A router has four interfaces addressed as 10.1.1.1/24, 10.1.2.1/24, 10.1.3.1/24, and 10.1.4.1/24
What is the smallest summary route that can be advertised covering these four subnets?
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A. B. C. D.C.
To determine the smallest summary route that can be advertised covering the four subnets, we need to perform a route summarization.
The first step is to convert the four subnets into binary form:
10.1.1.0 00001010.00000001.00000001.00000000 10.1.2.0 00001010.00000001.00000010.00000000 10.1.3.0 00001010.00000001.00000011.00000000 10.1.4.0 00001010.00000001.00000100.00000000
Next, we need to identify the common bits in the subnet mask. The subnet mask for all four subnets is /24, which means the first 24 bits are common:
00001010.00000001.00000001.xxxxxxxx 00001010.00000001.00000010.xxxxxxxx 00001010.00000001.00000011.xxxxxxxx 00001010.00000001.00000100.xxxxxxxx
The x's represent the bits that differ between the four subnets. We need to identify the first bit position at which the bits differ. In this case, the first bit position at which the bits differ is the 23rd bit position (counting from left to right).
We then take the common bits (the first 23 bits) and set all the bits to the right of the first bit position at which the bits differ to 0. This gives us the network address of the summary route:
00001010.00000001.000000xx.00000000 = 10.1.0.0
Finally, we need to calculate the summary route's subnet mask. Since the first bit position at which the bits differ is the 23rd bit position, the summary route's subnet mask will be /23:
10.1.0.0/23
Therefore, the answer is option C. 10.1.0.0/21 is the smallest summary route that can be advertised covering the four subnets.