1k views

We are given 9 tasks $T_1, T_2, \dots, T_9$. The execution of each task requires one unit of time. We can execute one task at a time. Each task $T_i$ has a profit $P_i$ and a deadline $d_i$. Profit $P_i$ is earned if the task is completed before the end of the $d_i^{th}$ unit of time.

Task $T_1$ $T_2$ $T_3$ $T_4$ $T_5$ $T_6$ $T_7$ $T_8$ $T_9$
Profit $15$ $20$ $30$ $18$ $18$ $10$ $23$ $16$ $25$
Deadline $7$ $2$ $5$ $3$ $4$ $5$ $2$ $7$ $3$

Are all tasks completed in the schedule that gives maximum profit?

2. $T_1$ and $T_6$ are left out

3. $T_1$ and $T_8$ are left out

4. $T_4$ and $T_6$ are left out

edited | 1k views

Question has been splitted. Please check the below URL for the B part https://gateoverflow.in/82514/gate2005-84b

Please refer the below URL for video lecture on this topic(Task Scheduling Problem).

Step -1 Sort the tasks in decreasing order of profit and if any conflict arises between two or more tasks,resolve them by sorting them on basis of having greater deadline first(Because we have more time to complete the task with greater deadline and same profit).

Step 2- Since Maximum deadline given is $7$, so we consider we have 7 time slots ranging from $0-7$ where a task $T_i$ having deadline say $2$ can be filled in slots either $0-1$ or $1-2$ and not beyond $2$ because this task has deadline of $2$ time units, so this task has to be completed by atmost time $T=2$.

Now according to question, since Each task completes in Unit time, so a single tasks takes only one slot as shown.

Now Take the first task in the list i.e. $T_3$ which has a deadline of $5$, so it can be completed in maximum $5$ time units, so place it in slot $4-5$ which is the maximum deadline by which this task can be completed.

Task $T_9$ with deadline $3$ is similarly placed in  slot $2-3$.

Task $T_7$ with deadline $2$ is placed in slot $1-2$.

Now for task $T_2$ having deadline $2$ can be placed in either $0-1$ or $1-2$ (Occupied by $T_7$). So $T_2$ will occupy slot $0-1$.

Task $T_5$ with deadline $4$ is placed in slot $3-4$.

Now comes task $T_4$ which has deadline $3$ can be put in slots $0-1$ or $1-2$ or $2-3$ and not beyond that.Unfortunately, all such slots are occupied so $T_4$ will be left out.

Task $T_8$ with deadline $7$ goes in slot $6-7$.

Task $T_1$ with deadline $7$ can be placed in  slot $5-6$.

Now all time slots are full.

So, Task $T_6$ will be left out.

So, option (d) is the answer.

edited by
Best Explanation. Thanks

The most important statement in question is

each task requires one unit of time

This shows that we can greedily choose the better task and that should give us the optimal solution. The best task would be the one with maximum profit. Thus we can sort the tasks based on deadline and then profit as follows:

Task T7 T2 T9 T4 T5 T3 T6 T8 T1
Deadline 2 2 3 3 4 5 5 7 7

0 ----T7 ----- 1----T2-----2-----T9-----3-----T5-----4-----T3-----5-----T8-----6-----T1------7

T4 and T6 left out

edited
Can you explain why T4 and T6 are left out

Since we have to be greedy about profit, so first we have to choose a task with max profit considering its deadline.

 Time 1 2 3 4 5 6 7 Order Of Task T7 T2 T9 T5 T3 T8 T1 Profit 23 20 25 18 30 16 15 deadline 2 2 3 4 5 7 7

TOTAL MAXIMUM PROFIT = (23 + 20 + 25 + 18 + 30 + 16 + 15) = 147.

@vijay can you provide some reference material to understand this ?
why this sequence is wrong ?

T2-------T7------T9------T5---------T3--------T1------T8?????

each task requires one unit of time,WHT DOES IT MEAN ?