CPU Performance Evaluation Cycles Per Instruction CPI Most

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CPU Performance Evaluation: Cycles Per Instruction (CPI) • Most computers run synchronously utilizing a

CPU Performance Evaluation: Cycles Per Instruction (CPI) • Most computers run synchronously utilizing a CPU clock running at a Clock cycle constant clock rate: where: • • Clock rate = 1 / clock cycle 1 cycle 2 cycle 3 The CPU clock rate depends on the specific CPU organization (design) and hardware implementation technology (VLSI) used A computer machine (ISA) instruction is comprised of a number of elementary or micro operations which vary in number and complexity depending on the instruction and the exact CPU organization (Design) – A micro operation is an elementary hardware operation that can be performed during one CPU clock cycle. – This corresponds to one micro-instruction in microprogrammed CPUs. – Examples: register operations: shift, load, clear, increment, ALU operations: add , subtract, etc. Thus a single machine instruction may take one or more CPU cycles to complete termed as the Cycles Per Instruction (CPI). Average CPI of a program: The average CPI of all instructions executed in the program on a given CPU design. (Chapter 4) Cycles/sec = Hertz = Hz EECC 550 - Shaaban #1 Lec # 3 Winter 2005 12 -6 -2005

Generic CPU Machine Instruction Processing Steps Instruction Fetch Instruction Decode Operand Fetch Execute Result

Generic CPU Machine Instruction Processing Steps Instruction Fetch Instruction Decode Operand Fetch Execute Result Store Next Instruction Obtain instruction from program memory Determine required actions and instruction size Locate and obtain operand data Compute result value or status Deposit results in storage (memory or register) for later use Determine successor or next instruction EECC 550 - Shaaban #2 Lec # 3 Winter 2005 12 -6 -2005

Computer Performance Measures: Program Execution Time • For a specific program compiled to run

Computer Performance Measures: Program Execution Time • For a specific program compiled to run on a specific machine (CPU) “A”, has the following parameters: – The total executed instruction count of the program. I – The average number of cycles per instruction (average CPI). CPI – Clock cycle of machine “A” C • How can one measure the performance of this machine (CPU) running this program? – Intuitively the machine (or CPU) is said to be faster or has better performance running this program if the total execution time is shorter. – Thus the inverse of the total measured program execution time is a possible performance measure or metric: Performance. A = 1 / Execution Time. A How to compare performance of different machines? What factors affect performance? How to improve performance? EECC 550 - Shaaban #3 Lec # 3 Winter 2005 12 -6 -2005

Comparing Computer Performance Using Execution Time • To compare the performance of two machines

Comparing Computer Performance Using Execution Time • To compare the performance of two machines (or CPUs) “A”, “B” running a given specific program: Performance. A = 1 / Execution Time. A Performance. B = 1 / Execution Time. B • Machine A is n times faster than machine B means (or slower? if n < 1) : Speedup = n = • Performance. A Performance. B Example: For a given program: Execution time on machine A: Execution time on machine B: Speedup= Performance. A / Performance. B = Execution Time. B Execution Time. A (i. e Speedup is ratio of performance, no units) Execution. A = 1 second Execution. B = 10 seconds = Execution Time. B / Execution Time. A = 10 / 1 = 10 The performance of machine A is 10 times the performance of machine B when running this program, or: Machine A is said to be 10 times faster than machine B when running this program. The two CPUs may target different ISAs provided the program is written in a high level language (HLL) EECC 550 - Shaaban #4 Lec # 3 Winter 2005 12 -6 -2005

CPU Execution Time: The CPU Equation • A program is comprised of a number

CPU Execution Time: The CPU Equation • A program is comprised of a number of instructions executed , I – Measured in: instructions/program • The average instruction executed takes a number of cycles per instruction (CPI) to be completed. Per Cycle (IPC): – Measured in: cycles/instruction, CPI Or Instructions IPC= 1/CPI • CPU has a fixed clock cycle time C = 1/clock rate – Measured in: seconds/cycle • CPU execution time is the product of the above three parameters as follows: Executed CPU time = Seconds Program T = execution Time per program in seconds = Instructions x Cycles Program I x Number of instructions executed x Seconds Instruction CPI x Average CPI for program (This equation is commonly known as the CPU performance equation) Cycle C CPU Clock Cycle EECC 550 - Shaaban #5 Lec # 3 Winter 2005 12 -6 -2005

