3. A benchmark program is run on a 3 GHz processor. The executed program consist

Question

3. A benchmark program is run on a 3 GHz processor. The executed program consists of 10,000,000 instruction executions, with the following instruction mix and clock cycle count: Instruction Type Instruction Count Cycles per Instruction Integer arithmetic 4,500,000 Data transfer Floating point Control transfer Determine: 3,200,000 1,500,000 800,000 a. Effective CP b. MIPS rate. c. FLOPS rate d. Program Execution time. e. If the processor is replaced by a 4.5 GHz processor, what is the Speed up, if any?

Explanation / Answer

The CPU clock rate depends on the specific CPU organization and hardware implementation technology used.
The average CPI of all instructions executed in the program in CPU design.
For a specific program compiled to run on a specific machine (CPU) “A”, has the following parameters:
The total executed instruction count of the program I
The average number of cycles per instruction effective CPI
– Clock cycle of machine C

1.Effective CPI
****************
CPI = Total program execution cycles / Instructions count Executed
4500000 + (2*3200000) + (3*1500000) + (800000*2) / (100 00000) = 1.7

2.Mips Rate
***********
MIPS Rating = Instruction count / (Execution Time x 106)
3G clocks / sec * (1/1.7 clocks per instruction) = 3 / 1.7 / 1000000 = 17.6 MIPs

3. Flops Rate
**************
FLOPS is "how many floating point operations per second". So it's a form of Hz, but with a specific value of "something"
A FLOP is a specific kind of instruction, and tends to take many more cycles than an integer op,
Flops = Number of floating-point operations / (Execution time x 106 )

4.Program Execution time
***************************
Execution time = (100 00000 instructions) * 1.7
CPI = 17000000 cyles * 1/3G sec = 0.003875 = 5.66 ms

5.Replaced by 4.5 ghz
*********************
CPI = 1.7;
MIPS rate = 26.5;
Execution time = 3.77 ns

Speedup = Performance for entire task using the enhancement / Performance for the entire task without using the enhancement

Speedup(E) = Execution Time without E Performance with E
________________________ = ____________________
Execution Time with E Performance without E

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