Performance Evaluation Theory

Question 1

How can we analyze if a processor X is faster than a processor Y?

Answer 1

We could either divide the execution time of Y by the execution time off X. or we could divide the performance of X by the performance of Y (performance is one above execution time)

Question 2

What is the the execution time of a program?

Answer 2

The execution time of a program or the CPU time corresponds to the number of clock cycles times the time each clock cycle takes to complete

Question 3

Based on how we calculate the CPU time, what are the means to reduce the execution time?

Answer 3

We could either reduce the number of clock cycles per program, reduce the clock period, and increase the clock frequency

Question 4

What is the other possibility to calculate the CPU time?

Answer 4

The CPU time corresponds to the amount of seconds the processor takes per program. Therefore in order to calculate it, we count the number of instructions per program multiply it by the clocks per instructions and finally by the time each clock cycle takes to complete

Question 5

How do we calculate the throughout of a processor?

Answer 5

The throughput or MIPS is calculated by the instruction count divided by the execution time times 10 to the 6.

Since the execution time corresponds to the instruction count times the CPI divided by the clock frequency the final way to calculate the MIPS is to divide the clock frequency by the CPI times 10 to 6

Question 6

How do we evaluate the speed up of a enhancement?

Answer 6

In order to calculate the speed up, we divide the execution time without the enhancement by the execution time with the enhancement or the performance with the enhancement by the performance without the enhancement

Question 7

How do we calculate the execution time of a processor with a certain enhancement?

Answer 7

The execution time of a possessor with any advancement is equal to the percentage of task that is not affected by enhancement times the previous execution time (the execution time without enhancement) plus the fraction of the task that is affected by the enhancement times the previous execution time divided by the speed up

Question 8

What is the Amdahl’s law?

Answer 8

The Amdahl’s law says that the performance improvement obtainable from using some faster execution modes is limited by the fraction of the time that the faster mode is used

Question 9

What is the impact of a pipeline in the throughput and execution time of processor?

Answer 9

Pipelining increases the CPU instruction throughput, but it does not reduce the execution time of a single instruction

Pipelining usually slightly increases the latency of each instructions due to imbalance among the pipeline stages and the overhead in the control of the pipeline

Question 10

Why does imbalances among by blind stages reduces performance?

Answer 10

Imbalance among pipeline stages, reduces performance since the clock period cannot be shorter than the time needed for these lowest pipe stage.

Question 11

What is the reason for pipeline overhead?

Answer 11

Pipeline overhead arises from pipeline register delay and clock skew

Question 12

In a five stage pipeline. How do we account for stall cycles in the throughout calculation?

Answer 12

We add the stall cycles +4 (the amount of cycles to finish the last operation) to the clock cycles in the CPI calculation

Question 13

What is the difference between calculating it throughout of a processor using the as asymptotic performance metric or not using it?

Answer 13

The difference is located on the CPI calculation, where the final clock cycles in order to finish the last operation execution, goes to zero as the number of instructions go to infinity, and therefore the CPI calculation relates to the amount of instructions and stall cycles

Question 14

What is the ideal CPI of a pipeline processor, why it is not achievable, and how do we calculate it?

Answer 14

The ideal CPI on a pipeline processor would be one, but still scores the pipeline performance to degrade from the ideal performance

The calculation is: The Ideal CPI plus the stall cycles per instruction

Question 15

Why does a processor stalls?

Answer 15

Due to structural, data or control hazards, or even due to memory stalls

Question 16

What is the rational behind the fact that a pipeline processor with Ideal CPI improves the performance by the depth of the pipeline

Answer 16

We calculate the speed up off a pipeline processor by dividing the unpipelined processor by pipeline one.

Since the CPI of a pipeline processor can be expressed as the ideal CPI plus the stall cycle per instruction, in the case of ideal CPI we would have no stalls

And if we assume that the unpipelined processor CPI is equal to the number of stages in the processor (pipeline depth). Our speed up is equal to the pipeline depth.

Question 17

What is the performance impact of conditional branches?

Answer 17

The pipeline speed up now is equal to the pipeline depth by one plus the stall cycles caused by the branches (which corresponds to the branch frequency times the branch penalty)

Question 18

What is the impact of memory hierarchy on the CPU time?

Answer 18

The CPU time can be calculated as the IC times the CPI divided by the clock frequency

Why the CPI is calculated as the amount of clocks to execute the instruction plus the store cycles

In order to relate the memory stalls to the stall cycles in a processor, we calculate the average misses per instruction, which corresponds to the percentage of memory accesses done in the program times the miss rate

Finally, we calculate the amount of clock stalls by multiplying the misses per instruction calculated previously by the miss penalty

Question 19

If machine A executes a program in 10 sec and machine B executes same program in 15 sec:

A is 50% faster than B or A is 33% faster than B?

Answer 19

The statement A is n%faster than B can be expressed as:
• “A is n% faster than B”
execution time (B)/execution time (A) = 1 +n/100 ⇒

n = (execution time (B)- execution time (A)) *100 /
execution time (A)

(15 -10)/10 *100 = 50 ⇒ A is 50% faster than B.

Question 20

Let us consider an enhancement for a CPU resulting 10 time faster on computation than the original one but the original CPU is busy with computation only 40% of the time. What is the overall speedup gained by introducing the enhancement?

Answer 20

Application of Amdahl’s Law where:
• FractionE = 0.4
• SpeedupE = 10
The overall speed up is given by:
1/((1-0.4) + (0.4/10)) = 1.56

Question 21

Let us consider a computer with a L1 cache and L2 cache memory hierarchy. Suppose that in 1000 memory references there are 40 misses in L1 and 20 misses in L2.
• What are the various miss rates?

Answer 21

Miss Rate L1 = 40 /1000 = 4% (either local or global)
Miss Rate L2 = 20 /40 = 50%
Global Miss Rate for Last Level Cache (L2): Miss Rate L1 . Miss Rate L2 = 2%