Native of a country whose natives qualify: In most cases, this means the country in which you were born. However, there are two other ways you may be able to qualify. First, if you were born in a country whose natives are ineligible but your spouse was born in a country whose natives are eligible, you can claim your spouse’s country of birth—provided that both you and your spouse are on the selected entry, are issued visas, and enter the United States simultaneously. Second, if you were born in a country whose natives are ineligible, but neither of your parents was born there or resided there at the time of your birth, you may claim nativity in one of your parents’ countries of birth if it is a country whose natives qualify for the DV-2011 program.
To enter the lottery, you must meet either the education or work experience requirement of the DV program: You must have either a high school education or its equivalent, defined as successful completion of a 12-year course of elementary and secondary education; OR two years of work experience within the past five years in an occupation requiring at least two years of training or experience to perform. The U.S. Department of Labor’s O*Net OnLine database will be used to determine qualifying work experience. For more information about qualifying work experience, see Frequently Asked Question #13.
If you cannot meet either of these requirements, you should NOT submit an entry to the DV program. PROCEDURES FOR SUBMITTING AN ENTRY TO DV-2011
The Department of State will only accept completed E-DV entry forms submitted electronically at www.dvlottery.state.gov during the registration period between noon, EDT (GMT-4), Friday, October 2, 2009 and noon, EST (GMT-5), Monday, November 30, 2009.
All entries by an individual will be disqualified if more than ONE entry for that individual is received, regardless of who submitted the entry. You may prepare and submit your own entry or have someone submit the entry for you.
A successfully registered entry will result in a confirmation screen containing your name and a unique confirmation number. You should print this confirmation screen for your records using the print function of your web browser. Starting July 1, 2010, you will be able to check the status of your DV-2011 entry by returning to the website and entering your unique confirmation number and personal information.
Paper entries will not be accepted.
It is very important that all required photographs be submitted. Your entry will be disqualified if all required photographs are not submitted. Recent photographs of the following people must be submitted electronically with the E-DV entry form:
You
Your spouse
Each unmarried child under 21 years of age at the time of your electronic entry, including all natural children as well as all legally adopted children and stepchildren, even if a child no longer resides with you or you do not intend for a child to immigrate under the DV program.
You do not need to submit a photo for a child who is already a U.S. citizen or a Lawful Permanent Resident. Group or family photographs will not be accepted; there must be a separate photograph for each family member. Failure to submit the required photographs for your spouse and each child listed will result in an incomplete entry to the E-DV system. The entry will not be accepted and must be resubmitted. Failure to enter the correct photograph of each individual into the E-DV system will result in disqualification of the principal applicant and refusal of all visas associated with the case at the time of the visa interview.
A digital photograph (image) of you, your spouse, and each child must be submitted online with the E-DV entry form. The image file can be produced either by taking a new digital photograph or by scanning a photographic print with a digital scanner.
Entries are subject to disqualification and visa refusal for cases in which the photographs are not recent or have been manipulated or fail to meet the specifications explained below.
3
Instructions for Submitting a Digital Photograph (Image) The image file must adhere to the following compositional and technical specifications and can be produced in one of the following ways: taking a new digital image or using a digital scanner to scan a photograph. Entrants may test their photos for suitability through the photo validator link on the e-DV website before submitting their entries. The photo validator provides additional technical advice on photo composition, along with examples of acceptable and unacceptable photos. Compositional Specifications: The submitted digital image must conform to the following compositional specifications or the entry will be disqualified.
Head Position
o The person being photographed must directly face the camera.
o The head of the person should not be tilted up, down, or to the side.
o The head height or facial region size (measured from the top of the head, including the hair, to the bottom of the chin) must be between 50 percent and 69 percent of the image's total height. The eye height (measured from the bottom of the image to the level of the eyes) should be between 56 percent and 69 percent of the image's height.
Background
o The person being photographed should be taken with the person in front of a neutral, light-colored background.
o Dark or patterned backgrounds are not acceptable.
Focus
o The photograph must be in focus.
Decorative Items
o Photographs in which the person being photographed is wearing sunglasses or other items that detract from the face will not be accepted.
Head Coverings and Hats
Photos of applicants wearing head coverings or hats are only acceptable if the head covering is worn for religious beliefs; even then, the head covering may not obscure any portion of the face of the applicant. Photographs of applicants with tribal or other headgear not specifically religious in nature will not be accepted; photographs of military, airline, or other personnel wearing hats will not be accepted.
Color photographs in 24-bit color depth are required. Color photographs may be downloaded from a camera to a file in the computer or they may be scanned onto a computer. If you are using a scanner, the settings must be for True Color or 24-bit color mode. See the additional scanning requirements below.
Technical Specifications
The submitted digital photograph must conform to the following specifications or the system will automatically reject the E-DV entry form and notify the sender.
Taking a New Digital Image. If a new digital image is taken, it must meet the following specifications:
Image File Format:
The image must be in the Joint Photographic Experts Group (JPEG) format.
Image File Size:
The maximum image file size is 240 kilobytes (240 KB).
Image Resolution and Dimensions:
Minimum acceptable dimensions are 600 pixels (width) x 600 pixels (height). Image pixel dimensions must be in a square aspect ratio (meaning the height must be equal to the width).
Image Color Depth:
24-bit color [Note: Color photographs are required. Black and white,
4
monochrome images (2-bit color depth), 8-bit color, or 8-bit grayscale will not be accepted.]
Scanning a Submitted Photograph. Before a photographic print is scanned, it must meet the compositional specifications listed above. If the photographic print meets the print color and compositional specifications, scan the print using the following scanner specifications:
Scanner Resolution:
Scanned at a resolution of at least 150 dots per inch (dpi).
Image File Format:
The image must be in the Joint Photographic Experts Group (JPEG) format.
Image File Size:
The maximum image file size is 240 kilobytes (240 KB).
Image Resolution:
600 by 600 pixels.
Image Color Depth:
24-bit color. [Note that black and white, monochrome, or grayscale images will not be accepted.]
INFORMATION REQUIRED FOR THE ELECTRONIC ENTRY
There is only one way to enter the DV-2011 lottery. You must submit the DS 5501, the Electronic Diversity Visa Entry Form (E-DV Entry Form), which is accessible only online at www.dvlottery.state.gov. Failure to complete the form in its entirety will disqualify the entry. Those who submit the E-DV entry will be asked to include the following information on the E-DV Entry Form.
NOTE: The Department of State strongly encourages applicants to complete the application without the assistance of “Visa Consultants,” “Visa Agents,” or other individuals who offer to submit an application on behalf of applicants. In many cases, these facilitators substitute their address for an applicant’s address and thereby receive the selection notification instead of it being received by the actual applicant. Subsequently, the visa facilitators extort money from the selectees in order to receive the notification information that should rightly have gone directly to the DV selectee.
1. FULL NAME – Last/Family Name, First Name, Middle name
2. DATE OF BIRTH – Day, Month, Year
3. GENDER – Male or Female
4. CITY WHERE YOU WERE BORN
5. COUNTRY WHERE YOU WERE BORN – The name of the country should be that which is currently in use for the place where you were born.
6. COUNTRY OF ELIGIBILITY OR CHARGEABILITY FOR THE DV PROGRAM – Your country of eligibility will normally be the same as your country of birth. Your country of eligibility is not related to where you live. If you were born in a country that is not eligible for the DV program, please review the instructions to see if there is another option for country chargeability available for you. For additional information on chargeability, please review ―Frequently Asked Question #1‖ of these instructions.
7. ENTRY PHOTOGRAPH(S) – See the technical information on photograph specifications. Make sure you include photographs of your spouse and all your children, if applicable. See Frequently Asked Question #3.
8. MAILING ADDRESS – In Care Of, Address Line 1, Address Line 2, City/Town, District/Country/Province/State, Postal Code/Zip Code, and Country
9. COUNTRY WHERE YOU LIVE TODAY
10. PHONE NUMBER (optional)
11. E-MAIL ADDRESS – provide an email address to which you have direct access. You will NOT receive an official selection letter at this address. However, if your entry is selected and you respond
5
to the official letter you receive from the Kentucky Consular Center (KCC), you may receive follow-up communication from them by email.
12. WHAT IS THE HIGHEST LEVEL OF EDUCATION YOU HAVE ACHIEVED, AS OF TODAY? You must indicate which one of the following represents your own highest level of educational achievement: (1) Primary school only, (2) High school, no degree, (3) High school degree, (4) Vocational school, (5) Some university courses, (6) University degree, (7) Some graduate level courses, (8) Master degree, (9) Some doctorate level courses, and (10) Doctorate degree
13. MARITAL STATUS – Unmarried, Married, Divorced, Widowed, or Legally Separated
14. NUMBER OF CHILDREN – Entries MUST include the name, date, and place of birth of your spouse and all natural children, as well as all legally adopted children and stepchildren who are unmarried and under the age of 21 on the date of your electronic entry (do not include children who are already U.S. citizens or Lawful Permanent Residents), even if you are no longer legally married to the child’s parent, and even if the spouse or child does not currently reside with you and/or will not immigrate with you. Note that married children and children 21 years or older are not eligible for the DV; however, U.S. law protects children from ―aging out‖ in certain circumstances. If your electronic DV entry is made before your unmarried child turns 21, and the child turns 21 before visa issuance, he/she may be protected from aging out by the Child Status Protection Act and be treated as though he/she were under 21 for visa-processing purposes. Failure to list all children who are eligible will result in disqualification of the principal applicant and refusal of all visas in the case at the time of the visa interview. See Frequently Asked Question #11. 15. SPOUSE INFORMATION – Name, Date of Birth, Gender, City/Town of Birth, Country of Birth, and Photograph. Failure to list your spouse will result in disqualification of the principal applicant and refusal of all visas in the case at the time of the visa interview. 16. CHILDREN INFORMATION – Name, Date of Birth, Gender, City/Town of Birth, Country of Birth, and Photograph: Include all children declared in question #14 above.
SELECTION OF APPLICANTS The computer will randomly select individuals from among all qualified entries. The selected individuals will be notified by mail between May and July 2010; the notification letters will provide further instructions, including information on fees connected with immigration to the United States. Those selected in the random drawing are NOT notified by e-mail. Those individuals NOT selected will NOT receive any notification. U.S. embassies and consulates will not provide a list of successful entrants. Successful entrants’ spouses and unmarried children under age 21 may also apply for visas to accompany or follow-to-join the principal applicant. DV-2011 visas will be issued between October 1, 2010, and September 30, 2011. Processing of entries and issuance of DVs to successful individuals and their eligible family members MUST occur by midnight on September 30, 2011. Under no circumstances can DVs be issued or adjustments approved after this date, nor can family members obtain DVs to follow-to-join the principal applicant in the United States after this date. In order to receive a DV to immigrate to the United States, those chosen in the random drawing must meet ALL eligibility requirements under U.S. law. These requirements may significantly increase the level of scrutiny required and time necessary for processing for natives of some countries listed in this notice including, but not limited to, countries identified as state sponsors of terrorism. Important Notice No fee is charged for the electronic lottery entry in the annual DV program. The U.S. Government employs no outside consultants or private services to operate the DV program. Any intermediaries or others who offer assistance to prepare DV entries do so without the authority or consent of the U.S. Government. Use of any outside intermediary or assistance to prepare a DV entry is entirely at the entrant’s discretion. A qualified electronic entry submitted directly by an applicant has an equal chance of being randomly selected by the computer at the KCC, as does a qualified electronic entry received from an outside intermediary on behalf of
6
the applicant. However, receipt of more than one entry per person will disqualify the person from registration, regardless of the source of the entry. FREQUENTLY ASKED QUESTIONS
1. WHAT DO THE TERMS “ELIGIBILITY”, “NATIVE” AND “CHARGEABILITY” MEAN? ARE THERE ANY SITUATIONS IN WHICH PERSONS WHO WERE NOT BORN IN A QUALIFYING COUNTRY MAY APPLY?
Your country of eligibility will normally be the same as your country of birth. Your country of eligibility is not related to where you live. ―Native‖ ordinarily means someone born in a particular country, regardless of the individual’s current country of residence or nationality. For immigration purposes, ―native‖ can also mean someone who is entitled to be ―charged‖ to a country other than the one in which he/she was born under the provisions of Section 202(b) of the Immigration and Nationality Act. For example, if you were born in a country that is not eligible for this year’s DV program, you may claim chargeability to the country where your derivative spouse was born, but you will not be issued a DV-1 unless your spouse is also eligible for and issued a DV-2, and both of you must enter the United States together with the DVs. In a similar manner, a minor dependent child can be ―charged‖ to a parent’s country of birth. Finally, if you were born in a country not eligible to participate in this year’s DV program, you can be ―charged‖ to the country of birth of either of your parents as long as neither parent was a resident of the ineligible country at the time of the your birth. In general, people are not considered residents of a country in which they were not born or legally naturalized if they are only visiting the country, studying in the country temporarily, or stationed temporarily in the country for business or professional reasons on behalf of a company or government from a country other than the country in which the applicant was born. If you claim alternate chargeability, you must indicate such information on the E-DV electronic online Entry Form, in question #6. Please be aware that listing an incorrect country of eligibility or chargeability (i.e., one to which you cannot establish a valid claim) may disqualify your entry.
2. ARE THERE ANY CHANGES OR NEW REQUIREMENTS IN THE APPLICATION PROCEDURES FOR THIS DV REGISTRATION?
Yes, you must provide an email address this year as part of your entry, it is no longer optional. If you are selected, you will still receive an official letter from KCC by regular mail, but KCC may send other communications to you by email. Please provide a personal email address that you can access rather than using someone else’s address or a standard company address. All other requirements for DV-2011 remain the same. The Entry Status Check will be available for DV-2011 beginning July 1, 2010. If you applied for the DV-2010 program, you may check the status of your entry until the end of June 2010.
3. ARE SIGNATURES AND PHOTOGRAPHS REQUIRED FOR EACH FAMILY MEMBER, OR ONLY FOR THE PRINCIPAL ENTRANT?
Signatures are not required on the E-DV Entry Form. Recent and individual photographs of you, your spouse, and all children under 21 years of age are required. Family or group photographs are not accepted. Refer to information on the photograph requirements located in this bulletin.
4. WHY DO NATIVES OF CERTAIN COUNTRIES NOT QUALIFY FOR THE DV PROGRAM?
DVs are intended to provide an immigration opportunity for persons from countries other than the countries that send large numbers of immigrants to the United States. The law states that no DVs shall be provided for natives of ―high-admission‖ countries. The law defines this to mean countries from which a total of 50,000 persons in the Family-Sponsored and Employment-Based visa categories immigrated to the United States during the previous five years. Each year, U.S. Citizenship and Immigration Services (USCIS) adds the family and employment immigrant admission figures for the previous five years to identify the countries whose
7
natives will be ineligible for the annual diversity lottery. Because there is a separate determination made before each annual E-DV entry period, the list of countries whose natives are not eligible may change from one year to the next.
5. WHAT IS THE NUMERICAL LIMIT FOR DV-2011?
By law, the U.S. DV program makes available a maximum of 55,000 permanent residence visas each year to eligible persons. However, the Nicaraguan Adjustment and Central American Relief Act (NACARA) passed by Congress in November 1997 stipulates that beginning as early as DV-1999, and for as long as necessary, up to 5,000 of the 55,000 annually-allocated DVs will be made available for use under the NACARA program. The actual reduction of the limit by up to 5,000 DVs began with DV-2000 and is likely to remain in effect through the DV-2011 program.
6. WHAT ARE THE REGIONAL DV LIMITS FOR DV-2011?
USCIS determines the regional DV limits for each year according to a formula specified in Section 203(c) of the INA. Once USCIS has completed these calculations, the regional visa limits will be announced.
7. WHEN WILL ENTRIES FOR THE DV-2011 PROGRAM BE ACCEPTED?
The DV-2011 entry period will run through the registration period listed above. Each year, millions of people apply for the program during the registration period. The massive volume of entries creates an enormous amount of work in selecting and processing successful individuals. Holding the entry period during October, November, and December will ensure that selectees are notified in a timely manner and gives both the visa applicants and our embassies and consulates time to prepare and complete cases for visa issuance. You are strongly encouraged to enter early in the registration period. Excessive demand at end of the registration period may slow the system down. No entries whatsoever will be accepted after noon EST Monday November 30, 2009.
8. MAY PERSONS WHO ARE IN THE UNITED STATES APPLY FOR THE PROGRAM?
Yes, an applicant may be in the United States or in another country, and the entry may be submitted from the United States or from abroad.
