Question

Problem 2. Probability of occurrence Program A Program B (commercial) (scientific) Instruction type Consider the instruction mixes in the given figure. Suppose that the system's clock frequency is 100 MHz, and all instructions except floating point instructions have an average execution time of 10 ns. (a) What is the average execution time of floating-point instructions if the overall average execution time per instruction for Program B is 18.1 ns? 15 pts. (b) What is the CPI for Program B? 20 pts NOTE: Show the details of your answer. 0.29 0.15 0.15 Memory load Memory store Fixed-point operations Floating-point operations Branch Other 0.24 0.12 0.27 0.00 0.17 0.20 0.19 0.10 0.12

Answer 1

Answer (a) :

Average execution time of each instruction (except floating point) = 10ns

In question we need to calculate the average execution time of floating point instruction and we have the average execution time of overall program B so we use simple average formula.

average execution time of overall program B =18.1ns

So average formula :

$\frac{\sum T_e}{n}$

where $T_e$ = Average execution time of each instruction

n=Number of instructions

10 + 10 + 10 + 1 + 10 + 10 = 18.1

x=Average execution time of floating point instruction

Answer) On calculation , x=2.172ns

Answer (b) :

Given system's clock frequency(clock rate)=100MHz

We need to calculate CPI in this question for program B. Now what is CPI ?

CPI : It is the average number of clock cycles per instruction. If for each instruction we are aware of its frequency and its number of clock cycles it will take for execution , then we can calculate the CPI of overall code.

CPI formula :

CPI = ΣF και Ν

where, F = frequency of each instruction (i.e the probability of occurence of each instruction as given in question

N = Number of clock cycles associated with each instruction iof the program

From the average execution time of each instructions we can take out the number of clock cycles per instruction by formula :

clock Rate - = noof Cycles * execution Time de

Clock rate is the clock frequency so   will give us the clock cycle time preferrably in (seconds).

clock Rate

Given, System's clock rate = 100MHz  

Execution time of floating point instruction = 2.172 ns

Execution time of all other instructions(except floating point) = 10 ns

  

For floating point instruction ,

100 * 106 .112* 10 ) * W

N (no of clock cycles) = 4.6

For all other instructions ,

100 * 106 = {10 * 10-9) *

N (no of clock cycles) = 1

Now CPI ,

CPI = (0.29*1)+(0.15*1)+(0.15*1)+(0.19*4.6)+(0.10*1)+(0.12*1) = 1.684clockcycles/instruction

Answer) CPI=1.684 clock cycles/instruction