How to Place Authenticating DC IN DFS Referral List

How to Place Authenticating DC IN DFS Referral List 

• KEY NAME: HKLM\System\CurrentControlSet\Services\Dfs
• Entry Name: PreferLogonDC
• Type: DWORD
• Value:1
  

How to Strict AD Replication To A Specific Port

How to Strict AD Replication To A Specific Port?

Method 2:-

• KEY NAME: HEKY_LOCAL_MACHINE\SYSTEM\CurrentControlSet\Services\NTDS\Parameters

• Entry Name: TCP/IP Port

• Value: Port No...

Method 1   :-

• KEY NAME: 

• HKEY_LOCAL_MACHINE\SYSTEM\CurrentControlSet\Services\Netlogon\Parameters

• Entry Name: DCTcpipPort

• Value: Port No...

How to Change Default DNS Interval For Domain Controllers

How to Change Default DNS Interval For Domain Controllers

• KEY Name: HKEY_LOCAL_MACHINE\SYSTEM\CurrentControlSet\Services\Netlogon\Parameters

• Entry Name: DnsRefreshInterval

• Type: REG_DWORD

• Value: value in seconds

How to Check DCs Register Records with DNS Dynamically

How to Check DCs Register Records with DNS Dynamically 

• KEY Name: HKEY_LOCAL_MACHINE\SYSTEM\CurrentControlSet\Services\Netlogon\Parameters

• Entry Name: UseDynamicDns

• Type: REG_DWORD 

• Value: 0(disabled) 1(enabled)

How to Netlogon Service on Domain Controller

How to Netlogon Service on Domain Controller 

This service is responsible for creating Secure Channel between Domain Controllers and client computers. Secure Channel is created to pass the authentication packets.

• Service performs the registration of SRV records, CNAME and other DC records in the DNS Server to advertise the availability of Domain Controllers in the domain.

• SRV Records registered by NetLogon Service are stored in C:\Windows\System32\Config\NetLogon.DNS File.

• Performs registration of SRV Records every 24 hours depending on the version of Operating System in use.

• Registers the SRV Records for a site where there is no Domain Controller. This is called Site Coverege. 

How to Prevent Registration of A and GC For Domains Controllers

How to Prevent Registration of A and GC For Domains Controllers 

Simple Steps Given Below:-

• DomainName.Com    A   IP_Address

• gc._msdcs.DomainName.Com     A    IP_Address

You can use the below mentioned registry entry on the domain controllers to dis-allow registration of these records:

• KEY Name: HKEY_LOCAL_MACHINE\SYSTEM\CurrentControlSet\Services\NetLogon\Parameters
• Entry Name: RegisterDnsARecords
• Data: DWORD
• Value: 1 (Register A Records), 0 (Do not register A Records)

Note: This regsitry entry does not exist by default.

How to Prevent Net Logon Service From Registering SRV & Domain Records

How to Prevent Net Logon Service From Registering SRV & Domain Records  

To disable this behaviour, use the below registry hack:

• KEY Name: HKEY_LOCAL_MACHINE\SYSTEM\CurrentControlSet\Services\NetLogon\Parameters
  

• Entry Name: UseDynamicDns
  

• Data: REG_DWORD
  

• Value: 0 (disabled registration of SRV records dynamically) 1 (enabled registration of SRV records dynamically)
  

Note: The default value is 1. So NetLogon performs the registration of SRV and other records dynamically. If you disable this then you must register the SRV and other Records found in C:\Windows\System32\Config\NetLogon.DNS file.

How to Change KCC Interval

How to Change KCC Interval

• KEY Name: HKLM\System\CurrentControlSet\Services\NTDS\Parameters

• Entry Name: Repl topology update period (secs)

• Type: REG_DWORD

• Value: 0x384 (this is the default value)  
  

ROSP AND FGPP

ROSP AND FGPP

However, you can try using one of the following methods to view the effect of FGPP on a user account

By using the LDIFDE command-line utility as follows:

Ldifde /d/l msds-resultantPSO /f con /p base /s

By using the Dsget command-line utility as follows:

dsget user-effectivepso

ADMT Migration & Networks Ports For Windows

ADMT Migration & Networks Ports For Windows 

If you are performing an ADMT migration from a computer which sits behind the firewall, it is important that you open the required network ports to allow ADMT computer to communicate with both source and target domain controllers

