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Paul [167]
4 years ago
10

Consider the parameter Δ used to define the working-set window inthe working-set model. When Δ is set to a low value, what is th

e effecton the page-fault frequency and the number of active (nonsuspended)processes currently executing in the system? What is the effect when Δis set to a very high value?
Computers and Technology
1 answer:
a_sh-v [17]4 years ago
3 0

Answer:

When Δ is set to large value, then a process's resident set is overestimated and this might prevent many processes from being scheduled even though their required pages are resident

Explanation:

When Δ is set to a small value, then the set of resident pages for a process might be underestimated, allowing a process to be scheduled even though all of its required pages are not resident.

This could result in a large number of page faults. When Δ is set to large value, then a process's resident set is overestimated and this might prevent many processes from being scheduled even though their required pages are resident.

Hoewever, once a process is scheduled, it is unlikely to generate page faults since its resident set has been overestimated.

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6 0
3 years ago
Read 2 more answers
Create a function called makePositive that accepts a single argument: an integer array. The function should walk through the arr
coldgirl [10]

Answer:

This solution is implemented in C++

void makePositive(int arr[]){

   int i =0;

   while(arr[i]!=0){

       if(arr[i]<0){

           arr[i] = abs(arr[i]);

       }

       i++;

   }

   i = 0;

   while(arr[i]!=0){

    cout<<arr[i]<<" ";

    i++;

}  

}

Explanation:

This defines the function makePositive

void makePositive(int arr[]){

This declares and initializes i to 0

   int i =0;

The following iteration is repeated until the last element in the array

   while(arr[i]!=0){

This checks if current array element is negative

       if(arr[i]<0){

If yes, it changes it to positive

           arr[i] = abs(arr[i]);

       }

The next element is then selected

       i++;

   }

This sets i to 0

   i = 0;

The following iteration prints the updated content of the array

<em>    while(arr[i]!=0){ </em>

<em>     cout<<arr[i]<<" "; </em>

<em>     i++; </em>

<em> }   </em>

}

See attachment for full program which includes the main

Download cpp
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nataly862011 [7]

Answer:

it is either c or d

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Answer:

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Explanation:

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Elena L [17]

Answer:

A. a series of actions that solve a particular problem.

Explanation:

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