1. Execution time Execution Time (processor-related) = IC x CPI x T
IC = instruction count
CPI = average number of system clock periods to execute an
instruction
T = clock period

2. Review

3. CS501 Advanced Computer Architecture
Lecture03
Dr.Noor Muhammad Sheikh

4. data transfer instructions
Example
Consider two SRC programs having three types of
instructions given as follows
Number of ..
Program 1
Program 2
data transfer instructions 2 1
control instructions 5
ALSU Instructions
Compare both the programs for the following parameters
Instruction count
Speed of execution

5. Example contd.. Instruction count IC. IC for program 1= 2+2+2=6
For execution time we can use the following SRC specifications.
ET = IC x CPI x T
ET1= (2x2)+(2x3)+(2x4)
= 18
ET2 =(5x2)+(1x3)+(1x4)
=17
Instruction Type
CPI
Control 2
ALSU 3
Data Transfer 4
Note: Since both programs are executing on the same machine, the T factor can be ignored while calculating ET.

6. lar r6,mpy ;load address of mpy lar r7, next ;load address of next
Problem: Consider the following SRC code segments for implementing the operation a=b+5c. Find which one is more efficient in terms of instruction count and execution time.
Program 1: Multiplication by using repeated addition in a for loop
org 0
a: .dw 1
b: .dw 1
c: .dw 1
.org 80
la r5, ; load value of loop
lar r6,mpy ;load address of mpy
lar r7, next ;load address of next
ld r2, b ; load contents of b
ld r3, c ; load contents of c
la r4, ;load 0 in r4
mpy:
brzr r7,r ; jump to next after 5
iterations
add r4,r4,r ;r4 contains r4+c
addi r5,r5, ; decrement index
br r ; loop again
next:
add r4,r4,r ; r4 contains sum of
b and 5c
st r4, a ;store at address a stop

7. Problem: Consider the following two SRC code segments for implementing the operation a=b+5c. Find which one is more efficient in terms of instruction count and execution time.
Program 2: Multiplication using sub-routine call
.org 0
a: .dw 1
b: .dw 1
c: .dw 1
.org 80
lar r1,mpy ;load address of mpy in r1
ld r2, b ; load contents of b in r2
la r3, ; load index in r3
ld r4,c ; load contents of c in r4
brl r5, r ; r5 contains PC
add r2,r2,r7 ; r2 contains sum b+5c
st r2, a
stop
mpy:
la r7, ;r7 contains zero
lar r8,again ;r8 contain again address
again:
brzr r5,r ;exit loop when index is
add r7,r7,r4 ; r7 contains r7+c
addi r3,r3, ; decrement index
br r8

8. Solution The instructions in both programs can be divided into 3
types and the respective count of each type is
Number of..
Program 1
Program 2
Data transfer instructions 7
Control instructions 3 4
ALSU instructions
IC for program 1 = = 13
IC for program 2 = = 14

9. Solution contd.. For execution time, consider the following SRC
specifications.
ET = IC x CPI x T
ET1= (7x4)+(3x2)+(3x3)
= 43T
ET2= (7x4)+(4x2)+(3x3)
= 45T
Conclusion:
Program 1 runs faster than program 2 as obvious from the execution time of both.
Instruction Type
CPI
Control 2
ALSU 3
Data Transfer 4

10. MIPS Millions of Instructions Per Second = IC / (ET x 106)
Capability of different instructions varies from machine to machine, eg. RISC machines have simpler instructions, so the same job will require more instructions
Was popular when the VAX 11/780 was treated as a reference – late 70s and early 80s

11. MIPS as a performance metric
MIPS is inversely proportional to execution time, ET= IC / (MIPS x 106 )

12. Example Consider a machine having a 100 MHz clock and three
instruction types with following
parameters.
Now suppose that two
different compilers generate
code for the same program.
The instruction count for each is given as follows
Instruction Type
CPI
Control 2
ALSU 3
Data Transfer 4
IC in millions
Code from compiler 1
Code from compiler 2
Control 5 10
ALSU 1
Data Transfer

13. Compare the two codes according to MIPS and according to execution time.
Solution:
First we find the CPI for both code sequences
Since CPI = clock cycles for each type of instruction / IC
CPI1= (5x2 + 1x3 + 1x4)/ 7 = 2.43
CPI2= (10x2 +1x3 + 1x4)/12 = 2.25
As MIPS= Clock Rate/ (CPI x 106 )
MIPS1= 100 x 106 / (2.43 x 106)
= 41.15
MIPS2=100 x 106 / (2.25 x 106)
= 44.44
Hence the code generated by compiler 2 has higher MIPS
Rating.

14. First we find the CPI for both code sequences
Compare the two codes according to MIPS and according to execution time.
Solution:
First we find the CPI for both code sequences
Since CPI = clock cycles for each type of instruction / IC
CPI1= (5x2 + 1x3 + 1x4)/ 7 = 2.43
CPI2= (10x2 +1x3 + 1x4)/12 = 2.25
As MIPS= Clock Rate/ (CPI x 106 )
MIPS1= 100 x 106 / (2.43 x 106)
= 41.15
MIPS2=100 x 106 / (2.25 x 106)
= 44.44
Hence the code generated by compiler 2 has higher MIPS
Rating.
As MIPS = IC / (ET x 106)
MIPS= (IC x clock rate)/
( IC x CPI x 106)
= Clock rate/(CPI x 106)

15. Solution contd.. Since ET = IC / (MIPS x 106)
ET1= (7 x 106) / (41.15 x 106)
= 0.17 seconds
ET2= (12 x 106) / ( x 106)
= 0.27 seconds
Hence code sequence 1 is much more efficient in terms of
execution time.

16. MFLOPS Millions of FLoating point Operations Per Second
Using FP operations makes more sense to some compared to using just any instructions
Results vary from FP op to FP op
Better compared to MIPS because of two reasons:

17. 2 reasons
FP ops are complex, and therefore, provide a better picture of the hardware capabilities on which they are run
Overheads (get operands, store results, etc. ) are effectively lumped with the FP ops they support

18. *** The name is a play on the word Whetstone
Dhrystones ***
Dhrystone is a general “integer performance” benchmark test originally developed by Reinhold Weicker in 1984.
Small program; less than 100 HLL statements
Compiles to about 1 to 1.5 Kb of code
*** The name is a play on the word Whetstone

19. Disadvantages of using Whetstones and Dhrystones
Both Whetstones and Dhrystones are now considered obsolete because of the following reasons.
Small, fit in cache
Obsolete instruction mix
Prone to compiler tricks
Difficult to reproduce results
Uncontrolled source code

20. SPEC System Performance Evaluation Cooperative
(SPEC) was founded in October, 1988, by Apollo, Hewlett-Packard, MIPS Computer Systems and SUN Microsystems
Latest version is SPEC CPU2000

21. SPEC The standard SPEC benchmark suite includes: A compiler
A Boolean minimization program
A spreadsheet program
A number of other programs that stress arithmetic processing speed
It uses a simple metric, elapsed time, to measure performance of competing machines
Machine independent code is used for fair comparison

22. Advantages It provides for ease of publication.
Each benchmark carries the same weight.
SPECratio is dimensionless.
It is not unduly influenced by long running programs.
It is relatively immune to performance variation on individual benchmarks.
It provides a consistent and fair metric.