Given ,No of instruction =$25000 $
No of branch instruction =$4500$
$\frac{2}{3}$ of the branches are taken So no of branch taken = $\frac{2}{3}*4500=3000$ .
If branch taken it will skip 5 instruction for this program .So ,with $3000$ branch instruction that taken , this program will skip =$5*3000=15000$ instruction .
So no of instruction it will execute =$25000-15000=10000$ .
Now the non pipeline and pipeline processor both will execute only $10000$ instruction .
Branch is determined in stage 3 so branch penalty=2 stall cycle.
So , no of clock cycle required to execute in pipeline =$10000*1+2*3000=16000$
No of clock cycle required to execute it in non pipeline $=5*10000=50000$
So, Speed up=$\frac{50000}{16000}=3.125$ [assuming clock cycle time for both pipeline and non pipeline are same]
