1. 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

2. 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

3. 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

4. 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