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I am Lyosha [343]
3 years ago
5

You are implementing a RAID storage system and have found a system with eight 100 GB drives. How much storage space will you hav

e available? How does the OS handle this RAID?
Computers and Technology
1 answer:
asambeis [7]3 years ago
4 0

Answer:

The space available will vary between 800 GB (100%) and 400 GB (50%) of the total disks, depending on the RAID level.

The OS will handle the RAID as a single disk.

Explanation:

Each RAID level implements parity and redundancy in a different way, so the amount of disks used for this extra information will reduce the space available for actual storage.

Usual RAID levels are:

<u>RAID 0:</u> does not implement any redundancy or parity, so you will have available 100% of the total storage: 8 x 100 GB = 800 GB

<u>RAID 1:</u> Duplicates all the information in one disk to a second disk. Space is reduced in half: 400 GB

<u>RAID 5:</u> Uses the equivalent of 1 disk of parity data distributed evenly on each disk, meaning the space available is \frac{n-1}{n} of the total disks: \frac{7}{8} of 800 GB = 700 GB

Writting and reading the information on a RAID storage is handled by a raid controller, either implemented in hardware or software. The OS will "see" a single disk and will read or write information as usual.

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Write the definition of a void function that takes as input two integer values, say n and m. The function returns the sum and av
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Answer:

void mn(int m, int n){

   int sum = 0;

int count = 0;

   if(m<n){

       for(int i = m;i<=n;i++){

           sum+=i;

       }

   }

   else{

       for(int i = n;i<=m;i++){

           sum+=i;

       }

   }

count = abs(m - n)+1;

   cout<<"Sum: "<<sum<<endl;

   cout<<"Average: "<<(float)sum/count;

}

Explanation:

This line defines the method

void mn(int m, int n){

This initializes sum and count to 0

   int sum = 0;

int count = 0;

This checks if m is less than n

   if(m<n){

This iterates from m to n and calculates the sum of numbers between this interval

<em>   for(int i = m;i<=n;i++){</em>

<em>            sum+=i;</em>

<em>        }</em>

<em>    }</em>

If otherwise,

   else{

This iterates from n to m and calculates the sum of numbers between this interval

<em>        for(int i = n;i<=m;i++){</em>

<em>            sum+=i;</em>

<em>        }</em>

<em>    }</em>

This calculates the range from m to n using absolute function

count = abs(m - n)+1;

This prints the calculated sum

   cout<<"Sum: "<<sum<<endl;

This calculates and prints the average

   cout<<"Average: "<<(float)sum/count;

}

<em>See attachment for complete program that includes the main (in c++)</em>

Download cpp
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