0 votes 0 votes Answer 3.12 CO and Architecture zeal pipelining numerical-answers test-series + – SKMAKM asked Oct 20, 2022 retagged Oct 20, 2022 by makhdoom ghaya SKMAKM 513 views answer comment Share Follow See 1 comment See all 1 1 comment reply Kabir5454 commented Dec 17, 2022 i moved by Kabir5454 Feb 1, 2023 reply Follow Share 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] 0 votes 0 votes Please log in or register to add a comment.