2. TEST 1 – Tuesday March 3 Lectures 1 - 8, Ch 1,2 HW Due Feb 24 –1.4.1 p.60 –1.4.4 p.60 –1.4.6 p.60 –1.5.2 p.60-61 –1.5.4 p.61 –1.5.5 p.61

3. CPI Invest Resources where time is Spent! CPI = Clock Cycles / Instruction Count = (CPU Time * Clock Rate) / Instruction Count “Average clock cycles per instruction” CPU Time = Instruction Count x CPI / Clock Rate = Instruction Count x CPI x Clock Cycle Time Average CPI = SUM of CPI (i) * I(i) for i=1, n Instruction Count Average CPI = SUM of CPI(i) * F(i) for i = 1, n F(i) is the Instruction Frequency

4. Example (RISC processor) Typical Mix Base Machine (Reg / Reg) OpFreqCyclesF(i)CPI(i)% Time ALU50%1 Load20%5 Store10%3 Branch20%2

5. Example (RISC processor) Typical Mix Base Machine (Reg / Reg) OpFreqCyclesF(i)CPI(i)% Time ALU50%1.5 Load20%5 1.0 Store10%3.3 Branch20%2.4 2.2 = CPI ave

6. Example (RISC processor) Typical Mix Base Machine (Reg / Reg) OpFreqCyclesF(i)CPI(i)% Time ALU50%1.5 Load20%5 1.0 Store10%3.3 Branch20%2.4 2.2 = CPI ave CPU Time(i) = Instr Cnt(i) * CPI(i) * Clk Cycle Time CPU Time Inst Cnt * CPI ave * Clk Cycle Time % Time = F(i) * CPI(i) / CPI ave

7. Example (RISC processor) Typical Mix Base Machine (Reg / Reg) OpFreqCyclesF(i)CPI(i)% Time ALU50%1.523% Load20%5 1.045% Store10%3.314% Branch20%2.418% 2.2 = CPI ave CPU Time(i) = Instr Cnt(i) * CPI(i) * Clk Cycle Time CPU Time Inst Cnt * CPI ave * Clk Cycle Time % Time = F(i) * CPI(i) / CPI ave

8. Example (RISC processor) Typical Mix Base Machine (Reg / Reg) OpFreqCyclesF(i)CPI(i)% Time ALU50%1.523% Load20%5 1.045% Store10%3.314% Branch20%2.418% 2.2 = CPI ave How much faster would the machine be if a better data cache reduced the average load time to 2 cycles?

9. Example (RISC processor) Typical Mix Base Machine (Reg / Reg) OpFreqCyclesF(i)CPI(i)% Time ALU50%1.523% Load20%5 (2) 1.0 (.4)45% Store10%3.314% Branch20%2.418% 2.2 (1.6) How much faster would the machine be if a better data cache reduced the average load time to 2 cycles? 2.2/1.6 = 1.375 CPU Time = Inst Cnt * CPI ave * Clk Cycle Time

10. Example (RISC processor) Typical Mix Base Machine (Reg / Reg) OpFreqCyclesF(i)CPI(i)% Time ALU50%1.523% Load20%5 1.045% Store10%3.314% Branch20%2.418% 2.2 How much faster would the machine be if a better data cache reduced the average load time to 2 cycles? CPI = 1.6 How does this compare with using branch prediction to shave a cycle off the branch time?

11. Example (RISC processor) Typical Mix Base Machine (Reg / Reg) OpFreqCyclesF(i)CPI(i)% Time ALU50%1.523% Load20%5 1.045% Store10%3.314% Branch20%2 (1).4 (.2)18% 2.2 (2.0) How much faster would the machine be if a better data cache reduced the average load time to 2 cycles? CPI = 1.6 How does this compare with using branch prediction to shave a cycle off the branch time? CPI = 2.0

12. Example (RISC processor) Typical Mix Base Machine (Reg / Reg) OpFreqCyclesF(i)CPI(i)% Time ALU50%1.523% Load20%5 1.045% Store10%3.314% Branch20%2.418% 2.2 How much faster would the machine be if a better data cache reduced the average load time to 2 cycles? CPI = 1.6 How does this compare with using branch prediction to shave a cycle off the branch time? CPI = 2.0 What if two ALU instructions could be executed at once?

13. Example (RISC processor) Typical Mix Base Machine (Reg / Reg) OpFreqCyclesF(i)CPI(i)% Time ALU50%1 (.5).5 (.25)23% Load20%5 1.045% Store10%3.314% Branch20%2.418% 2.2 (1.95) How much faster would the machine be if a better data cache reduced the average load time to 2 cycles? CPI = 1.6 How does this compare with using branch prediction to shave a cycle off the branch time? CPI = 2.0 What if two ALU instructions could be executed at once? CPI=1.95

14. A compiler designer is trying to decide between two code sequences for a particular machine. Based on the hardware implementation, there are three different classes of instructions: Class A has 1 cycle Class B has 2 cycles Class C has 3 cycles The first code sequence has 5 instructions: 2 of A, 1 of B, and 2 of C The second sequence has 6 instructions: 4 of A, 1 of B, and 1 of C. Which sequence will be faster? How much? What is the CPI for each sequence?

