Explain the three schemes via which the binding of instructions and data to memory addresses can be done. In each scheme, how th
1 answer:
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Answer:
p(x,n)
1. if(n==0) [if power is 0]
2. then result =1.
3.else
4. { result=1.
5. for i=1 to n.
6. { result = result * x. } [each time we multiply x once]
7. return result.
8. }
Let's count p(3,3)
30, so come to else part.
i=1: result = result *3 = 3
i=2: result = result *3 = 9
i=2: result = result *3 = 27
Explanation:
here the for loop at step 4 takes O(n) time and other steps take constant time. So overall time complexity = O(n)
Answer:
69.08265412 milliseconds
Explanation:
Lets first convert 7 MiB to bits
bits
Now convert bits to Gbits
Gbits
Queuing Delay = Total size/transmission link rate
Queuing Delay= seconds
Delay of packet number 3 = seconds
or milliseconds