Algorithms Dynamic programming suggest change

Weighted Job Scheduling Algorithm

Weighted Job Scheduling Algorithm can also be denoted as Weighted Activity Selection Algorithm.

The problem is, given certain jobs with their start time and end time, and a profit you make when you finish the job, what is the maximum profit you can make given no two jobs can be executed in parallel?

This one looks like Activity Selection using Greedy Algorithm, but there’s an added twist. That is, instead of maximizing the number of jobs finished, we focus on making the maximum profit. The number of jobs performed doesn’t matter here.

Let’s look at an example:

+-------------------------+---------+---------+---------+---------+---------+---------+
|          Name           |    A    |    B    |    C    |    D    |    E    |    F    |
+-------------------------+---------+---------+---------+---------+---------+---------+
|(Start Time, Finish Time)|  (2,5)  |  (6,7)  |  (7,9)  |  (1,3)  |  (5,8)  |  (4,6)  |
+-------------------------+---------+---------+---------+---------+---------+---------+
|         Profit          |    6    |    4    |    2    |    5    |    11   |    5    |
+-------------------------+---------+---------+---------+---------+---------+---------+

The jobs are denoted with a name, their start and finishing time and profit. After a few iterations, we can find out if we perform Job-A and Job-E, we can get the maximum profit of 17. Now how to find this out using an algorithm?

The first thing we do is sort the jobs by their finishing time in non-decreasing order. Why do we do this? It’s because if we select a job that takes less time to finish, then we leave the most amount of time for choosing other jobs. We have:

+-------------------------+---------+---------+---------+---------+---------+---------+
|          Name           |    D    |    A    |    F    |    B    |    E    |    C    |
+-------------------------+---------+---------+---------+---------+---------+---------+
|(Start Time, Finish Time)|  (1,3)  |  (2,5)  |  (4,6)  |  (6,7)  |  (5,8)  |  (7,9)  |
+-------------------------+---------+---------+---------+---------+---------+---------+
|         Profit          |    5    |    6    |    5    |    4    |    11   |    2    |
+-------------------------+---------+---------+---------+---------+---------+---------+

We’ll have an additional temporary array Acc_Prof of size n (Here, n denotes the total number of jobs). This will contain the maximum accumulated profit of performing the jobs. Don’t get it? Wait and watch. We’ll initialize the values of the array with the profit of each jobs. That means, Acc_Prof[i] will at first hold the profit of performing i-th job.

+-------------------------+---------+---------+---------+---------+---------+---------+
|         Acc_Prof        |    5    |    6    |    5    |    4    |    11   |    2    |
+-------------------------+---------+---------+---------+---------+---------+---------+

Now let’s denote position 2 with i, and position 1 will be denoted with j. Our strategy will be to iterate j from 1 to i-1 and after each iteration, we will increment i by 1, until i becomes n+1.

j        i

+-------------------------+---------+---------+---------+---------+---------+---------+
|          Name           |    D    |    A    |    F    |    B    |    E    |    C    |
+-------------------------+---------+---------+---------+---------+---------+---------+
|(Start Time, Finish Time)|  (1,3)  |  (2,5)  |  (4,6)  |  (6,7)  |  (5,8)  |  (7,9)  |
+-------------------------+---------+---------+---------+---------+---------+---------+
|         Profit          |    5    |    6    |    5    |    4    |    11   |    2    |
+-------------------------+---------+---------+---------+---------+---------+---------+
|         Acc_Prof        |    5    |    6    |    5    |    4    |    11   |    2    |
+-------------------------+---------+---------+---------+---------+---------+---------+

We check if Job[i] and Job[j] overlap, that is, if the finish time of Job[j] is greater than Job[i]‘s start time, then these two jobs can’t be done together. However, if they don’t overlap, we’ll check if Acc_Prof[j] + Profit[i] > Acc_Prof[i]. If this is the case, we will update Acc_Prof[i] = Acc_Prof[j] + Profit[i]. That is:

if Job[j].finish_time <= Job[i].start_time
    if Acc_Prof[j] + Profit[i] > Acc_Prof[i]
        Acc_Prof[i] = Acc_Prof[j] + Profit[i]
    endif
endif

Here Acc_Prof[j] + Profit[i] represents the accumulated profit of doing these two jobs toegther. Let’s check it for our example:

Here Job[j] overlaps with Job[i]. So these to can’t be done together. Since our j is equal to i-1, we increment the value of i to i+1 that is 3. And we make j = 1.

j                   i

+-------------------------+---------+---------+---------+---------+---------+---------+
|          Name           |    D    |    A    |    F    |    B    |    E    |    C    |
+-------------------------+---------+---------+---------+---------+---------+---------+
|(Start Time, Finish Time)|  (1,3)  |  (2,5)  |  (4,6)  |  (6,7)  |  (5,8)  |  (7,9)  |
+-------------------------+---------+---------+---------+---------+---------+---------+
|         Profit          |    5    |    6    |    5    |    4    |    11   |    2    |
+-------------------------+---------+---------+---------+---------+---------+---------+
|         Acc_Prof        |    5    |    6    |    5    |    4    |    11   |    2    |
+-------------------------+---------+---------+---------+---------+---------+---------+

Now Job[j] and Job[i] don’t overlap. The total amount of profit we can make by picking these two jobs is: Acc_Prof[j] + Profit[i] = 5 + 5 = 10 which is greater than Acc_Prof[i]. So we update Acc_Prof[i] = 10. We also increment j by 1. We get,

j         i

+-------------------------+---------+---------+---------+---------+---------+---------+
|          Name           |    D    |    A    |    F    |    B    |    E    |    C    |
+-------------------------+---------+---------+---------+---------+---------+---------+
|(Start Time, Finish Time)|  (1,3)  |  (2,5)  |  (4,6)  |  (6,7)  |  (5,8)  |  (7,9)  |
+-------------------------+---------+---------+---------+---------+---------+---------+
|         Profit          |    5    |    6    |    5    |    4    |    11   |    2    |
+-------------------------+---------+---------+---------+---------+---------+---------+
|         Acc_Prof        |    5    |    6    |    10   |    4    |    11   |    2    |
+-------------------------+---------+---------+---------+---------+---------+---------+

