Instruction Pipeline

Question

Consider an instruction pipeline with five stages ,  it allows overlapping of all instructions except branch type. Let there are 20% branch instructions and pipeline is operated with 800 megahertz.
1) what is the speedup?
2) what is average instruction time.
3) what is the time taken for 10 million instructions.
4) what is the throughput.

Comments

it allows overlapping of all instructions except branch type. Let there are 20% branch instructions 

Non branch instructions have CPI = 1 

Branch instructions have CPI = 5

% of branch instructions = 20%

% of non-branch instructions = 80%

Avg CPI = 0.8(1) + 0.2(5) = 1.8 

 

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Ohh got it

Thanks:)

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I think for branch instructions stall cycle =3 as it is a 5-stage pipeline branch target can be known after EX stage.
Therefore ,  

            speedup  =    K/(1+ stall frequency* stall cycle)

                              =     5/(1 + .2*3)

                              =     5/1.6

                              =     3.125

Let, me know if an wrong.....