Scheduling - Advanced Level: time slots ADVANCED

Step-by-step solution:

Conflict Detection Method:
1. Create doctor-time matrix:
Doctor | 10:00 | 10:30 | 11:00 | 11:30
------------|-------|-------|-------|-------
Dr. Patel | A | C | E | ---
Dr. Kumar | B | --- | --- | ---
Dr. Shah | D | --- | --- | ---

2. Check for conflicts:
- Dr. Patel at 10:00: Only Patient A (No conflict)
- Dr. Patel at 10:30: Only Patient C (No conflict)
- Dr. Patel at 11:00: Only Patient E (No conflict)
- Dr. Kumar at 10:00: Only Patient B (No conflict)
- Dr. Shah at 10:00: Only Patient D (No conflict)

3. Verification:
- No doctor has multiple patients in same slot
- All preferences can be honored

Answer: 0 patients need rescheduling

Key Strategy: Map all appointments to a doctor-time grid and identify slots where multiple patients request the same doctor.

Step-by-step solution:

1. Total matches in double round-robin: 9 × (9-1) = 72
2. Maximum matches per round: 4
3. Minimum rounds: 72 ÷ 4 = 9 rounds

Answer: 9 rounds

Step-by-step solution:

1. Total machines: 4
2. Maintenance duration: 2 day
3. Staggered schedule: Never have all machines down simultaneously
4. Maximum operational: 4 - 1 = 3

Answer: 3 machines

Step-by-step solution (Time Arithmetic):

1. Goal: To find the earliest start time for Event B, we must use the earliest possible schedule for Event A.
2. Calculate Earliest Finish Time for Event A:
- Earliest Start for A: 9:00 AM
- Duration of A: 90 minutes (1 hour 30 minutes)
- Earliest Finish for A: 9:00 AM + 1 hour 30 minutes = 10:30 AM.
3. Apply Minimum Gap:
- Earliest Start for B = (Earliest Finish A) + (Minimum Gap)
- Minimum Gap: 2 hours (120 minutes)
- Earliest Start for B: 10:30 AM + 2 hours = 12:30 PM.
4. Check Deadline for Event B:
- If B starts at 12:30 PM, its finish time is 12:30 PM + 60 minutes = 1:30 PM.
- The latest finish time for B is 3:00 PM. Since 1:30 PM is before 3:00 PM, the schedule is valid.
Answer: The earliest possible start time for Event B is 12:30 PM.
Key Strategy: To find the minimum time for the second event, use the minimum time for the first event, plus the mandatory gap.

Step-by-step solution:

1. Round Robin simulation:
2. Completion times:
- P1: 15
- P2: 25
- P3: 26
- P4: 30

3. Average: 96 ÷ 4 = 24.0

Answer: 24.0

Step-by-step solution (Fairness Scheduling):

1. Total shifts to assign:
- 7 days × 3 shifts = 21 shifts
2. Shifts per person: 21 ÷ 5 = 4 with 1 extra shifts
3. Undesirable weight distribution:
- Bob: 1 points
- Carol: 3 points
- David: 5 points
- Emma: 3 points
- Frank: 3 points

4. Fairness gap: 5 - 1 = 4

Key Strategy: Fair scheduling aims to minimize the maximum difference in undesirable shift assignments across all staff.

Step-by-step solution:

Set Intersection Method:
1. List all availability:
- Alex: {9:00 AM, 10:00 AM, 2:00 PM, 3:00 PM}
- Ben: {10:00 AM, 11:00 AM, 2:00 PM}
- Cara: {9:00 AM, 11:00 AM, 12:00 PM, 3:00 PM}
- Diana: {10:00 AM, 12:00 PM, 2:00 PM, 3:00 PM}

2. Find common slots (intersection):
- Common to all = Alex AND Ben AND Cara AND Diana
- Result: Empty set (No common time)

3. Conclusion: No common time slot available

Key Strategy: Use set intersection to find common availability, then choose the earliest time.

Step-by-step solution:

1. Patients: 12
2. Capacity per nurse: 3
3. Minimum nurses: ⌈12 ÷ 3⌉ = 4

Answer: 4 nurses

1. Timeline Approach & Constraint Application (Minimum Layover: 90 min):
The fastest total time is found by checking all 9 combinations and ensuring the layover time (F2 Dep Time - F1 Arr Time) is at least the minimum required.

