Network Address Interface 164.64.0.0/10 164.128.0.0/13 164.0.0.0/6 164.0.0.0/10

Question

Network Address Interface 164.64.0.0/10 164.128.0.0/13 164.0.0.0/6 164.0.0.0/10 167.0.0.0/8 otherwise 0 3 Table 1: IP forwarding table using CIDR notation [10 points]: Consider an IP router with four link interfaces numbered 0 through 3. Packets are to be forwarded according to the forwarding table given in Table (a) Using binary notation, list the IP address ranges corresponding to interfaces 0, 1, and 2 (b) Indicate how the IP forwarding table would determine the interface for datagrams with the following destination addresses: 164.132.81.85, 164.0.0.5, 164.225.64.195, 165.91.23.12, 164.133.128.17

Explanation / Answer

a. Range of IP address 232-10=4194304(interface 0)

232-13=524288(interface 1)

232-6=67108864(interface2)

b) Ans: 164.40.0.0
Taking the first 13 bits of 164.132.81.85 as network address, we have 164.40.0.0
The bit pattern of 164.132.85.81 is 10100100.00101100.01010001.01010101
When we perform the bit and operation with 13 leading bit 1s and 19 bit 0s, it is equivalent of making the last 19 bit zero. We get the following network address bit pattern: 10100100.00101000.00000000.00000000. The first byte is not changed. The 2nd type changes from 132 to 40 while the 3rd and 4th bytes become zero.
Match with network address in the routing table. The last row matches. The router will forward the packet to Interface 3.

bit pattern of 164.0.0.5=10100100.00000000.00000000..00000101

after making 1st 6bits 1 the address is 164.0.0.0 which matches to third and 4th row of the table.

so the interface will be either 2 or 3

bit pattern of 164.225.64.195=10100100.10111001.01000000.11000011

after making the first 10 bit 1 the address is164.128.0.0 which matches to 2nd row. so the interface will be 1

164.133.128.17=10100100.10000101.10000000.00010001

after making the first 13 bit 1 the address is 164.128.0.0

So,the interface will be 1

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