Question

.13 : 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. Recall that Amdahl’s law depends on the fraction of the original, unenhanced execution time that could make use of enhanced mode. Thus, we cannot directly use this 50% measurement to compute speedup with Amdahl’s law.

a. <1.9> What is the speedup we have obtained from fast mode?

b. <1.9> What percentage of the original execution time has been converted to fast mode?

1.14  When making changes to optimize part of a processor, it is often the case that speeding up one type of instruction comes at the cost of slowing down something else. For example, if we put in a complicated fast floatingpoint unit, that takes space, and something might have to be moved farther away from the middle to accommodate it, adding an extra cycle in delay to reach that unit. The basic Amdahl’s law equation does not take into account this trade-off.

a. <1.9> If the new fast floating-point unit speeds up floating-point operations by, on average, 2×, and floating-point operations take 20% of the original program’s execution time, what is the overall speedup (ignoring the penalty to any other instructions)?

b. <1.9> Now assume that speeding up the floating-point unit slowed down data cache accesses, resulting in a 1.5× slowdown (or 2/3 speedup). Data cache accesses consume 10% of the execution time. What is the overall speedup now?

1.15  Your company has just bought a new Intel Core i5 dualcore processor, and you have been tasked with optimizing your software for this processor. You will run two applications on this dual core, but the resource requirements are not equal. The first application requires 80% of the resources, and the other only 20% of the resources. Assume that when you parallelize a portion of the program, the speedup for that portion is 2.

a. <1.10> Given that 40% of the first application is parallelizable, how much speedup would you achieve with that application if run in isolation?

b. <1.10> Given that 99% of the second application is parallelizable, how much speedup would this application observe if run in isolation?

Answer 1

ANSWER :-

1.13

a) <1.9> What is the speedup we have obtained from fast mode?

We know that the accelerated execution time consisted of two halves: the unaccelerated phase (50%) and the accelerated phase (50%).

Without the enhancement, the unaccelerated phase would have taken just as long (50%), but the accelerated phase would take 10 times as long, i.e. 500%.

So the relative execution time without the enhancement would be 50% + 500% = 550%.

Overall speedup is execution time unaccelerated / execution time accelerated = 550% / 100% = 5.5

Thus the overall speedup = 5.5

b). <1.9> What percentage of the original execution time has been converted to fast mode?

To find the percentage of the original execution time which was accelerated, we plug these figures into Amdahl's Law again:

Fraction vectorised = speeduprm overall X speedup accelerated - speedup accelerated / speedup overll X speedup accelerated - speedup overall

fraction vectorised = 5.5 X 10 -10 / 5.5 X 10 - 5.5
=45 / 49.5
=90.90%

Thus the percentage of original time converted to fast mode is 90.90%,

1.14

a. <1.9> If the new fast floating-point unit speeds up floating-point operations by, on average, 2×, and floating-point operations take 20% of the original program’s execution time, what is the overall speedup (ignoring the penalty to any other instructions)?

The fraction is given as 20% Therefore the fraction can be taken as 0.20.The new fast floating point unit speed up floating point operations by the average is 2.

Overall speed up = 1 / ((1-x) + (x/N))

speed up = 1 / ((1-0.20) + (0.20 / 2))

Overall speed up = 1.11

<1.9> Now assume that speeding up the floating-point unit slowed down data cache accesses, resulting in a 1.5× slowdown (or 2/3 speedup). Data cache accesses consume 10% of the execution time. What is the overall speedup now?

That means it is able to speed up to 2/3 times.

Given the consumption of the data cache is 10% of the execution time.

Thus the overall speed up = 1 / (0.7+ 0.10 + 0.15)

Overall speed up = 1.05

1.15

a) <1.10> Given that 40% of the first application is parallelizable, how much speedup would you achieve with that application if run in isolation?

The maximum speedup for that application is given by 1 / ((1-x) + (x/N))

where N is 2 since it is dual core

If x is the fraction of the application that can be parallalized.

x = 0.4 since 40% is parallelized.

Maximum speedup = 1 / ((1-x) + (x/N))

speedup = 1 / ((1 - 0.4) + (0.4/2))

Maximum speedup = 1.25

b. <1.10> Given that 99% of the second application is parallelizable, how much speedup would this application observe if run in isolation?

x = 0.99 since 99% is parallelized.

Maximum speedup = 1 / ((1-x) + (x/N))

speedup = 1 / ((1 - 0.99) + (0.99/2))

Maximum speedup = 1.98