EC 413 Computer Organization

Transcription

1 EC 413 Computer Organzaton CPU Performance Evaluaton Prof. Mchel A. Knsy

2 Performance Measurement Processor performance: Executon tme Area Logc complexty Power Tme = Instructons Cycles Tme Program Program * Instructon * Cycle In ths class we wll focus on Executon tme

3 Compler Effects on Performance CPU tme = Instructon count x CPI / Clock rate A machne runnng at 100 MHz has these nstructon classes Instructon class CPI A 1 B 2 C 3 For a gven program, two complers produced the followng nstructon counts Instructon counts (n mllons) for each nstructon class Code from: A B C Compler Compler

4 Compler Effects on Performance CPU tme = Instructon count x CPI / Clock rate For compler 1: CPI 1 = (5 x x x 3) / ( ) = 10 / 7 = 1.43 CPU tme 1 = (( ) x 10 6 x 1.43) / (100 x 10 6 ) = 1 second For compler 2: CPI 2 = (10 x x x 3) / ( ) = 15 / 12 = 1.25 CPU tme 2 = (( ) x 10 6 x 1.25) / (100 x 10 6 ) = 1.5 seconds

5 Processor Performance Speed Up Equatons for Ppelnng CPI ppelned = Ideal CPI + Average Stall cycle per Instructon Speedup = Ideal CPI X Ppelne Depth Ideal CPI + Ppelne stall CPI X Clock Cycle Unppelned Clock Cycle Ppelned If Ideal CPI = 1 Speed Up <= Ppelne Depth Speedup = Ppelne Depth 1 + Ppelne stall CPI X Clock Cycle Unppelned Clock Cycle Ppelned

6 Illustratve Example We want to compare the performance of two machnes. Whch machne s faster? Machne A: Dual ported memory - so there are no memory stalls Machne B: Sngle ported memory, but ts ppelned mplementaton has a 1.05 tmes faster clock rate Assumptons Ideal CPI = 1 for both Loads are 40% of nstructons executed

7 Illustratve Example We want to compare the performance of two machnes. Whch machne s faster? Machne A: Dual ported memory - so there are no memory stalls Machne B: Sngle ported memory, but ts ppelned mplementaton has a 1.05 tmes faster clock rate Assumptons Ideal CPI = 1 for both Loads are 40% of nstructons executed Machne A speed = Ppelne Depth/(1 + 0) x (clock unppelne /clock ppelne ) = Ppelne Depth Machne B speed = Ppelne Depth/( x 1) x (clock unppelne /clock ppelne ) = (Ppelne Depth/1.4) x (clock unppelne /(1.05 *clock unppelne )) = 0.68 x Ppelne Depth A Speed/ B Speed = Ppelne Depth / (0.68 x Ppelne Depth) = 1.47

8 By Gene Amdahl Ths law answers the crtcal queston: How much of a speedup one can get for a gven archtectural mprovement/enhancement? The performance enhancement possble due to a gven desgn mprovement s lmted by the amount that the mproved feature s used Performance mprovement or speedup due to enhancement E Executon Tme wthout E Speedup(E) = = Executon Tme wth E Performance wth E Performance wthout E

9 By Gene Amdahl Ths law answers the crtcal queston: How much of a speedup one can get for a gven archtectural mprovement/enhancement? Suppose that enhancement E accelerates a fracton F of the executon tme by a factor S and the remander of the tme s unaffected then: Executon Tme wth E = ((1-F) + F/S) x Executon Tme wthout E Hence speedup s gven by: Executon Tme wthout E 1 Speedup(E) = = ((1 - F) + F/S) x Executon Tme wthout E (1 - F) + F/S

10 For the RISC machne wth the followng nstructon composton: Op Freq Cycles CPI() % Tme ALU 50% % Load 20% % Store 10% % Branch 20% % If a CPU desgn enhancement mproves the CPI of load nstructons from 5 to 2, what s the resultng performance mprovement from ths enhancement

11 For the RISC machne wth the followng nstructon composton: Op Freq Cycles CPI() % Tme ALU 50% % Load 20% % Store 10% % Branch 20% % If a CPU desgn enhancement mproves the CPI of load nstructons from 5 to 2, what s the resultng performance mprovement from ths enhancement Fracton enhanced = F = 45% or.45 Unaffected fracton = 100% - 45% = 55% or.55 Factor of enhancement = 5/2 = Speedup(E) = = = 1.37 (1 - F) + F/S /2.5

