Saturday, 12 November 2016

Active Directory

Active Directory (AD) is a directory service that Microsoft developed for Windows domain networks. A server running Active Directory Domain Services (AD DS) is called a domain controller. It authenticates and authorizes all users and computers in a Windows domain type network—assigning and enforcing security policies for all computers and installing or updating software. For example, when a user logs into a computer that is part of a Windows domain, Active Directory checks the submitted password and determines whether the user is a system administrator or normal user. Also, it allows management and storage of information at admin level and provides authentication and authorization mechanisms and a framework to deploy other related services (AD Certificate Services, AD Federated Services, etc.).

Active Directory is an integral part of the Windows 2000 architecture. Like other directory services, such as Novell Directory Services (NDS), Active Directory is a centralized and standardized system that automates network management of user data, security, and distributed resources, and enables interoperation with other directories. Active Directory is designed especially for distributed networking environments.

Active Directory provides a common interface for organizing and maintaining information related to resources connected to a variety of network directories. The directories may be systems-based (like Windows OS), application-specific or network resources, like printers. Active Directory serves as a single data store for quick data access to all users and controls access to users based on the directory's security policy.

Active Directory features include:

    • Support for the X.500 standard for global directories
    • The capability for secure extension of network operations to the Web
    • A hierarchical organization that provides a single point of access for system administration (management of user accounts, clients, servers, and applications, for example) to reduce redundancy and errors
    • An object-oriented storage organization, which allows easier access to information
    • Support for the Lightweight Directory Access Protocol (LDAP) to enable inter-directory operability

    • Designed to be both backward compatible and forward compatible.

    Active Directory (AD)

         Active Directory provides the following network services:
    • Lightweight Directory Access Protocol (LDAP) - An open standard used to access other directory services
    • Security service using the principles of Secure Sockets Layer (SSL) and Kerberos-based authentication
    • Hierarchical and internal storage of organizational data in a centralized location for faster access and better network administration
    • Data availability in multiple servers with concurrent updates to provide better scalability

    Active Directory is internally structured with a hierarchical framework. Each node in the tree-like structure is referred to as an object and associated with a network resource, such as a user or service. Like the database topic schema concept, the Active Directory schema is used to specify attribute and type for a defined Active Directory object, which facilitates searching for connected network resources based on assigned attributes. For example, if a user needs to use a printer with color printing capability, the object attribute may be set with a suitable keyword, so that it is easier to search the entire network and identify the object's location based on that keyword.
    A domain consists of objects stored in a specific security boundary and interconnected in a tree-like structure. A single domain may have multiple servers - each of which is capable of storing multiple objects. In this case, organizational data is stored in multiple locations, so a domain may have multiple sites for a single domain. Each site may have multiple domain controllers for backup and scalability reasons. Multiple domains may be connected to form a Domain Tree, which shares a common schema, configuration and global catalog (used for searching across domains). A Forest is formed by a set of multiple and trusted domain trees and forms the uppermost layer of the Active Directory.
    Novell's directory service - an Active Directory alternative - contains all server data within the directory itself, unlike Active Directory.

    References:
    https://en.wikipedia.org/wiki/Active_Directory
    https://www.techopedia.com/definition/25/active-directory
    http://serverfault.com/questions/402580/what-is-active-directory-domain-services-and-how-does-it-work

    Saturday, 16 July 2016

    Two circles have their centers 21 cm apart. The radii of the circles are 10 cm and 17 cm. Find the length (in cm) of the common chord of the two circles.

     if two circles having radii a and b with the distance between them as c,
    Length of common chord= 2* Sqrt [ a^2 - ((a^2+c^2-b^2)/2c)^2]

    Ans. 84

    If the HCF of two number is 23 and the LCM is 4186, one of the two numbers is one of the following: 

    a) 276 b) 300 c) 322 d) 345. Which one is it?


    Ans. c) 322 

     Both number should be divisible by its HCF and LCM should be divisible by  both the numbers whose  HCF and LCM are to determine. 
     a) 276%23=0
          4186%276=46

    b)300%23=1
      4186%300=286

    c)322%23=0
     4186%322=0

    d)345%23=0
     4186%345=46

    How do I find two numbers when I know that their LCM is 36 and HCF is 15?


    If the HCF of the two numbers say x and y is 15; the numbers can be represented as following:
    x = 15a and y = 15b where a and b are positive integers
    It is given that LCM of x and y is 36
    Now LCM (x,y) * HCF (x,y) = x * y
    Hence, 36 * 15 = 15a * 15b
    540 = 225*a*b
    or 2.4 = a*b
    The product of two integers cannot be 2.4 hence such two numbers do not exist.

    The sum of two numbers is 52 and their LCM is 168. What are the numbers?


    As LCM of two numbers is 168 that two numbers must be factors of 168
    now think about what are all factors of 168 that are less than 52(because sum of those two is 52)
    factors=(2,3,4,6,8,7,8,12,14,24,28,42)
    from above factors u have to select two numbers such that sum of two is 52
    only possibility is 24+28

    What is the difference between two numbers if their LCM is 630 and HCF is 9 and the sum of the two numbers is 153?


    Given GCD of two numbers is 9 and LCM is 630.
    Let a, b be two numbers
    As we know,
    LCM × HCF = The product of given numbers
    a×b=9×630
    Given a+b=153
    By using formula
    (a-b)^2=(a+b)^2-4×a×b
    (a-b)^2=(153^2)-(4×9×630)
    (a-b)^2=729
    x-y=27
    Ans.27

    Another Solution:

    Let's assume the numbers are 9*a and 9*b (as the hcf is 9).
    As we know,
    LCM × HCF = The product of given numbers
    So, (9*a)*(9*b) = 630*9
    So, a*b = 70
    Again 9*(a + b)= 153

    By solving those two equations, we get either a=10, b=7 or a=7 , b=10
    So either way, the two numbers are 90 and 63.
    So, (ab)=27 or the difference between the two numbers is 27.
    Ans.27

    The LCM of 1, 2, 3, ... 60 is N. What is the LCM of 1, 2, 3, ... 65?


    Answer: 61*2*n

    We are given LCM (1, 2, 3 ... 60) = n
    Now, we need to find out LCM (1, 2, 3... 65)

    = LCM (n, 61, 62, 63, 64, 65) 
    {The first 60 numbers can be replaced with 'n' as we already know their LCM}

    = LCM (n, 61,64) 
    {62, 63, and 65 can be broken down into prime factors which are in turn factors of n} 

    = 61*2*n  
    {61 is a prime number. 64 is 2^6. 2^5 occurs in 32. An extra 2 is required to accommodate it}

    122n

    Explanation:

    Lcm of first 60 numbers is n, so we need to consider 5 more numbers 61...65. Lets use prime factorization and try to figure out how inclusion of each of these numbers will affect lcm.
    61 - It is a prime number and so it has not been included in n till now. So Lcm of 1...61 become 61*n.
    62 - Prime factorization of 62 is 2*31 and both of these numbers are already included in n. So lcm doesn't change by inclusion of 62 and it remains 61*n.
    63 - Prime factorization of 63 is 3*3*7 and these numbers are already there in n. In fact 3 is there three times (for 27) and 7 is there two times (for 49). So inclusion of 63 also doesn't change lcm and lcm remain 61*n for numbers 1....63.
    64 - Factorization of 64 is 2^6. Now 2^5 is already there in n for 32. So we need to include only one more 2, and lcm of numbers 1..upto 64 becomes 61*2*n.
    65 - 65 factorizes as 5*13. Both of these numbers are there in n already and by inclusion of 65, lcm doesn't change. So it remains 61*2*n.