The Microsoft Certified Master: Exchange Server 2007 program provides the most in-depth and comprehensive training that is available today for Microsoft Exchange Server 2007. This three-week training program is delivered by recognized experts from Microsoft and Microsoft Partner organizations.
Training deepens and broadens the technical skills of experienced messaging professionals
Microsoft Certified Master training—a mixture of presentations, white-board discussions, and goal-based labs—covers the individual core Exchange Server 2007 technologies as well as broader architectural and operational areas, such as high availability of a messaging service, co-existence and migration, performance analysis, management, and operations. Candidates complete an independent lab exercise as a part of their training, which allows them to develop and apply their learning in an environment where instructor and peer support is readily available.
Certified individuals prove their technical mastery
Candidates who successfully complete the program can expect to have a greatly improved understanding of the Microsoft Exchange Server 2007 platform. Microsoft Certified Masters can design and build customer-specified messaging solutions, and they understand how design decisions affect the final solution. They can troubleshoot and diagnose configuration and performance issues, and they have the detailed knowledge and skills that are required to successfully operate and manage an enterprise-class Exchange Server 2007 infrastructure
Prerequisites
Applicants must meet or exceed the prerequisite requirements for their chosen program track in order to be accepted into a Microsoft Certified Master program. Required exams and certifications will be verified.
Experience
• Five or more years of hands-on experience with Microsoft Exchange Server 2003: installing, configuring, and troubleshooting
• One or more years of hands-on experience with Microsoft Exchange Server 2007: installing, configuring, and troubleshooting
• A thorough understanding of Microsoft Exchange Server design and architecture
• A thorough understanding of Active Directory, Domain Name System (DNS), and other core network services related to Exchange Server
• Ability to speak, understand, and write fluent English
Successful completion of the following exams and certifications
•
Exam 70-236: TS: Exchange Server 2007, Configuring
•
Exam 70-237: PRO: Designing Messaging Solutions with Microsoft Exchange Server 2007
•
Exam 70-238: PRO: Deploying Messaging Solutions with Microsoft Exchange Server 2007
Additionally required:
•
Microsoft Certified Systems Engineer (MCSE) on Windows Server 2003
or
Exam 70-640: TS: Windows Server 2008 Active Directory, Configuring
Wednesday, September 17, 2008
Exchange Server 2007 -EXAM DETAILS FOR FRESHERS
The Microsoft Certified Master: Exchange Server 2007 program provides the most in-depth and comprehensive training that is available today for Microsoft Exchange Server 2007. This three-week training program is delivered by recognized experts from Microsoft and Microsoft Partner organizations.
Training deepens and broadens the technical skills of experienced messaging professionals
Microsoft Certified Master training—a mixture of presentations, white-board discussions, and goal-based labs—covers the individual core Exchange Server 2007 technologies as well as broader architectural and operational areas, such as high availability of a messaging service, co-existence and migration, performance analysis, management, and operations. Candidates complete an independent lab exercise as a part of their training, which allows them to develop and apply their learning in an environment where instructor and peer support is readily available.
Certified individuals prove their technical mastery
Candidates who successfully complete the program can expect to have a greatly improved understanding of the Microsoft Exchange Server 2007 platform. Microsoft Certified Masters can design and build customer-specified messaging solutions, and they understand how design decisions affect the final solution. They can troubleshoot and diagnose configuration and performance issues, and they have the detailed knowledge and skills that are required to successfully operate and manage an enterprise-class Exchange Server 2007 infrastructure
Prerequisites
Applicants must meet or exceed the prerequisite requirements for their chosen program track in order to be accepted into a Microsoft Certified Master program. Required exams and certifications will be verified.
Experience
• Five or more years of hands-on experience with Microsoft Exchange Server 2003: installing, configuring, and troubleshooting
• One or more years of hands-on experience with Microsoft Exchange Server 2007: installing, configuring, and troubleshooting
• A thorough understanding of Microsoft Exchange Server design and architecture
• A thorough understanding of Active Directory, Domain Name System (DNS), and other core network services related to Exchange Server
• Ability to speak, understand, and write fluent English
Successful completion of the following exams and certifications
•
Exam 70-236: TS: Exchange Server 2007, Configuring
•
Exam 70-237: PRO: Designing Messaging Solutions with Microsoft Exchange Server 2007
•
Exam 70-238: PRO: Deploying Messaging Solutions with Microsoft Exchange Server 2007
Additionally required:
•
Microsoft Certified Systems Engineer (MCSE) on Windows Server 2003
or
Exam 70-640: TS: Windows Server 2008 Active Directory, Configuring
Training deepens and broadens the technical skills of experienced messaging professionals
Microsoft Certified Master training—a mixture of presentations, white-board discussions, and goal-based labs—covers the individual core Exchange Server 2007 technologies as well as broader architectural and operational areas, such as high availability of a messaging service, co-existence and migration, performance analysis, management, and operations. Candidates complete an independent lab exercise as a part of their training, which allows them to develop and apply their learning in an environment where instructor and peer support is readily available.
Certified individuals prove their technical mastery
Candidates who successfully complete the program can expect to have a greatly improved understanding of the Microsoft Exchange Server 2007 platform. Microsoft Certified Masters can design and build customer-specified messaging solutions, and they understand how design decisions affect the final solution. They can troubleshoot and diagnose configuration and performance issues, and they have the detailed knowledge and skills that are required to successfully operate and manage an enterprise-class Exchange Server 2007 infrastructure
Prerequisites
Applicants must meet or exceed the prerequisite requirements for their chosen program track in order to be accepted into a Microsoft Certified Master program. Required exams and certifications will be verified.
Experience
• Five or more years of hands-on experience with Microsoft Exchange Server 2003: installing, configuring, and troubleshooting
• One or more years of hands-on experience with Microsoft Exchange Server 2007: installing, configuring, and troubleshooting
• A thorough understanding of Microsoft Exchange Server design and architecture
• A thorough understanding of Active Directory, Domain Name System (DNS), and other core network services related to Exchange Server
• Ability to speak, understand, and write fluent English
Successful completion of the following exams and certifications
•
Exam 70-236: TS: Exchange Server 2007, Configuring
•
Exam 70-237: PRO: Designing Messaging Solutions with Microsoft Exchange Server 2007
•
Exam 70-238: PRO: Deploying Messaging Solutions with Microsoft Exchange Server 2007
Additionally required:
•
Microsoft Certified Systems Engineer (MCSE) on Windows Server 2003
or
Exam 70-640: TS: Windows Server 2008 Active Directory, Configuring
Storage Networks Explained: Basics and Application of Fibre Channel SAN, NAS iSCSI and InfiniBand
Storage Networks Explained has much to recommend it.
