Assume that we have a internet with a 9 bit address space. The addresses are divided between three networks (N0 to N2), with 64,192 and 256 addresses respectively. The internetwork communication is done through a router with03 interfaces (m0 to m2). S how the internet outline and forwarding table.
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Answer:
Step-by-step solution:
Step 1 of 4
• Given that the internet has 9-bit address space. Therefore N=9
• Therefore the number of addresses in the ISP is 29= 512 addresses.
• Given the addresses are divided into 3 networks, N0, N1, N2 and the number of addresses in each network are 64, 192, 256.
• That is the address range in N0 is 0 to 63.
• The address range in N1 is 64 to 255.
• The address range in N2 is 256 to 511.
• According CIDR rule, the number of addresses in a block or network should be a power of 2. All the networks except N1 do not satisfy this rule.
• In order to assign addresses to N1, network N1 is divided into two sub networks they are N11 with range from 64 to 127 and N12 with range from 128 to 255.
• Therefore the networks and their address ranges are:
Network N0: the number of addresses is 64. Range is 0 to 63
• The prefix n= N- log264= 9-log264 = 9-6 = 3
• The first address is 0 00000000
• The last address is 0 00111111
• The subnet mask or prefix is 000 or 3.
Network N 1: the number of addresses is 192. It is divided into two sub networks:
a) Network N11: the number addresses are 64. Range is 64 to 127
• The prefix n = N- log264= 9-log264 = 9-6 = 3
• The first address is 0 01000000
• The last address is 0 01111111
• The subnet mask or prefix is 001 or 3.
b) Network N12: the number of addresses is 128. Range is 128 to 255
• The prefix n = N- log2128= 9-log2128 = 9-7 = 2
• The first address is 0 10000000
• The last address is 0 11111111
• The subnet mask or prefix is 01 or 2.
Network N 2: the number of addresses is 256. Range is 256 to 511
• The prefix n= N- log2256= 9-log2256 = 9-8 = 1
• The first address is 1 00000000
• The last address is 1 11111111
• The subnet mask or prefix is 1.