9. IS EACH APPLICANT LIMITED TO ONLY ONE ENTRY DURING THE ANNUAL E-DV REGISTRATION PERIOD?
Yes, the law allows only one entry by or for each person during each registration period. Individuals for whom more than one entry is submitted will be disqualified. The Department of State will employ sophisticated technology and other means to identify individuals who submit multiple entries during the registration period. People submitting more than one entry will be disqualified, and an electronic record will be permanently maintained by the Department of State. Individuals may apply for the program each year during the regular registration period.
10. MAY A HUSBAND AND A WIFE EACH SUBMIT A SEPARATE ENTRY?
Yes, a husband and a wife may each submit one entry if each meets the eligibility requirements. If either is selected, the other is entitled to derivative status.
11. WHAT FAMILY MEMBERS MUST I INCLUDE ON MY E-DV ENTRY?
On your entry you must list your spouse (husband or wife) and all unmarried children under 21 years of age, with the exception of children who are already U.S. citizens or Lawful Permanent Residents. You must list your spouse even if you are currently separated from him/her, unless you are legally separated (i.e., there is a written agreement recognized by a court or a court order). If you are legally separated or divorced, you do not need to list your former spouse. You must list ALL your children who are unmarried and under 21 years of age at the time of your initial E-DV entry, whether they are your natural children, your spouse’s children, or children you have formally adopted in accordance with the laws of your country, unless such child is already a U.S. citizen or Lawful Permanent Resident. List all children under 21 years of age at
8
the time of your electronic entry, even if they no longer reside with you or you do not intend for them to immigrate under the DV program. The fact that you have listed family members on your entry does not mean that they must travel with you. They may choose to remain behind. However, if you include an eligible dependent on your visa application forms that you failed to include on your original entry, your case will be disqualified. This only applies to those who were family members at the time the original application was submitted, not those acquired at a later date. Your spouse may still submit a separate entry, even though he or she is listed on your entry, as long as both entries include details on all dependents in your family. See question #10 above.
12. MUST I SUBMIT MY OWN ENTRY, OR MAY SOMEONE ACT ON MY BEHALF?
You may prepare and submit your own entry, or have someone submit the entry for you. Regardless of whether an entry is submitted by the individual directly, or assistance is provided by an attorney, friend, relative, etc., only one entry may be submitted in the name of each person, and the entrant remains responsible for ensuring that information in the entry is correct and complete. If the entry is selected, the notification letter will be sent only to the mailing address provided on the entry. All entrants, including those not selected, will be able to check the status of their entry through the official DV website. Entrants should keep their own confirmation page information so that they may independently check the status of their entry.
13. WHAT ARE THE REQUIREMENTS FOR EDUCATION OR WORK EXPERIENCE?
The law and regulations require that every entrant must have at least a high school education or its equivalent or have, within the past five years, two years of work experience in an occupation requiring at least two years’ training or experience. A ―high school education or equivalent‖ is defined as successful completion of a twelve-year course of elementary and secondary education in the United States or successful completion in another country of a formal course of elementary and secondary education comparable to a high school education in the United States. Only formal courses of study meet this requirement; correspondence programs or equivalency certificates (such as the G.E.D.) are not acceptable. Documentary proof of education or work experience must be presented to the consular officer at the time of the visa interview. What Occupations qualify for the DV Program?
To determine eligibility based on work experience, definitions from the Department of Labor’s (DOL) O*Net OnLine database will be used. The O*Net Online Database groups job experience into five ―job zones.‖ While many occupations are listed on the DOL Website, only certain specified occupations qualify for the DV Program. To qualify for a DV on the basis of your work experience, you must have, within the past five years, two years of experience in an occupation that is designated as Job Zone 4 or 5, classified in a Specific Vocational Preparation (SVP) range of 7.0 or higher. How Do I Find the Qualifying Occupations on the Department of Labor Website? Qualifying DV Occupations are shown on the DOL O*Net Online Database. Follow these steps to find out if your occupation qualifies: Select ―Find Occupations‖ and then select a specific ―Job Family.‖ For example, select Architecture and Engineering and click ―GO.‖ Then click on the link for the specific Occupation. Following the same example, click Aerospace Engineers. After selecting a specific Occupation link, select the tab ―Job Zone‖ to find out the designated Job Zone number and Specific Vocational Preparation (SVP) rating range.
14. HOW WILL SUCCESSFUL ENTRANTS BE SELECTED?
At the KCC, all entries received from each region will be individually numbered. After the end of the registration period, a computer will randomly select entries from among all the entries received for each geographic region. Within each region, the first entry randomly selected will be the first case registered; the second entry selected the second registration, etc. All entries received during the registration period will have an equal chance of being selected within each region. When an entry has been selected, the entrant will be sent a notification letter by the KCC, which will provide visa application instructions. The KCC will continue to process the case until those selected to be visa applicants are instructed to appear for visa interviews at a
9
U.S. consular office or until those qualifying to change status in the United States apply at a domestic USCIS office. Important Note: Notifications to those selected in the random lottery are not sent by e-mail. Official notifications of selection are sent by regular postal mail from the KCC. Should you receive an email notification about your E-DV selection, be aware that the message is not legitimate. It is only after you are selected, notified by an official letter sent via postal mail, and processing begins on your case, that you may receive follow-up email communication from the KCC. The KCC will not ask you to send money to them by mail or by services such as Western Union.
15. MAY SELECTEES ADJUST THEIR STATUS WITH USCIS?
Yes, provided they are otherwise eligible to adjust status under the terms of Section 245 of the INA, selected individuals who are physically present in the United States may apply to USCIS for adjustment of status to permanent resident. Applicants must ensure that USCIS can complete action on their cases, including processing of any overseas derivatives, before September 30, 2011, since on that date registrations for the DV-2011 program expire. No visa numbers for the DV-2011 program will be available after midnight EST on September 30, 2011, under any circumstances.
16. WILL ENTRANTS WHO ARE NOT SELECTED BE INFORMED?
All entrants, including those NOT selected, may check the status of their entry through the E-DV website and find out if their entry was or was not selected. Entrants should keep their own confirmation page information from the time of their entry until they may check the status of their entry online. Status information for DV-2011 will be available online from July 1, 2010, until June 30, 2011. (Status information for the previous DV lottery, DV-2010, is available online from July 1, 2009, until June 30, 2010.) All official notification letters are sent to the address indicated on the entry within five to seven months from the end of the application period.
17. HOW MANY INDIVIDUALS WILL BE SELECTED?
There are 50,000 DV visas available for DV-2011, but because it is likely that some of the first 50,000 persons who are selected will not qualify for visas or pursue their cases to visa issuance, more than 50,000 entries will be selected by the KCC to ensure that all of the available DV visas are issued. However, this also means that there will not be a sufficient number of visas for all those who are initially selected. All applicants who are selected will be informed promptly of their place on the list. Interviews for the DV-2011 program will begin in October 2010. The KCC will send appointment letters to selected applicants four to six weeks before the scheduled interviews with U.S. consular officers at overseas posts. Each month, visas will be issued to those applicants who are ready for issuance during that month, visa-number availability permitting. Once all of the 50,000 DV visas have been issued, the program will end. In principle, visa numbers could be finished before September 2011. Selected applicants who wish to receive visas must be prepared to act promptly on their cases. Random selection by the KCC computer as a selectee does not automatically guarantee that you will receive a visa. You must qualify for the visa as well.
18. IS THERE A MINIMUM AGE FOR APPLICANTS TO APPLY FOR THE E-DV PROGRAM?
There is no minimum age to apply for the program, but the requirement of a high school education or work experience for each principal applicant at the time of application will effectively disqualify most persons who are under age 18.
19. ARE THERE ANY FEES FOR THE E-DV PROGRAM?
There is no fee for submitting an electronic lottery entry. DV applicants must pay all required visa fees at the time of visa application directly to the consular cashier at the embassy or consulate. Details of required DV and immigration visa application fees will be included with the instructions sent by the KCC to applicants who are selected.
20. DO DV APPLICANTS RECEIVE WAIVERS OF ANY GROUNDS OF VISA INELIGIBILITY OR RECEIVE SPECIAL PROCESSING FOR A WAIVER APPLICATION?
10
Applicants are subject to all grounds of ineligibility for immigrant visas specified in the INA. There are no special provisions for the waiver of any ground of visa ineligibility aside from those ordinarily provided in the INA, nor is there special processing for waiver requests. Some general waiver provisions for people with close relatives who are U.S. Citizens or Lawful Permanent Resident aliens may be available to DV applicants as well, but the time constraints in the DV program will make it difficult for applicants to benefit from such provisions.
21. MAY PERSONS WHO ARE ALREADY REGISTERED FOR AN IMMIGRANT VISA IN ANOTHER CATEGORY APPLY FOR THE DV PROGRAM?
Yes, such persons may apply for the DV program.
22. HOW LONG DO APPLICANTS WHO ARE SELECTED REMAIN ENTITLED TO APPLY FOR VISAS IN THE DV CATEGORY?
Persons selected in the DV-2011 lottery are entitled to apply for visa issuance only during fiscal year 2011, from October 1, 2010, through September 30, 2011. Applicants must obtain their visa or adjust status by the end of the fiscal year. There is no carry-over of DV benefits into the next year for persons who are selected but who do not obtain visas by September 30, 2010 (the end of the fiscal year.). Also, spouses and children who derive status from a DV-2011 registration can only obtain visas in the DV category between October 2010 and September 2011. Applicants who apply overseas will receive an appointment letter from the KCC four to six weeks before the scheduled appointment.
23. IF AN E-DV SELECTEE DIES, WHAT HAPPENS TO THE DV CASE?
The death of an individual selected in the lottery results in automatic revocation of the DV case. Any eligible spouse and/or children are no longer entitled to the DV visa for that entry.
24. WHEN WILL E-DV BE AVAILABLE ONLINE?
Online entry will be available during the registration period beginning at noon EDT (GMT-4) on October 2, 2009, and ending at noon EST (GMT-5) on November 30, 2009.
25. WILL I BE ABLE TO DOWNLOAD AND SAVE THE E-DV ENTRY FORM TO A MICROSOFT WORD PROGRAM (OR OTHER SUITABLE PROGRAM) AND THEN FILL IT OUT?
No, you will not be able to save the form into another program for completion and submission later. The E-DV Entry Form is a Web form only. This makes it more ―universal‖ than a proprietary word processor format. Additionally, it does require that the information be filled in and submitted while online.
26. IF I DON’T HAVE ACCESS TO A SCANNER, CAN I SEND PHOTOGRAPHS TO MY RELATIVE IN THE UNITED STATES TO SCAN THE PHOTOGRAPHS, SAVE THE PHOTOGRAPHS TO A DISKETTE, AND THEN MAIL THE DISKETTE BACK TO ME TO APPLY?
Yes, this can be done, as long as the photograph meets the photograph requirements in the instructions and the photograph is electronically submitted with, and at the same time as, the E-DV online entry. The applicants must already have the scanned photograph file when they submit the entry online. The photograph cannot be submitted separately from the online application. Only one online entry can be submitted for each person. Multiple submissions will disqualify the entry for that person for DV-2011. The entire entry (photograph and application together) can be submitted electronically from the United States or from overseas.
27. CAN I SAVE THE FORM ONLINE SO THAT I CAN FILL OUT PART AND THEN COME BACK LATER AND COMPLETE THE REMAINDER?
No, this cannot be done. The E-DV Entry Form is designed to be completed and submitted at one time. However, because the form is in two parts, and because of possible network interruptions and delays, the E-DV system is designed to permit up to sixty (60) minutes between the form’s download and when the entry is received at the E-DV website. If more than sixty minutes elapse and the entry has not been electronically
11
received, the information already received is discarded. This is done so that there is no possibility that a full entry could accidentally be interpreted as a duplicate of a previous partial entry. The DV-2011 instructions explain clearly and completely what information is required to fill in the form. Thus, you can be fully prepared, making sure you have all of the information needed before you start to complete the form online.
28. IF THE SUBMITTED DIGITAL IMAGES DO NOT CONFORM TO THE SPECIFICATIONS, THE PROCEDURES STATE THAT THE SYSTEM WILL AUTOMATICALLY REJECT THE E-DV ENTRY FORM AND NOTIFY THE SENDER. DOES THIS MEAN I WILL BE ABLE RE-SUBMIT MY ENTRY?
Yes, the entry can be resubmitted. Since the entry was automatically rejected, it was not actually considered as a submission to the E-DV website. It does not count as a submitted E-DV entry, and no confirmation notice of receipt is sent. If there are problems with the digital photograph sent, because it does not conform to the requirements, it is automatically rejected by the E-DV website. However, the amount of time it takes the rejection message to reach the sender is unpredictable, given the nature of the Internet. If the problem can be fixed by the applicant, and the Form Part One or Two is re-sent within sixty (60) minutes, there is no problem. Otherwise, the applicant will have to restart the submission process. An applicant can try to submit an application as many times as is necessary until a complete application is received and the confirmation notice sent.
29. WILL THE ELECTRONIC CONFIRMATION NOTICE THAT THE COMPLETED E-DV ENTRY FORM HAS BEEN RECEIVED THROUGH THE ONLINE SYSTEM BE SENT IMMEDIATELY AFTER SUBMISSION?
The response from the E-DV website which contains confirmation of the receipt of an acceptable E-DV Entry Form is sent by the E-DV website immediately. However, the amount of time it takes the response to reach the sender is unpredictable, given the nature of the Internet. If many minutes have elapsed since pressing the ―Submit‖ button, there is no harm in pressing the ―Submit‖ button a second time. The E-DV system will not be confused by a situation where the ―Submit‖ button is hit a second time, because no confirmation response has been received. An applicant can try to submit an application as many times as is necessary until a complete application is received and the confirmation notice sent. However, once you receive a confirmation notice, do not resubmit your information.
30. HOW WILL I KNOW IF THE NOTIFICATION OF SELECTION THAT I HAVE RECEIVED IS AUTHENTIC? HOW CAN I CONFIRM THAT I HAVE IN FACT BEEN CHOSEN IN THE RANDOM DV LOTTERY?
Keep your confirmation page. You will need it to check the status of your entry on the official DV website after the electronic lottery is conducted (usually March). If you lose your confirmation information, you will not be able to check your DV entry status by yourself, and we will not resend the confirmation page to you. If selected, you will also receive a letter from the KCC by mail sometime between May and July 2010 at the addresses listed on their E-DV entry. Only the randomly selected individuals will be notified by mail. Persons not selected may check their entry using their confirmation information through the official DV website, but they will not receive any additional official notification by email or by mail. We will NOT resend confirmation page information to you. If you lose your confirmation page information, you will only find out that you were selected if you receive an official letter by mail. U.S. Embassies and Consulates will NOT provide a list of those selected to continue the visa process. The KCC will send notification letters to those who are selected. These letters will contain instructions for the visa application process. The instructions say the selected applicants will pay all diversity and immigrant visa fees in person only at the U.S. Embassy or Consulate at the time of the visa application. The consular cashier or consular officer immediately gives the visa applicant a U.S. Government receipt for payment. You should never send money for DV fees through the mail, Western Union, or any other delivery service.
The E-DV lottery entries are submitted on the Internet, on the official U.S. Government E-DV website at www.dvlottery.state.gov. KCC sends only letters to the selected applicants. KCC, consular offices, or the U.S. Government have never sent e-mails to notify selected individuals, and there are no plans to use e-mail for this purpose for the DV-2011 program.
The Department of State’s Bureau of Consular Affairs advises the public that only Internet sites including the ―.gov‖ domain suffix are official government websites. Many other non-governmental websites (e.g., using the suffixes ―.com‖ or ―.org‖ or ―.net‖) provide immigration and visa related information and services. Regardless
12
of the content of non-governmental websites, the Department of State does not endorse, recommend, or sponsor any information or material shown at these other websites. Some websites may try to mislead customers and members of the public into thinking they are official websites and may contact you by e-mail to lure you to their offers. These websites may attempt to require you to pay for services such as forms and information about immigration procedures, which are otherwise free on the Department of State Visa Services website or through U.S. embassy consular sections’ websites. Additionally, these other websites may require you to pay for services you will not receive (such as fees for DV immigration applications and visas) in an effort to steal your money. If you send in money to one of these scams, you will never see it again. Also, you should be wary of sending any personal information to these websites that might be used for identity fraud/theft.
