GATE CSE 2015 Set 1 | Question: 46

Question

Consider a uniprocessor system executing three tasks $T_{1}, T_{2}$ and $T_{3}$ each of which is composed of an infinite sequence of jobs (or instances) which arrive periodically at intervals of $3, 7$ and $20$ milliseconds, respectively. The priority of each task is the inverse of its period, and the available tasks are scheduled in order of priority, which is the highest priority task scheduled first. Each instance of $T_{1}, T_{2}$ and $T_{3}$ requires an execution time of $1, 2$ and $4$ milliseconds, respectively. Given that all tasks initially arrive at the beginning of the $1^{\text{st}}$ millisecond and task preemptions are allowed, the first instance of $T_{3}$ completes its execution at the end of_____________________milliseconds.

Comments

I don’t understand why is there so much confusion about beginning and end. Suppose you consider beginning of 1st millisecond as all processes arrive at 0, then execution of T3 will be over at the range 11 to 12 so by your definition 11 is the beginning of 12th millisecond and 12 is the end of 12th millisecond.
And if suppose you belong to the other category which defines beginning of 1st millisecond as all processes arrive at 1, then execution of T3 will be over at the range 12 to 13 so by your definition 12 is beginning of 12th millisecond and 13 is the end of 12th millisecond.

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Answer is 11 if we start from 0 and 

the answer will be 12 if we start T1 execution  from 1 

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All processes available at the beginning of 1^st clock cycle.

Process T3 completes by the end of 12 clock cycle.

Gantt chart starting from 0 or 1 all that’s not matter.

Even if starting from 0 – total clock cycle is 12 i.e 12 msec

#Beautiful_Question

Answer 1

I think those who started with 1 ms are correct. The question asks "the first instance of T3 completes its execution at the  END of which millisecond and thus 12 is the right answer.

Answer 2

Solution:-

 

Comments

"arriving at beginning of 1ms"

it means 0th millisecond has already over & going to begin 1ms.

so, all jobs are executed from 1ms as the jobs are coming from 1ms.

Answer 3

For those who are getting answers as 13 please note that its 13 units on Gantt chart and all process arrive at 1ms. Hence final result of 13 has to be subtracted by 1.

Hence completion time of T3(in ms) is 13-1=12 ms

Hence answer is 12 ms. 

For clear understanding see the solution below where I have showed 1ms on Gantt chart. 

Answer 4

Tried to explain in simplest manner possible:-