2. Optimal Path Calculation:
The minimum elapsed time of 477 minutes is achieved by combining F1-1 (Arr: 11:58 AM) and F2-2 (Dep: 2:00 PM, Arr: 4:57 PM).
Total Elapsed Time = Final Arrival Time - Initial Departure Time.

3. Final Answer: The minimum elapsed time is 7 hours and 57 minutes.

Step-by-step solution:

1. Check for cycles: No circular dependencies
2. Check immediate constraints: Can be satisfied
3. Conclusion: Yes, a valid schedule exists

Answer: Yes, a valid schedule exists

Step-by-step solution:

1. Single elimination principle: Each match eliminates exactly one team
2. Teams to eliminate: 16 - 1 = 15 teams must be eliminated
3. Matches needed: 15 matches

Answer: 15 matches

Step-by-step solution (Johnson's Rule):

1. Apply Johnson's Rule:
- If M1 time < M2 time, schedule early
- If M2 time < M1 time, schedule late
2. Optimal sequence: Job E → Job A → Job D → Job C → Job B
3. Calculate makespan: 220

Answer: 220

Step-by-step solution (Breakdown Recovery):

1. Original SPT order: Job D → Job A → Job C → Job B
2. Simulate processing with breakdown:
- Job D: 0 → 54
- Job A: Starts at 54, breakdown at 107 (53 min completed), repair 22 min, resume 10 min → completes at 139
- Job C: 139 → 206
- Job B: 206 → 286

3. Total makespan: 286 minutes
4. Delay caused by breakdown: 22 minutes

Answer: 286 minutes

Key Strategy: Simulate the timeline, account for breakdown during active job processing.

Step-by-step solution:

Production Sequencing with Setup Times:
1. Calculate total production time (without setup):
- Product A: 1 x 4 = 4 hours
- Product C: 2 x 3 = 6 hours
- Product D: 3 x 1 = 3 hours
- Base production time: 13 hours

2. Minimize setup time by batching:
- Optimal sequence: Product C -> Product C -> Product A -> Product D -> Product D -> Product D
3. Total with setups:
- Product C: 3 hours
- Product C: 3 hours (no setup)
- Setup Product C→Product A: 1 hour + Product A: 4 hours
- Setup Product A→Product D: 1 hour + Product D: 1 hours
- Product D: 1 hours (no setup)
- Product D: 1 hours (no setup)

Total: 15 hours

Key Strategy: Batch identical products together to minimize setup changes.

Step-by-step solution:

Network Path Analysis:
1. Identify all possible routes from City A to City D:
- City A→City B -> City B→City D
- City A→City C -> City C→City D
- City A→City B -> City B→City C -> City C→City D

2. Best route found:
- T2: City A to City C (06:30 - 9:28 AM)
- Connection time: 122 minutes
- T5: City C to City D (11:30 - 2:29 PM)

Earliest arrival: 2:29 PM

Key Strategy: Enumerate all possible routes, verify connection times meet minimum requirements.

Step-by-step solution:

1. Total surgery time: 408 min = 6.8 hours
2. Available OR hours: 3 × 8 = 24 hours
3. Total ≤ Available → Can complete in one day

Answer: All surgeries can be scheduled within one day

Optimal sequence examined for all 24 permutations.
Minimum time: 266 minutes. Sequence: P4 - P3 - P1 - P2.
Calculation: P4 → P3 → P1 → P2.
Time: PT(P4) + (ST(P4→P3) + PT(P3)) + ...

Step-by-step solution:

1. Total slots available: (4 × 60) ÷ 15 = 16
2. Patients: 9
3. All patients can be scheduled

Answer: All 9 patients can be scheduled

Step-by-step solution:

1. Total machines: 5
2. Maintenance duration: 2 day
3. Staggered schedule: Never have all machines down simultaneously
4. Maximum operational: 5 - 1 = 4

Answer: 4 machines

Step-by-step solution (Deductive Logic):

1. Apply Consecutive Constraint: 'A immediately after C' -> (C, A)
2. Apply Before Constraint: 'D before E'
3. Apply Exclusion Constraint: 'B not first or last'
4. Apply Gap Constraints: 'B is 1 after E'

Final Sequence: C → A → D → B → E

Answer: The task in the fourth position is B.

Key Strategy: Use fixed pairs and gap constraints to anchor positions.