CPU Average CPI/Execution Time For a given program executed on a given machine (CPU):

CPU Average CPI/Execution Time For a given program executed on a given machine (CPU): CPI = Total program execution cycles / Instructions count (average) ® CPU clock cycles = Instruction count x CPI CPU execution time = T execution Time per program in seconds = CPU clock cycles x Clock cycle = Instruction count x CPI x Clock cycle = I x CPI x C Number of instructions executed Average CPI for program (This equation is commonly known as the CPU performance equation) CPU Clock Cycle EECC 550 - Shaaban #6 Lec # 3 Winter 2005 12 -6 -2005

CPU Execution Time: Example • A Program is running on a specific machine (CPU)

CPU Execution Time: Example • A Program is running on a specific machine (CPU) with the following parameters: – Total executed instruction count: 10, 000 instructions – Average CPI for the program: 2. 5 cycles/instruction. – CPU clock rate: 200 MHz. (clock cycle = 5 x 10 -9 seconds) • What is the execution time for this program: CPU time = Seconds Program = Instructions x Cycles Program x Seconds Instruction Cycle CPU time = Instruction count x CPI x Clock cycle = 10, 000 x 2. 5 x 1 / clock rate = 10, 000 x 2. 5 x 5 x 10 -9 =. 125 seconds T = I x CPI x C EECC 550 - Shaaban #7 Lec # 3 Winter 2005 12 -6 -2005

Factors Affecting CPU Performance CPU time = Seconds Program T = = Instructions x

Factors Affecting CPU Performance CPU time = Seconds Program T = = Instructions x Cycles Program x Seconds Instruction Cycle I x CPI x C Instruction Cycles per Clock Rate Count (1/C) Instruction Program Compiler Instruction Set Architecture (ISA) Organization (CPU Design) Technology (VLSI) EECC 550 - Shaaban #8 Lec # 3 Winter 2005 12 -6 -2005

Aspects of CPU Execution Time CPU Time = Instruction count x CPI x Clock

Aspects of CPU Execution Time CPU Time = Instruction count x CPI x Clock cycle Depends on: T = I x CPI x C Program Used Compiler ISA Instruction Count I (executed) Depends on: Program Used Compiler ISA CPU Organization CPI (Average CPI) Clock Cycle C Depends on: CPU Organization Technology (VLSI) EECC 550 - Shaaban #9 Lec # 3 Winter 2005 12 -6 -2005

Performance Comparison: Example • From the previous example: A Program is running on a

Performance Comparison: Example • From the previous example: A Program is running on a specific machine (CPU) with the following parameters: – Total executed instruction count, I: 10, 000 instructions – Average CPI for the program: 2. 5 cycles/instruction. – CPU clock rate: 200 MHz. • Using the same program with these changes: – A new compiler used: New executed instruction count, I: 9, 500, 000 New CPI: 3. 0 – Faster CPU implementation: New clock rate = 300 MHz • What is the speedup with the changes? Speedup = Old Execution Time = Iold x New Execution Time Inew x CPIold x Clock cycleold CPInew x Clock Cyclenew Speedup = (10, 000 x 2. 5 x 5 x 10 -9) / (9, 500, 000 x 3. 33 x 10 -9 ) =. 125 /. 095 = 1. 32 or 32 % faster after changes. Clock Cycle = 1/ Clock Rate T = I x CPI x C EECC 550 - Shaaban #10 Lec # 3 Winter 2005 12 -6 -2005

Instruction Types & CPI • Given a program with n types or classes of

Instruction Types & CPI • Given a program with n types or classes of instructions executed on given CPU with the following characteristics: Ci = Count of instructions of typei executed CPIi = Cycles per instruction for typei a i = 1, 2, …. n Then: CPI = CPU Clock Cycles / Instruction Count I Where: Executed Instruction Count I = S Ci T = I x CPI x C EECC 550 - Shaaban #11 Lec # 3 Winter 2005 12 -6 -2005

Instruction Types & CPI: An Example • An instruction set has three instruction classes:

Instruction Types & CPI: An Example • An instruction set has three instruction classes: Instruction class A B C CPI 1 2 3 For a specific CPU design • Two code sequences have the following instruction counts: Code Sequence 1 2 Instruction counts for instruction class A B C 2 1 2 4 1 1 • CPU cycles for sequence 1 = 2 x 1 + 1 x 2 + 2 x 3 = 10 cycles CPI for sequence 1 = clock cycles / instruction count = 10 /5 = 2 • CPU cycles for sequence 2 = 4 x 1 + 1 x 2 + 1 x 3 = 9 cycles CPI for sequence 2 = 9 / 6 = 1. 5 CPI = CPU Cycles / I EECC 550 - Shaaban #12 Lec # 3 Winter 2005 12 -6 -2005