31. HOW DO I REPORT INTERNET FRAUD OR UNSOLICITED EMAIL?
If you wish to file a complaint about Internet fraud, please see the econsumer.gov website, hosted by the Federal Trade Commission, in cooperation with consumer-protection agencies from 17 nations (http://www.econsumer.gov/english/). You may also report fraud to the Federal Bureau of Investigation (FBI) Internet Crime Complaint Center. To file a complaint about unsolicited email, contact the Department of Justice Contact Us page.
32. IF I AM SUCCESSFUL IN OBTAINING A VISA THROUGH THE DV PROGRAM, WILL THE U.S. GOVERNMENT ASSIST WITH MY AIRFARE TO THE UNITED STATES, PROVIDE ASSISTANCE TO LOCATE HOUSING AND EMPLOYMENT, PROVIDE HEALTHCARE, OR PROVIDE ANY SUBSIDIES UNTIL I AM FULLY SETTLED?
No, applicants who obtain a DV are not provided any type of assistance such as airfare, housing assistance, or subsidies. If you are selected to apply for a DV, you will be required to provide evidence that you will not become a public charge in the United States before being issued a visa. This evidence may be in the form of a combination of your personal assets, an Affidavit of Support (Form I-134) from a relative or friend residing in the United States, and/or an offer of employment from an employer in the United States.
13
LIST OF COUNTRIES BY REGION WHOSE NATIVES ARE ELIGIBLE FOR DV-2011 The list below shows the countries whose natives are eligible for DV-2011, grouped by geographic region. Dependent areas overseas are included within the region of the governing country. The countries whose natives are not eligible for the DV-2011 program were identified by USCIS, according to the formula in Section 203(c) of the INA. The countries whose natives are not eligible for the DV program (because they are the principal source countries of Family-Sponsored and Employment-Based immigration or ―high-admission‖ countries) are noted after the respective regional lists.
AFRICA
Algeria Angola Benin Botswana Burkina Faso Burundi Cameroon Cape Verde Central African Republic Chad Comoros Congo Congo, Democratic Republic of the Cote D’Ivoire (Ivory Coast) Djibouti Egypt Equatorial Guinea Eritrea Ethiopia Gabon Gambia, The Ghana Guinea Guinea-Bissau Kenya Lesotho Liberia
Libya Madagascar Malawi Mali Mauritania Mauritius Morocco Mozambique Namibia Niger Nigeria Rwanda Sao Tome and Principe Senegal Seychelles Sierra Leone Somalia South Africa Sudan Swaziland Tanzania Togo Tunisia Uganda Zambia Zimbabwe
Persons born in the Gaza Strip are chargeable to Egypt.
14
LIST OF COUNTRIES BY REGION WHOSE NATIVES ARE ELIGIBLE FOR DV-2011
ASIA
Afghanistan Bahrain Bangladesh Bhutan Brunei Burma Cambodia East Timor Hong Kong Special Administrative Region Indonesia
Iran
Iraq
Israel
Japan Jordan Kuwait Laos
Lebanon
Malaysia
Maldives
Mongolia
Nepal
North Korea
Oman
Qatar
Saudi Arabia
Singapore
Sri Lanka
Syria
Taiwan
Thailand
United Arab Emirates
Yemen
Natives of the following Asian countries are not eligible for this year’s diversity program: China (mainland-born), India, Pakistan, South Korea, Philippines, and Vietnam. Hong Kong S.A.R and Taiwan do qualify and are listed above. Macau S.A.R. also qualifies and is listed below (Europe). Persons born in the areas administered prior to June 1967 by Israel, Jordan, and Syria are chargeable, respectively, to Israel, Jordan, and Syria.
15
LIST OF COUNTRIES BY REGION WHOSE NATIVES ARE ELIGIBLE FOR DV-2011
EUROPE
Albania Andorra Armenia Austria Azerbaijan Belarus Belgium Bosnia and Herzegovina Bulgaria Croatia Cyprus Czech Republic Denmark (including components and dependent areas overseas) Estonia Finland France (including components and areas overseas) Georgia Germany Greece Hungary Iceland Ireland Italy Kazakhstan Kosovo Kyrgyzstan Latvia Liechtenstein
Lithuania
Luxembourg
Macau Special Administrative Region
Macedonia
Malta
Moldova
Monaco
Montenegro
Netherlands (including components and
dependent areas overseas)
Northern Ireland
Norway
Portugal (including components and
dependent areas overseas)
Romania
Russia
San Marino
Serbia Slovakia Slovenia Spain Sweden Switzerland Tajikistan Turkey Turkmenistan Ukraine Uzbekistan Vatican City
Natives of the following European countries are not eligible for this year’s diversity program: Great Britain (United Kingdom) and Poland. Great Britain (United Kingdom) includes the following dependent areas: Anguilla, Bermuda, British Virgin Islands, Cayman Islands, Falkland Islands, Gibraltar, Montserrat, Pitcairn, St. Helena, and Turks and Caicos Islands. Note that for purposes of the diversity program only, Northern Ireland is treated separately; Northern Ireland does qualify and is listed among the qualifying areas.
16
LIST OF COUNTRIES BY REGION WHOSE NATIVES ARE ELIGIBLE FOR DV-2011 NORTH AMERICA
The Bahamas
In North America, natives of Canada and Mexico are not eligible for this year’s diversity program.
OCEANIA
Australia (including components and dependent areas overseas) Fiji Kiribati Marshall Islands Micronesia, Federated States of Nauru
New Zealand (including components and dependent areas overseas) Palau Papua New Guinea Solomon Islands Tonga Tuvalu
Vanuatu
Samoa
SOUTH AMERICA, CENTRAL AMERICA, AND THE CARIBBEAN
Antigua and Barbuda
Argentina Barbados Belize Bolivia Chile Costa Rica Cuba Dominica Grenada Guyana Honduras Nicaragua Panama Paraguay Saint Kitts and Nevis Saint Lucia Saint Vincent and the Grenadines Suriname Trinidad and Tobago Uruguay Venezuela Countries in this region whose natives are not eligible for this year’s diversity program:
Brazil, Colombia, Dominican Republic, Ecuador, El Salvador, Guatemala, Haiti, Jamaica, Mexico, and Peru.
For more information pls click HERE
Wednesday, October 21, 2009
Tuesday, October 6, 2009
Router,Router Configuretion, 3 Router Configuretion
Router configuration:
Router A
Router Con0 is now available
Press RETURN to get started!
Router>en
Router#config t
Enter configuration commands, one per line. End with CNTL/Z
Router(config)#int f0/0
Router(config-if)#ip add 150.160.12.1 255.255.0.0
Router(config-if)#no shut
09:31:10 %LINK-3-UPDOWN: Interface Fastethernet0/0, changed state to up
09:31:10 %LINEPROTO-5-UPDOWN: Line protocol on Interface Fastethernet0/0, changed state to up
Router(config-if)#
Router#copy run start
Destination filename [startup-config]?
Building configuration...
[OK]
Router#config t
Enter configuration commands, one per line. End with CNTL/Z
Router(config)#int s0/0
Router(config-if)#ip add 10.10.10.1 255.0.0.0
Router(config-if)#no shut
09:32:09 %LINK-3-UPDOWN: Interface Serial0/0, changed state to up
09:32:09 %LINEPROTO-5-UPDOWN: Line protocol on Interface Serial0/0, changed state to up
Router(config-if)#
Router#copy run start
Destination filename [startup-config]?
Building configuration...
[OK]
Router#config t
Enter configuration commands, one per line. End with CNTL/Z
Router(config)#int s0/0
Router(config-if)#clock rate 64000
Router(config-if)#bandwidth 64
Router(config-if)#
Router#copy run start
Destination filename [startup-config]?
Building configuration...
[OK]
Router#
Router B
Router Con0 is now available
Press RETURN to get started!
Router>en
Router#config t
Enter configuration commands, one per line. End with CNTL/Z
Router(config)#int s0/0
Router(config-if)#ip add 10.10.10.2 255.0.0.0
Router(config-if)#
Router#config t
Enter configuration commands, one per line. End with CNTL/Z
Router(config)#int f0/0
Router(config-if)#ip add 200.200.150.4 255.255.255.0
Router(config-if)#no shut
09:44:10 %LINK-3-UPDOWN: Interface Fastethernet0/0, changed state to up
09:44:10 %LINEPROTO-5-UPDOWN: Line protocol on Interface Fastethernet0/0, changed state to up
Router(config-if)#
Router#copy run start
Destination filename [startup-config]?
Building configuration...
[OK]
Router#config t
Enter configuration commands, one per line. End with CNTL/Z
Router(config)#int s0/0
Router(config-if)#int s0/1
Router(config-if)#exit
Router(config)#int s0/1
Router(config-if)#ip add 11.10.10.3 255.0.0.0
Router(config-if)#no shut
09:46:03 %LINK-3-UPDOWN: Interface Serial0/1, changed state to up
09:46:03 %LINEPROTO-5-UPDOWN: Line protocol on Interface Serial0/1, changed state to up
Router(config-if)#
Router#config t
Enter configuration commands, one per line. End with CNTL/Z
Router(config)#int s0/1
Router(config-if)#clock rate 64000
Router(config-if)#bandwidth 64
Router(config-if)#
Router#copy run start
Destination filename [startup-config]?
Building configuration...
[OK]
Router#
Router C
Router Con0 is now available
Press RETURN to get started!
Router>en
Router#config t
Enter configuration commands, one per line. End with CNTL/Z
Router(config)#int f0/0
Router(config-if)#ip add 220.220.200.4 255.255.255.0
Router(config-if)#no shut
09:48:47 %LINK-3-UPDOWN: Interface Fastethernet0/0, changed state to up
09:48:47 %LINEPROTO-5-UPDOWN: Line protocol on Interface Fastethernet0/0, changed state to up
Router(config-if)#
Router#config t
Enter configuration commands, one per line. End with CNTL/Z
Router(config)#int s0/1
Router(config-if)#ip add 11.10.10.4 255.0.0.0
Router(config-if)#no shut
09:50:39 %LINK-3-UPDOWN: Interface Serial0/1, changed state to up
09:50:39 %LINEPROTO-5-UPDOWN: Line protocol on Interface Serial0/1, changed state to up
Router(config-if)#
Router#copy run start
Destination filename [startup-config]?
Building configuration...
[OK]
Router#
IP Routing:
Router A
Router Con0 is now available
Press RETURN to get started!
Router>en
Router#config t
Enter configuration commands, one per line. End with CNTL/Z
Router(config)#ip route
% Incomplete command.
Router(config)#ip route 200.200.150.0 255.255.255.0 10.10.10.2
Router(config)#ip route 0.0.0.0 255.255.255.255 11.10.10.4
Router(config)#
Router# copy run start
Destination filename [startup-config]?
Building configuration...
[OK]
Router#config t
Enter configuration commands, one per line. End with CNTL/Z
Router(config)#router rip
Router(config-router)#network 150.160.0.0
Router(config-router)#network 10.0.0.0
Router(config-router)#network 11.0.0.0
Router(config-router)#version 2
Router(config-router)#
Router#copy run start
Destination filename [startup-config]?
Building configuration...
[OK]
Router#
Router B
Router Con0 is now available
Press RETURN to get started!
Router>en
Router#config t
Enter configuration commands, one per line. End with CNTL/Z
Router(config)#ip route 150.160.0.0 255.255.0.0 10.10.10.1
Router(config)#ip route 0.0.0.0 255.255.255.255 11.10.10.4
Router(config)#
Router#config t
Enter configuration commands, one per line. End with CNTL/Z
Router(config)#router rip
Router(config-router)#network 200.200.150.0
Router(config-router)#network 10.0.0.0
Router(config-router)#network 11.0.0.0
Router(config-router)#version 2
Router(config-router)#
Router#copy run start
Destination filename [startup-config]?
Building configuration...
[OK]
Router#
Router C:
Router Con0 is now available
Press RETURN to get started!
Router>en
Router#config t
Enter configuration commands, one per line. End with CNTL/Z
Router(config)#ip route 150.160.0.0 255.255.0.0 10.10.10.1
Router(config)#ip route 0.0.0.0 255.255.255.255 11.10.10.3
Router(config)#router rip
Router(config-router)#network 220.220.200.0
Router(config-router)#network 11.0.0.0
Router(config-router)#version 2
Router(config-router)#
Router#copy run start
Destination filename [startup-config]?
Building configuration...
[OK]
Router#
Now,
Ping 220.220.200.2
Router Con0 is now available
Press RETURN to get started!
Router>ping 220.220.200.2
Type escape sequence to abort.
Sending 5, 100-byte ICMP Echos to 220.220.200.2, timeout is 2 seconds:
!!!!!
Success rate is 100 percent (5/5), round-trip min/avg/max = 4/4/4 ms
Router>
_______________
Router A
Router Con0 is now available
Press RETURN to get started!
Router>en
Router#config t
Enter configuration commands, one per line. End with CNTL/Z
Router(config)#int f0/0
Router(config-if)#ip add 150.160.12.1 255.255.0.0
Router(config-if)#no shut
09:31:10 %LINK-3-UPDOWN: Interface Fastethernet0/0, changed state to up
09:31:10 %LINEPROTO-5-UPDOWN: Line protocol on Interface Fastethernet0/0, changed state to up
Router(config-if)#
Router#copy run start
Destination filename [startup-config]?
Building configuration...
[OK]
Router#config t
Enter configuration commands, one per line. End with CNTL/Z
Router(config)#int s0/0
Router(config-if)#ip add 10.10.10.1 255.0.0.0
Router(config-if)#no shut
09:32:09 %LINK-3-UPDOWN: Interface Serial0/0, changed state to up
09:32:09 %LINEPROTO-5-UPDOWN: Line protocol on Interface Serial0/0, changed state to up
Router(config-if)#
Router#copy run start
Destination filename [startup-config]?
Building configuration...
[OK]
Router#config t
Enter configuration commands, one per line. End with CNTL/Z
Router(config)#int s0/0
Router(config-if)#clock rate 64000
Router(config-if)#bandwidth 64
Router(config-if)#
Router#copy run start
Destination filename [startup-config]?
Building configuration...
[OK]
Router#
Router B
Router Con0 is now available
Press RETURN to get started!
Router>en
Router#config t
Enter configuration commands, one per line. End with CNTL/Z
Router(config)#int s0/0
Router(config-if)#ip add 10.10.10.2 255.0.0.0
Router(config-if)#
Router#config t
Enter configuration commands, one per line. End with CNTL/Z
Router(config)#int f0/0
Router(config-if)#ip add 200.200.150.4 255.255.255.0
Router(config-if)#no shut
09:44:10 %LINK-3-UPDOWN: Interface Fastethernet0/0, changed state to up
09:44:10 %LINEPROTO-5-UPDOWN: Line protocol on Interface Fastethernet0/0, changed state to up
Router(config-if)#
Router#copy run start
Destination filename [startup-config]?
Building configuration...
[OK]
Router#config t
Enter configuration commands, one per line. End with CNTL/Z
Router(config)#int s0/0
Router(config-if)#int s0/1
Router(config-if)#exit
Router(config)#int s0/1
Router(config-if)#ip add 11.10.10.3 255.0.0.0
Router(config-if)#no shut
09:46:03 %LINK-3-UPDOWN: Interface Serial0/1, changed state to up
09:46:03 %LINEPROTO-5-UPDOWN: Line protocol on Interface Serial0/1, changed state to up
Router(config-if)#
Router#config t
Enter configuration commands, one per line. End with CNTL/Z
Router(config)#int s0/1
Router(config-if)#clock rate 64000
Router(config-if)#bandwidth 64
Router(config-if)#
Router#copy run start
Destination filename [startup-config]?
Building configuration...
[OK]
Router#
Router C
Router Con0 is now available
Press RETURN to get started!
Router>en
Router#config t
Enter configuration commands, one per line. End with CNTL/Z
Router(config)#int f0/0
Router(config-if)#ip add 220.220.200.4 255.255.255.0
Router(config-if)#no shut
09:48:47 %LINK-3-UPDOWN: Interface Fastethernet0/0, changed state to up
09:48:47 %LINEPROTO-5-UPDOWN: Line protocol on Interface Fastethernet0/0, changed state to up
Router(config-if)#
Router#config t
Enter configuration commands, one per line. End with CNTL/Z
Router(config)#int s0/1
Router(config-if)#ip add 11.10.10.4 255.0.0.0
Router(config-if)#no shut
09:50:39 %LINK-3-UPDOWN: Interface Serial0/1, changed state to up
09:50:39 %LINEPROTO-5-UPDOWN: Line protocol on Interface Serial0/1, changed state to up
Router(config-if)#
Router#copy run start
Destination filename [startup-config]?