…a rarity in the literature of digital data storage – a complete exposition of both the base subject matter and its applications, which at the same time offers a level of readability making it suitable as an introduction to the subject. Storage Networks Explained is also flexible. It can be read cover-to-cover, browsed, or used as a reference. I recommend Storage Networks Explained as an essential component of any active information technology library.”
Paul Massiglia, Technical Director, VERITAS Software Corporation
Storage networks will become a basic technology like databases or local area networks. According to market research, 700f external storage devices will be connected via storage networks in 2003. The authors have hands-on experience of network storage hardware and software, they teach customers about concrete network storage products, they understand the concepts behind storage networks, and show customers how storage networks address their business needs.
Storage networks provide shared access to stored data from multiple computers and servers, thus increasing storage efficiency and availability. They permit information management functions such as backup and recovery, data mirroring, disaster recovery, and data migration to be performed quickly and efficiently, with a minimum of system overhead.
This book explains how to use storage networks to fix malfunctioning business processes, covering the technologies as well as applications. A hot topic that will become increasingly important in the coming years.
* One of the first books to focus on using rather than building storage networks, and how to solve problems.
* Looking beyond technology and showing the true benefits of storage networks.
* Covers fibre channel SAN, Network Attached Storage, iSCSI and InfiniBand technologies.
* Contains several case studies (e.g. the example of a travel portal, protecting a critical database)
* Endorsed by the Storage Networking Industry Association.
* Written by very experienced professionals who tailored the book specifically to meet customer needs including support with supplementry material on Troppens website and Preface written by Tony Clark.
Provides basic application information key for systems administrators, database administrators and managers who need to know about the networking aspects of their systems. As well as systems architects, network managers, information management directors and decision makers.
This book also supports applications for graduate students and other relevant courses in the field.
Awarded Best System Administration Book 2005 by the Linux Journal
Download Description
Warehouses are often seen as a necessary evil: places that stop the flow of goods and thus increase costs without adding value. But the truth is that they have a critical part to play in supply chain management, and warehouse managers should be centrally involved in the strategic aspects of any business. Excellence in Warehouse Management covers everything you need to know to manage warehouse operations as part of a streamlined and holistic system, fine-tuned to serve the customer and drive the bottom-line. With thinking points, self-assessment exercises and case studies Stuart Emmett challenges you to consider your own operations in a new way, and plot a course into the future.
From the Back Cover
Stored data and information is the biggest asset of any company.
If stored data is lost, getting it back is expensive and time-consuming – if at all possible. Data networks must, therefore, be designed with built-in redundancy and for high availability. Storage networks permit information management functions such as backup and recovery, data mirroring, disaster recovery, and data migration to be performed quickly and efficiently, with a minimum of system overhead. Consequently, anyone who is involved in the planning or the operation of IT systems requires a basic knowledge of the principles and the use of storage networks.
After reading Storage Networks Explained you will be familiar with
* The basic concepts of storage networks like I/O-techniques, disk subsystems, virtualization, NAS, and SAN file systems.
* The design of storage networks to provide flexible, high-available, and scaleable IT-systems.
* The use of storage networks for data sharing, device sharing, and data protection.
* The management of storage networks using SNMP, SMI-S and IEEE 1244.
Storage Networks Explained provides system administrators and system architects as well as students with the tools they need for the optimal selection and the cost-effective use of storage networks, an indispensable technology in view of our ever-growing storage quantities.
“Storage Networks Explained has much to recommend it …a rarity in the literature of digital data storage – a complete exposition of both the base subject matter and its applications, which at the same time offers a level of readability making it suitable as an introduction to the subject. I recommend Storage Networks Explained as an essential component of any active information technology library.”
Paul Massiglia, Technical Director, VERITAS Software Corporation
“Storage Networks continue to proliferate, and newer technologies now enable their deployment in small business environments through enterprise data centers. This text is a ‘must read’ for all IT users involved in these exiting new technologies and solutions.”
Sheila Childs, Chair, Storage Networking Industry Association
“Comprehensive. Thorough. Easy to read. Well worth the time…”
Jai Menon, IBM Fellow, Director and Chief Technologist, Storage Systems Architecture and Design
About the Author
Ulf Troppens and Rainer Erkens are employed at IBM TotalStorage Interoperability Center in Mainz, a testing, development and demonstration laboratory for storage products and storage networks. Both authors work at the interface between technology and customers. Their duties include the testing of new products and the validation of concepts on the basis of customer environments set up in the laboratory. They present the latest hardware and software products in the field of storage networks to customers and discuss the set up test environment with them.
Ulf Tropes studied Computer Science at the University of Karlsruhe until 1995. Since 1989 he has been primarily involved in the administration of Unix systems, storage systems, data and storage networks and distributed applications. In April 2004 Ulf joined Wolfgang’s team to help with the roll-out of a software for tape library virtualization.
Rainer Erkens studied Mathematics at the University of Mainz until 2000. His experience in the management of computers and the management of distributed applications goes back to 1992. Since 2000 he has been working primarily with storage systems and storage networks. In February 2004 he was appointed to Chairman of the SNIA Europe Solutions Committee.
Wolfgang Müller is currently working as a software architect in the Storage Software Development Department at IBM in Mainz, Germany, where the focus is on software development projects supporting open standards such as SMI-S/CIM/WBEM and IEEE 1244. He received his Dipl.-Inform. (FH) degree in computer science from the University of Applied Sciences, Darmstadt, Germany, in 1993.
1 Introduction 1
1.1 Server-Centric IT Architecture and its Limitations 1
1.2 Storage-Centric IT Architecture and its Advantages 3
1.3 Case Study: Replacing a Server with Storage Networks 4
1.4 The Structure of the Book 7
PART I Technologies for Storage Networks 11
2 Intelligent Disk Subsystems 13
2.1 Architecture of Intelligent Disk Subsystems 14
2.2 Hard Disks and Internal I/O Channels 16
2.3 JBOD: Just a Bunch of Disks 19
2.4 Storage Virtualization using RAID 20
2.5 Different RAID Levels in Detail 22
2.5.1 RAID 0: block-by-block striping 23
2.5.2 RAID 1: block-by-block mirroring 24
2.5.3 RAID 0+1/RAID 10: striping and mirroring combined 24
2.5.4 RAID 4 and RAID 5: parity instead of mirroring 28
DOWNLOAD:
http://rapidshare.com/files/145977202/SNIA.pdf
Storage Networks Explained: Basics and Application of Fibre Channel SAN, NAS iSCSI and InfiniBand
Storage Networks Explained has much to recommend it.
…a rarity in the literature of digital data storage – a complete exposition of both the base subject matter and its applications, which at the same time offers a level of readability making it suitable as an introduction to the subject. Storage Networks Explained is also flexible. It can be read cover-to-cover, browsed, or used as a reference. I recommend Storage Networks Explained as an essential component of any active information technology library.”
