1 (C) . before starting the answer of this let us find the average CPI before new implementation
Type | CPI | PART | product |
---|---|---|---|
F | 5 | 0.25 | 1.25 |
I | 2 | 0.25 | 0.5 |
N | 2.4 | .50 | 1.2 |
Total Cpi = 2.95 with 600 MHz clock rate
now we will use the speed up factor to compare which new strategy will be good
old execution time = 2.95* 1/ (600 *10^-6)
1.) now just modify CPI of F to 2.5 so
Type | CPI | PART | product |
---|---|---|---|
F | 2.5 | 0.25 | 0.625 |
I | 2 | 0.25 | 0.5 |
N | 2.4 | .50 | 1.2 |
New Cpi = 2.325 new clock rate =600
new execution time = 2.325 * 1 / ( 600 * 10^-6)
Speed up S = 2.95 * 600 * 10^-6 / 2.325* 600 * 10^-6 = 1.27
2 .) now just modify CPI of I to 1.2 so
Type | CPI | PART | product |
---|---|---|---|
F | 5.0 | 0.25 | 1.25 |
I | 1.2 | 0.25 | 0.3 |
N | 2.4 | .50 | 1.2 |
New Cpi = 2.75 new clock rate =600
new execution time = 2.75 * 1 / ( 600 * 10^-6)
Speed up S = 2.95 * 600 * 10^-6 / 2.75* 600 * 10^-6 = 1.07
3.) now just update the clock speed so
Type | CPI | PART | product |
---|---|---|---|
F | 5 | 0.25 | 1.25 |
I | 2 | 0.25 | 0.5 |
N | 2.4 | .50 | 1.2 |
New Cpi = 2.95 with 750 MHz clock rate
new execution time = 2.95 * 1 / ( 750 * 10^-6)
Speed up S = 2.95 * 750 * 10^-6 / 2.95* 600 * 10^-6 = 1.25
so considering all the speedups we get largest speed up i.e. S =1.27 for the first approach why?
because in original the F class has largest CPI also the contribution in the CPI by F is 1.25 which is largest so if we anyhow modify this parameter then we may get the best possible result and which is done in the first approach that is why impact of first approach is maximum which leads us to a largest speedup
Also if we consider that in the 3rd approach we got 1.25 speedup which is near to the largest because the speed of all the instructions are increased .
I hope it'll help you so please give positive ratings :))))))))
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