1. Parameters that affect it How to improve it and by how much
Performance
Parameters that affect it
How to improve it and by how much

2. Performance
Response time/Execution time/Turn-around time/Latency/Wall-clock time/Elapsed time
Time between the start and completion of a task.
Users are normally interested in reducing this parameter.
Includes
CPU execution time for this task.
Time spent waiting for I/O.
Time spent waiting for CPU (in multi-programming systems)
Time due to OS overhead.

3. Increase performance implies decrease execution time
Throughput
Total amount of work done in a given time.
A system administrator would like to increase the throughput.
Increase performance implies
increase throughput,
decrease execution time

4. Performance vs Execution time
Performance (P) = 1/ tE
If X is ‘n’ times faster than Y, it implies:
Px/Py = n = tEy/tEx
i.e. Y times takes ‘n’ times longer than X.

5. Example 1
If a machine A runs a program in 10 seconds and machine B runs the same program in 15 seconds, how much faster is A than B?
n = PA/PB
= tB/tA
= 15/10
= 1.5
A is 1.5 times faster than B

6. CPU execution time/CPU time
Components of the execution time of a program:
CPU execution time
User CPU time
System CPU time
I/O time
Time spent running other programs

7. System performance refers to Response time.
CPU performance refers to CPU execution time.

8. CPU Performance equation
CPU time = CPU clock cycles * cycle time
= CPU clock cycles / clock rate
CPU clock cycles = IC * CPI
IC: instruction count (number of instructions per program)
CPI: average cycles per instruction
CPU time = IC * CPI * cycle time
seconds instructions clock cycles seconds
program program instruction clock cycle
CPU clock cycles = Σi (CPIi * ICi)
ICi : count of instructions of class i
CPIi : cycles that takes to execute instructions of class i = * *

9. Example 2
Computer A has a 400 MHz clock and runs a program in 10 seconds. Computer B has a 800 MHz clock and runs the same program in 6 seconds. The increase in the clock rate of B implies an increase in the number of clock cycles required by B. Determine by how much the number of clock cycles in computer B has increased to allow for the higher clock rate.
CPU time = CPU clock cycles * cycle time
= CPU clock cycles / clock rate
10 = CPU clock cycles for A * 1/400 MHz
= CPU clock cycles for B * 1/800 MHz
CPU clock cycles for B = 1/400 * = 1.2
CPU clock cycles for A /800 * 10

10. Example 3
Given a machine M1 with a clock rate 2.7GHz, how long will a program P1 take to run if there are 4,298 instructions and each instruction takes an average of 2.9 cycles.
seconds = instructions * clock cycles * seconds
program program instruction clock cycle
= x x 1/(2.7 x 109)
CPU time = 4.26 x 10-6 seconds

11. CPU time = IC * CPI * cycle time
Example 4
Given P1 and M1 from previous problem, what is the relative performance of M1 with respect to a machine M2 having clock rate 3.1 GHz running P1, where each instruction of P1 on M2 requires 3.2 cycles instead of 2.9 cycles?
CPU time = IC * CPI * cycle time
Perf of M1 / Perf of M2 = (IC2 * CP2 /Clock rate2)
(IC1 * CPI1 /Clock rate1)
= (CPI2 * Clock rate1)
(CPI1 * Clock rate2)
= (3.2 * 2.7GHz) / (2.9 * 3.1GHz) =
M1 is 1.04 times faster than M2.

12. Example 5
Given a machine M1 with clock rate 2.9 GHz, how long will a program P1 take to run that has 5,728 instructions and the following instruction mix and CPI?
Instruction Type % Incidence CPI
A | | 9
B | | 10
C | | 13
D | | 7
E | rem | 5

13. Example 5 Average CPI * Instruction Count of P1
Exec. Time of P1 on M1 = _______________________________________
Clock Rate of M1
CPU Clock cycles = (19*9+10*10+38*13+14*7+19*5) = 958
CPI = CPU Clock cycles/ Number of instructions = 958/100 = 9.58
9.58 * 5,728
Exec. Time of P1 on M1 = ______________
2.9 * 10^9
Exec. Time of P1 on M1 = * 10-5 seconds

14. Which code sequence executes the most instructions?
Example 6
Comparing two compiler code segments
Which code sequence executes the most instructions?
Which will be faster?
Instruction class i
CPIi for instruction class i A 1 B 2 C 3
Code sequence
Instruction counts (ICi) for instruction class A B C 1 2 4

15. Instruction count
S1 : = 5
S2 : = 6
S1 executes fewer instructions than S2
To determine CPI
CPU clock cycles = Σi (CPIi * ICi)
S1 : CPU clock cycles = (2 x 1) + (1 x 2) + (2 x 3) = 10 cycles
S2 : CPU clock cycles = (4 x 1) + (1 x 2) + (1 x 3) = = 9 cycles
S2 requires fewer clock cycles than S1.
b. CPI = CPU clock cycles
S1 : CPI = 10/5 = 2
S2 : CPI = 9/6 = 1.5

16. Scope of Performance Sources
CPU time = IC * CPI * Cycle time
Program
Compiler
ISA
Organization
Hardware

17. Evaluating performance
Ideal situation: known programs (workload)
Benchmarks
Real programs
Synthetic benchmarks
Risk: adjust design to benchmark requirements

18. Relative performance Total time: Σ exec timei
Arithmetic mean: AM = 1/n * Σ exec timei
Weighted mean: WM = Σ wi * exec timei
Geometric mean: GM = (Π exec time ratioi)1/n

19. Arithmetic Mean
Program
System A Execution Time
System B Execution Time
System C Execution Time V 50
100
500 W
200
400
600 X
250 Y
800 Z
5000
4100
3500
Average

20. Weighted Arithmetic Mean
Program
Execution Frequency
System A Execution Time
System C Execution Time V
50% 50
500 W
30%
200
600 X
10%
250 Y 5%
400
800 Z
5000
3500
Weighted Average
seconds seconds

21. Weighted Arithmetic Mean
Program
Execution Frequency
System A Execution Time
Execution Frequency (original) V
25% 50
50% W 5%
200
30% X
10%
250 Y
400 Z
55%
5000
Weighted Average
seconds
380 seconds

22. System B execution time System C execution time
Geometric Mean
Program
System A
Execution Time
Normalized to B
System B execution time
System C execution time V 50
100 1
500
0.2 W
200
400
600
0.667 X
250 Y
800 Z
5000
4100
3500
1.714
Geometric Mean
0.6898 2
0.82
1.6733

23. MIPS/MFLOPS
MIPS = Instruction Count
Execution Time x 106

24. Amdahl’s law
The performance improvement that can be obtained from using a faster mode of execution is limited by the fraction of time that the faster mode can be used.

25. Amdahl’s law Overall Speedup = PEnhanced/PUnenhanced
= tUnenhanced/tEnhanced
Depends on two factors:
f – fraction of execution time that can be enhanced.
s – speedup obtained for the fraction

26. Amdahl’s equation
Overall Speedup =

27. Example 7
Let’s say that your processes spend 70% of their time running in the CPU and 30% waiting for service from the disk. You have the option to upgrade to a CPU that is 50% faster than your current CPU or to a set of disk drives that promise to be two and a half times faster than your current drives. Which upgrade would you choose?

28. CPU upgrade :
f = 0.7 s = 1.5
Overall speedup =
Disk drive upgrade :
f = 0.3 s = 2.5
Overall speedup =
CPU upgrade cost = $10,000 Disk drives upgrade cost = $7,000 and if cost is a concern, which would you choose?
1% of CPU upgrade => $10000/30 = $333
1% of disk drive upgrade => $7000/22 = $318