4) We are interested in two implementations of a machine, one with and one witho

Question

4) We are interested in two implementations of a machine, one with and one without special floating-point hardware. Consider a program P with the following mix of operations: Instruction Type Floating-point multiply Floating-point add Floating-point divide Integer instructions Frequency in Mix 10% 15 % 5% 70 % Machine MFP (Machine with floating-point) has floating point hardware and can therefore implement the floating-point operations directly. It requires the following number of clock cycles for each instruction class: Instruction Type Floating-point multi Floating-point add Floating-point divide Integer instructions Number of Clock Cvcles 6 4 20 2 Machine MNFP (Machine with no floating-point) has no floating-point hardware and so must emulate the floating-point operations using integer instructions. The integer instructions all take 2 clock cycles. The number of integer instructions needed to implement each of the floating-point operations is as follows: Instruction Type Floating-point multipl Floating-point add Floating-point divide Number of Integer Operations 30 20 50 Assume that both machines have the same clock rate of 100 Mhz Find the MIPS ratings for both machines. If the machine MFP executes 300,000,000 instructions for this program, how many integer instructions does machine MNFP require for the same program? Determine the CPU time for each machine and compare the MIPS ratings with the CPU times for the same machines. Suppose that the research department says that the MFP machine in the previous problem can be re-engineered so that all floating-point instructions run 25% faster. There is no corresponding improvement in the rate of processing integer instruction. What performance gain for the MFP will this improvement produce? a. b. c. d.

Explanation / 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 MFP = 100 / 3.6 = 27.77 MIPS

MIPS ratings for MNFP = 100 / 9.9 = 10.1 MIPS

Number of instructions = 300 million

Number of integer instructions required for 300 million instructions = 300 x 9.9

= 2970 million.

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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