The Gateway to Computer Science Excellence

First time here? Checkout the FAQ!

x

+2 votes

Consider two processors P1 and P2 executing the same instruction set. Assume that under identical conditions, for the same input, a program running on P2 takes 25% less time but incurs 20% more CPI (clock cycles per instruction) as compared to the program running on P1. If the clock frequency of P1 is 1GHz, then the clock frequency of P2 (in GHz) is _________.

please do explain also

please do explain also

+4 votes

Total execution time, E = No. of instructions executed( #I) $\times$ CPI / clock frequency (F)

$E_{P2} = 0.75 \times E_{P1}$

$\implies 0.75 \times \#I_{P1} \times CPI_{P1} / F_{P1} = \#I_{P2} \times CPI_{P2}/F_{P2} $

$\implies 0.75 \times \#I_{P1} \times CPI_{P1} / F_{P1} = \#I_{P1} \times 1.2 \times CPI_{P1}/F_{P2} $

$\implies 0.75 \times F_{P1} = 1.2 /F_{P2} $

$\implies F_{P2} = 1.2 /(0.75 \times F_{P1}) $

$\implies F_{P2} = 1.2 /(0.75 \times 1) = 1.6$ GHz

$E_{P2} = 0.75 \times E_{P1}$

$\implies 0.75 \times \#I_{P1} \times CPI_{P1} / F_{P1} = \#I_{P2} \times CPI_{P2}/F_{P2} $

$\implies 0.75 \times \#I_{P1} \times CPI_{P1} / F_{P1} = \#I_{P1} \times 1.2 \times CPI_{P1}/F_{P2} $

$\implies 0.75 \times F_{P1} = 1.2 /F_{P2} $

$\implies F_{P2} = 1.2 /(0.75 \times F_{P1}) $

$\implies F_{P2} = 1.2 /(0.75 \times 1) = 1.6$ GHz

- All categories
- General Aptitude 1.2k
- Engineering Mathematics 4.9k
- Digital Logic 2k
- Programming & DS 3.6k
- Algorithms 3k
- Theory of Computation 3.9k
- Compiler Design 1.5k
- Databases 2.9k
- CO & Architecture 2.5k
- Computer Networks 2.9k
- Non GATE 949
- Others 1.3k
- Admissions 409
- Exam Queries 419
- Tier 1 Placement Questions 17
- Job Queries 55
- Projects 9

34,781 questions

41,758 answers

118,936 comments

41,400 users