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Internetworking and Addressing Schemes

   

Added on  2020-02-24

13 Pages2194 Words42 Views
Running Head: INTERNETWORKING AND ADDRESSING SCHEMESM2m Internetworking And Addressing Schemes[Student Names][Institution Name]

INTERNETWORKING AND ADDRESSINGSCHEMES2TASK 1 a.The designing of address solution.The M2M network requires an address scheme where it is using IP address named as 10.0.0.0 to come up with the subnets IP addresses.Below is the calculations involved in obtainingthe address of the various office as follows:1st Finance Office with 260 workstationsCalculations:This department has a total of 260 workstations therefore requires a total of 260 hosts Ip addresses this can be calculated as below:Hosts bits =2^k-2=2602^k=260+2 ---------> 2^k=262This gives the Host bits as 9 bits .Hence =8+8+8+8-9=23.The network subnet mask is given in binary format as 11111111.11111111.11111110.00000000.The network subnet mask is given in number format as i.e 255.255.254.000.Therefore the Block size is given as 2^9=512.Therefore the subnet IP address ranges from 10.0.0.0/23 to 10.0.0.511/23 with 260 workstations.

INTERNETWORKING AND ADDRESSINGSCHEMES32nd Transport Office with 130 workstationsCalculations:This department has a total of 130 workstations therefore requires a total of 130 hosts IP addresses this can be calculated as below:Hosts bits =2^k-2=1302^k=130+2 ---------> 2^k=132This gives the Host bits as 8 bits .Hence =8+8+8+8-8=24.The network subnet mask is given in binary format as 11111111.11111111.11111111.00000000The network subnet mask is given in number format as i.e 255.255.255.000.Therefore the Block size is given as 2^8=256Therefore the subnet IP address ranges from 10.0.1.0/24 to 10.0.1.255/24 with 130 workstations.3rd Research Office with 120 workstationsCalculations:This department has a total of 120 workstations therefore requires a total of 120 hosts IP addresses this can be calculated as below:Hosts bits =2^k-2=1202^k=120+2 ---------> 2^k=122This gives the Host bits as 7 bits .

INTERNETWORKING AND ADDRESSINGSCHEMES4Hence =8+8+8+8-7=25.The network subnet mask is given in binary format as 11111111.11111111.11111111.10000000The network subnet mask is given in number format as i.e 255.255.255.128.Therefore the Block size is given as 2^7=128Therefore the subnet IP address ranges from 10.0.2.0/25 to 10.0.2.127/25 with 120 workstations.4th Sales Office with 40 workstationsCalculations:This department has a total of 40 workstations therefore requires a total of 40 hosts IP addresses this can be calculated as below:Hosts bits =2^k-2=402^k=40+2 ---------> 2^k=42This gives the Host bits as 6 bits .Hence =8+8+8+8-7=26.The network subnet mask is given in binary format as 11111111.11111111.11111111.11000000The network subnet mask is given in number format as i.e 255.255.255.192.Therefore the Block size is given as 2^6=64Therefore the subnet IP address ranges from 10.0.3.0/26 to 10.0.3.63/26 with 40 workstations.

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