Building configuration...
[OK]
Router#
IP Routing:
Router A
Router Con0 is now available
Press RETURN to get started!
Router>en
Router#config t
Enter configuration commands, one per line. End with CNTL/Z
Router(config)#ip route
% Incomplete command.
Router(config)#ip route 200.200.150.0 255.255.255.0 10.10.10.2
Router(config)#ip route 0.0.0.0 255.255.255.255 11.10.10.4
Router(config)#
Router# copy run start
Destination filename [startup-config]?
Building configuration...
[OK]
Router#config t
Enter configuration commands, one per line. End with CNTL/Z
Router(config)#router rip
Router(config-router)#network 150.160.0.0
Router(config-router)#network 10.0.0.0
Router(config-router)#network 11.0.0.0
Router(config-router)#version 2
Router(config-router)#
Router#copy run start
Destination filename [startup-config]?
Building configuration...
[OK]
Router#
Router B
Router Con0 is now available
Press RETURN to get started!
Router>en
Router#config t
Enter configuration commands, one per line. End with CNTL/Z
Router(config)#ip route 150.160.0.0 255.255.0.0 10.10.10.1
Router(config)#ip route 0.0.0.0 255.255.255.255 11.10.10.4
Router(config)#
Router#config t
Enter configuration commands, one per line. End with CNTL/Z
Router(config)#router rip
Router(config-router)#network 200.200.150.0
Router(config-router)#network 10.0.0.0
Router(config-router)#network 11.0.0.0
Router(config-router)#version 2
Router(config-router)#
Router#copy run start
Destination filename [startup-config]?
Building configuration...
[OK]
Router#
Router C:
Router Con0 is now available
Press RETURN to get started!
Router>en
Router#config t
Enter configuration commands, one per line. End with CNTL/Z
Router(config)#ip route 150.160.0.0 255.255.0.0 10.10.10.1
Router(config)#ip route 0.0.0.0 255.255.255.255 11.10.10.3
Router(config)#router rip
Router(config-router)#network 220.220.200.0
Router(config-router)#network 11.0.0.0
Router(config-router)#version 2
Router(config-router)#
Router#copy run start
Destination filename [startup-config]?
Building configuration...
[OK]
Router#
Now,
Ping 220.220.200.2
Router Con0 is now available
Press RETURN to get started!
Router>ping 220.220.200.2
Type escape sequence to abort.
Sending 5, 100-byte ICMP Echos to 220.220.200.2, timeout is 2 seconds:
!!!!!
Success rate is 100 percent (5/5), round-trip min/avg/max = 4/4/4 ms
Router>
_______________
Tuesday, September 29, 2009
Virtual memory low solution
If your computer shows virtual memory too low then u can solve it by this way:
Go to :
My Computer>Properties>Advance>Performance>Setting>Advance(tab)
virtual memory>
Change
(All Drive
initial size = Ram x 2
Maximum size = Ram x 4)
Set
Ok
then restart your computer.
about IP Addressing
IP Addressing:
Short for Internet Protocol, IP is an address of a computer or other network device on a network using IP or TCP/IP . For example, the number "166.70.10.23" is an example of such an address. These addresses are similar to addresses used on houses and help data reach its appropriate destination on a network.
There are three classes of available IP ranges: Class A, Class B, and Class C. Each class allows for a range of valid IP addresses. Below is a listing of these addresses.
Class Address Range Supports
Class A 1.0.0.1 to 127.255.255.254 Supports 16 million hosts on each of 127 networks
Class B 128.1.0.1 to 191.255.255.254 Supports 65,000 hosts on each of 16,000 networks.
Class C 192.0.1.1 to 223.255.254.254 Supports 254 hosts on each of 2 million networks.
Ranges 127.x.x.x are reserved for loop back tests, for example, 127.0.0.1. Ranges 255.255.255.255 are used to broadcast to all hosts on the local network.
All IP addresses are broken down into 4 sets of octets that break down into binary to represent the actual IP address. The below chart is a basic example of the basic IP 255.255.255.255.
IP: 255. 255. 255. 255.
Binary value: 11111111. 11111111. 11111111. 11111111.
Octet value: 8 8 8 8
If we were to break down the IP "166.70.10.23", which is the IP address of Computer Hope, you would get the below value.
166. 70. 10. 23
10100110. 01000110. 00001010. 00010111
128+32+4+2=166 64+4+2=70 8+2=10 16+4+2+1=23
There are several IP addresses used or automatically assigned on a network. For example:
166.70.10.0 0 is the automatically assigned network address.
166.70.10.1 1 is the commonly used address used as the gateway.
166.70.10.2 2 is also a commonly used address used for a gateway.
166.70.10.255 255 is automatically assigned on most networks as the broadcast address.
IP Classes
Class A addresses always have the first bit of their IP addresses set to “0”. Since Class A networks have an 8-bit network mask, the use of a leading zero leaves only 7 bits for the network portion of the address, allowing for a maximum of 128 possible network numbers, ranging from 0.0.0.0 – 127.0.0.0. Number 127.x.x.x is reserved for loop back, used for internal testing on the local machine.
Class A addresses contain 7 bits in the network portion giving 27 - 2 = 126 possible networks since all 1's and all 0's are not allowed. Consequently 24 bits remain for the host portion allowing a total of 224 - 2 = 16,777,214 hosts.
Class B addresses always have the first bit set to “1” and their second bit set to “0”. Since Class B addresses have a 16-bit network mask, the use of a leading “10” bit-pattern leaves 14 bits for the network portion of the address.
Class B addresses contain 14 bits in the network portion allowing 214 - 2 = 16,384 possible networks, and 16 bits for the host portion allowing a possible total number of 216 - 2 = 65,534 hosts, ranging from 128.0.0.0 – 181.255.0.0.
Class C addresses have their first two bits set to “1” and their third bit set to “0”. Since Class C addresses have a 24-bit network mask, this leaves 21 bits for the network portion of the address.
Class C addresses contain 21 bits for the network portion giving a possible total of 221 - 2 = 2,097,152 networks, and 8 bits for the host portion giving a possible 28 - 2 = 254 hosts, ranging from 192.0.0.0 – 223.255.255.0.
Short for Internet Protocol, IP is an address of a computer or other network device on a network using IP or TCP/IP . For example, the number "166.70.10.23" is an example of such an address. These addresses are similar to addresses used on houses and help data reach its appropriate destination on a network.
There are three classes of available IP ranges: Class A, Class B, and Class C. Each class allows for a range of valid IP addresses. Below is a listing of these addresses.
Class Address Range Supports
Class A 1.0.0.1 to 127.255.255.254 Supports 16 million hosts on each of 127 networks
Class B 128.1.0.1 to 191.255.255.254 Supports 65,000 hosts on each of 16,000 networks.
Class C 192.0.1.1 to 223.255.254.254 Supports 254 hosts on each of 2 million networks.
Ranges 127.x.x.x are reserved for loop back tests, for example, 127.0.0.1. Ranges 255.255.255.255 are used to broadcast to all hosts on the local network.
All IP addresses are broken down into 4 sets of octets that break down into binary to represent the actual IP address. The below chart is a basic example of the basic IP 255.255.255.255.
IP: 255. 255. 255. 255.
Binary value: 11111111. 11111111. 11111111. 11111111.
Octet value: 8 8 8 8
If we were to break down the IP "166.70.10.23", which is the IP address of Computer Hope, you would get the below value.
166. 70. 10. 23
10100110. 01000110. 00001010. 00010111
128+32+4+2=166 64+4+2=70 8+2=10 16+4+2+1=23
There are several IP addresses used or automatically assigned on a network. For example:
166.70.10.0 0 is the automatically assigned network address.
166.70.10.1 1 is the commonly used address used as the gateway.
166.70.10.2 2 is also a commonly used address used for a gateway.
166.70.10.255 255 is automatically assigned on most networks as the broadcast address.
IP Classes
Class A addresses always have the first bit of their IP addresses set to “0”. Since Class A networks have an 8-bit network mask, the use of a leading zero leaves only 7 bits for the network portion of the address, allowing for a maximum of 128 possible network numbers, ranging from 0.0.0.0 – 127.0.0.0. Number 127.x.x.x is reserved for loop back, used for internal testing on the local machine.
Class A addresses contain 7 bits in the network portion giving 27 - 2 = 126 possible networks since all 1's and all 0's are not allowed. Consequently 24 bits remain for the host portion allowing a total of 224 - 2 = 16,777,214 hosts.
Class B addresses always have the first bit set to “1” and their second bit set to “0”. Since Class B addresses have a 16-bit network mask, the use of a leading “10” bit-pattern leaves 14 bits for the network portion of the address.
Class B addresses contain 14 bits in the network portion allowing 214 - 2 = 16,384 possible networks, and 16 bits for the host portion allowing a possible total number of 216 - 2 = 65,534 hosts, ranging from 128.0.0.0 – 181.255.0.0.
Class C addresses have their first two bits set to “1” and their third bit set to “0”. Since Class C addresses have a 24-bit network mask, this leaves 21 bits for the network portion of the address.
Class C addresses contain 21 bits for the network portion giving a possible total of 221 - 2 = 2,097,152 networks, and 8 bits for the host portion giving a possible 28 - 2 = 254 hosts, ranging from 192.0.0.0 – 223.255.255.0.
Subnet , subnet musk , subnetting
Subnet:
A subnet allows the flow of network traffic between hosts to be segregated based on a network configuration. By organizing hosts into logical groups, sub netting can improve network security and performance.
Subnet Mask:
Perhaps the most recognizable aspect of subnetting is the subnet mask. Like IP addresses, a subnet mask contains four bytes (32 bits) and is often written using the same "dotted-decimal" notation. For example, a very common subnet mask in its binary representation
11111111 11111111 11111111 00000000
is typically shown in the equivalent, more readable form 255.255.255.0
Applying a Subnet Mask:
IP addresses are broken down into three parts: the network part, the subnet part (now often considered part of the network part, although originally it was part of the rest part), and the host part. Even though classful networks are obsolete, both classful and classless networks are shown in the following table.
Class Leading bits Start End Default Subnet Mask in dotted decimal
A (CIDR /8) 0 0.0.0.0 127.255.255.255 255.0.0.0
B (CIDR /16) 10 128.0.0.0 191.255.255.255 255.255.0.0
C (CIDR /24) 110 192.0.0.0 223.255.255.255 255.255.255.0
Sub netting is the process of allocating bits from the host portion as a network portion. The above example shows the bitwise "AND" process being performed on a classful network. The following example shows bits being borrowed to turn a classful network into a subnet.
Example
Dot-decimal Address Binary
IP address 192.168.5.130 11000000.10101000.00000101.10000010
Subnet Mask 255.255.255.192 11111111.11111111.11111111.11000000
Network Portion 192.168.5.128 11000000.10101000.00000101.10000000
In this example two bits were borrowed from the original host portion. This is beneficial because it allows this network to be split into four smaller networks. A /24 suffix (Class C block) allows 254 hosts; split into four parts, the prefix is /26, each has 62 hosts.
Subnets and host count
It is possible to determine the number of hosts and sub networks available for any subnet mask. In the above example two bits were borrowed to create sub networks. Each bit can take the value 1 or 0, giving 4 possible subnets (22 = 4)
Network Network (binary) Broadcast address
192.168.5.0/26 11000000.10101000.00000101.00000000 192.168.5.63
192.168.5.64/26 11000000.10101000.00000101.01000000 192.168.5.127
192.168.5.128/26 11000000.10101000.00000101.10000000 192.168.5.191
192.168.5.192/26 11000000.10101000.00000101.11000000 192.168.5.255
The remaining bits after the subnet are used for addressing hosts within the subnet. In the above example the subnet mask consists of 26 bits, leaving 6 bits for the address (32 − 26). This allows for 64 possible combinations (26), however the all zeros value and all ones value are reserved for the network ID and broadcast address respectively, leaving 62 addresses.
In general the number of available hosts on a subnet can be calculated using the formula 2n − 2, where n is the number of bits used for the host portion of the address.
RFC 3021 specifies an exception to this rule when dealing with 31 bit subnet masks (i.e. 1 host bit). According to the above rule a 31 bit mask would allow for 21 − 2 = 0 hosts. The RFC makes allowances in this case for certain types of networks (point-to-point) to disregard the network and broadcast address, allowing two host addresses to be allocated.
Possible subnets for a /24 suffix (traditional Class C):
CIDR notation Network Mask Binary Available Networks Available Hosts per network Total usable hosts
/24 255.255.255.0 0000.0000 1 254 254
/25 255.255.255.128 1000.0000 2 126 252
/26 255.255.255.192 1100.0000 4 62 248
/27 255.255.255.224 1110.0000 8 30 240
/28 255.255.255.240 1111.0000 16 14 224
/29 255.255.255.248 1111.1000 32 6 192
/30 255.255.255.252 1111.1100 64 2 128
/31 255.255.255.254 1111.1110 128 2 * 256
* only applicable on point-to-point links
Once you have the two steps memorized, you can begin subnetting. Our first example will use the Class C mask of 255.255.255.192. Ask five simple questions to gather all the facts:
1. How many subnet bits are used in this mask?
2. How many host bits are available per subnet?
3. What are the subnet addresses?
4. What is the broadcast address of each subnet?
5. What is the valid host range of each subnet?
We already know how to answer questions one and two. To answer question three, use the formula 256-subnetmask to get the first subnet and our variable. Keep adding this number to itself until we get to the subnet mask value to determine the valid subnets. Once we verify all of the subnets, you can determine the broadcast address by looking at the next subnet’s value. The broadcast address is the number just before the next subnet number. Once we have the subnet number and broadcast address, the valid hosts are the numbers in between.
Here are the answers using 255.255.255.192:
1. How many subnet bits are used in this mask?
Answer: 2
22-2=2 subnets
2. How many host bits are available per subnet?
Answer: 6
26-2=62 hosts per subnet
3. What are the subnet addresses?
Answer: 256-192=64 (the first subnet)
64+64=128 (the second subnet)
64+128=192. However, although 192 is the subnet mask value, it’s not a valid subnet. The valid subnets are 64 and 128.
4. What is the broadcast address of each subnet?
Answer: 64 is the first subnet and 128 is the second subnet. The broadcast address is always the number before the next subnet. The broadcast address of the 64 subnet is 127. The broadcast address of the 128 subnet is 191.
5. What is the valid host range of each subnet?
Answer: The valid hosts are the numbers between the subnet number and the mask. For the 64 subnet, the valid host range is 64-126. For the 128 subnet, the valid host range is 129-190.
Let’s do a second example using the Class C mask of 255.255.255.224. Here are the answers:
1. How many subnet bits are used in this mask?
Answer: 3 bits or 23-2=6 subnets
2. How many host bits are available per subnet?
Answer: 5 bits or 25-2=30 hosts per subnet
3. What are the subnet addresses?
Answer: 256-224 =32, 64, 96, 128, 160 and 192 (Six subnets found by continuing to add 32 to itself.)
4. What is the broadcast address of each subnet?
Answer: The broadcast address for the 32 subnet is 63. The broadcast address for the 64 subnet is 95. The broadcast address for the 96 subnet is 127. The broadcast address for the 160 subnet is 191. The broadcast address for the 192 subnet is 223 (since 224 is the mask).
5. What is the valid host range of each subnet?
Answer: The valid hosts are the numbers in between the subnet and broadcast addresses. For example, the 32 subnet valid hosts are 33-62.
Let’s do a third example using the Class C mask of 255.255.255.240. Here are the answers:
1. How many subnet bits are used in this mask?
Answer: 4 bits or 24-2=14 subnets
2. How many host bits are available per subnet?
Answer: 4 bits or 24-2=14 hosts per subnet
3. What are the subnet addresses?
Answer: 256-240 =16, 32, 48, 64, 80, 96, 112, 128, 144. 160, 176, 192, 208 and 224 (14 subnets found by continuing to add 16 to itself.)
4. What is the broadcast address of each subnet?
Answer: Here are some examples of the broadcast address: The broadcast address for the 16 subnet is 31. The broadcast address for the 32 subnet is 47. The broadcast address for the 64 subnet is 79. The broadcast address for the 96 subnet is 111. The broadcast address for the 160 subnet is 175. The broadcast address for the 192 subnet is 207.
5. What is the valid host range of each subnet?
Answer: The valid hosts are the numbers in between the subnet and broadcast addresses. The 32 subnet valid hosts are 33-46.
