Scheduling - Beginner Level: timetable puzzles BEGINNER

Level up your scheduling skills with this entry level practice. 20 beginner-level problems await in Worksheet 4 of 30. Focus area: timetable puzzles. Learn timetable puzzles, appointment logic, calendar scheduling through systematic practice. Designed for entry-level learners seeking foundational concepts and basic patterns.

📝 Worksheet 4 of 30 • 20 questions • ⏱️ Estimated time: 20 minutes • 🎯 Beginner level

What you'll learn in this worksheet:
Your progress through Scheduling
Worksheet 4 of 30 (13% complete)

Question 1

Events need to be scheduled in rooms. Their time intervals are: - Event A: 1:00 to 5:00 - Event B: 15:00 to 16:00 - Event C: 19:00 to 26:00 - Event D: 12:00 to 16:00 - Event E: 20:00 to 23:00 - Event F: 16:00 to 21:00 - Event G: 14:00 to 21:00 What is the minimum number of rooms needed to schedule all events without overlap?
Step-by-step solution (Interval Graph):

1. Plot intervals on timeline:
Event A: ████ from 1 to 5
Event B: █ from 15 to 16
Event C: ███████ from 19 to 26
Event D: ████ from 12 to 16
Event E: ███ from 20 to 23
Event F: █████ from 16 to 21
Event G: ███████ from 14 to 21

2. Find maximum overlap:
Maximum 4 events overlap at once

Answer: 4 rooms needed

Question 2

In a round-robin tournament with 8 teams, each round consists of disjoint matches (no team plays twice in a round). What is the minimum number of rounds needed?
Step-by-step solution:

1. **Model as edge coloring of complete graph K_8
2. Vizing's theorem: χ'(K_n) = n-1 for even n, n for odd n
3. For 8 teams: 7 colors/rounds needed

Answer: 7 rounds

Question 3

Four colleagues need to schedule a meeting. Their available time slots are: - 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 What is the earliest time slot when all four can meet?
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.

Question 4

A factory has 3 production lines: Line 1, Line 2, Line 3. Three products require the following operations: **Product X:** - Cut: 30 min on Line 2 - Assemble: 45 min on Line 1 - Package: 15 min on Line 3 **Product Y:** - Cut: 20 min on Line 1 - Assemble: 60 min on Line 2 - Package: 20 min on Line 1 **Product Z:** - Cut: 40 min on Line 2 - Assemble: 30 min on Line 1 - Package: 25 min on Line 3 All products must be completed (all 3 operations each). Multiple operations can run in parallel on different lines. Which production line is the bottleneck, and what is its total load (in minutes)?
Step-by-step solution (Bottleneck Analysis):

1. Calculate total load per production line:
- Line 1: 115 minutes
- Line 2: 130 minutes
- Line 3: 40 minutes

2. Identify bottleneck: The line with maximum load = Line 2
3. Bottleneck load: 130 minutes

Answer: Line 2 (130 minutes)

Key Strategy: The bottleneck determines maximum throughput; optimize the bottleneck first for overall efficiency.

Question 5

Real-time tasks with Rate Monotonic Scheduling (shorter period = higher priority): - Task A: Execution 16, Period 50 - Task B: Execution 8, Period 30 - Task C: Execution 8, Period 30 - Task D: Execution 3, Period 10 Is the task set schedulable under RM?
Step-by-step solution:

1. Calculate utilization:
- Task A: 16/50 = 0.320
- Task B: 8/30 = 0.267
- Task C: 8/30 = 0.267
- