9. (20 pts). We are studying two implementations of a machine, one with and one
ID: 3748501 • Letter: 9
Question
9. (20 pts). We are studying two implementations of a machine, one with and one without special floating-point hardware. Consider a program P with the following mix of operations Floating-point multiply Floating-point add Floating-point divide Integer instructions 10% 15% 5% 70% Machine MFP (Machine with Floating-point) has floating-point hardware and can therefore implement the floating-point operations di for each instruction class: rect ly. It requires the following number of clock cycles Floating point Multiply Floating point Add Floating point Divide Integer instructions 6 cycles 4 cycles 20 cycles 2 cycles Machine MNFP (Machine with No Floating Point) has no floating-point hardware and so must emulate the floating-point operations using integer instructions. Each integer instruction on MNFP takes 2 clock cycles (so its CPI is 2). The number of integer instructions needed to implement each of the floating-point operations on MNFP is as follows:Explanation / Answer
Answer :
Convert the MNEP data into no. of cycles
Floating point multiply,
30 x 2 = 60 cycles
20 x 2 = 40 cycles
50 x 2 = 100 cycles
Using weighted AM method, find an average CPI for machine MFP and MNFP
Average CPI for MFP = (0.10 * 6) + (0.15 * 4) + (0.05 * 20) + (0.70 * 2) = 3.6 CPI
Average CPI for MNFP = (0.10 * 30) + (0.15 * 20) + (0.05 * 50) + (0.70 * 2) = 9.9 CPI
Clock rate = 100 MHZ
MIPS ratings for both machines
MIPS ratings for MFP = 100 / 3.6 = 27.77 MIPS
MIPS ratings for MNFP = 100 / 9.9 = 10.1 MIPS
b) For MNFP,
Number of instructions = 300 million
Number of integer instructions required for 300 million instructions = 300 x 9.9
= 2970 million.
c) CPU time for MFP and MNFP
CPU time for MFP = Number of instructions / MIPS = 300 / 27.77 = 10.8 sec.
CPU time for MNFP = Number of instructions / MIPS = 300 / 10.1 = 29.70 sec.
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