ADMT will fail to work if below network ports are not opened:

1. 389 - LDAP

2. 88 - Kerberos

3. 53 - DNS

4. 445 - SMB/CIFS

5. 3268 - GC

6. 135 - RPC

7. 1024-65535 - Dynamic Port Range   
   

layer model for osi

LAYER MODEL FOR OSI REFERENCE

The Open System Interconnect (OSI) reference model is a model, developed by the International Standards Organization (ISO), which describes how data from an application on one computer can be transferred to an application on another computer. The OSI reference model consists of seven conceptual layers which each specify different network functions. Each function of a network can be assigned to one, or perhaps a couple of adjacent layers, of these seven layers and is relatively independent of the other layers. This independence means that one layer does not need to be aware of what the implementation of an adjacent layer is, merely how to communicate with it. This is a major advantage of the OSI reference model and is one of the major reasons why it has become one of the most widely used architecture models for inter-computer communications.

The seven layers of the OSI reference model, as shown in Figure 1, are:

• Application
• Presentation
• Session
• Transport
• Network
• Data link
• Physical

Figure 1: Diagram of the OSI reference model layers, courtesy of catalyst.washington.edu

Over the next few articles I will be discussing each layer of the model and the networking hardware which relates to that layer. This article, as you have probably guessed from the title, will discuss layer 1; the physical layer.

While many people may simply state that all networking hardware belongs exclusively in the physical layer, they are wrong. Many networking hardware devices can perform functions belonging to the higher layers as well. For example, a network router performs routing functions which belong in the network layer.

What does the physical layer include? Well, the physical layer involves the actual transmission of signals over a medium from one computer to another. This layer includes specifications for the electrical and mechanical characteristics such as: voltage levels, signal timing, data rate, maximum transmission length, and physical connectors, of networking equipment. For a device to operate solely in the physical layer, it will not have any knowledge of the data which it transmits. A physical layer device simply transmits or receives data.

There are four general functions which the physical layer is responsible for. These functions are:

• Definitions of hardware specifications
• Encoding and signaling
• Data transmission and reception
• Topology and physical network design

Definitions of hardware specifications

Each piece of hardware in a network will have numerous specifications. If you read my previous article titled Copper and Glass: A Guide to Network Cables [link this title to my previous article of that title], you will learn about some of the more common specifications which apply to network cables. These specifications include things like the maximum length of a cable, the width of the cable, the protection from electromagnetic interference, and even the flexibility.

Another area of hardware specifications are the physical connectors. This includes both the shape and size of the connectors as well as the pin count and layout, if appropriate.

Encoding and signaling

Encoding and signaling is a very important part of the physical layer. This process can get quite complicated. For example, let's look at Ethernet. Most people learn that signals are sent in '1's and '0's using a high voltage level and a low voltage level to represent the two states. While this is useful for some teaching purposes, it is not correct. Signals over Ethernet are sent using Manchester encoding. This means that '1's and '0's are transmitted as rises and falls in the signal. Let me explain.

If you were to send signals over a cable where a high voltage level represents a '1' and a low voltage signal represents a '0' the receiver would also need to know when to sample that signal. This is usually done with a separate clock signal being transmitted. This method is called a Non-return to Zero (NRZ) encoding, and has some serious drawbacks. First, if you do include a separate clock signal you are basically transmitting two signals and doubling the work. If you don't want to transmit the clock signal, you could include an internal clock in the receiver but this must be in near perfect synchronization with the transmitter clock. Let's assume you can synchronize the clocks, which becomes much harder as the transmission speed increases, there is still the problem of keeping this synchronization when there is a long stretch of the same bit being transmitted; it is the transitions which help synchronize the clocks.

The limitations of the NRZ encoding can be overcome by technology developed in the 1940s at the University of Manchester [link University of Manchester to http://www.manchester.ac.uk/], in Manchester, UK. Manchester encoding combines the clock signal with the data signal. While this does increase the bandwidth of the signal, it also makes the successful transmission of the data much easier and reliable.

A Manchester encoded signal, transmits data as a rising or falling edge. Which edge represents the '1' and which represents the '0' must be decided first, but both are considered Manchester encoded signals. Ethernet and IEEE standards use the rising edge as a logical '1'. The original Manchester encoding used the falling edge as a '1'.

