Subject Computer Science Network Management and Data Communication

Question

Class C Subnetting: Given the original Network Address : 145.168.0.0 /24
Original mask : 255.255.255.0
Break this network into subnets using the subnet mask 255.255.255.240
New Network : 145.168.0.0 /28
Subnet mask: 255.255.2255.240
Your solution should look like the table below:
Network Address Starting Address Ending Address Broadcast Address
145.168.0.0/28 145.168.0.1 145.168.0.14 145.168.0.15
145.168.0.16/28           ……….     ………     ………..
………………….. ………………. …………………. ………………….
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However, we are already provided that initial address space is /24, thus by moving to /28, we get practically 2x2x2x2 =16 subnets. Each of these subnets has 14 usable host addresses (2 raised to 4 bits=16 and we lose two addresses per each subnet, because the first one is the network ID and the last one is the broadcast address....

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