Paul Massiglia, Technical Director, VERITAS Software Corporation
Storage networks will become a basic technology like databases or local area networks. According to market research, 700f external storage devices will be connected via storage networks in 2003. The authors have hands-on experience of network storage hardware and software, they teach customers about concrete network storage products, they understand the concepts behind storage networks, and show customers how storage networks address their business needs.
Storage networks provide shared access to stored data from multiple computers and servers, thus increasing storage efficiency and availability. They permit information management functions such as backup and recovery, data mirroring, disaster recovery, and data migration to be performed quickly and efficiently, with a minimum of system overhead.
This book explains how to use storage networks to fix malfunctioning business processes, covering the technologies as well as applications. A hot topic that will become increasingly important in the coming years.
* One of the first books to focus on using rather than building storage networks, and how to solve problems.
* Looking beyond technology and showing the true benefits of storage networks.
* Covers fibre channel SAN, Network Attached Storage, iSCSI and InfiniBand technologies.
* Contains several case studies (e.g. the example of a travel portal, protecting a critical database)
* Endorsed by the Storage Networking Industry Association.
* Written by very experienced professionals who tailored the book specifically to meet customer needs including support with supplementry material on Troppens website and Preface written by Tony Clark.
Provides basic application information key for systems administrators, database administrators and managers who need to know about the networking aspects of their systems. As well as systems architects, network managers, information management directors and decision makers.
This book also supports applications for graduate students and other relevant courses in the field.
Awarded Best System Administration Book 2005 by the Linux Journal
Download Description
Warehouses are often seen as a necessary evil: places that stop the flow of goods and thus increase costs without adding value. But the truth is that they have a critical part to play in supply chain management, and warehouse managers should be centrally involved in the strategic aspects of any business. Excellence in Warehouse Management covers everything you need to know to manage warehouse operations as part of a streamlined and holistic system, fine-tuned to serve the customer and drive the bottom-line. With thinking points, self-assessment exercises and case studies Stuart Emmett challenges you to consider your own operations in a new way, and plot a course into the future.
From the Back Cover
Stored data and information is the biggest asset of any company.
If stored data is lost, getting it back is expensive and time-consuming – if at all possible. Data networks must, therefore, be designed with built-in redundancy and for high availability. Storage networks permit information management functions such as backup and recovery, data mirroring, disaster recovery, and data migration to be performed quickly and efficiently, with a minimum of system overhead. Consequently, anyone who is involved in the planning or the operation of IT systems requires a basic knowledge of the principles and the use of storage networks.
After reading Storage Networks Explained you will be familiar with
* The basic concepts of storage networks like I/O-techniques, disk subsystems, virtualization, NAS, and SAN file systems.
* The design of storage networks to provide flexible, high-available, and scaleable IT-systems.
* The use of storage networks for data sharing, device sharing, and data protection.
* The management of storage networks using SNMP, SMI-S and IEEE 1244.
Storage Networks Explained provides system administrators and system architects as well as students with the tools they need for the optimal selection and the cost-effective use of storage networks, an indispensable technology in view of our ever-growing storage quantities.
“Storage Networks Explained has much to recommend it …a rarity in the literature of digital data storage – a complete exposition of both the base subject matter and its applications, which at the same time offers a level of readability making it suitable as an introduction to the subject. I recommend Storage Networks Explained as an essential component of any active information technology library.”
Paul Massiglia, Technical Director, VERITAS Software Corporation
“Storage Networks continue to proliferate, and newer technologies now enable their deployment in small business environments through enterprise data centers. This text is a ‘must read’ for all IT users involved in these exiting new technologies and solutions.”
Sheila Childs, Chair, Storage Networking Industry Association
“Comprehensive. Thorough. Easy to read. Well worth the time…”
Jai Menon, IBM Fellow, Director and Chief Technologist, Storage Systems Architecture and Design
About the Author
Ulf Troppens and Rainer Erkens are employed at IBM TotalStorage Interoperability Center in Mainz, a testing, development and demonstration laboratory for storage products and storage networks. Both authors work at the interface between technology and customers. Their duties include the testing of new products and the validation of concepts on the basis of customer environments set up in the laboratory. They present the latest hardware and software products in the field of storage networks to customers and discuss the set up test environment with them.
Ulf Tropes studied Computer Science at the University of Karlsruhe until 1995. Since 1989 he has been primarily involved in the administration of Unix systems, storage systems, data and storage networks and distributed applications. In April 2004 Ulf joined Wolfgang’s team to help with the roll-out of a software for tape library virtualization.
Rainer Erkens studied Mathematics at the University of Mainz until 2000. His experience in the management of computers and the management of distributed applications goes back to 1992. Since 2000 he has been working primarily with storage systems and storage networks. In February 2004 he was appointed to Chairman of the SNIA Europe Solutions Committee.
Wolfgang Müller is currently working as a software architect in the Storage Software Development Department at IBM in Mainz, Germany, where the focus is on software development projects supporting open standards such as SMI-S/CIM/WBEM and IEEE 1244. He received his Dipl.-Inform. (FH) degree in computer science from the University of Applied Sciences, Darmstadt, Germany, in 1993.
1 Introduction 1
1.1 Server-Centric IT Architecture and its Limitations 1
1.2 Storage-Centric IT Architecture and its Advantages 3
1.3 Case Study: Replacing a Server with Storage Networks 4
1.4 The Structure of the Book 7
PART I Technologies for Storage Networks 11
2 Intelligent Disk Subsystems 13
2.1 Architecture of Intelligent Disk Subsystems 14
2.2 Hard Disks and Internal I/O Channels 16
2.3 JBOD: Just a Bunch of Disks 19
2.4 Storage Virtualization using RAID 20
2.5 Different RAID Levels in Detail 22
2.5.1 RAID 0: block-by-block striping 23
2.5.2 RAID 1: block-by-block mirroring 24
2.5.3 RAID 0+1/RAID 10: striping and mirroring combined 24
2.5.4 RAID 4 and RAID 5: parity instead of mirroring 28
DOWNLOAD:
http://rapidshare.com/files/145977202/SNIA.pdf
Saturday, September 6, 2008
how to subnetting
Thanks to sadikov user -briano
====================================
1. What is an IP Address?
An IP address is made of 4 Octets, or 32 bits. It is usually represented in dotted decimal format like this: 131.107.2.205. Each number represents an octet. An octet is a group of 8 bits. Since we have 4 octets in an IP Address, there are 8*4=32 bits in an IP Address.
Computers do not understand dotted decimal notation, as they only function in binary. All computers understand is 1 and 0. Therefore, we must find a way to transfer an IP Address from dotted decimal format to binary. We do this octet by octet.
Each bit in an octet has an associated decimal value:
Bit 1 2 3 4 5 6 7 8
Bit value 128 64 32 16 8 4 2 1
So you can follow, here is a decimal to binary calculator. However, I advise you to do the math at first to better understand the concepts...