One situation which you may be thinking about is that if you need to transmit two '1's in a row the signal will already be high when you need to transmit the second '1'. This isn't the case because the rising or falling edge which represents data is transmitted in the middle of the bit boundaries; the edge of the bit boundaries either contain a transition or do not, which puts the signal in the right position for the next bit to be transmitted. The end result is that at the center of every bit is a transition, the direction of the transition represents either a '1' or a '0' and the timing of the transition is the clock.

While there are many other encoding schemes, many of which are much more advanced than NRZ or Manchester encoding, the simplicity and reliability of Manchester encoding has kept it a valuable standard still widely in use.

Data transmission and reception

Whether the network medium is an electrical cable, an optical cable, or radio frequency, there needs to be equipment that physically transmits the signal. Likewise, there also needs to be equipment that receives the signal. In the case of a wireless network, this transmission and reception is done by highly designed antennas which transmit, or receive, signals at predefined frequencies with predefined bandwidths.

Optical transmission lines use equipment which can produce and receive pulses of light, the frequency of which is used to determine the logical value of the bit. Equipment such as amplifiers and repeaters, which are commonly employed in long-haul optical transmissions, are also included in the physical layer of the OSI reference model.

preventing Network Pocket Collisions

Preventing Network Pocket Collisions 

Your 100-Mb network should be blazing along, but users complain that it's slow. Mismatched inter frame gap settings among the network adapters could be the problem. Learn how to fix this by adjusting the TCP/IP protocol and inter frame gap settings. 

Are you not getting the performance you’d expect from a network running at 100 Mb, even though you're using 100-Mb network adapters with TCP/IP as your network protocol? If so, 100-Mb network adapters not adhering to the IEEE 802.3 specification for Ethernet networks could be the cause for this slip in performance. The IEEE 802.3 specification calls for a 96-bit-times inter frame gap between the time the adapter switches from transmit mode to receive mode. 

When adapters that adhere to this specification are used on the same network with adapters that don’t, timing problems occur that eventually cause collisions on the network. This problem is compounded on networks that use the TCP/IP protocol because of the scheme it uses for acknowledging the receipt of data packets. In this Daily Drill Down, I’ll explain how you can solve this problem by adjusting the inter frame gap settings and the way that the TCP/IP protocol works with Windows 2000 Professional and Windows XP Professional clients. 

The inter frame problem in a nutshell 

The IEEE 802.3 specification calls for a 96-bit-times inter frame gap. Basically, this inter frame gap is a period of silence between the moment a particular station sends a data packet across the network and the time that the other station responds. For a 100-b network card, this 96-bit-times inter frame gap comes out to 960 nanoseconds. 

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Note 

For the sake of comparison, the 96-bit-times inter frame gap is 9.6 microseconds for a 10-Mb network card, 96 nanoseconds for a 1-Gb network card, and 9.6 nanoseconds for a 10-Gb network card. 

-------------------------------------------------------------------------------- 

The inter frame gap is designed to give the adapters involved in the communication time to switch from transmit mode to receive mode. In other words, once a network card has transmitted a data packet, the card needs a few nanoseconds before it is ready to receive a response. 

When it comes to the speed of today’s computer technology, you may think that 960 nanoseconds is a long time—and it is. However, when the specification was first hammered out, using a formula based on 96 bit times was necessary for the network cards of the day to properly switch from transmit mode to receive mode. 

While the 96-bit-times inter frame gap rule has remained set in stone, network cards have continued to evolve as the technology has improved and network speeds have increased. The newer 100-Mb cards no longer actually require a 96-bit-times inter frame gap to switch from transmit mode to receive mode. 

Considering this, some 100-Mb network card manufacturers are ignoring the standard and configuring their products with a much smaller inter frame gap to pump up the data transfer rate. However, other 100-Mb network card manufacturers continue to adhere to the 96-bit-times standard. 

If you're using 100-Mb network cards from the same manufacturer in all the systems on your network, you’ll be fine. But a more common scenario is an organization that uses 100-Mb network cards from multiple manufacturers. If you administer in this type of configuration, it's possible your network is suffering due to an inordinate amount of network collisions caused by differing inter frame gap settings. 

