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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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If the bending moment (M) is 4,176 ft-lb and the beam is an 1 beam, calculate the bending stress (psi) developed at a point with
SpyIntel [72]

Answer:

Bending stress at point 3.96 is \sigma_b = 1.37 psi

Explanation:

Given data:

Bending Moment M is 4.176 ft-lb = 50.12 in- lb

moment of inertia I = 144 inc^4

y = 3.96 in

\sigma_b = \frac{M}{I} \times y

putting all value to get bending stress

\sigma_b = \frac{50.112}{144} \times 3.96  

\sigma_b =  1.37 psi

Bending stress at point 3.96 is \sigma_b = 1.37 psi

3 0
4 years ago
Your family has asked you to estimate the operating costs of your clothes dryer for the year. The clothes dryer in your home has
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Answer:

The costs to run the dryer for one year are $ 9.03.

Explanation:

Given that the clothes dryer in my home has a power rating of 2250 Watts, and to dry one typical load of clothes the dryer will run for approximately 45 minutes, and in Ontario, the cost of electricity is $ 0.11 / kWh, to calculate the costs to run the dryer for one year the following calculation must be performed:

1 watt = 0.001 kilowatt

2250/45 = 50 watts per minute

45 x 365 = 16,425 / 60 = 273.75 hours of consumption

50 x 60 = 300 watt = 0.3 kw / h

0.3 x 273.75 = 82.125

82.125 x 0.11 = 9.03

Therefore, the costs to run the dryer for one year are $ 9.03.

8 0
3 years ago
A rectangular car-top carrier of 1.7-ft height, 5.0-ft length (front to back), and 4.2-ft width is attached to the top of a car.
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Answer:

\Delta P =1.2 \frac{1.3}{2}(26.822m/s)^2 (4.2*1.7*(0.3048)^2)=13.88 hp

Explanation:

We can assume that the general formula for the drag force is given by:

D= C_D \frac{\rho}{2}V^2 A

And we can see that is proportional to the area. On this case we can calculate the area with the product of the width and the height. And we can express the grad force like this:

D_1 = C_{D1} \frac{\rho}{2}V^2 (wh)

Where w is the width and h the height.

The last formula is without consider the area of the carrier, but if we use the area for the carrier we got:

D_2 = C_{D2} \frac{\rho}{2}V^2 (wh+ A_{carrier})

If we want to find the additional power added with the carrier we just need to take the difference between the multiplication of drag force by the velocity (assuming equal velocities for both cases) of the two cases, and we got:

\Delta P = C_{D2} \frac{\rho}{2}V^2 (wh+ A_{carrier}) V-  C_{D1} \frac{\rho}{2}V^2 (wh) V

We can assume the same drag coeeficient C_{D1}=C_{D2}=C_{D} and we got:

\Delta P = C_{D} \frac{\rho}{2}V^2 (wh+ A_{carrier}) V-  C_{D} \frac{\rho}{2}V^2 (wh) V

\Delta P = C_{D} \frac{\rho}{2}V^3 (A_{carrier})

1.7 ft =0.518 m

60 mph = 26.822 m/s

In order to find the drag coeffcient we ned to estimate the Reynolds number first like this:

R_E= \frac{Vl}{v}= \frac{26.822m/s*0.518 m}{1.58x10^{-4} Pa s}= 8.79 x10^{4}

And the value for the kinematic vicosity was obtained from the table of physical properties of the air under standard conditions.

Now we can find the aspect ratio like this:

\frac{l}{h}=\frac{5}{1.7}2.941

And we can estimate the calue of C_D = 1.2 from a figure.

And we can calculate the power difference like this:

\Delta P =1.2 \frac{1.3}{2}(26.822m/s)^2 (4.2*1.7*(0.3048)^2)=13.88 hp

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Answer:

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Explanation:

This type of form-fill can be described/developed based on the requirements or information needed by the organization in order to perfectly fulfill an order  and also to retain customers by making the form very easy and interactive attached below is an example of such form fill  interface

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Unless otherwise posted, the maximum speed limit is ______mph on a two-lane undivided highways and for vehicles towing trailers.
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Answer:

55 mph

Explanation:

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