Instruction Frequency & CPI • Given a program with n types or classes of

Instruction Frequency & CPI • Given a program with n types or classes of instructions with the following characteristics: i = 1, 2, …. n Ci = Count of instructions of typei CPIi = Average cycles per instruction of typei Fi = Frequency or fraction of instruction typei executed = Ci/ total executed instruction count = Ci/ I Then: Fraction of total execution time for instructions of type i = CPIi x Fi CPI EECC 550 - Shaaban #13 Lec # 3 Winter 2005 12 -6 -2005

Instruction Type Frequency & CPI: A RISC Example CPIi x Fi Program Profile or

Instruction Type Frequency & CPI: A RISC Example CPIi x Fi Program Profile or Executed Instructions Mix Given Base Machine (Reg / Reg) Op Freq, Fi CPIi ALU 50% 1 Load 20% 5 Store 10% 3 Branch 20% 2 Typical Mix CPIi x Fi. 5 1. 0. 3. 4 % Time 23% =. 5/2. 2 45% = 1/2. 2 14% =. 3/2. 2 18% =. 4/2. 2 Sum = 2. 2 CPI =. 5 x 1 +. 2 x 5 +. 1 x 3 +. 2 x 2 = 2. 2 =. 5 + 1 +. 3 +. 4 EECC 550 - Shaaban #14 Lec # 3 Winter 2005 12 -6 -2005

Metrics of Computer Performance (Measures) Execution time: Target workload, SPEC, etc. Application Programming Language

Metrics of Computer Performance (Measures) Execution time: Target workload, SPEC, etc. Application Programming Language Compiler ISA (millions) of Instructions per second – MIPS (millions) of (F. P. ) operations per second – MFLOP/s Datapath Control Megabytes per second. Function Units Transistors Wires Pins Cycles per second (clock rate). Each metric has a purpose, and each can be misused. EECC 550 - Shaaban #15 Lec # 3 Winter 2005 12 -6 -2005

Choosing Programs To Evaluate Performance Levels of programs or benchmarks that could be used

Choosing Programs To Evaluate Performance Levels of programs or benchmarks that could be used to evaluate performance: – Actual Target Workload: Full applications that run on the target machine. – Real Full Program-based Benchmarks: • Select a specific mix or suite of programs that are typical of targeted applications or workload (e. g SPEC 95, SPEC CPU 2000). – Small “Kernel” Benchmarks: • Key computationally-intensive pieces extracted from real programs. – Examples: Matrix factorization, FFT, tree search, etc. • Best used to test specific aspects of the machine. – Microbenchmarks: • Small, specially written programs to isolate a specific aspect of performance characteristics: Processing: integer, floating point, local memory, input/output, etc. EECC 550 - Shaaban #16 Lec # 3 Winter 2005 12 -6 -2005

Pros Types of Benchmarks • Representative • Portable. • Widely used. • Measurements useful

Pros Types of Benchmarks • Representative • Portable. • Widely used. • Measurements useful in reality. Actual Target Workload Full Application Benchmarks • Easy to run, early in the design cycle. • Identify peak performance and potential bottlenecks. Small “Kernel” Benchmarks Microbenchmarks Cons • Very specific. • Non-portable. • Complex: Difficult to run, or measure. • Less representative than actual workload. • Easy to “fool” by designing hardware to run them well. • Peak performance results may be a long way from real application performance EECC 550 - Shaaban #17 Lec # 3 Winter 2005 12 -6 -2005

SPEC: System Performance Evaluation Cooperative The most popular and industry-standard set of CPU benchmarks.

