Exercise 1.3.4 – 1.3.6 Given information in the following table: 1.3.4 [10] <1.4> Find the IPC (instructions per cycle) for each processor? 1.3.5 [5] <1.4> Find the clock rate for P2 that reduces its execution time to that of P1? 1.3.6 [5] <1.4> Find the number of instructions for P2 that reduces its execution time to that of P3? Processor Clock rate No. Instruction Time P1 3 GHz 20.00E+09 7 s P2 2.5 GHz 30.00E+09 10 s P3 4 GHz 90.00E+09 9 s

IPC means Instructions per Clock Cycle. From: Solution 1.3.4 IPC means Instructions per Clock Cycle. From: CPU-Time = IC x CPI x Clock Cycle Time = (IC x CPI)/Clock Rate Used for: IPC = 1/CPI = IC/(CPU-Time x Clock Rate) IPC(P1) = (20.00E+09)/(7s x 3.00E+09) = 20/21 = 0.95 IPC(P2) = (30.00E+09)/(10s x 2.50E+09) = 30/25 = 1.2 IPC(P3) = (90.00E+09)/(9s x 4.00E+09) = 90/36 = 2.5 Return

IPC means Instructions per Clock Cycle. From: Solution 1.3.5 IPC means Instructions per Clock Cycle. From: CPU-Time = IC x CPI x Clock Cycle Time = (IC x CPI)/Clock Rate Used for: CPU-Timeold = (ICold x CPIold)/(Clock Rateold) CPU-TimeNew = (ICNew x CPINew)/(Clock RateNew) Clock RateNew =(CPU-Timeold) x ICold x CPIold Clock Rateold = (CPU-TimeNew) x ICold x CPIold Clock RateNew = (10s/7s) x 2.5 GHz = 3.57 GHz Return

IPC means Instructions per Clock Cycle. From: Solution 1.3.6 IPC means Instructions per Clock Cycle. From: CPU-Time = IC x CPI x Clock Cycle Time = (IC x CPI)/Clock Rate Used for: CPU-Timeold = (ICold x CPIold)/(Clock Rateold) CPU-TimeNew = (ICNew x CPINew)/(Clock RateNew) and CPIold=CPINew ICNew = (9s/10s) x 30.00E+09 = 27.00E+9 Return