Let’s do a fourth example using the Class C mask of 255.255.255.248. Here are the answers:
1. How many subnet bits are used in this mask?
Answer: 5 bits or 25-2=30 subnets
2. How many host bits are available per subnet?
Answer: 3 bits or 23-2=6 hosts per subnet
3. What are the subnet addresses?
Answer 256-248 =8, 16, 24, 32, 40, 48, and so forth. The last subnet is 240 (30 subnets found by continuing to add 8 to itself).
4. What is the broadcast address of each subnet?
Answer: The broadcast address for the 8 subnet is 15. The broadcast address for the 16 subnet is 23. The broadcast address for the 48 subnet is 55.
5. What is the valid host range of each subnet?
Answer: The valid hosts are the numbers in between the subnet and broadcast addresses. For example, the 32 subnet valid hosts are 33-38.
Let’s do a fifth example using the Class C mask of 255.255.255.252. Here are the answers:
1. How many subnet bits are used in this mask?
Answer: 6 bits or 26-2=62 subnets
2. How many host bits are available per subnet?
Answer: 2 bits or 22-2=2 hosts per subnet
3. What are the subnet addresses?
Answer: 256-252 =4, 8, 12, 16, 20, and so forth. The last subnet is 248 (62 subnets found by continuing to add 4 to itself).
4. What is the broadcast address of each subnet?
Answer: The broadcast address for the 4 subnet is 7. The broadcast address for the 8 subnet is 11. The broadcast address for the 12 subnet is 15. The broadcast address for the 20 subnet is 23.
5. What is the valid host range of each subnet?
Answer: The valid hosts are the numbers in between the subnet and broadcast addresses. For example, the 16 subnet valid hosts are 17 and 18.
Conclusion
It is important to be able to subnet quickly and efficiently. After studying the examples presented in this Daily Drill Down, you should be familiar with this process with Class C addresses.
CIDR notation
The Classless Inter-Domain Routing address specification or CIDR notation, begins with the Internet Protocol address followed by a "/" character and a decimal number specifying the length, in bits, of the subnet mask or routing prefix.
In case of address block specifications the IP address is the starting address of the block.
For example
• 192.168.100.1/24 represents the given IP address and its associated routing prefix (192.168.100.0) or, equivalently, its subnet mask, 255.255.255.0.
• 192.168.0.0/22 represents the 1024 IP addresses from 192.168.0.0 through 192.168.3.255.
For IP networks, an alternative representation uses the network address followed by the network's subnet mask, written in dot-decimal notation:
• 192.168.0.0/24 could be written 192.168.0.0/255.255.255.0
• 192.168.0.0/22 could be written 192.168.0.0/255.255.252.0
The number of addresses of a subnet defined by the mask or prefix can be calculated as 2(address size-mask). For example, a mask of /29 gives: 2(32-29) = 23 = 8 addresses for an IP network. However, because at least one of these addresses is typically used for a gateway to other subnets, and because certain addresses are reserved as broadcast addresses, the number of addresses available for hosts is usually less.
Class B subnets:
In my last Daily Drill Down, we borrowed bits from the host portion of a Class C network address. Class C network addresses only have eight bits to manipulate into subnets. However, a Class B has 16 bits to play with. This will allow more subnets with more hosts per subnet than a Class C network ever could.
Table 1 lists all of the possible Class B subnets:
Table 1Mask Binary Subnets Hosts per subnet
255.255.128.0 10000000.00000000 2 32,766
255.255.192.0 11000000.00000000 2 16,382
255.255.224.0 11100000.00000000 6 8,190
255.255.240.0 11110000.00000000 14 4,094
255.255.248.0 11111000.00000000 30 2,046
255.255.252.0 11111100.00000000 62 1,022
255.255.254.0 11111110.00000000 126 510
255.255.255.0 11111111.00000000 254 254
255.255.255.128 11111111.10000000 510 126
255.255.255.192 11111111.11000000 1022 62
255.255.255.224 11111111.11100000 2,046 30
255.255.255.240 11111111.11110000 4,094 14
255.255.255.248 11111111.11111000 8,190 6
255.255.255.252 11111111.11111100 16,382 2
All possible Class B subnets
There are quite a few more masks we can use with a Class B network address than we can with a Class C network address. Remember that this is not harder than subnetting with Class C, but it can get confusing if you don’t pay attention to where the subnet bits and host bits are in a mask. This takes practice!
We will use the same techniques I used in the Class C article to subnet a network. We’ll start with the Class B subnet mask of 255.255.192.0 and figure out the subnets, broadcast address, and valid host range. We will answer the same five questions we answered for the Class C subnet masks:
1. How many subnets does this mask provide?
2. How many hosts per subnet does this mask provide?
3. What are the valid subnets?
4. What is the broadcast address for each subnet?
5. What is the host range of each subnet?
Before we answer these questions, there is one difference we need to be aware of when subnetting a Class B network address. When subnetting in the third octet, you need to add the fourth octet. For example, on the 255.255.192.0 mask, the subnetting will be done in the third octet. To create a valid subnet, you must add the fourth octet of all 0s and all 1s for the network and broadcast address (0 for all 0s and 255 for all 1s).
Example 1: Answers for the 255.255.192.0 mask
1. 2-2=2 subnets
2. 2-2=16,382 hosts per subnet
3. 256-192=64.0, 128.0
4. Broadcast for the 64.0 subnet is 127.255. Broadcast for the 128.0 subnet is 191.255.
5. The valid hosts are:
Subnet 64.0 128.0
first host 64.1 128.1
last host 127.254 191.254
broadcast 127.255 191.255
Notice that the numbers in the third octet are the same numbers we used in the fourth octet when subnetting the 192 mask. The only difference is that we add 0 and 255 in the fourth octet.
For the 64.0 subnet, all the hosts between 64.1 and 127.254 are in the 64 subnet. In the 128.0 subnet, the hosts are 128.1 through 191.254.
Example 2: 255.255.240.0
1. 2-2=14 subnets
2. 2-2=4094 hosts per subnet
3. 256-240=16.0, 32.0, 48.0, 64.0, etc.
4. Broadcast for the 16.0 subnet is 31.255. Broadcast for the 32.0 subnet is 47.255, etc.
5. The valid hosts are:
Subnet 16.0 32.0 48.0 64.0
first host 16.1 32.1 48.1 64.1
last host 31.254 47.254 63.254 79.254
broadcast 31.255 47.255 63.255 79.255
Example 3: 255.255.248.0
1. 2-2=30 subnets
2. 2-2=2,046 hosts per subnet
3. 256-248=8.0, 16.0, 24.0, 32.0, 40.0, 48.0, 56.0, 64.0, etc.
4. Broadcast for the 8.0 subnet is 15.255. Broadcast for the 16.0 subnet is 23.255, etc.
5. The valid hosts are:
Subnet 8.0 16.0 24.0 32.0 40.0 48.0 56.0 64.0
first host 8.1 16.1 24.1 32.1 40.1 48.1 56.1 64.1
last host 15.254 23.254 31.254 39.254 47.254 55.254 63.254 71.254
broadcast 15.255 23.255 31.255 39.255 47.255 55.255 63.255 71.255
Example 4: 255.255.252.0
1. 2-2=62 subnets
2. 2-2=1,022 hosts per subnet
3. 256-252=4.0, 8.0, 12.0, 16.0, 20.0, 24.0, 28.0, 32.0, etc.
4. Broadcast for the 4.0 subnet is 7.255. Broadcast for the 8.0 subnet is 11.255, etc.
5. The valid hosts are:
Subnet 4.0 8.0 12.0 16.0 20.0 24.0 28.0 32.0
first host 4.1 8.1 12.1 16.1 20.1 24.1 28.1 32.1
last host 7.254 11.254 15.254 19.254 23.254 27.254 31.254 35.254
broadcast 7.255 11.255 15.255 19.255 23.255 27.255 31.255 35.255
Example 5: 255.255.255.0
1. 2-2=254 subnets
2. 2-2=254 hosts per subnet
3. 256-255=1.0, 2.0, 3.0, 4.0, 5.0, 6.0, 7.0, 8.0, etc.
4. Broadcast for the 1.0 subnet is 1.255. Broadcast for the 2.0 subnet is 2.255, etc.
5. The valid hosts are:
Subnet 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0
first host 1.1 2.1 3.1 4.1 5.1 6.1 7.1 8.1
last host 1.254 21.254 3.254 4.254 5.254 6.254 7.254 8.254
broadcast 1.255 2.255 3.255 4.255 5.255 6.255 7.255 8.255
That last example was pretty simple. All the numbers were basically the same except we added the fourth octet into the address.
The more difficult process of subnetting a Class B network address is when you start using bits in the fourth octet for subnetting. For example, what happens when you use this mask with a Class B network address: 255.255.255.128? Is that valid? Absolutely! There are nine bits for subnetting and seven bits for hosts. That is 510 subnets, each with 126 hosts. However, it is the most difficult mask to figure out the valid hosts for.
Example 6: The Class B 255.255.255.128 subnet mask:
1. 2-2=510 subnets
2. 2-2=126 hosts per subnet
3. For the third octet, the mask would be 256-255=1, 2, 3, 4, 5, 6, etc.
4. For the fourth octet, the mask would be 256-128=128, which is one subnet if it is used. However, if you turn the subnet bit off, the value is 0. This means that for every subnet in the third octet, the fourth octet has two subnets: 0 and 128, for example 1.0 and 1.128.
5. Broadcast for the 0.128 subnet is 128.255; the broadcast for the 1.0 subnet is 1.127. Broadcast for the 1.128 subnet is 1.255, etc.
6. The valid hosts are:
Subnet 0.128 1.0 1.128 2.0 2.128 3.0 3.128 4.0
first host 0.129 1.1 1.129 2.1 2.129 3.1 3.129 4.1
last host 0.254 1.126 1.254 2.126 2.254 3.126 3.254 4.126
broadcast 0.255 1.127 1.255 2.127 2.255 3.127 3.255 4.127
The thing to remember is that for every subnet in the third octet, there are two in the fourth octet: 0 and 128. For the 0 subnet, the broadcast address is always 127. For the 128 subnet, the broadcast address is always 255.
Let’s continue with more subnetting into the fourth octet. This is exactly like subnetting a Class C network address, but the third octet is part of the subnet address.
Example 7: Class B network 255.255.255.192
1. 2-2=1022 subnets
2. 2-2=62 hosts per subnet
3. 256-255=1.0, 2.0, 3.0, etc. for the third octet. 256-192=64, 128, 192 for the fourth octet. For every valid subnet in the third octet, we get four subnets in the fourth octet: 0, 64, 128, and 192.
4. Broadcast for the 1.0 subnet is 1.63, since the next subnet is 1.64. Broadcast for the 1.64 subnet is 1.127, since the next subnet is 1.128. Broadcast for the 1.128 subnet is 1.191, since the next subnet is 1.192. Broadcast for the 1.192 subnet is 1.255.
5. The valid hosts are as follows:
Subnet 0.64 0.128 0.192 1.0 1.64 1.128 1.192 2.0
first host 0.65 0.129 0.193 1.1 1.65 1.129 1.193 2.1
last host 0.126 0.190 0.254 1.62 1.126 1.190 1.254 2.62
broadcast 0.127 0.191 0.255 1.63 1.127 1.191 1.255 2.63
On this one, the 0 and 192 subnets are valid, since we are using the third octet as well. The subnet range is 0.64 through 255.128. 0.0 is not valid since no subnet bits are on. 255.192 is not valid because then all subnet bits would be on.
Example 8: Class B network 255.255.255.224
1. 2-2=2046 subnets
2. 2-2=30 hosts per subnet
3. 256-255=1.0, 2.0, 3.0, etc. for the third octet. 256-224=32, 64, 96, 128, 160, 192 for the subnet value. (For every value in the third octet, we get eight subnets in the fourth octet: 0, 32, 64, 96, 128, 160, 192, 224.)
4. Broadcast for the 1.0 subnet is 1.63, since the next subnet is 1.64. Broadcast for the 1.64 subnet is 1.127, since the next subnet is 1.128. Broadcast for the 1.128 subnet is 1.191, since the next subnet is 1.192. Broadcast for the 1.192 subnet is 1.255.
5. The valid hosts are:
Subnet 0.32 0.64 0.96 0.128 0.160 0.192 0.224 1.0
first host 0.33 0.65 0.97 0.129 0.161 0.193 0.225 1.1
last host 0.62 0.94 0.126 0.158 0.190 0.222 0.254 1.30
broadcast 0.63 0.95 0.127 1.159 0.191 0.223 0.255 1.31
Subnetting a Class A network address
Class A subnetting is a form of network addressing typically reserved for the government and larger institutions. It can have a large number of unique nodes, which can prove quite a challenge. Here, Todd Lammle offers a simple approach to this addressing.
If we wanted to create smaller networks (subnetworks) out of a Class A network ID, you’d borrow bits from the host portion of the mask. The more bits you borrow, the more subnets we can have, but this means fewer hosts per subnet. However, with a Class A mask, you have 24 bits to manipulate, so this isn’t typically a problem.
Table A lists all the available Class A subnet masks:
Table AMask Prefix Subnets Hosts
255.0.0.0 (/8) 1 network with 16,777,214 hosts
255.128.0.0 (/9) 2 subnets with 8,388,606 hosts each
255.192.0.0 (/10) 4 subnets with 4,194,302 hosts each
255.224.0.0 (/11) 8 subnets with 2,097,150 hosts each
255.240.0.0 (/12) 16 subnets with 1,048,574 hosts each
255.248.0.0 (/13) 32 subnets with 524,286 hosts each
255.252.0.0 (/14) 64 subnets with 262,142 hosts each
255.254.0.0 (/15) 128 subnets with 131,070 hosts each
255.255.0.0 (/16) 256 subnets with 65,534 hosts each
255.255.128.0 (/17) 512 subnets with 32,766 hosts each
255.255.192.0 (/18) 1,024 subnets with 16,384 hosts each
255.255.224.0 (/19) 2,048 subnets with 8,190 hosts each
255.255.240.0 (/20) 4,096 subnets with 4,094 hosts each
255.255.248.0 (/21) 8,192 subnets with 2,046 hosts each
255.255.252.0 (/22) 16,384 subnets with 1,022 hosts each
255.255.254.0 (/23) 32,768 subnets with 510 hosts each
255.255.255.0 (/24) 65,536 subnets with 254 hosts each
255.255.255.128 (/25) 131,072 subnets with 126 hosts each
255.255.255.192 (/26) 262,144 subnets with 62 hosts each
255.255.255.224 (/27) 524,288 subnets with 30 hosts each
255.255.255.240 (/28) 1,048,576 subnets with 14 hosts each
255.255.255.248 (/29) 2,097,152 subnets with 6 hosts each
255.255.255.252 (/30) 4,194,304 subnets with 2 hosts each
Available Class A subnet masks
Subnet zero
This chart assumes you can use subnet zero. If you’re not using subnet zero, subtract two from each number in the Subnets column in Table A above.
Once you have an idea what your network will look like, write down the number of physical subnets you have and the number of hosts needed for each subnet. For example, on a WAN point-to-point link, you need only two IP addresses, so you can use a /30 mask.
/30
The slash (/) indicates the number of mask bits turned on. It saves you from typing, or pronouncing, the whole mask. For example, /8 means 255.0.0.0, /16 is 255.255.0.0, and /24 is 255.255.255.0. You pronounce it as "configure a slash 24 mask on that network." It's just an easier way of saying "configure a 255.255.255.0 mask on that network."
Subnet addresses
Once you have the mask assigned to each network, you must assign the valid subnet addresses and host ranges to each network. To determine the valid subnets and host addresses for each network, you need to answer three easy questions:
1. What is the valid subnet address?
2. What is the broadcast address?
3. What is the valid host range?
4. Here are some tips for finding the answers:
Valid subnet address:
To figure out the valid subnet address, simply subtract the subnet mask from 256. For example, if you had a Class A mask of 255.240.0.0, the equation would be 256-240=16. The number 16 is the first subnet and also your block size. Keep adding the block size (in this case 16) to itself until you reach the subnet mask value. The valid subnets in this example are 16, 32, 48, 64, 80, 96, 112, 128, 144, 160, 176, 192, 208, 224. As another example, if you had a Class A subnet mask of 255.255.240.0, you’d use the mask on the second and third octets minus 256. The second octet would be 256-255=1, 2, 3, etc., all the way to 254; the third octet would be 256-240=16, 32, 48, etc.
