Consider the set of 5 processes whose burst times are given below:| Process Id | Burst time ||------------|------------|| P1 | 5 || P2 | 3 || P3 | 1 || P4 | 2 || P5 | 3 |If the CPU scheduling policy is the Ideal Fair Scheduler algorithm with a time quantum of 5 units, calculate the average waiting time. Please explain.### Explanation:In the Ideal Fair Scheduler algorithm:1. **Time Quantum (Q):** 5 units2. **Process Burst Times:** - P1: 5 units - P2: 3 units - P3: 1 unit - P4: 2 units - P5: 3 unitsEach process gets a maximum of 5 units of CPU time per round. Here, since all the processes require less than or equal to 5 units, each process completes in one round.3. **Calculation of Waiting Time:** - P1 has no waiting time. - P2 needs to wait until P1 finishes (5 units), so waiting time = 5. - P3 needs to wait until P1 and P2 finish (5 + 3 units), so waiting time = 8. - P4 needs to wait until P1, P2, and P3 finish (5 + 3 + 1 units), so waiting time = 9. - P5 needs to wait until P1, P2, P3, and P4 finish (5 + 3 + 1 + 2 units), so waiting time = 11.4. **Average Waiting Time:** - \( \text{Average Waiting Time} = \frac{(0 + 5 + 8 + 9 + 11)}{5} = \frac{33}{5} = 6.6 \)The average waiting time for these processes under the Ideal Fair Scheduler algorithm with a time quantum of 5 units is 6.6 units.

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Consider the set of 5 processes whose burst times are given below Process Burst time Id P1 P2 P3 P4 P5 5 3 1 2 3

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A parallel storage file system is a sort of clustered file system. A clustered file system is a storage system shared by multiple devices simultaneously.

Question

Consider the set of 5 processes whose burst times are given below:

| Process Id | Burst time |
|------------|------------|
| P1 | 5 |
| P2 | 3 |
| P3 | 1 |
| P4 | 2 |
| P5 | 3 |

If the CPU scheduling policy is the Ideal Fair Scheduler algorithm with a time quantum of 5 units, calculate the average waiting time. Please explain.

### Explanation:

In the Ideal Fair Scheduler algorithm:

1. **Time Quantum (Q):** 5 units
2. **Process Burst Times:**
- P1: 5 units
- P2: 3 units
- P3: 1 unit
- P4: 2 units
- P5: 3 units

Each process gets a maximum of 5 units of CPU time per round. Here, since all the processes require less than or equal to 5 units, each process completes in one round.

3. **Calculation of Waiting Time:**
- P1 has no waiting time.
- P2 needs to wait until P1 finishes (5 units), so waiting time = 5.
- P3 needs to wait until P1 and P2 finish (5 + 3 units), so waiting time = 8.
- P4 needs to wait until P1, P2, and P3 finish (5 + 3 + 1 units), so waiting time = 9.
- P5 needs to wait until P1, P2, P3, and P4 finish (5 + 3 + 1 + 2 units), so waiting time = 11.

4. **Average Waiting Time:**
- \( \text{Average Waiting Time} = \frac{(0 + 5 + 8 + 9 + 11)}{5} = \frac{33}{5} = 6.6 \)

The average waiting time for these processes under the Ideal Fair Scheduler algorithm with a time quantum of 5 units is 6.6 units.

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