Insert Number:
Let’s take an example. I have an IP address of 131.107.2.4. What is that in Binary?
Let’s do this octet by octet:
131:
Bit Value 128 64 32 16 8 4 2 1
Binary 1 0 0 0 0 0 1 1
Column with a “1” in binary mean that we count the corresponding bit value. If you add up all the bit values marked with a one, you get 128+2+1=131.
107:
Bit Value 128 64 32 16 8 4 2 1
Binary 0 1 1 0 1 0 1 1
Column with a “1” in binary mean that we count the corresponding bit value. If you add up all the bit values marked with a one, you get 64+32+8+2+1=107
2:
Bit Value 128 64 32 16 8 4 2 1
Binary 0 0 0 0 0 0 1 0
Column with a “1” in binary mean that we count the corresponding bit value. If you add up all the bit values marked with a one, you get 2
4:
Bit Value 128 64 32 16 8 4 2 1
Binary 0 0 0 0 0 1 0 0
Column with a “1” in binary mean that we count the corresponding bit value. If you add up all the bit values marked with a one, you get 4
Now we know that another way of writing our example IP address of 131.107.2.4 is:
10000011.01101011.00000010.00000100
IP Addresses are made of two distinct parts: the Network ID and the Host ID. When you try to ping an IP address, IP at Layer 3 needs to determine whether the target IP address is local or remote to your subnet. To explain this, I like to ask my students the following question:
“Let’s say I live on Maple Tree Lane. You say that you also live on Maple Tree Lane. Are we neighbors? Well, we may be, or we may not be. We do not have enough information to answer that question. Specifically, we do not know if we live in the same town. If we do live in the same town, and our street addresses are similar, then we are neighbors. If we do not live in the same town, it does not matter whether our street addresses are similar or not: we are not neighbors.”
The same applies to IP addressing. Before I can find out what your Host ID is – i.e. your street address – I have to worry about what your Network ID is – i.e. your town.
So how does IP know what the Host ID and Network IDs are? That is the role of the Subnet Mask.
Please note that neither the network ID, the host ID nor the subnet bit can be all 0s or all ones. More to come about this later…
2. What is a Subnet Mask?
A Subnet Mask is used to enable IP at Layer 3 to find out whether the target IP address you are trying to contact is local or remote to your network. That is all it does. It helps you determine which part of an IP Address is the Network ID, and which is the Host ID. So how does it do that?
We have all seen a Subnet Mask before. It usually looks something like this:
255.255.255.0
This subnet mask is obviously displayed in dotted decimal format. As we already know, however, computers do not understand this format. They only understand binary. So let’s apply what we know about binary to our subnet mask:
255
Bit Value 128 64 32 16 8 4 2 1
Binary 1 1 1 1 1 1 1 1
Column with a “1” in binary mean that we count the corresponding bit value. If you add up all the bit values marked with a one, you get 128+64+32+16+8+4+2+1=255
So, 255 is “all ones”, as you may have heard before.
0:
Bit Value 128 64 32 16 8 4 2 1
Binary 0 0 0 0 0 0 0 0
Column with a “1” in binary mean that we count the corresponding bit value. If you add up all the bit values marked with a one, you get 0. So, 0 is “all zeros”, as you may have heard before.
In other words, our 255.255.255.0 subnet masks in binary looks like this:
11111111.11111111.11111111.00000000
Note that a subnet mask is also a 32 bit, 4 octet entity that matches the structure of our IP addresses…
If I were to superimpose the IP address and subnet mask we have translated so far, I would get this:
131.107.2.4 10000011. 01101011. 00000010. 00000100
255.255.255.0 11111111. 11111111. 11111111. 00000000
I highlighted the Network bits in Red, and the Host bits in Blue. Notice anything? All the network bits are the bits that had a value of “1” in the subnet mask, and all the host bits are the bits that had a value of “0” in the subnet mask. Easy, isn’t it? In our example, the Network ID is 131.107.2, and the host ID is 4. If I change the subnet mask to 255.255.0.0, what happens? Well, let’s apply the same logic:
131.107.2.4 10000011. 01101011. 00000010 00000100
255.255.0.0 11111111. 11111111. 00000000 00000000
What happens now? My network ID is now 131.107 and my host ID is 2.4! That is why an IP address on its own cannot exist! That is why a host on a network needs at least an IP address AND a subnet mask!
Let’s say that I tell you that I have 2 IP addresses:
131.107.2.4 and 131.107.5.6
Are they local to each other, or remote?
You cannot answer that question, because it is incomplete! I need to give you the subnet mask as well! Let’s see why…
Let’s say that the subnet mask is 255.255.255.0. Then we have:
131.107.2.4 10000011. 01101011. 00000010. 00000100
131.107.5.6 10000011. 01101011. 00000101. 00000110
255.255.255.0 11111111. 11111111. 11111111. 00000000
Do the Network IDs match? No! Look at the third octect:
0 0 0 0 0 0 1 0
0 0 0 0 0 1 0 1
If the network IDs do not match, then the two IP addresses are remote from each other. They are on different subnets, and you will need a router to go from one to the other.
Let’s take the same example with a different subnet mask of 255.255.0.0:
131.107.2.4 10000011. 01101011. 00000010. 00000100
131.107.5.6 10000011. 01101011. 00000101. 00000110
255.255.0.0 11111111. 11111111. 00000000. 00000000
Do the Network IDs match now? Yes!
If the network IDs match, then the two IP addresses will be on the same network. You do not need a router to go from one to the other in this scenario.
Let’s sum this up: we have seen how only having an IP address is not sufficient, and how the same two IP addresses can be either remote or local to each other depending on which subnet mask you are using. This is at the basis of IP troubleshooting.
3. What are IP classes?
You may have heard of the different classes for IP addresses, namely, class A, B and C. How does that work? Let’s look at the following table:
Class A 1-127* 0xxxxxxx.
Class B 128-191 10xxxxxx.
Class C 192-223 110xxxxx.
*127 is part of the class A range, but you cannot assign any 127.x.x.x IP addresses to hosts as the entire range is reserved by InterNIC for the loop back address…
What does this mean? When talking about a Class of IP addresses, you only look at the first octet to determine what class this IP address belongs to.
For class A, InterNIC decided that the 1st octet would start with a bit value of 0. Therefore, the lowest binary of the 1st octet is 0000001, and the highest is 01111111 (in decimal, that would be from 1 to 127)
For class B, InterNIC decided that the 1st octet would start with a bit value of 10.
Therefore, the lowest binary of the 1st octet is 1000000, and the highest is 10111111 (in decimal, that would be from 128 to 191)
For class C, InterNIC decided that the 1st octet would start with a bit value of 110.
