Ip Addressing and Subnetting

* Perform a vulnerability assessment scan on the targeted IP subnet to discover what the weakest link in the system.

2. Enterprise networks typically use LAN _____ at each site to connect devices using some type of media.

5 bits need to be borrowed with a total of 32 subnets. There will be a total of 2046 hosts per subnet. 211=2048 2048-2=2046

4) In IP subnet planning, you need to have at least 130 subnets. How large should your subnet part be?

Use the network diagram below and the additional information provided to answer the corresponding questions. [15 points]

135.46.52.2. In binary this is 10000111.00101110.00110100.00000010. Again, we are looking at one of the Interfaces, so we take the 22 MSB’s, which

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:

If we take the 172.16.0.0 range, that is /12. So that means that the network mask is 255.240.0.0. If we take 256-240 = 16 we get the block size which is 16. So that means that the range is from 172.16.0.0 to 172.31.255.255 because 172.32.0.0 is the next block and would be a public IP.

To meet the requirements given, the network was designed to be both scalable and secure. The networks required a specific number of hosts for the LAN each connection to the switches. For the EIGRP and OSPF companies, the devices used VLSM and were given /19, which provided many more addresses than the required 120 and 250. This addressing made sense, due to its scalability and because were seven devices on these networks. Had the network used a /24 an entire octet of addresses would have been wasted. For the RIP company, to account for its users a /24 was used because it allowed for 250 hosts and it could use vlsm and not waste addresses.

16. Consider the i-node shown in Fig. 4-13. If it contains 10 direct addresses and these

5. Administration departments spread over another 10 buildings. LAN IP’s: Building 1 (128.100.17.x), Building 2 (128.100.18.x), Building 3 (128.100.19.x), Building 4 (128.100.20.x), Building 5 (128.100.21.x), Building 6 (128.100.22.x), Building 7 (128.100.23.x), Building 8 (128.100.24.x), Building 9 (128.100.25.x), and Building 10 (128.100.26.x). Router IP’s will be respectively as follows: (128.100.254.11), (128.100.254.12), (128.100.254.13), (128.100.254.14), (128.100.254.15), (128.100.254.16), (128.100.254.17), (128.100.254.18),

Many organizations and companies ignore the importance of planning and administering a structured model approach to network layer addressing. Many problems result from such disorganization: there are duplicate network addresses; illegal addresses that routers cannot route to the Internet, running out of addresses for existing or future use; or have larger than required

What is IPv4? IPv4 is a 32-bit address used for communication between nodes using the TCP/IP protocol. The IPv4 address is broken into 4 octets each 8 bits long in binary and are broken into 5 different classes. Class A addresses start out where the first octet of the IP address ranged from 1-127 where 10 is reserved from private IP addresses. Class B addresses start out with 128 - 191 and has the 172.16.x.x to 172.31.x.x reserved from private IP addresses. Next is class C addresses that range from 192 - 223 in the first octet where 192.168.x.x is reserved from private IP addresses. Class D addresses range from 224 -239 in the first octet and are reserved from multicasting network traffic. Finally we have class E that range from 240 - 255 in the first octet and is reserved for experimental purposes ("IPv4," 2016).

Several classless routing protocol examples include OSPF, RIP version 2, Cisco 's EIGRP, BGP and IS-IS. We will deploy VLSM for this project. Use of a Class B address 172.17.0.0 is required to support a network that entails a total of 200 sites. The busiest LANs may support up to 100 hosts and there is a maximum projected total of 400 point-to-point WAN links. Hence there is a requirement for 600 subnets with a maximum of 100 hosts on any subnet. Even with a Class B address there is insufficient address space to meet this requirement without employing VLSM. When planning a VLSM solution you should start with the shortest mask in other words plan the subnets that support the most hosts. This is typically the mask that will be used on most or all of the LAN segments.

The proposed network is designed for the UMUC new building in Rockville. This new building will have some administrative offices, lecture rooms, library, and computer labs with an open Wi-Fi. This new building will have total six computer labs that will be used for instruction. Each of these labs will have 25 computers, 23 student computers and one instructor computer, one server in the closet and in addition to the six computer labs, there will also be a Student Computer Lab that will provide computer access to students for doing their own purpose like homework. In Student Computer Lab there will be 30 computers and a server in the closet. There will be 10 computers for student’s use and 5 computers for Library staff in the library so they can access to the library resources for staff use, with the exception of the admissions office, which will have five computers. There will be six class room which will have a computer for instructor’s use. Finally, there will be 7 offices in the building and each of these offices will have one computer. There will be two server rooms, one on the first floor and another one on the second floor. There should be at least 3 separate networks: Administrative, Instructional, and Public. All staff-accessed computers will be on the