12 For the RISC machne wth the followng nstructon composton: Op Freq Cycles CPI() % Tme ALU 50% % Load 20% % Store 10% % Branch 20% % If a CPU desgn enhancement mproves the CPI of load nstructons from 5 to 2, what s the resultng performance mprovement from ths enhancement

13 For the RISC machne wth the followng nstructon composton: Op Freq Cycles CPI() % Tme ALU 50% % Load 20% % Store 10% % Branch 20% % If a CPU desgn enhancement mproves the CPI of load nstructons from 5 to 2, what s the resultng performance mprovement from ths enhancement Old CPI = 2.2 New CPI =.5 x x x x 2 = 1.6 Orgnal Executon Tme Speedup(E) = = New Executon Tme Instructon count x old CPI x clock cycle Instructon count x new CPI x clock cycle old CPI 2.2 = = = 1.37 new CPI 1.6

14 A program takes 100 seconds to execute on a machne wth load operatons responsble for 80 seconds of ths tme. By how much must the load operaton be mproved to make the program four tmes faster?

15 A program takes 100 seconds to execute on a machne wth load operatons responsble for 80 seconds of ths tme. By how much must the load operaton be mproved to make the program four tmes faster? 100 Desred speedup = 4 = Executon Tme wth enhancement Executon tme wth enhancement = 100 * (1/4) = 25 seconds è 25 seconds = ( seconds) + 80 seconds / n è 25 seconds = 20 seconds + 80 seconds / n è 5 = 80 seconds / n è n = 80/5 = 16 Load operaton should be 16 tmes faster to get a speedup of 4!

16 A program takes 100 seconds to execute on a machne wth load operatons responsble for 80 seconds of ths tme. By how much must the load operaton be mproved to make the program fve tmes faster?

17 A program takes 100 seconds to execute on a machne wth load operatons responsble for 80 seconds of ths tme. By how much must the load operaton be mproved to make the program fve tmes faster? 100 Desred speedup = 5 = Executon Tme wth enhancement Executon tme wth enhancement = 100 * (1/5) = 20 seconds è 20 seconds = ( seconds) + 80 seconds / n è 20 seconds = 20 seconds + 80 seconds / n è 0 = 80 seconds / n No amount of load operaton mprovement wll be able acheve ths speed

18 Multple Enhancements Suppose that enhancement E accelerates a fracton F of the executon tme by a factor S and the remander of the tme s unaffected then: Speedup = ((1 Orgnal Executon Tme F )+ F ) S 1 Speedup = (( 1- ) + å F x Orgnal Executon Tme å F S )

19 Multple Enhancements Three CPU performance enhancements are proposed wth the followng speedups and percentage of the code executon tme affected: Speedup 1 = S 1 = 10 Percentage 1 = F 1 = 20% Speedup 2 = S 2 = 15 Percentage 1 = F 2 = 15% Speedup 3 = S 3 = 30 Percentage 1 = F 3 = 10% Whle all three enhancements are n place n the new desgn, each enhancement affects a dfferent porton of the code and only one enhancement can be used at a tme. What s the resultng overall speedup? 1 Speedup = (( 1- ) + å F å F S )

20 Multple Enhancements Three CPU performance enhancements are proposed wth the followng speedups and percentage of the code executon tme affected: Speedup 1 = S 1 = 10 Percentage 1 = F 1 = 20% Speedup 2 = S 2 = 15 Percentage 1 = F 2 = 15% Speedup 3 = S 3 = 30 Percentage 1 = F 3 = 10% Whle all three enhancements are n place n the new desgn, each enhancement affects a dfferent porton of the code and only one enhancement can be used at a tme. What s the resultng overall speedup? 1 Speedup = (( 1- ) + Speedup = 1 / [( ) +.2/ /15 +.1/30)] = 1 / [ ] = 1 /.5833 = 1.71 å å F F S )

21 Key Insghts The performance of any system s constraned by the speed or capacty of the slowest pont The mpact of an effort to mprove the performance of a program s prmarly constraned by the amount of tme that the program spends n parts of the program NOT TARGETED by the effort Amdahl's Law s a statement of the maxmum theoretcal speed-up you can ever hope to acheve The actual speed-ups are always less than the speedup predcted by Amdahl's Law

22 For software and hardware engneers MUST have a very deep understandng of Amdahl's Law f they are to avod havng unrealstc performance expectatons 1. For systems folks: ths law allows you to estmate the net performance beneft a new hardware feature wll add to program executons 1. For software folks: ths law allows you to estmate the amount of parallelsm your program/algorthm can acheve before you start wrtng your parallel code

23 Memory Prmtves Next Class