Therefore, the lowest binary of the 1st octet is 1100000, and the highest is 11011111 (in decimal, that would be from 192 to 223)
There are other classes, D and E, but they are not used right now and are illegal on the Internet.
Also, note that you can use any subnet mask classes with any IP address classes. Subnet mask classes are as follows:
Class A 255.0.0.0
Class B 255.255.0.0
Class C 255.255.255.0
In other words, it is totally possible to have a Class B IP address and a class C subnet mask… hang on… isn’t that what we have been taking as an example all along? Remember our 131.107.2.4 w/ a subnet mask of 255.255.255.0?
4. What is subnetting?
Subnetting is the action of taking an IP range where all IP addresses are local to each other and dividing it into different ranges – or subnets – where IP addresses from one range will be remote from any other IP addresses from any other range.
If you want to find out how many hosts you have in an IP range, first you need to determine how many host bits there are. Let’s take our previous example of 131.107.2.4 and 255.255.255.0. We already established previously that the Network ID was 131.107.2 and the Host ID was 4. In other words, we have 3 octets for the Network ID and one octet – or 8 bits – for the Host ID. Now that you have determined the amount of host bits you had, apply that number to the following formula:
(2^N)-2)=number of hosts, where N is your number of host bits
That gives us: ((2^8)-2)=254 hosts.
That means that in our example, we have the 131.107.2.x network, which contains 254 possible IP addresses, all local to each other.
What if I chose a class A subnet mask? I would then have:
(2^24)-2)=16,777,214 valid IP addresses in that range!
What if I do not need that many – heck, who does J -- and decided to divide that range in to several other, more manageable ranges? Well, I need to subnet.
Note: we have to subtract 2 because we lose the all ones and all zeroes values.
If I only had 2 bits to play with, in binary, I would have 2^2=4 possibilities:
00
01
10
11
However, all zeros in IP means the whole network, and all ones is the broadcast ID, neither one being valid IP addresses that can be assigned to a host. That is why we always lose 2 and have to subtract two from (2^N)…
Let’s look at the following table:
Bit Value 128 64 32 16 8 4 2 1
Subnet Mask 192 224 240 248 252 254 255
Number of Subnets 2 6 14 30 62 126 254
This table is the only table you need to learn to understand IP subnetting and IP addressing! Three little lines!
Let’s look at it line by line. The first line we are already acquainted with, so I will not spend more time on it.
The second line tells you what your subnet mask will be. How do we get those results? If you look at the bit value line, and add up the bit values, you will see that:
128+64=192
192+32=224
224+16=240
240+8=248
248+4=252
252+2=254
254+1=255
Easy? J
Now to the third line. This one tells you how many subnets will get if you use the corresponding subnet mask. In other words: if you use the 192 in your subnet mask, you will get 2 subnets. If you use 224, you will get 6 subnets, etc… So how did I get those numbers? Well, let’s take 192 as an example. How many bits did I use to get 192? Well, I added 128 and 64, so that would be two bits, right? Let’s look at our well-known formula:
((2^2)-2)= 2
That is where the values in the third line come from. Want to check another one? Let’s look at 248. How many bits did I use to get to 248? 128+64+32+16+8=248, or a total of 5 bits. ((2^5)-2)=30
So now, we know how to build that table, and we understand how it is built. We are almost done with subnetting!
Let’s say my boss shows up one morning and says:
“Dave, I have an IP range of 131.107.0.0 and a subnet mask of 255.255.0.0. I want 6 subnets out of this.” What do I do now?
I look at my chart, and I see that to get 6 subnets, I need a subnet mask of 224. I already have an existing subnet of 255.255.0.0, which in binary looks like this:
11111111.11111111.00000000.00000000
I cannot “borrow” any more bits from the first 2 octets, as they are already network bits – represented by a binary value of 1. So the only I can add this “224” to my subnet mask is to borrow from the host ID. I take the next available octec – the third in our example – and end up with a subnet mask of 255.255.224.0, which looks like this in binary:
11111111.11111111.11100000.00000000
Note that we “borrowed” 3 host bits – binary value of 0 -- in the third octet and made them network bits – binary value of 1. Note as well that 3 bits is what I need to make 224: 128+64+32=224.
Now that I have a subnet mask, I can tell that I have 13 host bits, meaning that I will have ((2^13)-2)=8190 valid IP addresses per range J
So I can go back to my boss and say “Boss, your new subnet mask for the 131.107.x.x network will be 255.255.224.0, and you will have 6 subnets with 8190 IP addresses in each range.” What’s my boss’ next question?
“Dave, what are those ranges?” Oh yeah… maybe I forgot that part J
5. Calculating IP ranges in subnetted environments
Bit Value 128 64 32 16 8 4 2 1
Subnet Mask 192 224 240 248 252 254 255
Number of Subnets 2 6 14 30 62 126 254
Here is our little subnetting chart again. In this section, we will learn how to derive the actual IP ranges from the network address and subnet mask.
In our example, we had:
Original IP range: 131.107.x.x
Original Subnet Mask: 255.255.0.0
Subnetted subnet mask: 255.255.224.0
Amount of subnets: 6
The first valid range will be 131.107.32.1 to 131.107.63.254. How did I get there? I used the above chart.
Once I determined that 224 was my subnet bit, I asked myself: what is the lowest bit value needed to make 224?” The answer is: 32 (128+64+32=224, and 32 is the lowest of all 3 bit values.) If you look at the chart, you can see this:
Bit value 32
Subnet Mask 224
Number of subnets 6
Visually, it is easy to see that to have 6 subnets, I will use 224 as a subnet bit and will start my first range at 32. I will then increment each range by that same value of 32. I love this chart!
My 6 ranges will be:
131.107.32.1 to 131.107.63.254
131.107.64.1 to 131.107.95.254
131.107.96.1 to 131.107.127.254
131.107.128.1 to 131.107.159.254
131.107.160.1 to 131.107.191.254
131.107.192.1 to 131.107.223.254
As you can see, to get the next range, I simply incremented my original value by the same value (look at the numbers in RED) I started at 32, then incremented 5 times by that same value of 32.
If you look at the values in BLUE you will see that they always are the next red value minus 1. For example, if I look at the first range x.x.32.1 to x.x.63.254, I looked at the next line’s red value, here 64, and subtracted 1 to get to 63. That is all there is to it!
By the way, note that my last octet on the start of range is always 1 – it cannot be 0, or else my host ID would be all 0s – and that the last octet is always 254 in the last IP of each range – it cannot be 255, or else it would be all 1s, which we know is the broadcast address, and is therefore not available for hosts to have.
In this example, we subnetted a Class B network into 6 subnets. Note that it is somewhat different if you subnet a Class A and a Class C. Let’s do a Class A example.
Original IP range: 10.x.x.x
Original Subnet mask: 255.0.0.0
Amount of subnets needed: 14
To get 14 subnets I need a subnet bit of 240, therefore my new subnet mask becomes 255.240.0.0. Note that my subnet bit is now the second octet, not the third as in the previous example. This will have its importance when we create our ranges.
