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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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Base course aggregate has a target density of 121.8 lb/ft3 in place It will be laid down and compacted in a rectangular street r
irakobra [83]

Answer:

total weight of the aggregate = 3594878.28 lbs

Explanation:

given data

density = 121.8 lb/ft³

area =  1000 ft x 60 ft x 6 in = 1000 ft x 60 ft x 0.5 ft

moisture = 3.5 %

compaction = 95%

solution

we get here first volume of the space that is filled with the aggregate that is

volume = 1000 ft x 60 ft x 0.5 ft  = 30,000 cu ft

now we get fill space with aggregate that compact to 95% of dry density.

so we fill space with aggregate of density that is = 95% of 121.8

= 115.71 lb/ cu ft

so now dry weight of aggregate is

dry weight of aggregate = 30,000  × 115.71 = 3471300 lb

when we assume that moisture percentage is by weight

then weight of the moisture in aggregate will be

weight of the moisture in aggregate = 3.56 % of 3471300 lb  

weight of the moisture in aggregate = 123578.28 lbs

and

we get total weight of the aggregate to fill space that is

total weight of the aggregate = 3471300 lb +123578.28 lb

total weight of the aggregate = 3594878.28 lbs

4 0
3 years ago
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