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Alexeev081 [22]
3 years ago
12

Given a program with execution times broken down shown below. Assume that techniques can only be applied to accelerate the integ

er instructions. What is the theoretical upper bound (theoretical maximum) of the speedup that can be achieved? Show your derivation Continued from the previous question, now assume our goal is to redesign the machine to achieve an overall speedup of 2 times. A technique was found to accelerate the floating point instructions by 4x. To achieve our design goal, how much speedup is needed from optimizing the integer instructions?
Engineering
1 answer:
Alex17521 [72]3 years ago
7 0

<u>Solution and Explanation:</u>

Floating point and Integer consumes 1000 seconds of 1600 seconds.

Since we shall be talking of theoretical speed-up, please do not try to relate this with any practical scenario :P

To achieve maximum speed up, we need to reduce the time consumed by FP and Integer as large as possible. So we consider a system that does not at all consume time to perform FP and Integer operations. With such specifications, we can say that our new system will consume 600 seconds.

 speed up = 1600 divide by 600 = 2.66

<u>part b: </u>

We need a System with enhancement that will result in speedup of 2.

So the time required for the new system would be 800 sec

 Required time = 1600 divide by = 800 seconds

It is now given that Floating Point can now be accelerated by 5 times, so our enhanced system will consume 40 seconds to perform Floating point operations

We know that Load/Store and branch operations cannot be enhanced and hence they will consume 600 seconds.

Therefore to attend speedup of 2, Integer operations must be completed in 160 seconds (160 = 800 – (600 + 40))

So speed up required for Integer Operations is 800/160 = 5

So Integer Operation should need 5 times less time to achieve speed up of 2

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Answer:

203.0160

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Durante el segundo trimestre de 2001, Tiger Woods fue el golfista que más dinero ganó en el PGATour. Sus ganancias sumaron un to
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Answer: a. 0.4667

b. 0.4667 and C 0.0667

Explanation:

Given Data:

N = population size (10)

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Therefore

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f(1) = ( 7/1 ) ( 10 - 7 / 2 -1 ) / ( 10/2 )

= 7 / 15

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When y = 2

f(2) = ( 7/2 ) ( 10 - 7 / 2 -2 ) / ( 10/2 )

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When y = 0

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2 years ago
A completely reversible heat pump produces heat ata rate of 300 kW to warm a house maintained at 24°C. Theexterior air, which is
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Answer:

Change in entropy S = 0.061

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T2 = 24°C = 297 K

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Change in entropy =

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21.Why are throttling devices commonly used in refrigeration and air-conditioning<br> applications?
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Answer is given below

Explanation:

we know that some common types of throttling devices are

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so here throttling devices commonly used in refrigeration and air-conditioning because

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