What is the smallest bit value needed to make 240? 16. Therefore I will start my ranges at 16 and will increment by that same value of 16.
My ranges are:
10.16.0.1 to 10.31.255.254 10.128.0.1 to 10.143.255.254
10.32.0.1 to 10.47.255.254 10.144.0.1 to 10.159.255.254
10.48.0.1 to 10.63.255.254 10.160.0.1 to 10.175.255.254
10.64.0.1 to 10.79.255.254 10.176.0.1 to 10.191.255.254
10.80.0.1 to 10.95.255.254 10.192.0.1 to 10.207.255.254
10.96.0.1 to 10.111.255.254 10.208.0.1 to 10.223.255.254
10.112.0.1 to 10.127.255.254 10.224.0.1 to 10.239.255.254
Now let's do this with a class C. Remember, class C is the hardest, so follow this carefully!
Original IP range: 192.168.2.x
Original Subnet mask: 255.255.255.0
Amount of subnets needed: 6
To get 6 subnets I need a subnet bit of 224, therefore my new subnet mask becomes 255.255.255.224. Note that my subnet bit is now the 4th octet, not the third or second as in the previous examples. This will have its importance when we create our ranges.
What is the smallest bit value needed to make 224? 32. Therefore I will start my ranges at 32 and will increment by that same value of 32. (Are you getting the hang of it by now?)
My ranges are:
192.168.2.33 to 192.168.2.62 192.168.2.129 to 192.168.2.158
192.168.2.65 to 192.168.2.94 192.168.2.161 to 192.168.2.190
192.168.2.97 to 192.168.2.126 192.168.2.193 to 192.168.2.222
But, you say, I thought I was starting at 32?!? My first range is starting at 33! Well, remember the other examples? We always started at .1, didn't we? Except that here, because we are already subnetting at the fourth octet, we don't have room for a fifth to add the .1, so we have to incorporate it in the last octet. So why can we not use 192.168.2.32 w/ a subnet mask of 255.255.255.224? Let's see why:
192.168.2.32 11000000. 10101000. 00000010. 00100000
255.255.255.224 11111111. 11111111. 11111111. 11100000
Looking at it in binary, it is obvious that the Host ID is all zeroes, which we know is not possible...
That is it. If you know and understand this, you can subnet! Enjoy practicing...
how to subnetting
Thanks to sadikov user -briano
====================================
1. What is an IP Address?
An IP address is made of 4 Octets, or 32 bits. It is usually represented in dotted decimal format like this: 131.107.2.205. Each number represents an octet. An octet is a group of 8 bits. Since we have 4 octets in an IP Address, there are 8*4=32 bits in an IP Address.
Computers do not understand dotted decimal notation, as they only function in binary. All computers understand is 1 and 0. Therefore, we must find a way to transfer an IP Address from dotted decimal format to binary. We do this octet by octet.
Each bit in an octet has an associated decimal value:
Bit 1 2 3 4 5 6 7 8
Bit value 128 64 32 16 8 4 2 1
So you can follow, here is a decimal to binary calculator. However, I advise you to do the math at first to better understand the concepts...
Insert Number:
Let’s take an example. I have an IP address of 131.107.2.4. What is that in Binary?
Let’s do this octet by octet:
131:
Bit Value 128 64 32 16 8 4 2 1
Binary 1 0 0 0 0 0 1 1
Column with a “1” in binary mean that we count the corresponding bit value. If you add up all the bit values marked with a one, you get 128+2+1=131.
107:
Bit Value 128 64 32 16 8 4 2 1
Binary 0 1 1 0 1 0 1 1
Column with a “1” in binary mean that we count the corresponding bit value. If you add up all the bit values marked with a one, you get 64+32+8+2+1=107
2:
Bit Value 128 64 32 16 8 4 2 1
Binary 0 0 0 0 0 0 1 0
Column with a “1” in binary mean that we count the corresponding bit value. If you add up all the bit values marked with a one, you get 2
4:
Bit Value 128 64 32 16 8 4 2 1
Binary 0 0 0 0 0 1 0 0
Column with a “1” in binary mean that we count the corresponding bit value. If you add up all the bit values marked with a one, you get 4
Now we know that another way of writing our example IP address of 131.107.2.4 is:
10000011.01101011.00000010.00000100
IP Addresses are made of two distinct parts: the Network ID and the Host ID. When you try to ping an IP address, IP at Layer 3 needs to determine whether the target IP address is local or remote to your subnet. To explain this, I like to ask my students the following question:
“Let’s say I live on Maple Tree Lane. You say that you also live on Maple Tree Lane. Are we neighbors? Well, we may be, or we may not be. We do not have enough information to answer that question. Specifically, we do not know if we live in the same town. If we do live in the same town, and our street addresses are similar, then we are neighbors. If we do not live in the same town, it does not matter whether our street addresses are similar or not: we are not neighbors.”
The same applies to IP addressing. Before I can find out what your Host ID is – i.e. your street address – I have to worry about what your Network ID is – i.e. your town.
So how does IP know what the Host ID and Network IDs are? That is the role of the Subnet Mask.
Please note that neither the network ID, the host ID nor the subnet bit can be all 0s or all ones. More to come about this later…
2. What is a Subnet Mask?
A Subnet Mask is used to enable IP at Layer 3 to find out whether the target IP address you are trying to contact is local or remote to your network. That is all it does. It helps you determine which part of an IP Address is the Network ID, and which is the Host ID. So how does it do that?
We have all seen a Subnet Mask before. It usually looks something like this:
255.255.255.0
This subnet mask is obviously displayed in dotted decimal format. As we already know, however, computers do not understand this format. They only understand binary. So let’s apply what we know about binary to our subnet mask:
255
Bit Value 128 64 32 16 8 4 2 1
Binary 1 1 1 1 1 1 1 1
Column with a “1” in binary mean that we count the corresponding bit value. If you add up all the bit values marked with a one, you get 128+64+32+16+8+4+2+1=255
So, 255 is “all ones”, as you may have heard before.
0:
Bit Value 128 64 32 16 8 4 2 1
Binary 0 0 0 0 0 0 0 0
Column with a “1” in binary mean that we count the corresponding bit value. If you add up all the bit values marked with a one, you get 0. So, 0 is “all zeros”, as you may have heard before.
