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S=(1-alpha)P+alpha Gustafson's law states that the speed up, S, of a computation on P processors is given by the equation where 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 1 ? Choose 1 answer: (A) 1-alpha (B) (alpha)/(alpha-1) (c) (1)/(1-alpha) (D) alpha

S=(1−α)P+α S=(1-\alpha) P+\alpha \newlineGustafson's law states that the speed up, S S , of a computation on P P 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 ?\newlineChoose 11 answer:\newline(A) 1−α 1-\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+α S=(1-\alpha) P+\alpha \newlineGustafson's law states that the speed up, S S , of a computation on P P 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 ?\newlineChoose 11 answer:\newline(A) 1−α 1-\alpha \newline(B) αα−1 \frac{\alpha}{\alpha-1} \newline(C) 11−α \frac{1}{1-\alpha} \newline(D) α \alpha
  1. Start with Gustafson's Law: We start with Gustafson's law, which is given by the equation 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 SS and the new speedup as S+1S + 1. We can set up two equations based on Gustafson's law for these two scenarios.
  2. Set up Equations: For the initial speedup SS, we have the equation S=(1−α)P+αS = (1 - \alpha)P + \alpha.
  3. Find ΔP\Delta P: For the new speedup S+1S + 1, we have the equation S+1=(1−α)(P+δP)+αS + 1 = (1 - \alpha)(P + \delta P) + \alpha.
  4. Subtract Equations: Now we want to find the value of δP\delta P that satisfies the increase of 11 in speedup. We subtract the first equation from the second to eliminate α\alpha and get an equation for δP\delta P.\newlineS+1−S=(1−α)(P+δP)+α−((1−α)P+α)S + 1 - S = (1 - \alpha )( P + \delta P) + \alpha - ((1 - \alpha)P + \alpha)
  5. Simplify Equation: Simplifying the equation, we get:\newline1=(1−α)P+(1−α)δP+α−(1−α)P−α1 = (1 - \alpha)P + (1 - \alpha)\delta P + \alpha - (1 - \alpha)P - \alpha
  6. Cancel Terms: The terms (1−α)P(1 - \alpha)P and α\alpha cancel out on both sides, leaving us with: 1=(1−α)δP1 = (1 - \alpha)\delta P
  7. Divide to Find Delta P: To find delta P, we divide both sides of the equation by (1−α)(1 - \alpha):\newlineΔP=1(1−α)\Delta P = \frac{1}{(1 - \alpha)}
  8. Match with Choices: We look at the given choices to find the one that matches our expression for δP\delta P. The correct choice is (C) 11−α\frac{1}{1-\alpha}.

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