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Please assist with problem below. Please label answer with associated number. Pl

ID: 3856590 • Letter: P

Question

Please assist with problem below. Please label answer with associated number. Please use the layout provided in your answer.

Please answer accurately and claarly....

#1.

Analyze the performance of three scheduling mechanisms: Exponential Averaging, MLFQ, and true ShortestBurstFirst.   Calculate the Average Completion Time AND count the number of context switches.   The workload consists of the following:

P1: 10(3,2,4); P2: 4(2,6); P3: 16(2,4,5,6).      I/O between bursts for 4 time quantums.   Exp Ave default =2, alpha=0.6

MLFQ: 3 queues of 2,3 and 4 quantums for the queues. Processes are executed in FIFO order within each queue.

Executing (initial pred=2, alpha =0.6)

Time

0

Exec

Ready

1,2,3

P1

2

P2

2

P3

2

I/O:4

Ave Ave Averagecompletion time = Context switches =

MLFQ: Executing in Queues

Time

0

Q1:1

Q2:2

Q3:4

MLFQ: In Queues

Q1

1,2,3

Q2

Q3

IO:4

Ave AverageCompletion Time =        Context switches =

True SJF:

Time

0

Executing

ReadyQ

1,2,3

I/O:4

Completion time ave =                                                            .   Context switches =     .

Executing (initial pred=2, alpha =0.6)

Time

0

Exec

Ready

1,2,3

P1

2

P2

2

P3

2

I/O:4

Explanation / Answer

Alright, so after reviewing with the professor, this is how this actually works.

The default of 2 means they all have an equal priority and just get run in order; process P1, P2, and P3.

Now, once we see P1 run for 3 quantums after the first burst, then we calculate its new prediction.

(0.6) * 2 [original default] + (0.4) * 3 [last run] = 1.2 + 1.2 = Prediction of 2.4

Same thing with P2 & P3 (both have a first burst of 2 quantums.)

(0.6) * 2 [original default] + (0.4)* 2 [last run] = 1.2 + 0.8 = Prediction of 2.0

So for the next quantums' order, it would be P2, P3, THEN P1.

Then after that run you would just do;

(0.6) * [2nd to last burst] + (0.4) * [last burst] = new prediction to figure out which proces would run next.

So the actual order would be;

Quantum: Process

Q3: P1:3

Q5: P2:2

Q7: P3:2

Q8: P2:2 (FINISHES)

Q12: P3:3

Q14: P1:2

Q19: P3:5

Q24: P1:5 (FINISHES)

Q30: P3:6 (FINISHES)