Greedy Algorithm

In this tutorial we will learn about Job Sequencing Problem with Deadline. This problem consists of n jobs each associated with a deadline and profit and our objective is to earn maximum profit. We will earn profit only when job is completed on or before deadline. We assume that each job will take unit time to complete.

- In this problem we have n jobs j1, j2, … jn each has an associated deadline d1, d2, … dn and profit p1, p2, ... pn.
- Profit will only be awarded or earned if the job is completed on or before the deadline.
- We assume that each job takes unit time to complete.
- The objective is to earn maximum profit when only one job can be scheduled or processed at any given time.

Consider the following 5 jobs and their associated deadline and profit.

index | 1 | 2 | 3 | 4 | 5 |

JOB | j1 | j2 | j3 | j4 | j5 |

DEADLINE | 2 | 1 | 3 | 2 | 1 |

PROFIT | 60 | 100 | 20 | 40 | 20 |

Note! If two or more jobs are having the same profit then sort them as per their entry in the job list.

index | 1 | 2 | 3 | 4 | 5 |

JOB | j2 | j1 | j4 | j3 | j5 |

DEADLINE | 1 | 2 | 2 | 3 | 1 |

PROFIT | 100 | 60 | 40 | 20 | 20 |

Looking at the jobs we can say the max deadline value is 3.

So, **dmax** = 3

As dmax = 3 so we will have THREE slots to keep track of free time slots. Set the time slot status to EMPTY

time slot | 1 | 2 | 3 |

status | EMPTY | EMPTY | EMPTY |

Total number of jobs is 5.

So we can write **n** = 5

Note!

If we look at job j2, it has a deadline 1. This means we have to complete job j2 in time slot 1 if we want to earn its profit.

Similarly, if we look at job j1 it has a deadline 2. This means we have to complete job j1 on or before time slot 2 in order to earn its profit.

Similarly, if we look at job j3 it has a deadline 3. This means we have to complete job j3 on or before time slot 3 in order to earn its profit.

Our objective is to select jobs that will give us higher profit.

```
for i = 1 to n do
Set k = min(dmax, DEADLINE(i)) //where DEADLINE(i) denotes deadline of ith job
while k >= 1 do
if timeslot[k] is EMPTY then
timeslot[k] = job(i)
break
endif
Set k = k - 1
endwhile
endfor
```

In the code we are saving jobs from 0 index.

```
#include <stdio.h>
#define MAX 100
typedef struct Job {
char id[5];
int deadline;
int profit;
} Job;
void jobSequencingWithDeadline(Job jobs[], int n);
int minValue(int x, int y) {
if(x < y) return x;
return y;
}
int main(void) {
//variables
int i, j;
//jobs with deadline and profit
Job jobs[5] = {
{"j1", 2, 60},
{"j2", 1, 100},
{"j3", 3, 20},
{"j4", 2, 40},
{"j5", 1, 20},
};
//temp
Job temp;
//number of jobs
int n = 5;
//sort the jobs profit wise in descending order
for(i = 1; i < n; i++) {
for(j = 0; j < n - i; j++) {
if(jobs[j+1].profit > jobs[j].profit) {
temp = jobs[j+1];
jobs[j+1] = jobs[j];
jobs[j] = temp;
}
}
}
printf("%10s %10s %10s\n", "Job", "Deadline", "Profit");
for(i = 0; i < n; i++) {
printf("%10s %10i %10i\n", jobs[i].id, jobs[i].deadline, jobs[i].profit);
}
jobSequencingWithDeadline(jobs, n);
return 0;
}
void jobSequencingWithDeadline(Job jobs[], int n) {
//variables
int i, j, k, maxprofit;
//free time slots
int timeslot[MAX];
//filled time slots
int filledTimeSlot = 0;
//find max deadline value
int dmax = 0;
for(i = 0; i < n; i++) {
if(jobs[i].deadline > dmax) {
dmax = jobs[i].deadline;
}
}
//free time slots initially set to -1 [-1 denotes EMPTY]
for(i = 1; i <= dmax; i++) {
timeslot[i] = -1;
}
printf("dmax: %d\n", dmax);
for(i = 1; i <= n; i++) {
k = minValue(dmax, jobs[i - 1].deadline);
while(k >= 1) {
if(timeslot[k] == -1) {
timeslot[k] = i-1;
filledTimeSlot++;
break;
}
k--;
}
//if all time slots are filled then stop
if(filledTimeSlot == dmax) {
break;
}
}
//required jobs
printf("\nRequired Jobs: ");
for(i = 1; i <= dmax; i++) {
printf("%s", jobs[timeslot[i]].id);
if(i < dmax) {
printf(" --> ");
}
}
//required profit
maxprofit = 0;
for(i = 1; i <= dmax; i++) {
maxprofit += jobs[timeslot[i]].profit;
}
printf("\nMax Profit: %d\n", maxprofit);
}
```

```
Job Deadline Profit
j2 1 100
j1 2 60
j4 2 40
j3 3 20
j5 1 20
dmax: 3
Required Jobs: j2 --> j1 --> j3
Max Profit: 180
```

The time complexity of this problem is O(n^{2}).