15. A compiler designer is trying to decide between two code sequences for a particular machine. Based on the hardware implementation, there are three different classes of instructions: Class A has 1 cycle Class B has 2 cycles Class C has 3 cycles The first code sequence has 5 instructions: 2 of A, 1 of B, and 2 of C 2*1+1*2+2*3 = 10 The second sequence has 6 instructions: 4 of A, 1 of B, and 1 of C. 4*1+1*2+1*3 = 9 Which sequence will be faster? How much? What is the CPI for each sequence?

16. A compiler designer is trying to decide between two code sequences for a particular machine. Based on the hardware implementation, there are three different classes of instructions: Class A has 1 cycle Class B has 2 cycles Class C has 3 cycles The first code sequence has 5 instructions: 2 of A, 1 of B, and 2 of C 2*1+1*2+2*3 = 10 The second sequence has 6 instructions: 4 of A, 1 of B, and 1 of C. 4*1+1*2+1*3 = 9 Which sequence will be faster? How much? 10 / 9 = 1.11 What is the CPI for each sequence?

17. A compiler designer is trying to decide between two code sequences for a particular machine. Based on the hardware implementation, there are three different classes of instructions: Class A has 1 cycle Class B has 2 cycles Class C has 3 cycles The first code sequence has 5 instructions: 2 of A, 1 of B, and 2 of C 2*1+1*2+2*3 = 10 The second sequence has 6 instructions: 4 of A, 1 of B, and 1 of C. 4*1+1*2+1*3 = 9 Which sequence will be faster? How much? 10 / 9 = 1.11 What is the CPI for each sequence? 10/5 = 2 9/6 = 1.5

18. MIPS = Instruction Count Execution time x 10 6 A popular performance metric is MIPS, the number of millions of instructions per second. For a given program,

19. MIPS = Instruction Count Execution time x 10 6 A popular performance metric is MIPS, the number of millions of instructions per second. For a given program, 1.Cannot compare if instruction set is different

20. MIPS = Instruction Count Execution time x 10 6 A popular performance metric is MIPS, the number of millions of instructions per second. For a given program, 1.Cannot compare if instruction set is different 2.Highly dependent on the program

21. MIPS = Instruction Count Execution time x 10 6 A popular performance metric is MIPS, the number of millions of instructions per second. For a given program, 1.Cannot compare if instruction set is different 2.Highly dependent on the program 3.Can be inversely proportional to performance

22. Two different compilers are being tested for a 100 MHz. machine with three different classes of instructions: Class A has 1 cycle,Class B has 2 cycles, Class C has 3 cycles Instruction counts ( billions) Code fromABC Compiler 1511 Compiler 21011 Which sequence will be faster according to MIPS? Which sequence will be faster according to execution time? MIPS example

23. Two different compilers are being tested for a 100 MHz. machine with three different classes of instructions: Class A Class B Class C CPI 1 2 3 Instruction counts ( billions) Code fromABCTotal Compiler 1 511 7 Compiler 21011 12 CPU cyclesExec TimeMIPS Compiler 1 5+1x2+1x3=10 billion Compiler 2 MIPS example

24. Two different compilers are being tested for a 100 MHz. machine with three different classes of instructions: Class A Class B Class C CPI 1 2 3 Instruction counts ( billions) Code fromABCTotal Compiler 1 511 7 Compiler 21011 12 CPU cyclesExec TimeMIPS Compiler 1 10 billion Compiler 2 15 billion MIPS example

25. Two different compilers are being tested for a 100 MHz. machine with three different classes of instructions: Class A Class B Class C CPI 1 2 3 Instruction counts ( billions) Code fromABCTotal Compiler 1 511 7 Compiler 21011 12 CPU cyclesExec TimeMIPS Compiler 1 10 billion10 10 x10 -8 =100 Compiler 2 15 billion MIPS example

26. Two different compilers are being tested for a 100 MHz. machine with three different classes of instructions: Class A Class B Class C CPI 1 2 3 Instruction counts ( billions) Code fromABCTotal Compiler 1 511 7 Compiler 21011 12 CPU cyclesExec TimeMIPS Compiler 1 10 billion 100 sec Compiler 2 15 billion 150 sec MIPS example

