A certain program takes 26.67 seconds to run on 3 processors and 16 s to run on 7 processors. Find the execution time on one processor, the fraction of the program that can be parallelized, the theoretical execution time on an infinite number of processors, and P1/2.

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Answer:

The explanation for the question is listed in the section below on Explanation.

Explanation:

The time needed for a program to execute in 26.67 seconds on 3 processors. Time expected to run the program, therefore, should be as follows:

The time to run on one processor = [tex]\frac{26.67}{3}[/tex] = 8.89 seconds.

The time needed for a program to execute in 26.67 seconds on 7 processors, should be:

Time = [tex]\frac{16}{7}[/tex] = 2.2857 seconds.

Whenever the amount of processors increases the performance or speed should increase.

The time is taken in percentage = [tex]\frac{26.67}{16}[/tex]×100% = 170 %

As the number of processors increases from 3 to 7, the time required according to each processor increases by 70%.

In one processor the execution time should be as follows:

26.67+(26.67×0.7) = 45.339

                              = 45 seconds

Therefore. the execution period for a single processor is about 45 seconds.

The work is performed in parallel, whereas the program is already being parallelized. The overall function ought to be shared between the processor. The time is shortened to around half. Hence the duration taken will be 8 seconds after parallelization.

Theoretical execution time would be Tp > T infinity.

Thus, T infinity<45

The theoretical processing interval on processor count P1/2 would therefore be less than 8 seconds. Because, the time required during parallelization was just that.