SPEC: System Performance Evaluation Cooperative The most popular and industry-standard set of CPU benchmarks. • SPECmarks, 1989: – 10 programs yielding a single number (“SPECmarks”). • SPEC 92, 1992: – SPECInt 92 (6 integer programs) and SPECfp 92 (14 floating point programs). • SPEC 95, 1995: – SPECint 95 (8 integer programs): • go, m 88 ksim, gcc, compress, li, ijpeg, perl, vortex – SPECfp 95 (10 floating-point intensive programs): • tomcatv, swim, su 2 cor, hydro 2 d, mgrid, applu, turb 3 d, apsi, fppp, wave 5 – Performance relative to a Sun Super. Spark I (50 MHz) which is given a score of SPECint 95 = SPECfp 95 = 1 • SPEC CPU 2000, 1999: – CINT 2000 (11 integer programs). CFP 2000 (14 floating-point intensive programs) – Performance relative to a Sun Ultra 5_10 (300 MHz) which is given a score of SPECint 2000 = SPECfp 2000 = 100 All based on execution time and give speedup over a reference CPU EECC 550 - Shaaban #18 Lec # 3 Winter 2005 12 -6 -2005

SPEC 95 Programs application domain: Engineering and scientific computation Integer Floating Point Resulting Performance

SPEC 95 Programs application domain: Engineering and scientific computation Integer Floating Point Resulting Performance relative to a Sun Super. Spark I (50 MHz) which is given a score of SPECint 95 = SPECfp 95 = 1 EECC 550 - Shaaban #19 Lec # 3 Winter 2005 12 -6 -2005

Sample SPECint 95 (Integer) Results Source URL: http: //www. macinfo. de/bench/specmark. html Sun Super.

Sample SPECint 95 (Integer) Results Source URL: http: //www. macinfo. de/bench/specmark. html Sun Super. Spark I (50 MHz) score = 1 EECC 550 - Shaaban #20 Lec # 3 Winter 2005 12 -6 -2005

Sample SPECfp 95 (Floating Point) Results Source URL: http: //www. macinfo. de/bench/specmark. html Sun

Sample SPECfp 95 (Floating Point) Results Source URL: http: //www. macinfo. de/bench/specmark. html Sun Super. Spark I (50 MHz) score = 1 EECC 550 - Shaaban #21 Lec # 3 Winter 2005 12 -6 -2005

SPEC CPU 2000 Programs Benchmark Language Descriptions CINT 2000 (Integer) 164. gzip 175. vpr

SPEC CPU 2000 Programs Benchmark Language Descriptions CINT 2000 (Integer) 164. gzip 175. vpr 176. gcc 181. mcf 186. crafty 197. parser 252. eon 253. perlbmk 254. gap 255. vortex 256. bzip 2 300. twolf C C C C++ C C Compression FPGA Circuit Placement and Routing C Programming Language Compiler Combinatorial Optimization Game Playing: Chess Word Processing Computer Visualization PERL Programming Language Group Theory, Interpreter Object-oriented Database Compression Place and Route Simulator CFP 2000 (Floating Point) 168. wupwise 171. swim 172. mgrid 173. applu 177. mesa 178. galgel 179. art 183. equake 187. facerec 188. ammp 189. lucas 191. fma 3 d 200. sixtrack 301. apsi Fortran 77 C Fortran 90 Fortran 77 Physics / Quantum Chromodynamics Shallow Water Modeling Multi-grid Solver: 3 D Potential Field Parabolic / Elliptic Partial Differential Equations 3 -D Graphics Library Computational Fluid Dynamics Image Recognition / Neural Networks Seismic Wave Propagation Simulation Image Processing: Face Recognition Computational Chemistry Number Theory / Primality Testing Finite-element Crash Simulation High Energy Nuclear Physics Accelerator Design Meteorology: Pollutant Distribution Programs application domain: Engineering and scientific computation Source: http: //www. spec. org/osg/cpu 2000/ EECC 550 - Shaaban #22 Lec # 3 Winter 2005 12 -6 -2005

