GATE | Gate IT 2007 | Question 36

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 unit 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)
(A) D1 > D > D2
(B) D2 > D > D1
(C) D > D2 > D1
(D) D > D1 > D2


Answer: (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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