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 1.4k
- Engineering Mathematics 5.9k
- Digital Logic 2.3k
- Programming & DS 4.3k
- Algorithms 3.7k
- Theory of Computation 4.6k
- Compiler Design 1.7k
- Databases 3.4k
- CO & Architecture 2.9k
- Computer Networks 3.4k
- Non GATE 1.2k
- Others 1.3k
- Admissions 506
- Exam Queries 482
- Tier 1 Placement Questions 22
- Job Queries 64
- Projects 15

40,903 questions

47,558 answers

146,289 comments

62,306 users