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Answer:
2^11
Explanation:
Physical Memory Size = 32 KB = 32 x 2^10 B
Virtual Address space = 216 B
Page size is always equal to frame size.
Page size = 16 B. Therefore, Frame size = 16 B
If there is a restriction, the number of bits is calculated like this:
number of page entries = 2^[log2(physical memory size) - log2(n bit machine)]
where
physical memory size = 32KB which is the restriction
n bit machine = frame size = 16
Hence, we have page entries = 2^[log2(32*2^10) - log2(16)] = 2ˆ[15 - 4 ] = 2ˆ11
Answer:
Code is too large , i attached a source file below and also a text file from where i get Questions
Explanation:
Answer:
anycast
Explanation:
Anycast -
It refers to as the routing methodology and the networking address where only one destination address has more than two pathway for the end destination , is referred to as an anycast .
In this case , the router has the capability to select the pathway to reach the destination , depending on the latency measurement , cost , distance and hops .
Hence , from the given information of the question ,
The correct answer is anycast .