The TCP/IP headache 

While the differences in the inter frame gap settings are a big enough performance problem to deal with, the situation can get worse when you’re using TCP/IP as your main protocol. 

The TCP/IP protocol specification calls for sending an ACK (acknowledgment) for every two data packets that a network station receives. If the network cards in the two systems are using different inter frame gap settings, it’s possible that a network station that has received two TCP/IP data packets could be sending its ACK at the same time the sender station is sending another data packet. This would compound the network collision problem already initiated by the mismatched inter frame gap settings. 

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Use a protocol analyzer to detect collisions 

If you’re encountering network performance problems, you can easily test your network with a protocol analyzer, or a similar device, to measure collisions, runt packets, or CRC (cyclic redundancy check) errors. Keep in mind that under regular operating conditions, some collisions are normal. But a high percentage of collisions, runt packets, or CR errors indicates a serious performance problem that could very well be related to the differing inter frame gap settings of the various network cards on your network. 

Adjusting inter frame gap settings 

To begin, you need to have on hand a list of all the brands and versions of 100-Mb network cards installed on your network. (This is where a network inventory will come in very handy.) Next, check the network card documentation for a specification on the inter frame gap setting. You may need to contact the manufacturer directly, or via the company's Web site, to track down the inter frame gap setting. Also find out if the network card provides a method of adjusting the inter frame gap setting. 

The goal here is to have all the cards on your network use the same inter frame gap setting. So, if those cards that have nonstandard inter frame gap settings provide a mechanism for adjusting the setting, you’ll want to bring them in line with all of the other cards on your network. 

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Other terms 

As you’re searching for information relating to your specific network card’s inter frame gap setting, keep in mind that this setting has several aliases. Some manufacturers call it the inter frame space or inter-frame space. I’ve also seen it referred to as the inter packet gap. It also goes by the acronyms IFG, IFS, and IPG. 

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Adjusting the TCP/IP protocol 

You can also alleviate the problem by adjusting the way the TCP/IP protocol works. As I mentioned, the default configuration for the TCP/IP protocol calls for sending an ACK for every two data packets that a network station receives, which can cause a higher volume of network collisions on a network using network adapters with mismatched inter frame gap settings. The adjustment I’ll explain here will configure the TCP/IP protocol to send an ACK for every data packet received. To do so, you’ll need to edit the registry and add a special parameter called the TcpWindowSize to the TCP/IP key. 

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Note 

Since editing the registry can be dangerous, consider performing a full backup before you attempt this operation. 

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To begin, select the Run command on the Start menu and launch the Registry Editor by typing regedit.exe in the Open text box. Once you have the Registry Editor up and running, double-click on the key HKEY_LOCAL_MACHINE. When this subtree is visible, open each of the following subtrees in succession: 

SYSTEM 

CurrentControlSet 

Services 

Tcpip 

Parameters 

Pull down the Edit menu and select New | DWORD Value. Then, assign the new value the name TcpWindowSize and press [Enter] twice. When you see the Edit DWORD Value dialog box, type 2920 in the Value Data text box. Finally, select the Decimal option in the Base panel. Your system will be set to send an ACK with every data packet received. 

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Note 

Keep in mind that adding this TcpWindowsSize parameter to the registry can degrade network performance for Wide Area Networks (WANs) because there is already some extra latency introduced by this type of connection. 

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Conclusion 

When you use 100-Mb network adapters from various manufacturers and are using TCP/IP as your network protocol, you might not be getting the performance you’d expect from a network running at 100 Mb due to mismatched interframe gap settings. I’ve explained how to solve the problem by tracking down information on your network cards' interframe gap settings and how to adjust the setting, if needed. In addition, I’ve explained how to alleviate the problem by adjusting the way the TCP/IP protocol works.

Networking Essentials

Networking Essentials Notes - Section 1

Network Orientation:

Peer to Peer Networks

• No dedicated server or hierarchy, also called a workgroup.
• Usually 10 or fewer workstations.
• Users act as their own administrator and security.
• Computers are in same general area.
• Limited growth.

Server Based Networks

• 10 or more users.
• Employs specialized servers.
1. File and Print
2. Application
3. Mail
4. Fax
5. Communications (gateways)
• Central administration.
• Greater security.
• Centralized backup.
• Data Redundancy.
• Supports many users

• Combination Networks

• Combines the features of both Peer to Peer and Server based networks
• Users can share resources among themselves as well as access server-based resources.