27. Two different compilers are being tested for a 100 MHz. machine with three different classes of instructions: Class A Class B Class C CPI 1 2 3 Instruction counts ( billions) Code fromABCTotal Compiler 1 511 7 Compiler 21011 12 CPU cyclesExec TimeMIPS Compiler 1 10 billion 100 sec7x10 3 /100 Compiler 2 15 billion 150 sec MIPS example

28. Two different compilers are being tested for a 100 MHz. machine with three different classes of instructions: Class A Class B Class C CPI 1 2 3 Instruction counts ( billions) Code fromABCTotal Compiler 1 511 7 Compiler 21011 12 CPU cyclesExec TimeMIPS Compiler 1 10 billion 100 sec 70 Compiler 2 15 billion 150 sec 12x10 3 /150 MIPS example

29. Two different compilers are being tested for a 100 MHz. machine with three different classes of instructions: Class A Class B Class C CPI 1 2 3 Instruction counts ( billions) Code fromABCTotal Compiler 1 511 7 Compiler 21011 12 CPU cyclesExec TimeMIPS Compiler 1 10 billion 100 sec 70 Compiler 2 15 billion 150 sec 80 MIPS example

30. Performance best determined by running a real application –Use programs typical of expected workload –Or, typical of expected class of applications e.g., compilers/editors, scientific applications, graphics, etc. Benchmarks

31. Performance best determined by running a real application –Use programs typical of expected workload –Or, typical of expected class of applications e.g., compilers/editors, scientific applications, graphics, etc. Small benchmarks –nice for architects and designers –easy to standardize –can be abused Benchmarks

32. SPEC CPU 2006 SPEC - System Performance Evaluation Cooperative 12 Integer Benchmarks 17 Floating Point Benchmarks Fig 1.20 P.49

33. Execution Time After Improvement = Execution Time Unaffected + ( Execution Time Affected / Amount of Improvement ) Amdahl's Law

34. Execution Time After Improvement = Execution Time Unaffected + ( Execution Time Affected / Amount of Improvement ) Example: Suppose a program runs in 100 seconds on a machine, with multiply responsible for 80 seconds of this time. How much do we have to improve the speed of multiplication if we want the program to run 4 times faster? Amdahl's Law

35. Execution Time After Improvement = Execution Time Unaffected + ( Execution Time Affected / Amount of Improvement ) Example: Suppose a program runs in 100 seconds on a machine, with multiply responsible for 80 seconds of this time. How much do we have to improve the speed of multiplication if we want the program to run 4 times faster? Improved Time = (100 – 80) + 80/n = 100/4 Amdahl's Law

36. Execution Time After Improvement = Execution Time Unaffected + ( Execution Time Affected / Amount of Improvement ) Example: Suppose a program runs in 100 seconds on a machine, with multiply responsible for 80 seconds of this time. How much do we have to improve the speed of multiplication if we want the program to run 4 times faster? Improved Time = (100 – 80) + 80/n = 100/4 20 + 80/n = 25 80 = 5n ; n = 16 Amdahl's Law

37. Execution Time After Improvement = Execution Time Unaffected + ( Execution Time Affected / Amount of Improvement ) Example: Suppose a program runs in 100 seconds on a machine, with multiply responsible for 80 seconds of this time. How much do we have to improve the speed of multiplication if we want the program to run 4 times faster? How about making it 5 times faster? Amdahl's Law

38. Execution Time After Improvement = Execution Time Unaffected + ( Execution Time Affected / Amount of Improvement ) Example: Suppose a program runs in 100 seconds on a machine, with multiply responsible for 80 seconds of this time. How much do we have to improve the speed of multiplication if we want the program to run 4 times faster? How about making it 5 times faster? Improved Time = (100 –80) + 80/n = 100/5 Amdahl's Law

39. Execution Time After Improvement = Execution Time Unaffected + ( Execution Time Affected / Amount of Improvement ) Example: Suppose a program runs in 100 seconds on a machine, with multiply responsible for 80 seconds of this time. How much do we have to improve the speed of multiplication if we want the program to run 4 times faster? How about making it 5 times faster? Improved Time = (100 –80) + 80/n = 100/5 20 + 80/n = 20 80/n = 0 Impossible! Amdahl's Law

40. Performance is specific to a particular program/s –Total execution time is a consistent summary of performance Remember

41. Performance is specific to a particular program/s –Total execution time is a consistent summary of performance For a given architecture performance increases come from: –increases in clock rate (without adverse CPI affects) Remember

42. Performance is specific to a particular program/s –Total execution time is a consistent summary of performance For a given architecture performance increases come from: –increases in clock rate (without adverse CPI affects) –improvements in processor organization that lower CPI Remember

43. Performance is specific to a particular program/s –Total execution time is a consistent summary of performance For a given architecture performance increases come from: –increases in clock rate (without adverse CPI affects) –improvements in processor organization that lower CPI –compiler enhancements that lower CPI and/or instruction count Remember