Top 20 SPEC CPU 2000 Results (As of March 2002) Top 20 SPECint 2000

Top 20 SPEC CPU 2000 Results (As of March 2002) Top 20 SPECint 2000 Top 20 SPECfp 2000 # MHz Processor int peak int base MHz 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 1300 2200 1667 1000 1400 1050 1533 750 833 1400 833 600 675 900 552 750 700 800 400 POWER 4 Pentium 4 Xeon Athlon XP Alpha 21264 C Pentium III Ultra. SPARC-III Cu Athlon MP PA-RISC 8700 Alpha 21264 B Athlon Alpha 21264 A MIPS R 14000 SPARC 64 GP Ultra. SPARC-III PA-RISC 8600 POWER RS 64 -IV Pentium III Xeon Itanium MIPS R 12000 814 811 810 724 679 664 610 609 604 571 554 533 500 478 467 441 439 438 365 353 790 788 697 621 648 537 587 568 497 495 511 483 449 438 417 409 431 358 328 1300 1050 2200 833 800 833 1667 750 1533 600 675 900 1400 500 450 500 400 Processor POWER 4 Alpha 21264 C Ultra. SPARC-III Cu Pentium 4 Xeon Pentium 4 Alpha 21264 B Itanium Alpha 21264 A Athlon XP PA-RISC 8700 Athlon MP MIPS R 14000 SPARC 64 GP Ultra. SPARC-III Athlon Pentium III PA-RISC 8600 POWER 3 -II Alpha 21264 MIPS R 12000 Performance relative to a Sun Ultra 5_10 (300 MHz) which is given a score of SPECint 2000 = SPECfp 2000 = 100 Source: http: //www. aceshardware. com/SPECmine/top. jsp fp peak 1169 960 827 802 801 784 701 644 642 581 547 529 509 482 458 456 440 433 422 407 fp base 1098 776 701 779 643 701 571 596 526 504 499 371 427 426 437 397 426 383 382 EECC 550 - Shaaban #23 Lec # 3 Winter 2005 12 -6 -2005

 • Computer Performance Measures : MIPS (Million Instructions Per Second) Rating For a

• Computer Performance Measures : MIPS (Million Instructions Per Second) Rating For a specific program running on a specific CPU the MIPS rating is a measure of how many millions of instructions are executed per second: MIPS Rating = Instruction count / (Execution Time x 106) = Instruction count / (CPU clocks x Cycle time x 106) = (Instruction count x Clock rate) / (Instruction count x CPI x 106) = Clock rate / (CPI x 106) • Major problem with MIPS rating: As shown above the MIPS rating does not account for the count of instructions executed (I). – A higher MIPS rating in many cases may not mean higher performance or better execution time. i. e. due to compiler design variations. • In addition the MIPS rating: – Does not account for the instruction set architecture (ISA) used. • Thus it cannot be used to compare computers/CPUs with different instruction sets. – Easy to abuse: Program used to get the MIPS rating is often omitted. • Often the Peak MIPS rating is provided for a given CPU which is obtained using a program comprised entirely of instructions with the lowest CPI for the given CPU design which does not represent real programs. EECC 550 - Shaaban #24 Lec # 3 Winter 2005 12 -6 -2005

Computer Performance Measures : MIPS (Million Instructions Per Second) Rating • Under what conditions

Computer Performance Measures : MIPS (Million Instructions Per Second) Rating • Under what conditions can the MIPS rating be used to compare performance of different CPUs? • The MIPS rating is only valid to compare the performance of different CPUs provided that the following conditions are satisfied: 1 The same program is used (actually this applies to all performance metrics) 2 The same ISA is used 3 The same compiler is used Þ (Thus the resulting programs used to run on the CPUs and obtain the MIPS rating are identical at the machine code level including the same instruction count) EECC 550 - Shaaban #25 Lec # 3 Winter 2005 12 -6 -2005

Compiler Variations, MIPS & Performance: An Example • For a machine (CPU) with instruction

Compiler Variations, MIPS & Performance: An Example • For a machine (CPU) with instruction classes: Instruction class A B C CPI 1 2 3 • For a given high-level language program, two compilers produced the following executed instruction counts: Code from: Compiler 1 Compiler 2 Instruction counts (in millions) for each instruction class A B C 5 1 1 10 1 1 • The machine is assumed to run at a clock rate of 100 MHz. EECC 550 - Shaaban #26 Lec # 3 Winter 2005 12 -6 -2005

Compiler Variations, MIPS & Performance: An Example (Continued) MIPS = Clock rate / (CPI

Compiler Variations, MIPS & Performance: An Example (Continued) MIPS = Clock rate / (CPI x 106) = 100 MHz / (CPI x 106) CPI = CPU execution cycles / Instructions count CPU time = Instruction count x CPI / Clock rate • For compiler 1: – CPI 1 = (5 x 1 + 1 x 2 + 1 x 3) / (5 + 1) = 10 / 7 = 1. 43 – MIPS Rating 1 = 100 / (1. 428 x 106) = 70. 0 MIPS – CPU time 1 = ((5 + 1) x 106 x 1. 43) / (100 x 106) = 0. 10 seconds • For compiler 2: – CPI 2 = (10 x 1 + 1 x 2 + 1 x 3) / (10 + 1) = 15 / 12 = 1. 25 – MIPS Rating 2 = 100 / (1. 25 x 106) = 80. 0 MIPS – CPU time 2 = ((10 + 1) x 106 x 1. 25) / (100 x 106) = 0. 15 seconds MIPS rating indicates that compiler 2 is better while in reality the code produced by compiler 1 is faster EECC 550 - Shaaban #27 Lec # 3 Winter 2005 12 -6 -2005

