Evaluation of computer performance

Explained response time and CPU time, difference between them, Explained clock cycle time, clock frequency, clock rate, cycle per instruction(CPI), million instruction per second(MIPS), etc. Describe Amdahl's law with numerical examples

1. Evaluation of Computer performance Dr. Prasenjit Dey

2. Performance The performance of a computer is defined by its speed of processing of the instructions Faster the machine, better the performance What are the factors influences the performance of a computer?

3. Performance Metrics Response Time The total time required by an instruction to complete From start time of the instruction to the finish time It is also known as elapsed time Response time = Memory access time + waiting time + CPU time Throughput : Number of instructions completed in a specific time

4. Execution Time CPU time It is the time when an instruction utilizes the CPU Time duration when an instruction is in running state It is also known as execution time time spent to execute the lines of codes in a program CPU time = user CPU time + system CPU time elapsed time = user CPU time + system CPU time + waiting time Computer performance is measured on the basis of CPU time

5. Comparison of Performance The performance of a machine is inversely proportional to the CPU time Performance = 1/CPU time If machine A is n times faster than machine B then Performance(A)/performance(B) = n

6. Clock Cycles CPU time is measured with the help of CPU cycles Each instruction uses a certain number of CPU cycles to execute an instruction The time taken by an instruction is (number of CPU cycles) x (time to execute 1 CPU cycle) If all instructions take equal number of cycles then, The time required to execute a program of N instructions is N x (number of CPU cycle) x (time to execute 1 CPU cycle) Avg. program execution time (cycles/program) x (seconds/cycle) seconds/cycle  cycle time Clock rate/frequency  cycle/seconds , used more often

7. Performance of a Program CPU execution time of a program is (No . of CPU cycles) x (cycle time) (No . of CPU cycles)/(clock rate) An efficient program should require less number of CPU cycles and/or high clock rate clock rate  1 Hz. = 1 cycle/sec, If clock rate = 200 Mhz ? Time to execute a program = 1/(200*106) = 5*10 -3 *10 -6 = 5*10 -9 sec

8. Performance of a Program In computer, different instructions use different amounts of CPU cycles Bitwise instructions require less number of CPU cycles, whereas multiplicative instructions, floating point instructions require more number of CPU cycles, To compute the execution time of a program, one should compute the avg . CPU cycle time or clock rate

9. CPI: Cycles Per Instruction For a given program Compute number of instructions Compute total number of cycles to execute all instructions Divide total number of cycles by number of instructions cycles per instruction (CPI) The avg. amount of time required to execute an instruction in a program Measured in terms of MIPS (millions of instructions per second)

10. CPU Execution Time CPU execution time of a program is N x ( )/N x cycle time , Here N is the total number of instructions in a program Average CPI x instruction count x cycle time ( Average CPI x instruction count ) / clock rate  

11. Problem 1 Let there are 2 machines, machine A and machine B, which execute the sample program. Where, Machine A clock cycle is 10ns and CPI is 2 Machine B clock cycle is 20ns and CPI is 1.2 Which one has better performance? Computation for machine A Avg. CPI x clock cycle time = 2 x 10 = 20ns Computation for machine B Avg. CPI x clock cycle time = 1.2 x 20 = 24ns Machine A has better performance for this program

12. Problem 2 Let us consider that two programs that contain three different types of instructions: type A, type B, and type C Type A, type B, and type C instructions require 4, 3, 5 cycles respectively. Suppose program 1, uses 1 type A instructions, 2 type B instructions, and 2 type C instructions Suppose program 2, uses 1 type A instructions, 4 type B instructions, and 1 type C instructions Then which program is faster? Computation for program 1 1*4 + 2*3 + 2*5 = 20cycles Computation for program 2 1*4 + 4*3 + 1*5 = 21cycles Program is 1 faster

13. Problem 3 Suppose your program consists of 2500 instructions . The proportion of different kinds of instructions in the program is as follow: Data transfer instruction 50% , arithmetic instruction 30% and branching related instructions 20%. The cycles consumed by these types of instructions are 2, 5, and 10 respectively. What will be the execution time for a 4 GHz processor to execute your program? Avg CPI = 0.5*2 + 0.3*5 + 0.2*10 = 4.5 Avg execution time = 2500* (4.5/4 *10 6 )Sec = 2500*1.125 *10 -6 Sec = 2.8125ms

14. Amdahl's Law I t computes the overall performance enhancement when the performance of a fraction of code(program) is enhanced Overall performance enhancement  Overall Speedup = old execution time/new execution time = Fraction_enhanced T he sub part of the code/program which has been enhanced by using some hardware or compiler Speedup The performance gain in the enhanced fraction of code  

15. Amdahl's Law: example Let a program contains 10 multiplicative instructions and 10 additive instructions. Each multiplicative instruction takes 50ns and each additive instructions take 10ns. Now by adding some hardware, we enhanced the performance of multiplicative instructions and complete a multiplicative instructions in 20ns. What will be the overall speedup? Old execution time = 50*10 + 10*10 =600ns New execution time = = = = =  

16. Problems on Amdahl's Law With the use of Amdahl’s law, conclude among the given options which possible improvement is the best one Possible improvement Branch CPI can be decreased from 4 to 3 Increase clock frequency from 2 to 2.3GHz Store CPI can be decreased from 3 to 2 Instruction type Frequency CPI ALU 40% 1 Branch 20% 4 Load 30% 2 Store 10% 3

17. Solution Avg. CPI = ( 0.4 *1 + 0.2 *4 + 0.3 *2 + 0.1 *3) = 2.1 Clock rate = 2GHz Avg. instruction execution time = Avg. CPI/clock rate = 2.1/2 = 1.05*10 -6 sec Case A Current execution time = (0.4*1 + 0.2*3 + 0.3*2 + 0.1*3 )/ 2 = 0.95*10 -6 sec Case B Increase clock frequency from 2 to 2.3 Current execution time = (0.4*1 + 0.2*4 + 0.3*2 + 0.1*3 )/ (2.3) = 0.91*10 -6 sec Case C Current execution time = (0.4*1 + 0.2*4 + 0.3*2 + 0.1*2 )/ 2 = 1*10 -6 sec

18. Conclusion Performance of a machine measured with the help of clock cycles Each instructions require different clock cycles, need to compute average clock cycle per instruction(CPI) Overall performance gain can be achieved by enhancing a fraction of a program

19. Thank you