Broadcast address:
To determine the broadcast address of each subnet, just subtract 1 from the next subnet value. For example, within the 16 subnet, the next subnet is 32, so the broadcast address of the 16 subnet is 31. The broadcast address for the 32 subnet is 47, because the next subnet is 48. The broadcast address for the 48 subnet is 63, because the next subnet is 64.
Valid host range:
The valid hosts are the numbers between the subnet address and the broadcast address. For the 16 subnet, the valid host range you can assign on a network is 17-30 because the subnet number is 16 and the broadcast address is 31. For the 32 subnet, the valid host range is 33 to 46 because the subnet number is 32 and the broadcast address is 47. You can’t use the subnet number and broadcast addresses as valid host addresses.
Assigning valid host addresses
Using Figure A as an example, let’s assign networks and valid host ranges to each network. First, create another table of usable block sizes. Figure B shows how to assign block sizes to your network on a granular basis.
Assigning other Class A masks
Figure B is an example of variable length subnet masks and is typically the easiest way you’ll assign IP addresses on an internetwork. However, let’s go through a few practice examples of subnetting Class A masks, to make sure you really understand the process.
Practice Class A mask 255.240.0.0 (/12)
This mask provides you with only four subnet bits, or 16 subnets (14 if you’re not using subnet zero) with 1,048,574 hosts each. The valid subnets are 256-240=16, 32, 48, 64, 80, etc., all the way to 224. (Subnets 0 and 240 are available if you’re using subnet zero.)
The first subnet, assuming subnet zero, is:
• Subnet: 10.0.0.0
• Broadcast: 10.15.255.255
• Valid host range: 10.0.0.1 through 10.15.255.254
The last subnet, assuming subnet zero, is:
• Subnet: 10.240.0.0
• Broadcast: 10.255.255.255
• Valid host range: 10.240.0.1 through 10.255.255.254
Practice Class A mask 255.255.128.0
This mask provides you with nine bits of subnetting and 15 host bits (/17). This gives you 512 subnets with 32,766 hosts each. The second octet is 256-255=1, 2, 3, etc., all the way to 255. Zero is available in the second octet if you have either a subnet bit on in the third octet or are, of course, using subnet zero.
The first available subnet is:
• Subnet: 10.0.0.0
• Broadcast: 10.0.127.255
• Valid host range: 10.0.0.1 through 10.0.127.254
You must remember that the third octet is using only one subnet bit. This bit can be either off or on; if it is off, the subnet is 0. If it is on, the subnet is 128.
Here’s an example of the 10.0.128.0 subnet:
• Subnet: 10.0.128.0
• Broadcast: 10.0.255.255
• Valid host range: 10.0.128.1 through 10.0.255.254
The last available subnet is:
• Subnet: 10.255.128.0
• Broadcast: 10.255.255.255
• Valid host range: 10.255.128.1 through 10.255.255.254
Practice Class A mask 255.255.255.252
This mask is the easiest to subnet. Even if it weren’t a Class A mask, and you used this mask with a Class B or Class C mask, you’d always have only two available host IDs. The reason you would use this with a Class A mask is because it can give you up to 4,194,304 subnets with two hosts each. This is a perfect mask for a point-to-point link, so I suggest always saving a few block sizes of four (/30) masks for use on WANs and point-to-point LAN connections.
If you use the 10.2.3.0 network, your subnets are always 2.3 in the second and third octets, respectively. But the fourth octet is where it changes, as in 256-252=4, 8, 12, 16, 20, 24, 28, etc., all the way to 248. If you use subnet zero, your first subnets are 0, and your last subnet is 255.
An example of the 10.2.3.0 subnet is:
• Subnet: 10.2.3.0
• Broadcast: 10.2.3.3
• Valid hosts: 10.2.3.1 and 10.2.3.2
An example of the 10.2.3.252 subnet is:
• Subnet: 10.2.3.252
• Broadcast: 10.2.3.255
• Valid hosts: 10.2.3.253 and 10.2.3.254
A subnet allows the flow of network traffic between hosts to be segregated based on a network configuration. By organizing hosts into logical groups, sub netting can improve network security and performance.
Subnet Mask:
Perhaps the most recognizable aspect of subnetting is the subnet mask. Like IP addresses, a subnet mask contains four bytes (32 bits) and is often written using the same "dotted-decimal" notation. For example, a very common subnet mask in its binary representation
11111111 11111111 11111111 00000000
is typically shown in the equivalent, more readable form 255.255.255.0
Applying a Subnet Mask:
IP addresses are broken down into three parts: the network part, the subnet part (now often considered part of the network part, although originally it was part of the rest part), and the host part. Even though classful networks are obsolete, both classful and classless networks are shown in the following table.
Class Leading bits Start End Default Subnet Mask in dotted decimal
A (CIDR /8) 0 0.0.0.0 127.255.255.255 255.0.0.0
B (CIDR /16) 10 128.0.0.0 191.255.255.255 255.255.0.0
C (CIDR /24) 110 192.0.0.0 223.255.255.255 255.255.255.0
Sub netting is the process of allocating bits from the host portion as a network portion. The above example shows the bitwise "AND" process being performed on a classful network. The following example shows bits being borrowed to turn a classful network into a subnet.
Example
Dot-decimal Address Binary
IP address 192.168.5.130 11000000.10101000.00000101.10000010
Subnet Mask 255.255.255.192 11111111.11111111.11111111.11000000
Network Portion 192.168.5.128 11000000.10101000.00000101.10000000
In this example two bits were borrowed from the original host portion. This is beneficial because it allows this network to be split into four smaller networks. A /24 suffix (Class C block) allows 254 hosts; split into four parts, the prefix is /26, each has 62 hosts.
Subnets and host count
It is possible to determine the number of hosts and sub networks available for any subnet mask. In the above example two bits were borrowed to create sub networks. Each bit can take the value 1 or 0, giving 4 possible subnets (22 = 4)
Network Network (binary) Broadcast address
192.168.5.0/26 11000000.10101000.00000101.00000000 192.168.5.63
192.168.5.64/26 11000000.10101000.00000101.01000000 192.168.5.127
192.168.5.128/26 11000000.10101000.00000101.10000000 192.168.5.191
192.168.5.192/26 11000000.10101000.00000101.11000000 192.168.5.255
The remaining bits after the subnet are used for addressing hosts within the subnet. In the above example the subnet mask consists of 26 bits, leaving 6 bits for the address (32 − 26). This allows for 64 possible combinations (26), however the all zeros value and all ones value are reserved for the network ID and broadcast address respectively, leaving 62 addresses.
In general the number of available hosts on a subnet can be calculated using the formula 2n − 2, where n is the number of bits used for the host portion of the address.
RFC 3021 specifies an exception to this rule when dealing with 31 bit subnet masks (i.e. 1 host bit). According to the above rule a 31 bit mask would allow for 21 − 2 = 0 hosts. The RFC makes allowances in this case for certain types of networks (point-to-point) to disregard the network and broadcast address, allowing two host addresses to be allocated.
Possible subnets for a /24 suffix (traditional Class C):
CIDR notation Network Mask Binary Available Networks Available Hosts per network Total usable hosts
/24 255.255.255.0 0000.0000 1 254 254
/25 255.255.255.128 1000.0000 2 126 252
/26 255.255.255.192 1100.0000 4 62 248
/27 255.255.255.224 1110.0000 8 30 240
/28 255.255.255.240 1111.0000 16 14 224
/29 255.255.255.248 1111.1000 32 6 192
/30 255.255.255.252 1111.1100 64 2 128
/31 255.255.255.254 1111.1110 128 2 * 256
* only applicable on point-to-point links
Once you have the two steps memorized, you can begin subnetting. Our first example will use the Class C mask of 255.255.255.192. Ask five simple questions to gather all the facts:
1. How many subnet bits are used in this mask?
2. How many host bits are available per subnet?
3. What are the subnet addresses?
4. What is the broadcast address of each subnet?
5. What is the valid host range of each subnet?
We already know how to answer questions one and two. To answer question three, use the formula 256-subnetmask to get the first subnet and our variable. Keep adding this number to itself until we get to the subnet mask value to determine the valid subnets. Once we verify all of the subnets, you can determine the broadcast address by looking at the next subnet’s value. The broadcast address is the number just before the next subnet number. Once we have the subnet number and broadcast address, the valid hosts are the numbers in between.
Here are the answers using 255.255.255.192:
1. How many subnet bits are used in this mask?
Answer: 2
22-2=2 subnets
2. How many host bits are available per subnet?
Answer: 6
26-2=62 hosts per subnet
3. What are the subnet addresses?
Answer: 256-192=64 (the first subnet)
64+64=128 (the second subnet)
64+128=192. However, although 192 is the subnet mask value, it’s not a valid subnet. The valid subnets are 64 and 128.
4. What is the broadcast address of each subnet?
Answer: 64 is the first subnet and 128 is the second subnet. The broadcast address is always the number before the next subnet. The broadcast address of the 64 subnet is 127. The broadcast address of the 128 subnet is 191.
5. What is the valid host range of each subnet?
Answer: The valid hosts are the numbers between the subnet number and the mask. For the 64 subnet, the valid host range is 64-126. For the 128 subnet, the valid host range is 129-190.
Let’s do a second example using the Class C mask of 255.255.255.224. Here are the answers:
1. How many subnet bits are used in this mask?
Answer: 3 bits or 23-2=6 subnets
2. How many host bits are available per subnet?
Answer: 5 bits or 25-2=30 hosts per subnet
3. What are the subnet addresses?
Answer: 256-224 =32, 64, 96, 128, 160 and 192 (Six subnets found by continuing to add 32 to itself.)
4. What is the broadcast address of each subnet?
Answer: The broadcast address for the 32 subnet is 63. The broadcast address for the 64 subnet is 95. The broadcast address for the 96 subnet is 127. The broadcast address for the 160 subnet is 191. The broadcast address for the 192 subnet is 223 (since 224 is the mask).
5. What is the valid host range of each subnet?
Answer: The valid hosts are the numbers in between the subnet and broadcast addresses. For example, the 32 subnet valid hosts are 33-62.
Let’s do a third example using the Class C mask of 255.255.255.240. Here are the answers:
1. How many subnet bits are used in this mask?
Answer: 4 bits or 24-2=14 subnets
2. How many host bits are available per subnet?
Answer: 4 bits or 24-2=14 hosts per subnet
3. What are the subnet addresses?
Answer: 256-240 =16, 32, 48, 64, 80, 96, 112, 128, 144. 160, 176, 192, 208 and 224 (14 subnets found by continuing to add 16 to itself.)
4. What is the broadcast address of each subnet?
Answer: Here are some examples of the broadcast address: The broadcast address for the 16 subnet is 31. The broadcast address for the 32 subnet is 47. The broadcast address for the 64 subnet is 79. The broadcast address for the 96 subnet is 111. The broadcast address for the 160 subnet is 175. The broadcast address for the 192 subnet is 207.
5. What is the valid host range of each subnet?
Answer: The valid hosts are the numbers in between the subnet and broadcast addresses. The 32 subnet valid hosts are 33-46.
Let’s do a fourth example using the Class C mask of 255.255.255.248. Here are the answers:
1. How many subnet bits are used in this mask?
Answer: 5 bits or 25-2=30 subnets
2. How many host bits are available per subnet?
Answer: 3 bits or 23-2=6 hosts per subnet
3. What are the subnet addresses?
Answer 256-248 =8, 16, 24, 32, 40, 48, and so forth. The last subnet is 240 (30 subnets found by continuing to add 8 to itself).
4. What is the broadcast address of each subnet?
Answer: The broadcast address for the 8 subnet is 15. The broadcast address for the 16 subnet is 23. The broadcast address for the 48 subnet is 55.
5. What is the valid host range of each subnet?
Answer: The valid hosts are the numbers in between the subnet and broadcast addresses. For example, the 32 subnet valid hosts are 33-38.
Let’s do a fifth example using the Class C mask of 255.255.255.252. Here are the answers:
1. How many subnet bits are used in this mask?
Answer: 6 bits or 26-2=62 subnets
2. How many host bits are available per subnet?
Answer: 2 bits or 22-2=2 hosts per subnet
3. What are the subnet addresses?
Answer: 256-252 =4, 8, 12, 16, 20, and so forth. The last subnet is 248 (62 subnets found by continuing to add 4 to itself).
4. What is the broadcast address of each subnet?
Answer: The broadcast address for the 4 subnet is 7. The broadcast address for the 8 subnet is 11. The broadcast address for the 12 subnet is 15. The broadcast address for the 20 subnet is 23.
5. What is the valid host range of each subnet?
Answer: The valid hosts are the numbers in between the subnet and broadcast addresses. For example, the 16 subnet valid hosts are 17 and 18.
Conclusion
It is important to be able to subnet quickly and efficiently. After studying the examples presented in this Daily Drill Down, you should be familiar with this process with Class C addresses.
CIDR notation
The Classless Inter-Domain Routing address specification or CIDR notation, begins with the Internet Protocol address followed by a "/" character and a decimal number specifying the length, in bits, of the subnet mask or routing prefix.
In case of address block specifications the IP address is the starting address of the block.
For example
• 192.168.100.1/24 represents the given IP address and its associated routing prefix (192.168.100.0) or, equivalently, its subnet mask, 255.255.255.0.
• 192.168.0.0/22 represents the 1024 IP addresses from 192.168.0.0 through 192.168.3.255.
For IP networks, an alternative representation uses the network address followed by the network's subnet mask, written in dot-decimal notation:
• 192.168.0.0/24 could be written 192.168.0.0/255.255.255.0
• 192.168.0.0/22 could be written 192.168.0.0/255.255.252.0
The number of addresses of a subnet defined by the mask or prefix can be calculated as 2(address size-mask). For example, a mask of /29 gives: 2(32-29) = 23 = 8 addresses for an IP network. However, because at least one of these addresses is typically used for a gateway to other subnets, and because certain addresses are reserved as broadcast addresses, the number of addresses available for hosts is usually less.
Class B subnets:
In my last Daily Drill Down, we borrowed bits from the host portion of a Class C network address. Class C network addresses only have eight bits to manipulate into subnets. However, a Class B has 16 bits to play with. This will allow more subnets with more hosts per subnet than a Class C network ever could.
Table 1 lists all of the possible Class B subnets:
Table 1Mask Binary Subnets Hosts per subnet
255.255.128.0 10000000.00000000 2 32,766
255.255.192.0 11000000.00000000 2 16,382
255.255.224.0 11100000.00000000 6 8,190
255.255.240.0 11110000.00000000 14 4,094
255.255.248.0 11111000.00000000 30 2,046
255.255.252.0 11111100.00000000 62 1,022
255.255.254.0 11111110.00000000 126 510
255.255.255.0 11111111.00000000 254 254
255.255.255.128 11111111.10000000 510 126
255.255.255.192 11111111.11000000 1022 62
255.255.255.224 11111111.11100000 2,046 30
255.255.255.240 11111111.11110000 4,094 14
255.255.255.248 11111111.11111000 8,190 6
255.255.255.252 11111111.11111100 16,382 2
All possible Class B subnets
There are quite a few more masks we can use with a Class B network address than we can with a Class C network address. Remember that this is not harder than subnetting with Class C, but it can get confusing if you don’t pay attention to where the subnet bits and host bits are in a mask. This takes practice!
We will use the same techniques I used in the Class C article to subnet a network. We’ll start with the Class B subnet mask of 255.255.192.0 and figure out the subnets, broadcast address, and valid host range. We will answer the same five questions we answered for the Class C subnet masks:
1. How many subnets does this mask provide?
2. How many hosts per subnet does this mask provide?
3. What are the valid subnets?
4. What is the broadcast address for each subnet?
5. What is the host range of each subnet?
Before we answer these questions, there is one difference we need to be aware of when subnetting a Class B network address. When subnetting in the third octet, you need to add the fourth octet. For example, on the 255.255.192.0 mask, the subnetting will be done in the third octet. To create a valid subnet, you must add the fourth octet of all 0s and all 1s for the network and broadcast address (0 for all 0s and 255 for all 1s).
Example 1: Answers for the 255.255.192.0 mask
1. 2-2=2 subnets
2. 2-2=16,382 hosts per subnet
3. 256-192=64.0, 128.0
4. Broadcast for the 64.0 subnet is 127.255. Broadcast for the 128.0 subnet is 191.255.
5. The valid hosts are:
Subnet 64.0 128.0
first host 64.1 128.1
last host 127.254 191.254
broadcast 127.255 191.255
Notice that the numbers in the third octet are the same numbers we used in the fourth octet when subnetting the 192 mask. The only difference is that we add 0 and 255 in the fourth octet.
