Min Max Algorithm in Grid Computing with code in C

Here you will find the easiest explanation for min max algorithm in grid computing.

Let us first know why this algorithm is needed?

This is a static task scheduling algorithm used for load balancing.

The Min-Max algorithm first finds the maximum execution time of all tasks. Then it chooses the task with the minimum execution time among all the tasks. The same procedure is repeated by Min Max until all tasks are scheduled.

Why we call it Min-Max when actually we are finding the maximum first then minimum?

This is because the semantics of saying Min-Max means that the task with minimum execution time (lightest process ) is allocated to the machine with maximum completion time (worst machine).

Let us take an example before moving to the code

Here Task 1 will execute in 140ms in VM1 and in 100 ms in VM2. Task 2 will execute for 20ms in VM1 and in 100ms in VM2 . Task 3 will run for 60 ms in VM1 and for 70 ms in VM2.

What is a Task ?

Let’s take an example. Suppose you have developed a social media app called fakebook. Now user one wants to change her DP , another user wants to create her account and the third user wants to like a picture.

All these are the tasks and involve manipulation in the database. The entity responsible for that change is the backend server. For effective use of the server you have made virtual partitions of the server making some virtual machines good for light computation and some virtual machines for heavy computation.

Now for sure you will not let a weak virtual machine to do heavy computation and a strong machine to do a trivial task, but this is the optimal condition, therefore we need task scheduling algorithm to make this decision for us.

Algorithm to solve the problem

1. Find maximum time for each task , in the above image maximum time for each task is (140 ms ,T1 ,VM1), (100 ms, T2, VM2) ,(70 ms, T3, VM2)
2. Now find the set with minimum time and that is (70 ms, T3, VM2)
3. Execute T3 in VM2 for 70ms
4. Since 70ms has been elapsed we need to update our data. New data is:

5. Here we impute T3 as INT_MIN to let our code know that this task is over and it denotes negative infinite. Now repeat step 1

6. Find maximum time for each task , in the above image maximum time for each task is (170 ms ,T1 ,VM2), (170 ms, T2, VM2)

7. Now find the set with minimum time and that is (170 ms, T2, VM2)

8. Execute T2 in VM2 for 170ms

9. Since 170ms has been elapsed we need to update our data. New data is:

10. Find maximum time for each task , in the above image maximum time for each task is (340 ms ,T1 ,VM2)

11. Exit Since all task has been scheduled.

Makespan produced by any algorithm for a schedule can be calculated as follows:

makespan = max (CT (ti, mj))

CTij = Rj+ETij

Where CT completion time of machines ,ETij expected execution time of job i on resource j , Rj ready time or availability time of resource j after completing the previously assigned jobs.

Here makespan = max(70,170,340) = 340 ms

Finally the Code

#include<stdio.h>
#include <limits.h>
int main(){

int nT,nM;//number of tasks , number of machines
printf(“\nEnter number of machines and tasks\n”);
scanf(“%d%d”,&nM,&nT);

/*
Declare a 2d-array of size nM x nT
Data should be in the following format :

T1 T2 T3
M1 | 140 | 20 | 60 |
M2 | 100 | 100 | 70 |

*/
int minMax[nM][nT];
int tmp[nM][nT];
int makespan=0;
printf(“\nFill Data\n”);
for(int i=0;i<nM;i++)
for(int j=0;j<nT;j++){
scanf(“%d”,&minMax[i][j]);
tmp[i][j]=minMax[i][j];
}

// visualise data
printf(“\nOriginal Data\n”);

for(int i=0;i<nM;i++){
for(int j=0;j<nT;j++)
printf(“%d “,minMax[i][j]);
printf(“\n”);
}

//This array will hold the answer
int resultTask[nT];
int resultMachine[nT];
int resultTime[nT];

int ptr=-1; //Indicates if result set is full or not

while(ptr<nT-1){
int time[nT],machine[nT]; //stores minimum time w.r.t machine of each task
for(int j=0;j<nT;j++){
int maximum = INT_MIN;
int pos=-1;
for(int i=0;i<nM;i++){
if(minMax[i][j]>maximum){
maximum=minMax[i][j];
pos=i;
}
}
time[j]=maximum;
machine[j]=pos;
}

// Now we find task with minimum time

int minimum=INT_MAX;
int pos=-1;

for(int j=0;j<nT;j++){
if(time[j]<minimum && time[j] != INT_MIN){
minimum=time[j];
pos=j;
}
}

resultTask[++ptr]=pos;
resultMachine[ptr]=machine[pos];
resultTime[ptr]=tmp[machine[pos]][pos];
if(minimum>makespan)
makespan=minimum;
// resetting states

for(int i=0;i<nM;i++){
for(int j=0;j<nT;j++){
if(j==resultTask[ptr])
minMax[i][j]=INT_MIN;
else if(i==resultMachine[ptr] && minMax[i][j]!=INT_MIN)
minMax[i][j]+=minimum;
else
continue;
}
}

}

//printing answer
printf(“\nScheduled Task are :\n”);
for(int i=0;i<nT;i++){
printf(“\nTask %d Runs on Machine %d with Time %d units\n”,resultTask[i]+1,resultMachine[i]+1,resultTime[i]);
}

printf(“\nMakeSpan time : %d units\n”,makespan);
return 0;
}

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