Scheduling - Beginner Level: calendar scheduling BEGINNER

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. **Model as edge coloring of complete graph K_4
2. Vizing's theorem: χ'(K_n) = n-1 for even n, n for odd n
3. For 4 teams: 3 colors/rounds needed

Answer: 3 rounds

Step-by-step solution:

Timeline Approach:
1. Convert all times to 24-hour format for easy comparison
- Evening Walk: 6:00 PM
- Lunch Break: 1:00 PM
- Morning Yoga: 6:00 AM
- Office Work: 9:00 AM

2. Arrange in chronological order:
1. Lunch Break at 1:00 PM
2. Morning Yoga at 6:00 AM
3. Evening Walk at 6:00 PM
4. Office Work at 9:00 AM

Final Schedule: Lunch Break -> Morning Yoga -> Evening Walk -> Office Work

Key Strategy: Convert all times to 24-hour format and arrange from earliest to latest.

Step-by-step solution:

1. Total surgery time: 381 min = 6.3 hours
2. Available OR hours: 2 × 8 = 16 hours
3. Total ≤ Available → Can complete in one day

Answer: All surgeries can be scheduled within one day

Step-by-step solution:

1. Jobs per batch: 3
2. Number of batches: ⌈14 ÷ 3⌉ = 5
3. Total time: 5 × 32 = 160 minutes

Answer: 160 minutes

No valid schedule found given the constraints. The only guaranteed assignment is: Alice must handle Testing.
If the constraints cannot all be satisfied, fallback is to force rule 1's assignment.

Step-by-step solution:

1. Rotation cycle: 5 employees × 7 days = 35 days
2. Verification: Each employee cycles through all shifts

Answer: 35 days

Step-by-step solution:

1. **Model as edge coloring of complete graph K_6
2. Vizing's theorem: χ'(K_n) = n-1 for even n, n for odd n
3. For 6 teams: 5 colors/rounds needed

Answer: 5 rounds

Step-by-step solution:

Priority-Deadline Scheduling Algorithm:
1. Assign priority weights:
- High = 3, Medium = 2, Low = 1

2. Create priority-deadline table:
Task | Priority | Deadline | Duration
--------------|----------|----------|----------
Report | High | 5 | 3
Email | Low | 6 | 1
Presentation | High | 4 | 2
Analysis | Medium | 7 | 2

3. Sorting criteria:
- Primary: Highest priority first
- Secondary: Earliest deadline (if priority is same)

4. Sorted order:
1. Presentation (Priority: High, Deadline: 4)
2. Report (Priority: High, Deadline: 5)
3. Analysis (Priority: Medium, Deadline: 7)
4. Email (Priority: Low, Deadline: 6)

Answer: Presentation should be completed first

Key Strategy: Sort by priority first (descending), then by deadline (ascending) for tasks with equal priority.

Step-by-step solution (Interval Graph):

1. Plot intervals on timeline:
Event A: ██████ from 15 to 21
Event B: ██████ from 15 to 21
Event C: █ from 19 to 20
Event D: █ from 20 to 21
Event E: ██ from 19 to 21
Event F: ████████ from 5 to 13
Event G: ███ from 8 to 11
Event H: ████████ from 1 to 9

2. Find maximum overlap:
Maximum 5 events overlap at once

Answer: 5 rooms needed

Step-by-step solution (Redundancy Detection):

1. List all constraints:
1. Task 3 must be before Task 2
2. Task 2 must be before Task 4
3. Task 3 must be before Task 4
4. Task 1 must be after Task 4

2. Check for transitive relationships:
- From Constraint 1: Task 3 before Task 2
- From Constraint 2: Task 2 before Task 4
- By transitivity: Task 3 before Task 4
- This makes Constraint 3 unnecessary (redundant)

3. Verify other constraints are independent:
- Constraint 4 (Task 1 after Task 4) adds unique information

Answer: Constraint 3 is redundant

Key Strategy: Look for transitive relationships (A→B, B→C implies A→C).

Step-by-step solution:

1. Total demand: 24
2. Vehicle capacity: 13
3. Minimum vehicles: ⌈24 ÷ 13⌉ = 2

Answer: 2 vehicles

Step-by-step solution:

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

Answer: 2 machines

Step-by-step solution:

1. Round trip time: 38 × 2 = 76 minutes
2. Headway: 15 minutes
3. Buses needed: ⌈76 ÷ 15⌉ = 6

Answer: 6 buses

Step-by-step solution:

1. Pareto dominance: Schedule X dominates Y if X is better in at least one objective and not worse in others
2. Pareto frontier schedules: Schedule A, Schedule B, Schedule C, Schedule D, Schedule E

Answer: Schedule A, Schedule B, Schedule C, Schedule D, Schedule E

Step-by-step solution:

1. **Model as edge coloring of complete graph K_6
2. Vizing's theorem: χ'(K_n) = n-1 for even n, n for odd n
3. For 6 teams: 5 colors/rounds needed

Answer: 5 rounds

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 396 minutes is achieved by combining F1-1 (Arr: 11:30 AM) and F2-1 (Dep: 1:00 PM, Arr: 4:06 PM).
Total Elapsed Time = Final Arrival Time - Initial Departure Time.

3. Final Answer: The minimum elapsed time is 6 hours and 36 minutes.

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. Total invigilator slots per time: 2
2. Minimum teachers needed: At least 2 (one per invigilator slot)
3. Same teachers can invigilate multiple slots

Answer: 2 teachers

Step-by-step solution:

1. Total matches in double round-robin: 7 × (7-1) = 42
2. Maximum matches per round: 3
3. Minimum rounds: 42 ÷ 3 = 7 rounds

Answer: 7 rounds