In other words, our 255.255.255.0 subnet masks in binary looks like this:
11111111.11111111.11111111.00000000
Note that a subnet mask is also a 32 bit, 4 octet entity that matches the structure of our IP addresses…
If I were to superimpose the IP address and subnet mask we have translated so far, I would get this:
131.107.2.4 10000011. 01101011. 00000010. 00000100
255.255.255.0 11111111. 11111111. 11111111. 00000000
I highlighted the Network bits in Red, and the Host bits in Blue. Notice anything? All the network bits are the bits that had a value of “1” in the subnet mask, and all the host bits are the bits that had a value of “0” in the subnet mask. Easy, isn’t it? In our example, the Network ID is 131.107.2, and the host ID is 4. If I change the subnet mask to 255.255.0.0, what happens? Well, let’s apply the same logic:
131.107.2.4 10000011. 01101011. 00000010 00000100
255.255.0.0 11111111. 11111111. 00000000 00000000
What happens now? My network ID is now 131.107 and my host ID is 2.4! That is why an IP address on its own cannot exist! That is why a host on a network needs at least an IP address AND a subnet mask!
Let’s say that I tell you that I have 2 IP addresses:
131.107.2.4 and 131.107.5.6
Are they local to each other, or remote?
You cannot answer that question, because it is incomplete! I need to give you the subnet mask as well! Let’s see why…
Let’s say that the subnet mask is 255.255.255.0. Then we have:
131.107.2.4 10000011. 01101011. 00000010. 00000100
131.107.5.6 10000011. 01101011. 00000101. 00000110
255.255.255.0 11111111. 11111111. 11111111. 00000000
Do the Network IDs match? No! Look at the third octect:
0 0 0 0 0 0 1 0
0 0 0 0 0 1 0 1
If the network IDs do not match, then the two IP addresses are remote from each other. They are on different subnets, and you will need a router to go from one to the other.
Let’s take the same example with a different subnet mask of 255.255.0.0:
131.107.2.4 10000011. 01101011. 00000010. 00000100
131.107.5.6 10000011. 01101011. 00000101. 00000110
255.255.0.0 11111111. 11111111. 00000000. 00000000
Do the Network IDs match now? Yes!
If the network IDs match, then the two IP addresses will be on the same network. You do not need a router to go from one to the other in this scenario.
Let’s sum this up: we have seen how only having an IP address is not sufficient, and how the same two IP addresses can be either remote or local to each other depending on which subnet mask you are using. This is at the basis of IP troubleshooting.
3. What are IP classes?
You may have heard of the different classes for IP addresses, namely, class A, B and C. How does that work? Let’s look at the following table:
Class A 1-127* 0xxxxxxx.
Class B 128-191 10xxxxxx.
Class C 192-223 110xxxxx.
*127 is part of the class A range, but you cannot assign any 127.x.x.x IP addresses to hosts as the entire range is reserved by InterNIC for the loop back address…
What does this mean? When talking about a Class of IP addresses, you only look at the first octet to determine what class this IP address belongs to.
For class A, InterNIC decided that the 1st octet would start with a bit value of 0. Therefore, the lowest binary of the 1st octet is 0000001, and the highest is 01111111 (in decimal, that would be from 1 to 127)
For class B, InterNIC decided that the 1st octet would start with a bit value of 10.
Therefore, the lowest binary of the 1st octet is 1000000, and the highest is 10111111 (in decimal, that would be from 128 to 191)
For class C, InterNIC decided that the 1st octet would start with a bit value of 110.
Therefore, the lowest binary of the 1st octet is 1100000, and the highest is 11011111 (in decimal, that would be from 192 to 223)
There are other classes, D and E, but they are not used right now and are illegal on the Internet.
Also, note that you can use any subnet mask classes with any IP address classes. Subnet mask classes are as follows:
Class A 255.0.0.0
Class B 255.255.0.0
Class C 255.255.255.0
In other words, it is totally possible to have a Class B IP address and a class C subnet mask… hang on… isn’t that what we have been taking as an example all along? Remember our 131.107.2.4 w/ a subnet mask of 255.255.255.0?
4. What is subnetting?
Subnetting is the action of taking an IP range where all IP addresses are local to each other and dividing it into different ranges – or subnets – where IP addresses from one range will be remote from any other IP addresses from any other range.
If you want to find out how many hosts you have in an IP range, first you need to determine how many host bits there are. Let’s take our previous example of 131.107.2.4 and 255.255.255.0. We already established previously that the Network ID was 131.107.2 and the Host ID was 4. In other words, we have 3 octets for the Network ID and one octet – or 8 bits – for the Host ID. Now that you have determined the amount of host bits you had, apply that number to the following formula:
(2^N)-2)=number of hosts, where N is your number of host bits
That gives us: ((2^8)-2)=254 hosts.
That means that in our example, we have the 131.107.2.x network, which contains 254 possible IP addresses, all local to each other.
What if I chose a class A subnet mask? I would then have:
(2^24)-2)=16,777,214 valid IP addresses in that range!
What if I do not need that many – heck, who does J -- and decided to divide that range in to several other, more manageable ranges? Well, I need to subnet.
Note: we have to subtract 2 because we lose the all ones and all zeroes values.
If I only had 2 bits to play with, in binary, I would have 2^2=4 possibilities:
00
01
10
11
However, all zeros in IP means the whole network, and all ones is the broadcast ID, neither one being valid IP addresses that can be assigned to a host. That is why we always lose 2 and have to subtract two from (2^N)…
Let’s look at the following table:
Bit Value 128 64 32 16 8 4 2 1
Subnet Mask 192 224 240 248 252 254 255
Number of Subnets 2 6 14 30 62 126 254
This table is the only table you need to learn to understand IP subnetting and IP addressing! Three little lines!
Let’s look at it line by line. The first line we are already acquainted with, so I will not spend more time on it.
The second line tells you what your subnet mask will be. How do we get those results? If you look at the bit value line, and add up the bit values, you will see that:
128+64=192
192+32=224
224+16=240
240+8=248
248+4=252
252+2=254
254+1=255
Easy? J
Now to the third line. This one tells you how many subnets will get if you use the corresponding subnet mask. In other words: if you use the 192 in your subnet mask, you will get 2 subnets. If you use 224, you will get 6 subnets, etc… So how did I get those numbers? Well, let’s take 192 as an example. How many bits did I use to get 192? Well, I added 128 and 64, so that would be two bits, right? Let’s look at our well-known formula:
((2^2)-2)= 2
That is where the values in the third line come from. Want to check another one? Let’s look at 248. How many bits did I use to get to 248? 128+64+32+16+8=248, or a total of 5 bits. ((2^5)-2)=30
So now, we know how to build that table, and we understand how it is built. We are almost done with subnetting!
Let’s say my boss shows up one morning and says:
“Dave, I have an IP range of 131.107.0.0 and a subnet mask of 255.255.0.0. I want 6 subnets out of this.” What do I do now?
