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TCP/IP IP Address Management Assignment

   

Added on  2020-09-17

14 Pages2343 Words114 Views
TCP/IPIP Address Management

To achieve efficient IP address management, businesses must first formulatea strategy that caters to their specifications. While chalking out this plan,administrators need to focus on 3 major components of IP management: IPAddress Inventory Administration, Domain Name Service (DNS)Management, and Dynamic Host Settings Protocol (DHCP) Management.TCP/IP ModelWhile the OSI strategies slowly developed, due to their official committeebased mainlyEngineering, technique, the TCP/IP protocol package deal evolved andmatured. Using its public Request for Comments (RFC) policy of improvingplus updating the protocol selection, it has established itself because theprotocol of choice for many data communication networks.

TCP/IP is the internationally accepted normal for local networks, networkingand internet. TCP (Transmission Control Protocol) + Internet Protocol (IP)TCP = the connection process.Internet Protocol = the specific routing protocol.Network Design StepsBelow is a typical rule-of-thumb approach to IP program design. It presentsthe structured approach to analyzing in addition developing a networkdesign to suit the needs of a company.Designing IP addresses using FLSM: IP addressing for the following six locationsSolution:

From the given network address, the subletting must be done using the last octet (last 8 bits). -------------- network portion ---------  host portion192168 .100Decimal form1100 0000 .1010 1000 .0000 1010 .0000 0000Binary formAs the company has 4 departments, you need to have 4 subnets and each ofthose subnets must accommodate all the PCs and printers. Each network willneed an address for the gateway. So for each subnet we need 20 PCs + 3 printers + 1 gateway = 24 IPs. We have 8 bits in the last octet of the network address (host portion in the given network address). Among these we have to decide how many bits for host part and how many bits for subnet part. To find the number of bits needed to accommodate m hosts the formula is: 2n - 2 ≥ m Where n is the number of host bits and m is the total number of host IPs needed. Using this formula for 24 hosts, we need 5 bits. (25 - 2 = 30 ≥ 23) If the last octet is considered, 0 0 00 0 0 0 0|-sub net part-|-----host part--------| The new subnet mask can be found after deciding on the number of host bits, by the following formula. Host bits are the last few bits. So (32 - n) will be the subnet mask number which is 32 - 5 = 27 here. Subnet mask is /27 and in decimal form it is 27 ones among 32 bits. Subnet mask in prefix notation:/27Subnet mask in binary form:1111 1111. 1111 1111. 1111 1111. 1110 0000Subnet Mask in decimal form:255255 .255 .224

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