Answer:
1 PROCESSOR :
(1 × 2.56 × 10^9) + (12 × 1.28 × 10^9) + (5 × 2.56 × 10^8) / 2 GHz = 9.6 s
2 PROCESSORS :
(1×2.56×10^9)+(12×1.28×10^9)/0.7×2 + (5 × 2.56 × 10^8) / 2 GHz = 7.04 s
Speed -up is 1.36
4 PROCESSORS :
(1×2.56×10^9)+(12×1.28×10^9)/0.7×4 + (5 × 2.56 × 10^8) / 2 GHz = 3.84 s
Speed -up is 2.5
5 PROCESSORS :
(1×2.56×10^9)+(12×1.28×10^9)/0.7×8 + (5 × 2.56 × 10^8) / 2 GHz = 2.24 s
Speed -up is 4.29
Explanation:
The following formula is used in this answer:
EXECUTION TIME = CLOCK CYCLES / CLOCK RATE
Execution Time is equal to the clock cycle per clock rate
Answer:
yes yes yes yes yes yes yes
Answer:
The answer to this question is given below in the explanation section.
Explanation:
The answer is the seek time.
The seek time is the time for the disk arm to move the heads to the cylinder containing the desired sector.
<u>Answer:</u>
<em>void main ( ) </em>
<em> { </em>
<em> int counter; </em>
<em> cout<<""Even numbers between 1 to 20 are:""<<endl ; </em>
<em> //Method 1
</em>
<em> for (counter = 1; counter <= 20; counter++) </em>
<em> { </em>
<em> if ( counter%2 == 0) </em>
<em> {
</em>
<em> cout<<counter<<""\t""<<endl ; </em>
<em> } </em>
<em> } </em>
<em>//Method 2 – simplest one
</em>
<em>for (counter = 2; counter <= 20;) </em>
<em> { </em>
<em> cout<<counter<<""\t""<<endl ; </em>
<em>counter = counter + 2;
</em>
<em> </em>
<em> }
</em>
<em>
</em>
<em> return 0; </em>
<em>}
</em>
<u>Explanation:</u>
In this, Method 1 runs a for loop and check whether each number is divided by 2. If yes, then printed otherwise it is skipped.
In the second method, it runs for loop only for even numbers. <em>This is obtained by incrementing the counter by 2.
</em>
Answer:
Please see the attached file for the complete answer.
Explanation:
Please Read the flow chart very carefully and try to read it as one Picture, Because Picture size is large.
