no of instructions

Question

Comments

22.22%

---

22.2??

---

Yes..how?

Answer 1

Let the number of instructions be 'n'.

Speedup = $\frac{Time_{non-pipeline}}{Time_{pipeline}}$

Time_non-pipeline = 5 * n * T_clk

Now, let 'x' instructions incur 0 stall cycles, so CPI of these 'x' instrutions =1

Then, 'n-x' instructions incur 3 stall cycles, so CPI of these 'n-x' instructions = 4

Time_pipeline = [(x * 1) + [(n-x) * 4]] * T_clk = (4n - 3x) * T_clk

Speedup = $\frac{5n *T_{clk}}{(4n - 3x) * T_{clk}}$ = 3.

$\therefore$ x = $\frac{7n}{9}$

So, number of instructions which incurred 3 stall cycles = n - $\frac{7n}{9}$ = $\frac{2n}{9}$

Percentage = $\frac{\frac{2n}{9}}{n}$ * 100 = 22.2%

Comments

I did this way

5 / ((4x+y) / (x+y)) = 3

x = no of instructions with 3 stalls

y = remaining instructions

 

What is wrong in this approach?

---

Didn't understand the denominator part!!

---

(4x+y) / (x+y)

This is the avg cycles for a single instruction.

---

speedup =totaql time taken in non-pipleine/total time taken in pipeline

here,time taken by non-pipeline will be 5 considering each stage takes one cycle

and pipeline will take (CPI + percentage of instruction*stall cycles)

so,3 =5/(1+ (x/100)*3)

which comes to be 22.2 %