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Pipeline and Vector Processing Q.25

0. The floating point unit of a processor using a design D takes 2t cycles compared to t cycles taken by the fixed point unit. There are two more design suggestions D1 and D2. D1 uses 30% more cycles for fixed point but 30% less cycles for floating point unit as compared to design D. D2 uses 40% less cycles for fixed point unit but 10% more cycles for floating point unit as compared to design D. For a given program which has 80% fixed point operations and 20% floating point operations, which of the following ordering reflects the relative performances of three designs? (Di > Dj denotes that Di is faster than Dj)

  • Option : A
  • Explanation : 0.8 * (time taken in fixed point) + 0.2 (time taken in floating point) Say, t = 1
    D = 0.8(1) + 0.2(2)
    = 1.2
    D1 = 0.8(1.3) + 0.2(1.4)
    = 1.04 +.28 = 1.32
    D2 = 0.8(1 – 0.04) + 0.2(2 – 2*0.1)
    = 0.8 * 0.96 + 0.2 * 1.8
    = 0.768 + 0.36 = 1.128
    D1 > D > D2
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