768 Words4 Pages

IP Addressing Scenario
Unit 1 Exercise 1
ITT Technical Institute
Floyd Washington Jr.
April 4, 2015
When setting up a network that will consist of many host computers, one of the first things that an administrator must do is to determine what class of networks that they must administer to a given business. This is the point where every administrator must know how to implement classful and classless IP addressing. A classful network is a network addressing architecture used in the internet from 1981 until the introduction of Classless Inter-Domain Routing (CIDR) in 1993. Classful IP addressing divides the addtess space on the internet into five address classes. Each class is coded in the first four bits of the address. Today*…show more content…*

For the second half of this assignment I will set up a network LAN for 145 host with an allowance for 50% growth over the next two years. When I do the math the number that I came up with is 281 address spaces needed for this company. To get 218, multiply the original 145 hosts by .50, which is equal to 72.5. Then add the 72.5 to the original 145 and the result is 217.5 and I rounded up to 218. So I will need at least 218 host addresses on my LAN. I now know that from the number of addresses needed I will be in the C Class of Classful networks, because there are 218 network addresses needed and Class C networks will allow up to 256 addresses per network. From the start I would break this down into at least five or more networks depending on how the company is divided. A company with this many host computers will have at least four or more divisions (IT, Administrations, Accounting, and Human Resources), and each division will need its own network. In order to get my five networks I will have to make room for eight networks, because the next lowest number for networks is 4. Remembering that I must use the n-2 notation for networks the number 8-2 = 6 which covers my 4 networks needed. The subnet mask for this will be 255.255.255.224 because 128 + 64 +32 = 224. The CIDR notation is /27 because of the bit value. Now that I have this information I know that my subnet mask is 255.255.255.224/27. The range of my four networks are as follows:

For the second half of this assignment I will set up a network LAN for 145 host with an allowance for 50% growth over the next two years. When I do the math the number that I came up with is 281 address spaces needed for this company. To get 218, multiply the original 145 hosts by .50, which is equal to 72.5. Then add the 72.5 to the original 145 and the result is 217.5 and I rounded up to 218. So I will need at least 218 host addresses on my LAN. I now know that from the number of addresses needed I will be in the C Class of Classful networks, because there are 218 network addresses needed and Class C networks will allow up to 256 addresses per network. From the start I would break this down into at least five or more networks depending on how the company is divided. A company with this many host computers will have at least four or more divisions (IT, Administrations, Accounting, and Human Resources), and each division will need its own network. In order to get my five networks I will have to make room for eight networks, because the next lowest number for networks is 4. Remembering that I must use the n-2 notation for networks the number 8-2 = 6 which covers my 4 networks needed. The subnet mask for this will be 255.255.255.224 because 128 + 64 +32 = 224. The CIDR notation is /27 because of the bit value. Now that I have this information I know that my subnet mask is 255.255.255.224/27. The range of my four networks are as follows:

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