Network Topologies

There are 4 basic topologies with variations

Bus Topology

• Bus consists of a single linear cable called a trunk.
• Data is sent to all computers on the trunk. Each computer examines EVERY packet on the wire to determine who the packet is for and accepts only messages addressed to them.
• Bus is a passive topology.
• Performance degrades as more computers are added to the bus.
• Signal bounce is eliminated by a terminator at each end of the bus.
• Barrel connectors can be used to lengthen cable.
• Repeaters can be used to regenerate signals.
• Usually uses Thinnet or Thicknet
• both of these require 50 ohm terminator
• good for a temporary, small (fewer than 10 people) network
• But its difficult to isolate malfunctions and if the backbone goes down, the entire network goes down.

Star Topology

• Computers are connected by cable segments to a centralized hub.
• Signal travels through the hub to all other computers.
• Requires more cable.
• If hub goes down, entire network goes down.
• If a computer goes down, the network functions normally.
• most scalable and reconfigurable of all topologies

Ring Topology

• Computers are connected on a single circle of cable.
• usually seen in a Token Ring or FDDI (fiber optic) network
• Each computer acts as a repeater and keeps the signal strong => no need for repeaters on a ring topology
• No termination required => because its a ring

• Token passing is used in Token Ring networks. The token is passed from one computer to the next, only the computer with the token can transmit. The receiving computer strips the data from the token and sends the token back to the sending computer with an acknowledgment. After verification, the token is regenerated.
• relatively easy to install, requiring ;minimal hardware

Mesh

• The mesh topology connects each computer on the network to the others
• Meshes use a significantly larger amount of network cabling than do the other network topologies, which makes it more expensive.
• The mesh topology is highly fault tolerant.
• Every computer has multiple possible connection paths to the other com-puters on the network, so a single cable break will not stop network communications between any two computers.

Star Bus Topology

• Several star topologies linked with a linear bus.
• No single computer can take the whole network down. If a single hub fails, only the computers and hubs connected to that hub are affected.

Star Ring Topology

• Also known as star wired ring because the hub itself is wired as a ring. This means it's a physical star, but a logical ring.
• This topology is popular for Token Ring networks because it is easier to implement than a physical ring, but it still provides the token passing capabilities of a physical ring inside the hub.
• Just like in the ring topology, computers are given equal access to the network media through
• the passing of the token.
• A single computer failure cannot stop the entire network, but if the hub fails, the ring that the hub controls also fails.

Hybrid Mesh

• most important aspect is that a mesh is fault tolerant
• a true mesh is expensive because of all the wire needed
• another option is to mesh only the servers that contain information that everyone has to get to. This way the servers (not all the workstations) have fault tolerance at the cabling level.

CCNA Number System

Introduction

A number system defines a set of values used to represent quantity. You may regard each digit as a box that can hold a number. In the binary system, there can be only two choices for this number -- either a "0" or a "1". In the octal system, there can be eight possibilities:

"0", "1", "2", "3", "4", "5", "6", "7".

In the decimal system, there are ten different numbers that can enter the digit box:

"0", "1", "2", "3", "4", "5", "6", "7", "8", "9".

In the hexadecimal system, we allow 16 numbers:

"0", "1", "2", "3", "4", "5", "6", "7", "8", "9", "A", "B", "C", "D", "E", and "F".

As demonstrated by the following table, there is a direct correspondence between the binary system and the Hexa system, with four binary digits translate directly into one hexadecimal digit. In computer usage, hexadecimal notation is especially common because it easily replaces the binary notation, which is too long and human mistakes in transcribing the binary numbers are too easily made.

