[EE_CSIE] Computer Architecture Chapter01 Notes

1.2 Quiz1 : Disscuss the critical system design issues of desktop, server and embbeded systems.

Ans1 : 1.Desktop : price-performance, graphics performance. 2.Server : throughput, availability, scalability. 3.Embedded : price, power consumption, application specific performance.

--- 1.4 Cost of an integrated circuit :

Cost of IC = [ ( Cost of die ) + ( Cost of testing ) + ( Cost of packaging ) ] / ( Final test yeild )

Cost of die = ( Cost of Wafer ) / [ ( Dies per wafer ) * ( Die yield ) ]

Dies per wafer = π * [ ( Wafer diameter / 2 ) ]2 / (Die area) - π * ( Wafer diameter ) / [ 2 * ( Die area ) ] 1/2

--- 1.6 Amdahl's Law (亞當斯定理) : the performance improvement to be gained from using some faster mode of execution is limited by the fraction of the time the faster mode can be used. 藉由使用某種較快的執行方式所增進的效能，會受限於可採用此種執行方式所佔的時間比例。

--- 公式： Speedup = ( Execution time before omprovement ) / ( Execution time after improvement ) => Make the common cast fast.

ExTime new = ExTime old * [ ( 1 - Fraction enhanced ) + ( Fraction enhanced / Speedup enhanced ) ]

Speedup overall = ExTime old / ExTime new

= 1 / [ (1 - Fraction enhanced ) + ( Fraction enhanced / Speedup enhanced )]

--- Amdahl's Law expresses the law of diminishing returns (報酬遞減法則): The incremental improvement in speedup gained by an additional improvement in the performance of just a portion of the computation diminishes as improvements are added. An important corollary of Amdahl's Law is that if an enhancement is only usable for a fraction of a task, we can't speed up the task by more than the reciprocal of 1 minus that fraction.

CPU Time = Seconds / Program = ( Instructions / Program) * ( Cycles / Instruction ) * ( Seconds / Cycle )

--- Q1.2 Assume that we make an enhancement to a computer that improves some mode of execution by a factor of 10. Enhanced mode is used 50% of the time, measured as a percentage of the execution time when the enhanced mode is in use (rather than as defined in the notes, where the percentage of the running time without the enhancement is used). 1. What is the speedup we have obtained from fast mode? 2. What percentage of the original execution time has been converted to fast mode? Ans: 1. What is the speedup we have obtained from fast mode? Assume that the time taken to execute some program P is 100t seconds when the enhancement is in use. The amount of time during which enhanced mode is in use is 50% of this time, i.e. 50t seconds. Now the enhanced mode gives us a speedup factor of 10. This means the original time (in unenhanced mode) to execute these 50t seconds would be 500t seconds. The total original time would therefore be 500t + 50t = 550t seconds. Thus Speedup = Original time / Enhanced time = 550t / 100t = 5.5

2. What percentage of the original execution time has been converted to fast mode? The amount of original time converted to fast mode is 500 seconds. The total original time is 550t seconds. Thus the percentage of original time converted to fast mode is 500t / 550t = 90.91%

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