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 ( 𝐢=𝟏
𝐍
𝐜𝐲𝐜𝐥𝐞 𝐭𝐢𝐦𝐞 𝐟𝐨𝐫 𝐢𝐧𝐬𝐭𝐫𝐜𝐭𝐢𝐨𝐧 𝐢)/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 *106 )Sec
= 2500*1.125 *10-6 Sec
= 2.8125ms

14. Amdahl's Law
 It 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
=
1
(1 – fraction_enhanced) + fraction_enhanced
speedup
 Fraction_enhanced
 The 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 =
1
(1 – fraction_enhanced) + fraction_enhanced
speedup
=
1
1−
500
600
+
500
600
500
200
=
1
1
6
+(
1
3
)
=
1
3
6
=
3
6
= 2

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
A. Branch CPI can be decreased from 4 to 3
B. Increase clock frequency from 2 to 2.3GHz
C. 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