A. intentionally or recklessly disrupt, degrade, or destroy information or services on the computer
Big-O notation is a way to describe a function that represents the n amount of times a program/function needs to be executed.
(I'm assuming that := is a typo and you mean just =, by the way)
In your case, you have two loops, nested within each other, and both loop to n (inclusive, meaning, that you loop for when i or j is equal to n), and both loops iterate by 1 each loop.
This means that both loops will therefore execute an n amount of times. Now, if the loops were NOT nested, our big-O would be O(2n), because 2 loops would run an n amount of times.
HOWEVER, since the j-loop is nested within i-loop, the j-loop executes every time the i-loop <span>ITERATES.
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As previously mentioned, for every i-loop, there would be an n amount of executions. So if the i-loop is called an n amount of times by the j loop (which executes n times), the big-O notation would be O(n*n), or O(n^2).
(tl;dr) In basic, it is O(n^2) because the loops are nested, meaning that the i-loop would be called n times, and for each iteration, it would call the j-loop n times, resulting in n*n runs.
A way to verify this is to write and test program the above. I sometimes find it easier to wrap my head around concepts after testing them myself.
<u>Answer:</u>
a) First, we need to determine the pipeline stage amounting to the maximum time. In the given case, the maximum time required is 2ns for MEM. In addition, the pipeline register delay=0.1 ns.
Clock cycled time of the pipelined machine= max time+delay
=2ns+0.1 ns
=2.1 ns
b) For any processor, ideal CPI=1. However, since there is a stall after every four instructions, the effective CPI of the new machine is specified by:
c) The speedup of pipelined machine over the single-cycle machine=avg time per instruction of single cycle/avg time per instruction of pipelined.
Single cycle processor:
CPI=1
Clock period=7 ns
Pipelined processor:
Clock period=2.1 ns
CPI=1.25
Therefore, speedup=
=7/2.625
= 2.67
d) As the number of stages approach infinity, the speedup=k where k is the number of stages in the machine.
Answer:
For text and table in common: Keep Text Only
And hence, B. Paste as a text item is the correct option.
The above is the common option available in case we are pasting the Text or you are pasting the Table. If you copy a text and try to paste it, and when you will try to paste the table, then you will find this option in common, and hence it is the option that is generally available. Hence, B is the correct option.
Explanation:
Please check the answer section.
