Computer Architecture Fundamentals Slides

Front 1

Performance

Back 1

1 / Execution time

Front 2

Two methods of calculating execution time for multi-testing programs.

Back 2

Average of execution times


Weighted average

Front 3

Amdahl's Law

Back 3

A fundamental law used to quantify performance improvements in computer design.

Front 4

Speed up

Back 4

Performance with enchancment /
Performance without enchancement

OR

execution time without enchancement /
execution time with enchancement

speedup >= 1

Front 5

Fractional Enhancement

Back 5

fraction of thecomputation time in the original machine thatcan be converted to take advantage of theenhancement.

FE <= 1

Example:
An original program takes 60 secs as execution time,and within the total 60 secs, only 20 secs of theprogram can be enhanced. Therefore, FE = 20/60

Front 6

What is speedup enhancement (SE)?

Back 6

How much faster the program would run if the enhanced mode was used for the entire program

SE >= 1

Example:
If an enhanced mode takes 2 secs for a program thatrequires 20 secs without the enhanced mode,then SE = 20/2=10

Front 7

How do you calculate the new execution time?

Back 7

ET(new) = ET(old) x [(1-FE) + FE/SE]

Front 8

How do you calculate the overall speedup?

Back 8

ET(old)/ET(new) = 1/ (1 - FE) + FE/SE

Front 9

Example

A CPU is usable 40% of the total execution time;• The enhanced CPU runs 10 times faster;

• What is the overall speedup of the new machine?FE = 0.4, SE = 10

Back 9

1 / (1-0.4) + 0.4/10

Speedup = 1.56

Front 10

CPU time

Back 10

CPU clock cycles for a program / clock rate

Front 11

CPI:

Back 11

CPU clock cycles for a program / IC

Front 12

Clock rate

Back 12

1/clock cycle time

Front 13

CPU time

Back 13

IC x CPI / clock rate

OR

IC x CPI x clock cycle time

Front 14

What three things is CPU performance dependant on?

Back 14

- Clock rate
Hardware technology and organisation
- CPI
Organisational an instruction set architecture
- IC
Instruction set architecture and compiler technology

Front 15

Total CPU clock cycles

Back 15

SUMOF = CPI x IC

Front 16

Instruction count

Back 16

CPI = (SUMOF CPI x IC) / IC

Front 17

Considering two alternatives for conditional branchinstructions

CPU A: a condition code is set by a compare instruction; abranch tests the condition code.

- CPU B: a compare instruction is included in the branch
- Known: conditional branch takes 2 cycles on both CPUs; allother instructions take 1 cycle. To CPU A, 20% of allinstructions are conditional branches.

- Suppose: clock cycle time of CPU A is (1) 1.25 times faster (2)1.1 times faster than that of CPU B.

Back 17

CPI(a) = (0.2 x 2) + (0.8 x 1) / 1= 1.2
CPU time(a) = IC(a) x 1.2 x clock cycle time