For the 64.0 subnet, all the hosts between 64.1 and 127.254 are in the 64 subnet. In the 128.0 subnet, the hosts are 128.1 through 191.254.
Example 2: 255.255.240.0
1. 2-2=14 subnets
2. 2-2=4094 hosts per subnet
3. 256-240=16.0, 32.0, 48.0, 64.0, etc.
4. Broadcast for the 16.0 subnet is 31.255. Broadcast for the 32.0 subnet is 47.255, etc.
5. The valid hosts are:
Subnet 16.0 32.0 48.0 64.0
first host 16.1 32.1 48.1 64.1
last host 31.254 47.254 63.254 79.254
broadcast 31.255 47.255 63.255 79.255
Example 3: 255.255.248.0
1. 2-2=30 subnets
2. 2-2=2,046 hosts per subnet
3. 256-248=8.0, 16.0, 24.0, 32.0, 40.0, 48.0, 56.0, 64.0, etc.
4. Broadcast for the 8.0 subnet is 15.255. Broadcast for the 16.0 subnet is 23.255, etc.
5. The valid hosts are:
Subnet 8.0 16.0 24.0 32.0 40.0 48.0 56.0 64.0
first host 8.1 16.1 24.1 32.1 40.1 48.1 56.1 64.1
last host 15.254 23.254 31.254 39.254 47.254 55.254 63.254 71.254
broadcast 15.255 23.255 31.255 39.255 47.255 55.255 63.255 71.255
Example 4: 255.255.252.0
1. 2-2=62 subnets
2. 2-2=1,022 hosts per subnet
3. 256-252=4.0, 8.0, 12.0, 16.0, 20.0, 24.0, 28.0, 32.0, etc.
4. Broadcast for the 4.0 subnet is 7.255. Broadcast for the 8.0 subnet is 11.255, etc.
5. The valid hosts are:
Subnet 4.0 8.0 12.0 16.0 20.0 24.0 28.0 32.0
first host 4.1 8.1 12.1 16.1 20.1 24.1 28.1 32.1
last host 7.254 11.254 15.254 19.254 23.254 27.254 31.254 35.254
broadcast 7.255 11.255 15.255 19.255 23.255 27.255 31.255 35.255
Example 5: 255.255.255.0
1. 2-2=254 subnets
2. 2-2=254 hosts per subnet
3. 256-255=1.0, 2.0, 3.0, 4.0, 5.0, 6.0, 7.0, 8.0, etc.
4. Broadcast for the 1.0 subnet is 1.255. Broadcast for the 2.0 subnet is 2.255, etc.
5. The valid hosts are:
Subnet 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0
first host 1.1 2.1 3.1 4.1 5.1 6.1 7.1 8.1
last host 1.254 21.254 3.254 4.254 5.254 6.254 7.254 8.254
broadcast 1.255 2.255 3.255 4.255 5.255 6.255 7.255 8.255
That last example was pretty simple. All the numbers were basically the same except we added the fourth octet into the address.
The more difficult process of subnetting a Class B network address is when you start using bits in the fourth octet for subnetting. For example, what happens when you use this mask with a Class B network address: 255.255.255.128? Is that valid? Absolutely! There are nine bits for subnetting and seven bits for hosts. That is 510 subnets, each with 126 hosts. However, it is the most difficult mask to figure out the valid hosts for.
Example 6: The Class B 255.255.255.128 subnet mask:
1. 2-2=510 subnets
2. 2-2=126 hosts per subnet
3. For the third octet, the mask would be 256-255=1, 2, 3, 4, 5, 6, etc.
4. For the fourth octet, the mask would be 256-128=128, which is one subnet if it is used. However, if you turn the subnet bit off, the value is 0. This means that for every subnet in the third octet, the fourth octet has two subnets: 0 and 128, for example 1.0 and 1.128.
5. Broadcast for the 0.128 subnet is 128.255; the broadcast for the 1.0 subnet is 1.127. Broadcast for the 1.128 subnet is 1.255, etc.
6. The valid hosts are:
Subnet 0.128 1.0 1.128 2.0 2.128 3.0 3.128 4.0
first host 0.129 1.1 1.129 2.1 2.129 3.1 3.129 4.1
last host 0.254 1.126 1.254 2.126 2.254 3.126 3.254 4.126
broadcast 0.255 1.127 1.255 2.127 2.255 3.127 3.255 4.127
The thing to remember is that for every subnet in the third octet, there are two in the fourth octet: 0 and 128. For the 0 subnet, the broadcast address is always 127. For the 128 subnet, the broadcast address is always 255.
Let’s continue with more subnetting into the fourth octet. This is exactly like subnetting a Class C network address, but the third octet is part of the subnet address.
Example 7: Class B network 255.255.255.192
1. 2-2=1022 subnets
2. 2-2=62 hosts per subnet
3. 256-255=1.0, 2.0, 3.0, etc. for the third octet. 256-192=64, 128, 192 for the fourth octet. For every valid subnet in the third octet, we get four subnets in the fourth octet: 0, 64, 128, and 192.
4. Broadcast for the 1.0 subnet is 1.63, since the next subnet is 1.64. Broadcast for the 1.64 subnet is 1.127, since the next subnet is 1.128. Broadcast for the 1.128 subnet is 1.191, since the next subnet is 1.192. Broadcast for the 1.192 subnet is 1.255.
5. The valid hosts are as follows:
Subnet 0.64 0.128 0.192 1.0 1.64 1.128 1.192 2.0
first host 0.65 0.129 0.193 1.1 1.65 1.129 1.193 2.1
last host 0.126 0.190 0.254 1.62 1.126 1.190 1.254 2.62
broadcast 0.127 0.191 0.255 1.63 1.127 1.191 1.255 2.63
On this one, the 0 and 192 subnets are valid, since we are using the third octet as well. The subnet range is 0.64 through 255.128. 0.0 is not valid since no subnet bits are on. 255.192 is not valid because then all subnet bits would be on.
Example 8: Class B network 255.255.255.224
1. 2-2=2046 subnets
2. 2-2=30 hosts per subnet
3. 256-255=1.0, 2.0, 3.0, etc. for the third octet. 256-224=32, 64, 96, 128, 160, 192 for the subnet value. (For every value in the third octet, we get eight subnets in the fourth octet: 0, 32, 64, 96, 128, 160, 192, 224.)
4. Broadcast for the 1.0 subnet is 1.63, since the next subnet is 1.64. Broadcast for the 1.64 subnet is 1.127, since the next subnet is 1.128. Broadcast for the 1.128 subnet is 1.191, since the next subnet is 1.192. Broadcast for the 1.192 subnet is 1.255.
5. The valid hosts are:
Subnet 0.32 0.64 0.96 0.128 0.160 0.192 0.224 1.0
first host 0.33 0.65 0.97 0.129 0.161 0.193 0.225 1.1
last host 0.62 0.94 0.126 0.158 0.190 0.222 0.254 1.30
broadcast 0.63 0.95 0.127 1.159 0.191 0.223 0.255 1.31
Subnetting a Class A network address
Class A subnetting is a form of network addressing typically reserved for the government and larger institutions. It can have a large number of unique nodes, which can prove quite a challenge. Here, Todd Lammle offers a simple approach to this addressing.
If we wanted to create smaller networks (subnetworks) out of a Class A network ID, you’d borrow bits from the host portion of the mask. The more bits you borrow, the more subnets we can have, but this means fewer hosts per subnet. However, with a Class A mask, you have 24 bits to manipulate, so this isn’t typically a problem.
Table A lists all the available Class A subnet masks:
Table AMask Prefix Subnets Hosts
255.0.0.0 (/8) 1 network with 16,777,214 hosts
255.128.0.0 (/9) 2 subnets with 8,388,606 hosts each
255.192.0.0 (/10) 4 subnets with 4,194,302 hosts each
255.224.0.0 (/11) 8 subnets with 2,097,150 hosts each
255.240.0.0 (/12) 16 subnets with 1,048,574 hosts each
255.248.0.0 (/13) 32 subnets with 524,286 hosts each
255.252.0.0 (/14) 64 subnets with 262,142 hosts each
255.254.0.0 (/15) 128 subnets with 131,070 hosts each
255.255.0.0 (/16) 256 subnets with 65,534 hosts each
255.255.128.0 (/17) 512 subnets with 32,766 hosts each
255.255.192.0 (/18) 1,024 subnets with 16,384 hosts each
255.255.224.0 (/19) 2,048 subnets with 8,190 hosts each
255.255.240.0 (/20) 4,096 subnets with 4,094 hosts each
255.255.248.0 (/21) 8,192 subnets with 2,046 hosts each
255.255.252.0 (/22) 16,384 subnets with 1,022 hosts each
255.255.254.0 (/23) 32,768 subnets with 510 hosts each
255.255.255.0 (/24) 65,536 subnets with 254 hosts each
255.255.255.128 (/25) 131,072 subnets with 126 hosts each
255.255.255.192 (/26) 262,144 subnets with 62 hosts each
255.255.255.224 (/27) 524,288 subnets with 30 hosts each
255.255.255.240 (/28) 1,048,576 subnets with 14 hosts each
255.255.255.248 (/29) 2,097,152 subnets with 6 hosts each
255.255.255.252 (/30) 4,194,304 subnets with 2 hosts each
Available Class A subnet masks
Subnet zero
This chart assumes you can use subnet zero. If you’re not using subnet zero, subtract two from each number in the Subnets column in Table A above.
Once you have an idea what your network will look like, write down the number of physical subnets you have and the number of hosts needed for each subnet. For example, on a WAN point-to-point link, you need only two IP addresses, so you can use a /30 mask.
/30
The slash (/) indicates the number of mask bits turned on. It saves you from typing, or pronouncing, the whole mask. For example, /8 means 255.0.0.0, /16 is 255.255.0.0, and /24 is 255.255.255.0. You pronounce it as "configure a slash 24 mask on that network." It's just an easier way of saying "configure a 255.255.255.0 mask on that network."
Subnet addresses
Once you have the mask assigned to each network, you must assign the valid subnet addresses and host ranges to each network. To determine the valid subnets and host addresses for each network, you need to answer three easy questions:
1. What is the valid subnet address?
2. What is the broadcast address?
3. What is the valid host range?
4. Here are some tips for finding the answers:
Valid subnet address:
To figure out the valid subnet address, simply subtract the subnet mask from 256. For example, if you had a Class A mask of 255.240.0.0, the equation would be 256-240=16. The number 16 is the first subnet and also your block size. Keep adding the block size (in this case 16) to itself until you reach the subnet mask value. The valid subnets in this example are 16, 32, 48, 64, 80, 96, 112, 128, 144, 160, 176, 192, 208, 224. As another example, if you had a Class A subnet mask of 255.255.240.0, you’d use the mask on the second and third octets minus 256. The second octet would be 256-255=1, 2, 3, etc., all the way to 254; the third octet would be 256-240=16, 32, 48, etc.
Broadcast address:
To determine the broadcast address of each subnet, just subtract 1 from the next subnet value. For example, within the 16 subnet, the next subnet is 32, so the broadcast address of the 16 subnet is 31. The broadcast address for the 32 subnet is 47, because the next subnet is 48. The broadcast address for the 48 subnet is 63, because the next subnet is 64.
Valid host range:
The valid hosts are the numbers between the subnet address and the broadcast address. For the 16 subnet, the valid host range you can assign on a network is 17-30 because the subnet number is 16 and the broadcast address is 31. For the 32 subnet, the valid host range is 33 to 46 because the subnet number is 32 and the broadcast address is 47. You can’t use the subnet number and broadcast addresses as valid host addresses.
Assigning valid host addresses
Using Figure A as an example, let’s assign networks and valid host ranges to each network. First, create another table of usable block sizes. Figure B shows how to assign block sizes to your network on a granular basis.
Assigning other Class A masks
Figure B is an example of variable length subnet masks and is typically the easiest way you’ll assign IP addresses on an internetwork. However, let’s go through a few practice examples of subnetting Class A masks, to make sure you really understand the process.
Practice Class A mask 255.240.0.0 (/12)
This mask provides you with only four subnet bits, or 16 subnets (14 if you’re not using subnet zero) with 1,048,574 hosts each. The valid subnets are 256-240=16, 32, 48, 64, 80, etc., all the way to 224. (Subnets 0 and 240 are available if you’re using subnet zero.)
The first subnet, assuming subnet zero, is:
• Subnet: 10.0.0.0
• Broadcast: 10.15.255.255
• Valid host range: 10.0.0.1 through 10.15.255.254
The last subnet, assuming subnet zero, is:
• Subnet: 10.240.0.0
• Broadcast: 10.255.255.255
• Valid host range: 10.240.0.1 through 10.255.255.254
Practice Class A mask 255.255.128.0
This mask provides you with nine bits of subnetting and 15 host bits (/17). This gives you 512 subnets with 32,766 hosts each. The second octet is 256-255=1, 2, 3, etc., all the way to 255. Zero is available in the second octet if you have either a subnet bit on in the third octet or are, of course, using subnet zero.
The first available subnet is:
• Subnet: 10.0.0.0
• Broadcast: 10.0.127.255
• Valid host range: 10.0.0.1 through 10.0.127.254
You must remember that the third octet is using only one subnet bit. This bit can be either off or on; if it is off, the subnet is 0. If it is on, the subnet is 128.
Here’s an example of the 10.0.128.0 subnet:
• Subnet: 10.0.128.0
• Broadcast: 10.0.255.255
• Valid host range: 10.0.128.1 through 10.0.255.254
The last available subnet is:
• Subnet: 10.255.128.0
• Broadcast: 10.255.255.255
• Valid host range: 10.255.128.1 through 10.255.255.254
Practice Class A mask 255.255.255.252
This mask is the easiest to subnet. Even if it weren’t a Class A mask, and you used this mask with a Class B or Class C mask, you’d always have only two available host IDs. The reason you would use this with a Class A mask is because it can give you up to 4,194,304 subnets with two hosts each. This is a perfect mask for a point-to-point link, so I suggest always saving a few block sizes of four (/30) masks for use on WANs and point-to-point LAN connections.
If you use the 10.2.3.0 network, your subnets are always 2.3 in the second and third octets, respectively. But the fourth octet is where it changes, as in 256-252=4, 8, 12, 16, 20, 24, 28, etc., all the way to 248. If you use subnet zero, your first subnets are 0, and your last subnet is 255.
An example of the 10.2.3.0 subnet is:
• Subnet: 10.2.3.0
• Broadcast: 10.2.3.3
• Valid hosts: 10.2.3.1 and 10.2.3.2
An example of the 10.2.3.252 subnet is:
• Subnet: 10.2.3.252
• Broadcast: 10.2.3.255
• Valid hosts: 10.2.3.253 and 10.2.3.254
Saturday, August 15, 2009
Flynn's taxonomy
Flynn's taxonomy is a classification of computer architectures, proposed by Michael J. Flynn in 1996.
Parallel Computing Platforms
In order to achieve parallel execution in software, hardware must provide a platform that supports the simultaneous execution of multiple threads. Generally speaking, computer architectures can be classified by two different dimensions. The first dimension is the number of instruction streams that a particular computer architecture may be able to process at a single point in time. The second dimension is the number of data streams that can be processed at a single point in time. In this way, any given computing system can be described in terms of how instructions and data are processed. This classification system is known as Flynn’s taxonomy (Flynn, 1972), and is graphically depicted in the following Figure.
Figure : Flynn’s Taxonomy
Flynn’s taxonomy places computing platforms in one of four categories:
��A single instruction, single data (SISD) machine is a traditional sequential computer that provides no parallelism in hardware. Instructions are executed in a serial fashion. Only one data stream is processed by the CPU during a given clock cycle. Examples of
these platforms include older computers such as the original IBM PC, older mainframe computers, or many of the 8-bit home computers such as the Commodore 64 that were popular in the early 1980s.
��A multiple instruction, single data (MISD) machine is capable of processing a single data stream using multiple instruction streams simultaneously. In most cases, multiple instruction streams need multiple data streams to be useful, so this class of parallel
computer is generally used more as a theoretical model than a practical, mass-produced computing platform.