MIPS (The ISA not the metric) Loop Performance Example High Memory $6 points here

MIPS (The ISA not the metric) Loop Performance Example High Memory $6 points here For the loop: X[999] X[998] . . for (i=0; i<1000; i=i+1){ x[i] = x[i] + s; } $2 initially MIPS assembly code is given by: loop: lw addi lw add sw addi bne $3, $6, $4, $5, $2, $6, 8($1) $2, 4000 0($2) $4, $3 0($2) $2, 4 $2, loop Last element to compute points here ; ; ; ; X[0] Low Memory First element to compute load s in $3 $6 = address of last element + 4 load x[i] in $4 $5 has x[i] + s store computed x[i] increment $2 to point to next x[ ] element last loop iteration reached? The MIPS code is executed on a specific CPU that runs at 500 MHz (clock cycle = 2 ns = 2 x 10 -9 seconds) with following instruction type CPIs : For this MIPS code running on this CPU find: Instruction type ALU Load Store Branch CPI 4 5 7 3 1 - Fraction of total instructions executed for each instruction type 2 - Total number of CPU cycles 3 - Average CPI 4 - Fraction of total execution time for each instructions type 5 - Execution time 6 - MIPS rating , peak MIPS rating for this CPU X[ ] array of words in memory, base address in $2 , s a constant word value in memory, address in $1 EECC 550 - Shaaban #28 Lec # 3 Winter 2005 12 -6 -2005

MIPS (The ISA) Loop Performance Example (continued) • • The code has 2 instructions

MIPS (The ISA) Loop Performance Example (continued) • • The code has 2 instructions before the loop and 5 instructions in the body of the loop which iterates 1000 times, Thus: Total instructions executed, I = 5 x 1000 + 2 = 5002 instructions 1 Number of instructions executed/fraction Fi for each instruction type: – – ALU instructions = 1 + 2 x 1000 = 2001 Load instructions = 1 + 1 x 1000 = 1001 Store instructions = 1000 Branch instructions = 1000 CPIALU = 4 Fraction. ALU = FALU = 2001/5002 = 0. 4 = 40% CPILoad = 5 Fraction. Load = FLoad = 1001/5002= 0. 2 = 20% CPIStore = 7 Fraction. Store = FStore = 1000/5002 = 0. 2 = 20% CPIBranch = 3 Fraction. Branch= FBranch = 1000/5002= 0. 2 = 20% 2 3 4 = 2001 x 4 + 1001 x 5 + 1000 x 7 + 1000 x 3 = 23009 cycles Average CPI = CPU clock cycles / I = 23009/5002 = 4. 6 Instruction type Fraction of execution time for each instruction type: – – CPI ALU Load Store Branch Fraction of time for ALU instructions = CPIALU x FALU / CPI= 4 x 0. 4/4. 6 = 0. 348 = 34. 8% Fraction of time for load instructions = CPIload x Fload / CPI= 5 x 0. 2/4. 6 = 0. 217 = 21. 7% Fraction of time for store instructions = CPIstore x Fstore / CPI= 7 x 0. 2/4. 6 = 0. 304 = 30. 4% Fraction of time for branch instructions = CPIbranch x Fbranch / CPI= 3 x 0. 2/4. 6 = 0. 13 = 13% 4 5 7 3 5 Execution time = I x CPI x C = CPU cycles x C = 23009 x 2 x 10 -9 = = 4. 6 x 10 -5 seconds = 0. 046 msec = 46 usec 6 MIPS rating = Clock rate / (CPI x 106) = 500 / 4. 6 = 108. 7 MIPS – The CPU achieves its peak MIPS rating when executing a program that only has instructions of the type with the lowest CPI. In this case branches with CPIBranch = 3 – Peak MIPS rating = Clock rate / (CPIBranch x 106) = 500/3 = 166. 67 MIPS EECC 550 - Shaaban #29 Lec # 3 Winter 2005 12 -6 -2005

Computer Performance Measures : MFLOPS (Million FLOating-Point Operations Per Second) • • A floating-point