Base Conversion Table

     BIN    HEX   DEC
  ----------------
  0000    0     0
  0001    1     1
  0010    2     2
  0011    3     3
  0100    4     4
  0101    5     5
  0110    6     6
  0111    7     7
  ----------------
  1000    8     8
  1001    9     9
  1010    A    10
  1011    B    11
  1100    C    12
  1101    D    13
  1110    E    14
  1111    F    15

Convert From Any Base To Decimal

Let's think more carefully what a decimal number means. For example, 1234 means that there are four boxes (digits); and there are 4 one's in the right-most box (least significant digit), 3 ten's in the next box, 2 hundred's in the next box, and finally 1 thousand's in the left-most box (most significant digit). The total is 1234:

Original Number: 1 2 3 4 | | | | How Many Tokens: 1 2 3 4 Digit/Token Value: 1000 100 10 1 Value: 1000 + 200 + 30 + 4 = 1234 or simply, 1*1000 + 2*100 + 3*10 + 4*1 = 1234

Thus, each digit has a value: 10^⁰=1 for the least significant digit, increasing to 10^¹=10, 10^²=100, 10^³=1000, and so forth.

Likewise, the least significant digit in a hexadecimal number has a value of 16^⁰=1 for the least significant digit, increasing to 16^¹=16 for the next digit, 16^²=256 for the next, 16^³=4096 for the next, and so forth. Thus, 1234 means that there are four boxes (digits); and there are 4 one's in the right-most box (least significant digit), 3 sixteen's in the next box, 2 256's in the next, and 1 4096's in the left-most box (most significant digit). The total is:

1*4096 + 2*256 + 3*16 + 4*1 = 4660

Example. Convert the hexadecimal number 4B3 to decimal notation.

Solution:

     Original Number  :   4    B    3
                       |    |    |
  How Many Tokens  :   4   11    3
  Digit/Token Value: 256   16    1
  Value:            1024 +176  + 3 = 1203

Another way is to think of a cash register with different slots, each holding bills of a different denomination

Convert From Decimal to Any Base

Again, let's think about what you do to obtain each digit. As an example, let's start with a decimal number 1234 and convert it to decimal notation. To extract the last digit, you move the decimal point left by one digit, which means that you divide the given number by its base 10.

1234/10 = 123 + 4/10

The remainder of 4 is the last digit. To extract the next last digit, you again move the decimal point left by one digit and see what drops out.

123/10 = 12 + 3/10

The remainder of 3 is the next last digit. You repeat this process until there is nothing left. Then you stop. In summary, you do the following:

Quotient Remainder ----------------------------- 1234/10 = 123 4 --------+ 123/10 = 12 3 ------+ | 12/10 = 1 2 ----+ | | 1/10 = 0 1 --+ | | | (Stop when the quotient is 0) | | | | 1 2 3 4 (Base 10)

Now, let's try a nontrivial example. Let's express a decimal number 1341 in binary notation. Note that the desired base is 2, so we repeatedly divide the given decimal number by 2.

Quotient Remainder ----------------------------- 1341/2 = 670 1 ----------------------+ 670/2 = 335 0 --------------------+ | 335/2 = 167 1 ------------------+ | | 167/2 = 83 1 ----------------+ | | | 83/2 = 41 1 --------------+ | | | | 41/2 = 20 1 ------------+ | | | | | 20/2 = 10 0 ----------+ | | | | | | 10/2 = 5 0 --------+ | | | | | | | 5/2 = 2 1 ------+ | | | | | | | | 2/2 = 1 0 ----+ | | | | | | | | | 1/2 = 0 1 --+ | | | | | | | | | | (Stop when the | | | | | | | | | | | quotient is 0) 1 0 1 0 0 1 1 1 1 0 1 (BIN; Base 2)

Let's express the same decimal number 1341 in hexadecimal notation.

Quotient Remainder ----------------------------- 1341/16 = 83 13 ------+ 83/16 = 5 3 ----+ | 5/16 = 0 5 --+ | | (Stop when the quotient is 0) | | | 5 3 D (HEX; Base 16)

Example. Convert the decimal number 3315 to hexadecimal notation.

Solution:

Quotient Remainder ----------------------------- 3315/16 = 207 3 ------+ 207/16 = 12 15 ----+ | 12/16 = 0 12 --+ | | (Stop when the quotient is 0) | | | C F 3 (HEX; Base 16) Thus, 3315 (DEC) --> CF3 (HEX)

Note that from the Base Conversion Table, you can easily get the binary notation from the hexadecimal number by grouping four binary digits per hexadecimal digit, or from or the octal number by grouping three binary digits per octal digit, and vice versa.

HEX 5 3 D BIN 0101 0011 1101 OCT 2 4 7 5 BIN 010 100 111 101

Note: In 640-801 exam there is one question to convert a number from one base to others.