��A single instruction, multiple data (SIMD) machine is one in which a single
instruction stream has the ability to process multiple data streams simultaneously. These machines are useful in applications such as general digital signal processing, image
processing, and multimedia applications such as audio and video. Originally, supercomputers known as array processors or vector processors such as the Cray-1 provided SIMD processing capabilities. Almost all computers today implement some form of SIMD instruction set. Intel processors implement the MMX™,
Streaming SIMD Extensions (SSE), Streaming SIMD Extensions 2 (SSE2), and Streaming SIMD Extensions 3 (SSE3) instructions that are capable of processing multiple data elements in a single clock. The multiple data elements are stored in the floating point registers. PowerPC† processors have implemented the AltiVec
instruction set to provide SIMD support.
��A multiple instruction, multiple data (MIMD) machine is capable of is executing multiple instruction streams, while working on a separate and independent data stream. This is the most common parallel computing platform today. New multi-core platforms
such as the Intel® Core™ Duo processor fall into this category. Given that modern computing machines are either the SIMD or MIMD machines, software developers have the ability to exploit data-level and task level parallelism in software.
Flynn’s Taxonomy falls short in a number of ways:
• First, there appears to be no need for MISD machines.
• Second, parallelism is not homogeneous. This assumption ignores the contribution of specialized processors.
• Third, it provides no straightforward way to distinguish architectures of the MIMD category.
– One idea is to divide these systems into those that share memory, and those that don’t, as well as whether the interconnections are bus-based or switch-based.
• Symmetric multiprocessors (SMP) and massively parallel processors (MPP) are MIMD architectures that differ in how they use memory.
• SMP systems share the same memory and MPP do not.
• An easy way to distinguish SMP from MPP is: MPP ⇒ many processors + distributed memory + communication via network SMP ⇒ fewer processors + shared memory + communication via memory
• Other examples of MIMD architectures are found in distributed computing, where processing takes place collaboratively among networked computers.
– A network of workstations (NOW) uses otherwise idle systems to solve a problem.
– A collection of workstations (COW) is a NOW where one workstation coordinates the actions of the others.
– A dedicated cluster parallel computer (DCPC) is a group of workstations brought together to solve a specific problem.
– A pile of PCs (POPC) is a cluster of (usually) heterogeneous systems that form a dedicated parallel system
• Flynn’s Taxonomy has been expanded to include SPMD (single program, multiple data) architectures.
• Each SPMD processor has its own data set and program memory. Different nodes can execute different instructions within the same program using
instructions similar to: If myNodeNum = 1 do this, else do that
• Yet another idea missing from Flynn’s is whether the architecture is instruction driven or data driven.
Parallel Computing Platforms
In order to achieve parallel execution in software, hardware must provide a platform that supports the simultaneous execution of multiple threads. Generally speaking, computer architectures can be classified by two different dimensions. The first dimension is the number of instruction streams that a particular computer architecture may be able to process at a single point in time. The second dimension is the number of data streams that can be processed at a single point in time. In this way, any given computing system can be described in terms of how instructions and data are processed. This classification system is known as Flynn’s taxonomy (Flynn, 1972), and is graphically depicted in the following Figure.
Figure : Flynn’s Taxonomy
Flynn’s taxonomy places computing platforms in one of four categories:
��A single instruction, single data (SISD) machine is a traditional sequential computer that provides no parallelism in hardware. Instructions are executed in a serial fashion. Only one data stream is processed by the CPU during a given clock cycle. Examples of
these platforms include older computers such as the original IBM PC, older mainframe computers, or many of the 8-bit home computers such as the Commodore 64 that were popular in the early 1980s.
��A multiple instruction, single data (MISD) machine is capable of processing a single data stream using multiple instruction streams simultaneously. In most cases, multiple instruction streams need multiple data streams to be useful, so this class of parallel
computer is generally used more as a theoretical model than a practical, mass-produced computing platform.
��A single instruction, multiple data (SIMD) machine is one in which a single
instruction stream has the ability to process multiple data streams simultaneously. These machines are useful in applications such as general digital signal processing, image
processing, and multimedia applications such as audio and video. Originally, supercomputers known as array processors or vector processors such as the Cray-1 provided SIMD processing capabilities. Almost all computers today implement some form of SIMD instruction set. Intel processors implement the MMX™,
Streaming SIMD Extensions (SSE), Streaming SIMD Extensions 2 (SSE2), and Streaming SIMD Extensions 3 (SSE3) instructions that are capable of processing multiple data elements in a single clock. The multiple data elements are stored in the floating point registers. PowerPC† processors have implemented the AltiVec
instruction set to provide SIMD support.
��A multiple instruction, multiple data (MIMD) machine is capable of is executing multiple instruction streams, while working on a separate and independent data stream. This is the most common parallel computing platform today. New multi-core platforms
such as the Intel® Core™ Duo processor fall into this category. Given that modern computing machines are either the SIMD or MIMD machines, software developers have the ability to exploit data-level and task level parallelism in software.
Flynn’s Taxonomy falls short in a number of ways:
• First, there appears to be no need for MISD machines.
• Second, parallelism is not homogeneous. This assumption ignores the contribution of specialized processors.
• Third, it provides no straightforward way to distinguish architectures of the MIMD category.
– One idea is to divide these systems into those that share memory, and those that don’t, as well as whether the interconnections are bus-based or switch-based.
• Symmetric multiprocessors (SMP) and massively parallel processors (MPP) are MIMD architectures that differ in how they use memory.
• SMP systems share the same memory and MPP do not.
• An easy way to distinguish SMP from MPP is: MPP ⇒ many processors + distributed memory + communication via network SMP ⇒ fewer processors + shared memory + communication via memory
• Other examples of MIMD architectures are found in distributed computing, where processing takes place collaboratively among networked computers.
– A network of workstations (NOW) uses otherwise idle systems to solve a problem.
– A collection of workstations (COW) is a NOW where one workstation coordinates the actions of the others.
– A dedicated cluster parallel computer (DCPC) is a group of workstations brought together to solve a specific problem.
– A pile of PCs (POPC) is a cluster of (usually) heterogeneous systems that form a dedicated parallel system
• Flynn’s Taxonomy has been expanded to include SPMD (single program, multiple data) architectures.
• Each SPMD processor has its own data set and program memory. Different nodes can execute different instructions within the same program using
instructions similar to: If myNodeNum = 1 do this, else do that
• Yet another idea missing from Flynn’s is whether the architecture is instruction driven or data driven.
New features that are added to microprocessor generation by generation.
Microprocessor:
A device that integrates the functions of the central processing unit (CPU) of a computer onto one semiconductor chip or integrated circuit (IC). In essence, the microprocessor contains the core elements of a computer system, its computation and control engine. Only a power supply, memory, peripheral interface ICs, and peripherals (typically input/output and storage devices) need be added to build a complete computer system.
A microprocessor consists of multiple internal function units. A basic design has an arithmetic logic unit (ALU), a control unit, a memory interface, an interrupt or exception controller, and an internal cache. More sophisticated microprocessors might also contain extra units that assist in floating-point match calculations, program branching, or vector processing
A microprocessor consists of multiple independent function units. The memory interface fetches instructions from, and writes data to, external memory. The control unit issues one or more instructions to other function units. These units process the instructions in parallel to boost performance.
The ALU performs all basic computational operations: arithmetic, logical, and comparisons.
The History of Intel
In 1968, Bob Noyce and Gordon Moore were two unhappy engineers working for the Fairchild Semiconductor Company who decided to quit and create their own company at a time when many Fairchild employees were leaving to create start-ups. People like Noyce and Moore were nicknamed the "Fairchildren".
Bob Noyce typed himself a one page idea of what he wanted to do with his new company, and that was enough to convince San Francisco venture capitalist Art Rock to back Noyce's and Moore's new venture. Rock raised $2.5 million dollars in less than 2 days.
A device that integrates the functions of the central processing unit (CPU) of a computer onto one semiconductor chip or integrated circuit (IC). In essence, the microprocessor contains the core elements of a computer system, its computation and control engine. Only a power supply, memory, peripheral interface ICs, and peripherals (typically input/output and storage devices) need be added to build a complete computer system.
A microprocessor consists of multiple internal function units. A basic design has an arithmetic logic unit (ALU), a control unit, a memory interface, an interrupt or exception controller, and an internal cache. More sophisticated microprocessors might also contain extra units that assist in floating-point match calculations, program branching, or vector processing
A microprocessor consists of multiple independent function units. The memory interface fetches instructions from, and writes data to, external memory. The control unit issues one or more instructions to other function units. These units process the instructions in parallel to boost performance.
The ALU performs all basic computational operations: arithmetic, logical, and comparisons.
The History of Intel
In 1968, Bob Noyce and Gordon Moore were two unhappy engineers working for the Fairchild Semiconductor Company who decided to quit and create their own company at a time when many Fairchild employees were leaving to create start-ups. People like Noyce and Moore were nicknamed the "Fairchildren".
Bob Noyce typed himself a one page idea of what he wanted to do with his new company, and that was enough to convince San Francisco venture capitalist Art Rock to back Noyce's and Moore's new venture. Rock raised $2.5 million dollars in less than 2 days.
computer protidin 12-08-09
ਊগলাইজার িদেয় তথઘ োখঁাজা
ওেয়বসাইট োদখার মઓਡઙ সফটওয়ઘার ফায়ারফেਈ োখালা োকােনা ওেয়বসাইট োথেকোসখানকার োকােনা শਲ਼ সઃপেকગ িবએ੪ািরত জানেত ਊগেল তথઘ োখঁাজার (সাচગ করা) পਤিত খઓবই সহজ৷ শਲ਼িট িনবગাচন কের মাউেস ডান বাটন িকઆক কের োমনઓઘ োথেক Search Google for অপশনিট োবেছ িনেত হেব৷ একিট নতઓন উইেਫােত ফলাফল োদখা যােব৷
হয়েতা অাপিন মઓਡઙ অপােরিটং িসেઍটম উবઓ੯টઓেত োকােনা িপিডএফ ফাইল বা টারিমনােল োকােনা মઘানઓয়াল পড়েছন, োসখােন মাউেসর ডান বাটেনর োমনઓઘেত এমন োকােনা অপশন োনই৷ িকੰ੫ অাপনার োয তথઘ োখঁাজার পઝেয়াজন হেব না, এমনিট নয়৷ োখঁাজার এই সઓিবধািট যઓਡઙ করেত পােরন ਊগলাইজার সফটওয়ઘারিট বઘবহার কের৷ িসনઘাপিটক পઘােকজ মઘােনজার বা Add/Remove োথেক ইনઍটল কের িনেত পরেবন সফট্ওয়ઘারিট৷ ইনઍটেলর পর ApplicationsInternet োমনઓઘেত পাওয়া যােব সফটওয়ઘারিট৷ পઝেয়াজেন পઘােনেল যઓਡઙ কের িনেত পােরন৷
এিট বઘবহার করার পਤিত খઓবই সহজ৷ োকােনা োলখা িনবગাচন কের ਊগলাইজার অাইকেন িকઆক করেত হেব৷ সফট্ওয়ઘারিট বઘবহার কের উবઓ੯টઓর োয োকােনা োপઝাগઝাম োথেক োলখা ਊগল বা ਊগলাইজােরর সাচગ ইিਙেন োখঁাজা যােব৷ যিদ োকােনা ফায়ারফਈ উইেਫা োখালা থােক, তেব োসখােন নতઓন একিট টઘােব অার যিদ ফায়ারফਈ োখালা না থােক, তেব ফায়ারফਈ চালઓ হেয় ফলাফল োদখােব৷
ਊগলাইজার বઘবহার কের www.google.com ওেয়বসাইেট তথઘ খઓঁেজ ফলাফল োদখােনা হয়৷ তেব োখঁাজার জনઘ এ সাইেটর পিরবেতગ ਊগেলর িবিভনੱ એઐানীয় (োলাকালাইজড) সাইট ও অনઘানઘ সাচગ ইিਙন বઘবহার করা যােব৷ অােগ োথেক িনধગািরত (িডফઈট) সাচગ ইিਙন পিরবতગন করার জনઘ পઘােনেলর অাইকনিটেত ডান িকઆক কের োপઝাপািটગজ িনবગাচন করেল নতઓন একিট উইেਫা খઓলেব হেব৷ োসখােন command-এর পােশর জায়গায় োয িঠকানা োলখা হেব, িডফઈট সাচગ ইিਙন িহেসেব োসিট কাজ করেব৷ োযমন- www.google.com.bd সাইটিট বઘবহার করেত চাইেল োসখােন িলখেত হেব
googlizer �url http://www.google.com.bd/search? hl=bn&q=
www.bing.com, www.yahoo.com বা www.ask.com এর জনઘ িলখেত হেব
googlizer �url http://www.bing.com/search?q=
googlizer �url http://search.yahoo.com/search?p=
googlizer �url http://www.ask.com/web?q=
Properties োথেক ই੧ছা করেল োলােগািটও পিরবতગন কের বઘবহার করা যায়৷
ওেয়বসাইট োদখার মઓਡઙ সফটওয়ઘার ফায়ারফেਈ োখালা োকােনা ওেয়বসাইট োথেকোসখানকার োকােনা শਲ਼ সઃপেকગ িবએ੪ািরত জানেত ਊগেল তথઘ োখঁাজার (সাচગ করা) পਤিত খઓবই সহজ৷ শਲ਼িট িনবગাচন কের মাউেস ডান বাটন িকઆক কের োমনઓઘ োথেক Search Google for অপশনিট োবেছ িনেত হেব৷ একিট নতઓন উইেਫােত ফলাফল োদখা যােব৷
হয়েতা অাপিন মઓਡઙ অপােরিটং িসেઍটম উবઓ੯টઓেত োকােনা িপিডএফ ফাইল বা টারিমনােল োকােনা মઘানઓয়াল পড়েছন, োসখােন মাউেসর ডান বাটেনর োমনઓઘেত এমন োকােনা অপশন োনই৷ িকੰ੫ অাপনার োয তথઘ োখঁাজার পઝেয়াজন হেব না, এমনিট নয়৷ োখঁাজার এই সઓিবধািট যઓਡઙ করেত পােরন ਊগলাইজার সফটওয়ઘারিট বઘবহার কের৷ িসনઘাপিটক পઘােকজ মઘােনজার বা Add/Remove োথেক ইনઍটল কের িনেত পরেবন সফট্ওয়ઘারিট৷ ইনઍটেলর পর ApplicationsInternet োমনઓઘেত পাওয়া যােব সফটওয়ઘারিট৷ পઝেয়াজেন পઘােনেল যઓਡઙ কের িনেত পােরন৷
এিট বઘবহার করার পਤিত খઓবই সহজ৷ োকােনা োলখা িনবગাচন কের ਊগলাইজার অাইকেন িকઆক করেত হেব৷ সফট্ওয়ઘারিট বઘবহার কের উবઓ੯টઓর োয োকােনা োপઝাগઝাম োথেক োলখা ਊগল বা ਊগলাইজােরর সাচગ ইিਙেন োখঁাজা যােব৷ যিদ োকােনা ফায়ারফਈ উইেਫা োখালা থােক, তেব োসখােন নতઓন একিট টઘােব অার যিদ ফায়ারফਈ োখালা না থােক, তেব ফায়ারফਈ চালઓ হেয় ফলাফল োদখােব৷
ਊগলাইজার বઘবহার কের www.google.com ওেয়বসাইেট তথઘ খઓঁেজ ফলাফল োদখােনা হয়৷ তেব োখঁাজার জনઘ এ সাইেটর পিরবেতગ ਊগেলর িবিভনੱ એઐানীয় (োলাকালাইজড) সাইট ও অনઘানઘ সাচગ ইিਙন বઘবহার করা যােব৷ অােগ োথেক িনধગািরত (িডফઈট) সাচગ ইিਙন পিরবতગন করার জনઘ পઘােনেলর অাইকনিটেত ডান িকઆক কের োপઝাপািটગজ িনবગাচন করেল নতઓন একিট উইেਫা খઓলেব হেব৷ োসখােন command-এর পােশর জায়গায় োয িঠকানা োলখা হেব, িডফઈট সাচગ ইিਙন িহেসেব োসিট কাজ করেব৷ োযমন- www.google.com.bd সাইটিট বઘবহার করেত চাইেল োসখােন িলখেত হেব
googlizer �url http://www.google.com.bd/search? hl=bn&q=
www.bing.com, www.yahoo.com বা www.ask.com এর জনઘ িলখেত হেব
googlizer �url http://www.bing.com/search?q=
googlizer �url http://search.yahoo.com/search?p=
googlizer �url http://www.ask.com/web?q=
Properties োথেক ই੧ছা করেল োলােগািটও পিরবতગন কের বઘবহার করা যায়৷
Subscribe to:
Posts (Atom)