CODEWITHSUNDEEP
c#randompercentage
user3245565
user3245565

Reputation: 43

Random with a specific amount of a specific value

my problem is following: I need to create some random values for scheduling. For example process times are given. Lets say a job j on a machine i gets a random value between (1,99). thats the time the jobs needs on this machine.

Now, I need to manipulate this random values. I like to say of all random process times, 20% of them are zero process times. So does anybody know how it is possible to give an array with integers a specific amount with a specific time??

here the normal random:

p_machine_job_completionTime = new int[Constants.numberMachines][];
        for (i = 0; i < Constants.numberMachines; i++)
        {
            p_machine_job_completionTime[i] = new int[Constants.numberJobs];
            for (j = 0; j < Constants.numberJobs; j++)
            {
                p_machine_job_completionTime[i][j] = random.Next(1, 99);
            }
        }

Now, jobs may skip a machine and consequently have a processing time of 0. Is it possible to limit the random values, with guaranteeing that x% of all my random values has the value 0 ?? e.g.: 

20% of p_machine_job_completionTime[i][j] = 0
80% of p_machine_job_completionTIme[i][j] = random (1,99)

I am very thankful for any small any tiny advice.

Upvotes: 4

Views: 144

Answers (4)

aquinas
aquinas

Reputation: 23796

I think what the existing answers are missing is this important point:

"Is it possible to limit the random values, with guaranteeing that x% of all my random values has the value 0"

If you need to guarantee that at the end of the day some random exactly x% of the items are given a value of zero, then you can't use random as in the answer from @Douglas. As @Douglas says, "To get 0 with 20% probability." But as stated in the question we don't want 20% probability, we want EXACTLY 20%, and the other exactly 80% to have random values. I think the code below does what you want. Filled in some values for numberMachines and numberJobs so the code can be run. 

int numberMachines = 5;
int numberJobs = 20;
Random random = new Random();
var p_machine_job_completionTime = new int[numberMachines][];
var theTwentyPercent = new HashSet<int>(Enumerable.Range(0,(numberJobs * numberMachines) -1 ).OrderBy(x => Guid.NewGuid()).Take(Convert.ToInt32((numberMachines * numberJobs) * 0.2)));

for (int i = 0; i < numberMachines; i++) {
    p_machine_job_completionTime[i] = new int[numberJobs];
    for (int j = 0; j < numberJobs; j++) {
        int index = (i * numberJobs) + j;
        if (theTwentyPercent.Contains(index)) {
            p_machine_job_completionTime[i][j] = 0;
        }
        else {
            p_machine_job_completionTime[i][j] = random.Next(1, 99);
        }
    }
}

Debug.Assert( p_machine_job_completionTime.SelectMany(x => x).Count(val => val==0) == (numberMachines * numberJobs) * 0.2 );

Upvotes: 0

Douglas
Douglas

Reputation: 54897

One way to do this would be by generating two random values: One for determining whether to use 0, and (possibly) another to generate non-zero values. However, you can combine both randoms into one by increasing the range of your random values by the appropriate amount, and converting any results above your limit to 0:

int val = random.Next(1, 123);
if (val >= 99)
    val = 0;

In this case, your target range contains 98 possible values (1 to 98, since the upper bound is exclusive). To get 0 with 20% probability, you need to extend the range of your random generator to 1 / (1 - 20%), or 125% of its present value, which would be 123.

Upvotes: 1

Paweł Bejger
Paweł Bejger

Reputation: 6366

I think that you do not need to include this within Random in your case. You could simply use modulo as a part of your for loop: 

    p_machine_job_completionTime = new int[Constants.numberMachines][];
    int counter = 0;  
    for (i = 0; i < Constants.numberMachines; i++)
    {
        p_machine_job_completionTime[i] = new int[Constants.numberJobs];
        for (j = 0; j < Constants.numberJobs; j++)
        {
              if( counter%5 == 0)
              {
                 p_machine_job_completionTime[i][j] = 0;
              }
              else
              {
                 p_machine_job_completionTIme[i][j] = random (1,99);                         
              }
              counter++;
        }
    }

Upvotes: 0

Dmitrii Bychenko
Dmitrii Bychenko

Reputation: 186803

Just separate two cases: 20% when 0 should be returned and 80% when 1..99 is the outcome

Random random;

...

int value = random.Next(5) == 0 ? 0 : random.Next(99) + 1;

Upvotes: 1

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