The floating point unit of a processor using a design D takes 2t cycles…
2007
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
Attempted by 15 students.
Show answer & explanation
Correct answer: B
Step-by-step calculation of average cycles per operation for each design (program mix: 80% fixed-point, 20% floating-point):
Baseline design D: fixed = t, floating = 2t → average = 0.8*t + 0.2*(2t) = 1.2t.
Design D1: fixed = 30% more = 1.3t; floating = 30% less = 0.7*(2t) = 1.4t → average = 0.8*1.3t + 0.2*1.4t = 1.32t.
Design D2: fixed = 40% less = 0.6t; floating = 10% more = 1.1*(2t) = 2.2t → average = 0.8*0.6t + 0.2*2.2t = 0.92t.
Interpretation:
Lower average cycles per operation means better (faster) performance.
Comparing averages: D2 = 0.92t (fastest) < D = 1.2t < D1 = 1.32t (slowest).
Conclusion: The correct relative performance ordering is D2 > D > D1.