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Drupady [299]
3 years ago
6

Explain the purpose of the checkpoint mechanism. How often should checkpoints be performed?How does the frequency of checkpoints

affect:
System performance when no failure occurs?
The time it takes to recover from a system crash?
The time it takes to recover from a disk crash?
Engineering
1 answer:
Olegator [25]3 years ago
3 0

Explanation:

Answer:

1) Check point mechanism is a system, point or mechanism within the database management system that allows for data, information, transactions and general logs passed through the system be retrieved and stored into a separate storage system and talking about how often the check point should be performed, it is advised that checkpoint should be done as often as possible because the more often the checkpoints is done, the lower the probability that obsolete data will be updated or retrieved during the checkpoint process.

How does the frequency of check point affects system performance when no failure occurs?

Answer: When failure does not occur, the process of checkpoint mechanism have only succeeded in accruing unnecessary cost of performing the process which at this time would have been necessary.

How does the frequency of check point affects the time it takes to recover from a system crash ?

Answer: It is important to note that the purpose of a check point mechanism in the first place is for us to be able to recover our data in the database before a time when the database became corrupt. So if we constantly run the process of check point mechanism, it will drastically reduce the time it takes to recover from a system crash because we would have previously detected obsolete data and updated the system before the crash.

How does the frequency of check point affects the time it takes to recover from a disk crash?

Answer: This can also be related to the same time of recovery as that of the system crash.  We can substitute the answer for time to recover from a system crash for that of a disc crash also. The only difference here is that it applies to a disc.

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• Build upon the results of problem 3-85 to determine the minimum factor of safety for fatigue based on infinite life, using the
Rudik [331]

Answer:

minimum factor of safety for fatigue is = 1.5432

Explanation:

given data

AISI 1018 steel cold drawn as table

ultimate strength Sut = 63.800 kpsi

yield strength Syt = 53.700 kpsi

modulus of elasticity E = 29.700 kpsi

we get here

\sigma a = \sqrt{(\sigma a \times kb)^2+3\times (za\times kt)^2}    ...........1

here kb and kt = 1 combined bending and torsion fatigue factor

put here value and we get

\sigma a =  \sqrt{(12 \times 1)^2+3\times (0\times 1)^2}  

\sigma a = 12 kpsi

and

\sigma m = \sqrt{(\sigma m \times kb)^2+3\times (zm\times kt)^2}     ...........2

put here value and we get

\sigma m = \sqrt{(-0.9 \times 1)^2+3\times (10\times 1)^2}  

\sigma m = 17.34 kpsi

now we apply here goodman line equation here that is

\frac{\sigma m}{Sut} +  \frac{\sigma a}{Se} = \frac{1}{FOS}     ...................3

here Se = 0.5 × Sut

Se = 0.5 × 63.800 = 31.9 kspi

put value in equation 3 we get

\frac{17.34}{63.800} +  \frac{12}{31.9} = \frac{1}{FOS}  

solve it we get

FOS = 1.5432

6 0
3 years ago
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