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Fed [463]
3 years ago
10

4. Assume a multiple level queue system with a variable time quantum per queue, and that the incoming job needs 50ms to run to c

ompletion. If the first queue has a time quantum of 5ms and each queue thereafter has a time quantum that is twice as large as the previous one, how many times will the job be interrupted, and on which queue will it finish its execution? Explain how much time it spends in each queue.
Physics
1 answer:
Troyanec [42]3 years ago
6 0

Answer:

Explanation:

For the completion of incoming job it will take 50ms

First queue takes 5ms quantum time and the subsequent queue takes double of the previous question

So,

First queue T_1 = 5ms

Second queue T_2 = 2 × T_1 = 2 × 5 = 10ms

Third queue T_3 = 2 × T_2 = 2 × 10 = 20ms

Fourth queue T_4 = 2 × T_3 = 2 × 20 = 40ms

Fifth queue T_5= 2 × T_4 = 2 × 40 = 80ms.

Now, the job will be done after the fifth queue.

So, after the first queue, the job is not completed, so, we have first interruption

After the second queue, the job is not completed, so, we have second interruption

After the third queue, the job is not completed, so we have third interruption.

After the fourth queue, the job is not yet completed, so we have the fourth interruption

And in the fifth queue the job is completed, so we don't have any interruption here.

So, the job will be interrupted 4 times and it will finished on the fifth queue.

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Rapid gas exchange can be accomplished more easily in air than in water because A. the O2 content of water is higher than that o
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C. water is more dense and viscous

Explanation:

Rapid gas exchange can be accomplished more easily in air than in water because water is more dense and viscous.

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How does the Earth's rotation affect our view of stars?
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5 0
2 years ago
A paper clip is made of wire 0.5 mm in diameter. If the original material from which the wire is made is a rod 25 mm in diameter
gizmo_the_mogwai [7]

Answer:

longitudinal engineering strain = 624.16

true strain is 6.44

Explanation:

given data

diameter d1 = 0.5 mm

diameter d2 = 25 mm

to find out

longitudinal engineering and true strains

solution

we know both the volume is same

so

volume 1 = volume 2

A×L(1) = A×L(2)

( π/4 × d1² )×L(1) = ( π/4 × d2² )×L(2)

( π/4 × 0.5² )×L(1) = ( π/4 × 25² )×L(2)

0.1963 ×L(1) = 122.71 ×L(2)

L(1) / L(2) = 122.71 / 0.1963 = 625.16

and we know longitudinal engineering strain is

longitudinal engineering strain = L(1) / L(2)  - 1

longitudinal engineering strain = 625.16  - 1

longitudinal engineering strain = 624.16

and

true strain is

true strain = ln ( L(1) / L(2))

true strain = ln ( 625.16)

true strain is 6.44

3 0
3 years ago
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