In the case of parallelization, Amdahl's law states that if \(𝑃\) is the…

2015

In the case of parallelization, Amdahl's law states that if \(𝑃\) is the proportion of a program that can be made parallel and \((1−𝑃)\) is the proportion that cannot be parallelized, then the maximum speed-up that can be achieved by using \(𝑁\) processors is:

Answer: C. \(\dfrac{1}{(1-P)+\frac{P}{N}} \\\)Correct formula: S(N) = 1 / ((1 - P) + P/N) Derivation: Normalize the total work so that execution time on one processor is 1. The serial (non-parallelizable)…

  1. A.

    \(\dfrac{1}{(1-P)+N.P} \\\)

  2. B.

    \(\dfrac{1}{(N-1)P+P} \\\)

  3. C.

    \(\dfrac{1}{(1-P)+\frac{P}{N}} \\\)

  4. D.

    \(\dfrac{1}{P+\frac{(1-P)}{N}}\)

Attempted by 61 students.

Show answer & explanation

Correct answer: C

Correct formula: S(N) = 1 / ((1 - P) + P/N)

Derivation:

  1. Normalize the total work so that execution time on one processor is 1.

  2. The serial (non-parallelizable) portion takes time (1 − P).

  3. The parallelizable portion P is divided evenly across N processors, so its time becomes P/N.

  4. Total execution time with N processors = (1 − P) + P/N.

  5. Speed-up S(N) is the ratio of the original time to the new time, so S(N) = 1 / ((1 − P) + P/N).

Limiting case: As N → ∞ the P/N term goes to 0, so the maximum possible speed-up is 1/(1 − P).

Quick example: If P = 0.9 and N = 10, total time = 0.1 + 0.9/10 = 0.19, so S(10) ≈ 1/0.19 ≈ 5.26.

Explore the full course: Mca Entrance Exam

Loading lesson…