Computer Performance Measures : MFLOPS (Million FLOating-Point Operations Per Second) • • A floating-point operation is an addition, subtraction, multiplication, or division operation applied to numbers represented by a single or a double precision floating-point representation. MFLOPS, for a specific program running on a specific computer, is a measure of millions of floating point-operation (megaflops) per second: MFLOPS = Number of floating-point operations / (Execution time x 106 ) • • • MFLOPS rating is a better comparison measure between different machines (applies even if ISAs are different) than the MIPS rating. – Applicable even if ISAs are different Program-dependent: Different programs have different percentages of floating -point operations present. i. e compilers have no floating- point operations and yield a MFLOPS rating of zero. Dependent on the type of floating-point operations present in the program. – Peak MFLOPS rating for a CPU: Obtained using a program comprised entirely of the simplest floating point instructions (with the lowest CPI) for the given CPU design which does not represent real floating point programs. EECC 550 - Shaaban #30 Lec # 3 Winter 2005 12 -6 -2005

Quantitative Principles of Computer Design • Amdahl’s Law: The performance gain from improving some

Quantitative Principles of Computer Design • Amdahl’s Law: The performance gain from improving some portion of a computer is calculated by: Speedup = Performance for entire task using the enhancement Performance for the entire task without using the enhancement or Speedup = Execution time without the enhancement Execution time for entire task using the enhancement EECC 550 - Shaaban #31 Lec # 3 Winter 2005 12 -6 -2005

Performance Enhancement Calculations: Amdahl's Law • The performance enhancement possible due to a given

Performance Enhancement Calculations: Amdahl's Law • The performance enhancement possible due to a given design improvement is limited by the amount that the improved feature is used • Amdahl’s Law: Performance improvement or speedup due to enhancement E: Execution Time without E Speedup(E) = -------------------Execution Time with E Performance with E = ----------------Performance without E – Suppose that enhancement E accelerates a fraction F of the execution time by a factor S and the remainder of the time is unaffected then: Execution Time with E = ((1 -F) + F/S) X Execution Time without E Hence speedup is given by: Execution Time without E 1 Speedup(E) = ----------------------------- = ----------((1 - F) + F/S) X Execution Time without E (1 - F) + F/S F (Fraction of execution time enhanced) refers to original execution time before the enhancement is applied EECC 550 - Shaaban #32 Lec # 3 Winter 2005 12 -6 -2005

Pictorial Depiction of Amdahl’s Law Enhancement E accelerates fraction F of original execution time

Pictorial Depiction of Amdahl’s Law Enhancement E accelerates fraction F of original execution time by a factor of S Before: Execution Time without enhancement E: (Before enhancement is applied) • shown normalized to 1 = (1 -F) + F =1 Unaffected fraction: (1 - F) Affected fraction: F Unchanged Unaffected fraction: (1 - F) F/S After: Execution Time with enhancement E: Execution Time without enhancement E 1 Speedup(E) = --------------------------- = ---------Execution Time with enhancement E (1 - F) + F/S EECC 550 - Shaaban #33 Lec # 3 Winter 2005 12 -6 -2005

Performance Enhancement Example • For the RISC machine with the following instruction mix given

Performance Enhancement Example • For the RISC machine with the following instruction mix given earlier: Op ALU Load Store Freq 50% 20% 10% Cycles 1 5 3 CPI(i). 5 1. 0. 3 % Time 23% 45% 14% CPI = 2. 2 Branch 20% 2. 4 18% • If a CPU design enhancement improves the CPI of load instructions from 5 to 2, what is the resulting performance improvement from this enhancement: Fraction enhanced = F = 45% or. 45 Unaffected fraction = 1 - F = 100% - 45% = 55% or. 55 Factor of enhancement = S = 5/2 = 2. 5 Using Amdahl’s Law: 1 1 Speedup(E) = --------------------- = (1 - F) + F/S. 55 +. 45/2. 5 1. 37 EECC 550 - Shaaban #34 Lec # 3 Winter 2005 12 -6 -2005

