evolution of computers

May 13th, 2015
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When using the non-VLSM method of assigning addresses, all subnets have the same number of addresses assigned to them. In order to provide each network with an adequate number of addresses, we base the number of addresses for all networks on the addressing requirements for the largest network.

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evolution of computers3 LANNetworks: 41 WANHost RequirementsStudent: 481Computers: 460Router (LAN Gateway): 1Switches: 20Instructor: 69Computers: 64Router (LAN Gateway): 1Switches:4Administration: 23Computers: 20Server:1Router (LAN Gateway): 1Switch:1WAN:2Router Router:2Given IP Address: 172.16.0.0 /21When using the non-VLSM method of assigning addresses, all subnets have the same number of addresses assigned to them. In order to provide each network with an adequate number of addresses, we base the number of addresses for all networks on the addressing requirements for the largest network.In the above topology, the Student LAN is the largest network, requiring 481 addresses. Formula= 2^n-2; where n=9= 2^9-2= 512-2= 510 usable hostsBecause there are four networks in the internetwork, we will need four blocks of 512 addresses each, for a total of 2048 addresses. We will use the address block 172.16.0.0 /23. This provides addresses in the range from 172.16.0.0 to 172.16.7.255. Let's examine the address calculations for the networks:Address: 172.16.0.0Mask: 255.255.254.010101100.00010000.00000000.0000000011111111.11111111.11111110.00000000This mask will provide the four address ranges shown in the figure. Address ranges:172.16.0.0/23172.16.1.255172.16.2.0/23172.16.3.255172.16.4.0/23172.16.5.255172.16.6.0/23172.16.7.25510101100.00010000.00000000.00000000+ 1.11111111 = 9 bits10101100.00010000.00000001.111

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