Branch instruction

Question

Consider two machines M1 and M2. For both machines, all instructions except for mispredicted branches take one cycle. Mispredicted branches take one cycle plus an additional misprediction penalty.

 Machine A has a clock rate of 1.0 GHz and a misprediction penalty of 5 cycles.

  Machine B has a clock rate of 2.0 GHz and a misprediction penalty of 20 cycles.

 Branches are 25% of all instructions and 80% of branches are predicted correctly. Which of the following statement is correct?

 (a) Machine A is faster than Machine B

 (b) Machine B is faster than Machine A

 (c) Both are same

 (d) None of these

Comments

I'm solving like that: For M/C A

Out of 25% ,80% prediction correctly and 20% misprediction right then

and 75% don't have any branch int then only 1 cycle

Overall cycle : 0.25*0.8*1 + 0.25*0.2*5 + 0.75*1 = 1.2 cycle or 1.2 ns  (converting Freq. into T 1GHZ ->1ns)

but correct ans for machine A is 1.3 ns how ??

or what's wrong in my concept plz corrrect it if any.

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@ASUR penalty should be considered as (5+1) as it's mentioned in question that "Mispredicted branches take one cycle plus an additional misprediction penalty.'

Answer 1

Answer: opt b) Machine B is faster

Cycles required for A = 0.75*1+0.25*0.80*1+0.25*0.20*(1+5)=1.25

Cycles required for B=

0.75*1+0.25*0.80*1+0.25*0.20*(1+20)=2

Execution time = cycles / clock rate

Execution time of A=1.25/1 GHz =1.25 nsec

Execution time of B = 2/2 GHz = 1 nsec

Hence machine B is faster than machine A as execution time of B is less than A

Comments

OOOO understood.

 i missed it    "Mispredicted branches take one cycle plus an additional misprediction penalty."

Thankyou for clearification.