I look at my chart, and I see that to get 6 subnets, I need a subnet mask of 224. I already have an existing subnet of 255.255.0.0, which in binary looks like this:
11111111.11111111.00000000.00000000
I cannot “borrow” any more bits from the first 2 octets, as they are already network bits – represented by a binary value of 1. So the only I can add this “224” to my subnet mask is to borrow from the host ID. I take the next available octec – the third in our example – and end up with a subnet mask of 255.255.224.0, which looks like this in binary:
11111111.11111111.11100000.00000000
Note that we “borrowed” 3 host bits – binary value of 0 -- in the third octet and made them network bits – binary value of 1. Note as well that 3 bits is what I need to make 224: 128+64+32=224.
Now that I have a subnet mask, I can tell that I have 13 host bits, meaning that I will have ((2^13)-2)=8190 valid IP addresses per range J
So I can go back to my boss and say “Boss, your new subnet mask for the 131.107.x.x network will be 255.255.224.0, and you will have 6 subnets with 8190 IP addresses in each range.” What’s my boss’ next question?
“Dave, what are those ranges?” Oh yeah… maybe I forgot that part J
5. Calculating IP ranges in subnetted environments
Bit Value 128 64 32 16 8 4 2 1
Subnet Mask 192 224 240 248 252 254 255
Number of Subnets 2 6 14 30 62 126 254
Here is our little subnetting chart again. In this section, we will learn how to derive the actual IP ranges from the network address and subnet mask.
In our example, we had:
Original IP range: 131.107.x.x
Original Subnet Mask: 255.255.0.0
Subnetted subnet mask: 255.255.224.0
Amount of subnets: 6
The first valid range will be 131.107.32.1 to 131.107.63.254. How did I get there? I used the above chart.
Once I determined that 224 was my subnet bit, I asked myself: what is the lowest bit value needed to make 224?” The answer is: 32 (128+64+32=224, and 32 is the lowest of all 3 bit values.) If you look at the chart, you can see this:
Bit value 32
Subnet Mask 224
Number of subnets 6
Visually, it is easy to see that to have 6 subnets, I will use 224 as a subnet bit and will start my first range at 32. I will then increment each range by that same value of 32. I love this chart!
My 6 ranges will be:
131.107.32.1 to 131.107.63.254
131.107.64.1 to 131.107.95.254
131.107.96.1 to 131.107.127.254
131.107.128.1 to 131.107.159.254
131.107.160.1 to 131.107.191.254
131.107.192.1 to 131.107.223.254
As you can see, to get the next range, I simply incremented my original value by the same value (look at the numbers in RED) I started at 32, then incremented 5 times by that same value of 32.
If you look at the values in BLUE you will see that they always are the next red value minus 1. For example, if I look at the first range x.x.32.1 to x.x.63.254, I looked at the next line’s red value, here 64, and subtracted 1 to get to 63. That is all there is to it!
By the way, note that my last octet on the start of range is always 1 – it cannot be 0, or else my host ID would be all 0s – and that the last octet is always 254 in the last IP of each range – it cannot be 255, or else it would be all 1s, which we know is the broadcast address, and is therefore not available for hosts to have.
In this example, we subnetted a Class B network into 6 subnets. Note that it is somewhat different if you subnet a Class A and a Class C. Let’s do a Class A example.
Original IP range: 10.x.x.x
Original Subnet mask: 255.0.0.0
Amount of subnets needed: 14
To get 14 subnets I need a subnet bit of 240, therefore my new subnet mask becomes 255.240.0.0. Note that my subnet bit is now the second octet, not the third as in the previous example. This will have its importance when we create our ranges.
What is the smallest bit value needed to make 240? 16. Therefore I will start my ranges at 16 and will increment by that same value of 16.
My ranges are:
10.16.0.1 to 10.31.255.254 10.128.0.1 to 10.143.255.254
10.32.0.1 to 10.47.255.254 10.144.0.1 to 10.159.255.254
10.48.0.1 to 10.63.255.254 10.160.0.1 to 10.175.255.254
10.64.0.1 to 10.79.255.254 10.176.0.1 to 10.191.255.254
10.80.0.1 to 10.95.255.254 10.192.0.1 to 10.207.255.254
10.96.0.1 to 10.111.255.254 10.208.0.1 to 10.223.255.254
10.112.0.1 to 10.127.255.254 10.224.0.1 to 10.239.255.254
Now let's do this with a class C. Remember, class C is the hardest, so follow this carefully!
Original IP range: 192.168.2.x
Original Subnet mask: 255.255.255.0
Amount of subnets needed: 6
To get 6 subnets I need a subnet bit of 224, therefore my new subnet mask becomes 255.255.255.224. Note that my subnet bit is now the 4th octet, not the third or second as in the previous examples. This will have its importance when we create our ranges.
What is the smallest bit value needed to make 224? 32. Therefore I will start my ranges at 32 and will increment by that same value of 32. (Are you getting the hang of it by now?)
My ranges are:
192.168.2.33 to 192.168.2.62 192.168.2.129 to 192.168.2.158
192.168.2.65 to 192.168.2.94 192.168.2.161 to 192.168.2.190
192.168.2.97 to 192.168.2.126 192.168.2.193 to 192.168.2.222
But, you say, I thought I was starting at 32?!? My first range is starting at 33! Well, remember the other examples? We always started at .1, didn't we? Except that here, because we are already subnetting at the fourth octet, we don't have room for a fifth to add the .1, so we have to incorporate it in the last octet. So why can we not use 192.168.2.32 w/ a subnet mask of 255.255.255.224? Let's see why:
192.168.2.32 11000000. 10101000. 00000010. 00100000
255.255.255.224 11111111. 11111111. 11111111. 11100000
Looking at it in binary, it is obvious that the Host ID is all zeroes, which we know is not possible...
That is it. If you know and understand this, you can subnet! Enjoy practicing...
Friday, August 29, 2008
Pass4sure CCNA 640-802, New ver 3.20--394Q_Updated: 2008-08-25
Exam Number/Code: 640-802
Exam Name: Cisco Certified Network Associate
QUOTE
Cisco Certified Network Associate", also known as 640-802 exam, is a Cisco certification.Preparing for the 640-802 exam? Searching 640-802 Test Questions, 640-802 Practice Exam, 640-802 Dumps?With the complete collection of questions and answers, Pass4sure has assembled to take you through 347 Q&As to your 640-802 Exam preparation. In the 640-802 exam resources, you will cover every field and category in CCNA helping to ready you for your successful Cisco Certification.Questions and Answers : 394 Q&AsUpdated: 2008-08-25
Cisco Certified Network Associate", also known as 640-802 exam, is a Cisco certification.Preparing for the 640-802 exam? Searching 640-802 Test Questions, 640-802 Practice Exam, 640-802 Dumps?With the complete collection of questions and answers, Pass4sure has assembled to take you through 347 Q&As to your 640-802 Exam preparation. In the 640-802 exam resources, you will cover every field and category in CCNA helping to ready you for your successful Cisco Certification.Questions and Answers : 394 Q&AsUpdated: 2008-08-25
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