Here, Job[j] overlaps with Job[i] and j is also equal to i-1. So we increment i by 1, and make j = 1. We get,

j                             i

+-------------------------+---------+---------+---------+---------+---------+---------+
|          Name           |    D    |    A    |    F    |    B    |    E    |    C    |
+-------------------------+---------+---------+---------+---------+---------+---------+
|(Start Time, Finish Time)|  (1,3)  |  (2,5)  |  (4,6)  |  (6,7)  |  (5,8)  |  (7,9)  |
+-------------------------+---------+---------+---------+---------+---------+---------+
|         Profit          |    5    |    6    |    5    |    4    |    11   |    2    |
+-------------------------+---------+---------+---------+---------+---------+---------+
|         Acc_Prof        |    5    |    6    |    10   |    4    |    11   |    2    |
+-------------------------+---------+---------+---------+---------+---------+---------+

Now, Job[j] and Job[i] don’t overlap, we get the accumulated profit 5 + 4 = 9, which is greater than Acc_Prof[i]. We update Acc_Prof[i] = 9 and increment j by 1.

j                   i

+-------------------------+---------+---------+---------+---------+---------+---------+
|          Name           |    D    |    A    |    F    |    B    |    E    |    C    |
+-------------------------+---------+---------+---------+---------+---------+---------+
|(Start Time, Finish Time)|  (1,3)  |  (2,5)  |  (4,6)  |  (6,7)  |  (5,8)  |  (7,9)  |
+-------------------------+---------+---------+---------+---------+---------+---------+
|         Profit          |    5    |    6    |    5    |    4    |    11   |    2    |
+-------------------------+---------+---------+---------+---------+---------+---------+
|         Acc_Prof        |    5    |    6    |    10   |    9    |    11   |    2    |
+-------------------------+---------+---------+---------+---------+---------+---------+

Again Job[j] and Job[i] don’t overlap. The accumulated profit is: 6 + 4 = 10, which is greater than Acc_Prof[i]. We again update Acc_Prof[i] = 10. We increment j by 1. We get:

j         i

+-------------------------+---------+---------+---------+---------+---------+---------+
|          Name           |    D    |    A    |    F    |    B    |    E    |    C    |
+-------------------------+---------+---------+---------+---------+---------+---------+
|(Start Time, Finish Time)|  (1,3)  |  (2,5)  |  (4,6)  |  (6,7)  |  (5,8)  |  (7,9)  |
+-------------------------+---------+---------+---------+---------+---------+---------+
|         Profit          |    5    |    6    |    5    |    4    |    11   |    2    |
+-------------------------+---------+---------+---------+---------+---------+---------+
|         Acc_Prof        |    5    |    6    |    10   |    10   |    11   |    2    |
+-------------------------+---------+---------+---------+---------+---------+---------+

If we continue this process, after iterating through the whole table using i, our table will finally look like:

+-------------------------+---------+---------+---------+---------+---------+---------+
|          Name           |    D    |    A    |    F    |    B    |    E    |    C    |
+-------------------------+---------+---------+---------+---------+---------+---------+
|(Start Time, Finish Time)|  (1,3)  |  (2,5)  |  (4,6)  |  (6,7)  |  (5,8)  |  (7,9)  |
+-------------------------+---------+---------+---------+---------+---------+---------+
|         Profit          |    5    |    6    |    5    |    4    |    11   |    2    |
+-------------------------+---------+---------+---------+---------+---------+---------+
|         Acc_Prof        |    5    |    6    |    10   |    14   |    17   |    8    |
+-------------------------+---------+---------+---------+---------+---------+---------+

* A few steps have been skipped to make the document shorter.

If we iterate through the array Acc_Prof, we can find out the maximum profit to be 17! The pseudo-code:

Procedure WeightedJobScheduling(Job)
sort Job according to finish time in non-decreasing order
for i -> 2 to n
    for j -> 1 to i-1
        if Job[j].finish_time <= Job[i].start_time
            if Acc_Prof[j] + Profit[i] > Acc_Prof[i]
                Acc_Prof[i] = Acc_Prof[j] + Profit[i]
            endif
        endif
    endfor
endfor

maxProfit = 0
for i -> 1 to n
    if maxProfit < Acc_Prof[i]
        maxProfit = Acc_Prof[i]
return maxProfit

The complexity of populating the Acc_Prof array is O(n2). The array traversal takes O(n). So the total complexity of this algorithm is O(n2).

Now, If we want to find out which jobs were performed to get the maximum profit, we need to traverse the array in reverse order and if the Acc_Prof matches the maxProfit, we will push the name of the job in a stack and subtract Profit of that job from maxProfit. We will do this until our maxProfit > 0 or we reach the beginning point of the Acc_Prof array. The pseudo-code will look like:

Procedure FindingPerformedJobs(Job, Acc_Prof, maxProfit):
S = stack()
for i -> n down to 0 and maxProfit > 0
    if maxProfit is equal to Acc_Prof[i]
        S.push(Job[i].name
        maxProfit = maxProfit - Job[i].profit
    endif
endfor

The complexity of this procedure is: O(n).

One thing to remember, if there are multiple job schedules that can give us maximum profit, we can only find one job schedule via this procedure.

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