An Alternative Solution Using CPU Equation Op ALU Load Store Freq 50% 20% 10%

An Alternative Solution Using CPU Equation Op ALU Load Store Freq 50% 20% 10% Cycles 1 5 3 CPI(i). 5 1. 0. 3 % Time 23% 45% 14% CPI = 2. 2 Branch 20% 2. 4 18% • If a CPU design enhancement improves the CPI of load instructions from 5 to 2, what is the resulting performance improvement from this enhancement: Old CPI = 2. 2 New CPI =. 5 x 1 +. 2 x 2 +. 1 x 3 +. 2 x 2 = 1. 6 Original Execution Time Speedup(E) = -----------------New Execution Time Instruction count x old CPI x clock cycle = --------------------------------Instruction count x new CPI x clock cycle old CPI = ------ = new CPI 2. 2 ----1. 6 = 1. 37 Which is the same speedup obtained from Amdahl’s Law in the first solution. EECC 550 - Shaaban #35 Lec # 3 Winter 2005 12 -6 -2005

Performance Enhancement Example • A program runs in 100 seconds on a machine with

Performance Enhancement Example • A program runs in 100 seconds on a machine with multiply operations responsible for 80 seconds of this time. By how much must the speed of multiplication be improved to make the program four times faster? Desired speedup = 4 = ® ® ® 100 --------------------------Execution Time with enhancement Execution time with enhancement = 100/4 = 25 seconds = (100 - 80 seconds) + 80 seconds / S 25 seconds = 20 seconds + 80 seconds / S 5 = 80 seconds / S S = 80/5 = 16 Alternatively, it can also be solved by finding enhanced fraction of execution time: F = 80/100 =. 8 and then solving Amdahl’s speedup equation for desired enhancement factor S 1 Speedup(E) = --------- = 4 = (1 - F) + F/S 1 1 --------- = -------(1 -. 8) +. 8/S. 2 +. 8/s Hence multiplication should be 16 times faster to get an overall speedup of 4. Solving for S gives S= 16 EECC 550 - Shaaban #36 Lec # 3 Winter 2005 12 -6 -2005

Performance Enhancement Example • For the previous example with a program running in 100

Performance Enhancement Example • For the previous example with a program running in 100 seconds on a machine with multiply operations responsible for 80 seconds of this time. By how much must the speed of multiplication be improved to make the program five times faster? Desired speedup = 5 = ® 100 --------------------------Execution Time with enhancement Execution time with enhancement = 100/5 = 20 seconds = (100 - 80 seconds) + 80 seconds / s 20 seconds = 20 seconds + 80 seconds / s ® 0 = 80 seconds / s No amount of multiplication speed improvement can achieve this. EECC 550 - Shaaban #37 Lec # 3 Winter 2005 12 -6 -2005

Extending Amdahl's Law To Multiple Enhancements • Suppose that enhancement Ei accelerates a fraction

Extending Amdahl's Law To Multiple Enhancements • Suppose that enhancement Ei accelerates a fraction Fi of the original execution time by a factor Si and the remainder of the time is unaffected then: Unaffected fraction Note: All fractions Fi refer to original execution time before the enhancements are applied. EECC 550 - Shaaban #38 Lec # 3 Winter 2005 12 -6 -2005

Amdahl's Law With Multiple Enhancements: Example • Three CPU performance enhancements are proposed with

Amdahl's Law With Multiple Enhancements: Example • Three CPU performance enhancements are proposed with the following speedups and percentage of the code execution time affected: Speedup 1 = S 1 = 10 Speedup 2 = S 2 = 15 Speedup 3 = S 3 = 30 • • • Percentage 1 = F 1 = 20% Percentage 1 = F 2 = 15% Percentage 1 = F 3 = 10% While all three enhancements are in place in the new design, each enhancement affects a different portion of the code and only one enhancement can be used at a time. What is the resulting overall speedup? Speedup = 1 / [(1 -. 2 -. 15 -. 1) +. 2/10 +. 15/15 +. 1/30)] = 1/ [. 55 +. 0333 ] = 1 /. 5833 = 1. 71 EECC 550 - Shaaban #39 Lec # 3 Winter 2005 12 -6 -2005

Pictorial Depiction of Example Before: Execution Time with no enhancements: 1 Unaffected, fraction: .

Pictorial Depiction of Example Before: Execution Time with no enhancements: 1 Unaffected, fraction: . 55 S 1 = 10 F 1 =. 2 S 2 = 15 S 3 = 30 F 2 =. 15 F 3 =. 1 / 15 / 10 / 30 Unchanged Unaffected, fraction: . 55 After: Execution Time with enhancements: . 55 +. 02 +. 01 +. 00333 =. 5833 Speedup = 1 /. 5833 = 1. 71 Note: All fractions refer to original execution time. EECC 550 - Shaaban #40 Lec # 3 Winter 2005 12 -6 -2005