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S=(1−α)P+αS=(1-\alpha)P+\alpha Gustafson's law states that the speed up, SS, of a computation on PP processors is given by the equation where α\alpha is a known constant related to the parallelizability. Which of the following expressions equals the increase in the number of processors needed for the speedup to increase by 11? Choose 11 answer:\newline(A) 1−α1-\alpha\newline(B) αα−1\frac{\alpha}{\alpha-1}\newline(C) 11−α\frac{1}{1-\alpha}\newline(D) α\alpha

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Q. S=(1−α)P+αS=(1-\alpha)P+\alpha Gustafson's law states that the speed up, SS, of a computation on PP processors is given by the equation where α\alpha is a known constant related to the parallelizability. Which of the following expressions equals the increase in the number of processors needed for the speedup to increase by 11? Choose 11 answer:\newline(A) 1−α1-\alpha\newline(B) αα−1\frac{\alpha}{\alpha-1}\newline(C) 11−α\frac{1}{1-\alpha}\newline(D) α\alpha
  1. Given Gustafson's Law: We are given Gustafson's law in the form S=(1−α)P+αS = (1-\alpha)P + \alpha. We want to find the increase in the number of processors (ΔP\Delta P) needed for the speedup (SS) to increase by 11. Let's denote the initial speedup as S1S_1 and the increased speedup as S2S_2, where S2=S1+1S_2 = S_1 + 1.
  2. Expressing Speedups: First, we express S1S_1 and S2S_2 using Gustafson's law. S1=(1−α)P+αS_1 = (1-\alpha)P + \alpha and S2=(1−α)(P+ΔP)+αS_2 = (1-\alpha)(P + \Delta P) + \alpha.
  3. Simplifying Equation: Since S2S_2 is 11 greater than S1S_1, we can write S2=S1+1S_2 = S_1 + 1. Substituting the expressions for S1S_1 and S2S_2, we get (1−α)(P+ΔP)+α=(1−α)P+α+1(1-\alpha)(P + \Delta P) + \alpha = (1-\alpha)P + \alpha + 1.
  4. Distributing Terms: Now, we simplify the equation. The α\alpha terms on both sides cancel out, leaving us with (1−α)(P+ΔP)=(1−α)P+1(1-\alpha)(P + \Delta P) = (1-\alpha)P + 1.
  5. Canceling Terms: We distribute (1−α)(1-\alpha) on the left side of the equation: (1−α)P+(1−α)ΔP=(1−α)P+1(1-\alpha)P + (1-\alpha)\Delta P = (1-\alpha)P + 1.
  6. Finding ΔP\Delta P: The (1−α)P(1-\alpha)P terms on both sides of the equation cancel out, leaving us with (1−α)ΔP=1(1-\alpha)\Delta P = 1.
  7. Matching Answer Choice: To find ΔP\Delta P, we divide both sides of the equation by (1−α)(1-\alpha): ΔP=1(1−α)\Delta P = \frac{1}{(1-\alpha)}.
  8. Matching Answer Choice: To find ΔP\Delta P, we divide both sides of the equation by (1−α)(1-\alpha): ΔP=1(1−α)\Delta P = \frac{1}{(1-\alpha)}.We look at the answer choices to find the one that matches our expression for ΔP\Delta P. The correct answer is (C) 1(1−α)\frac{1}{(1-\alpha)}.

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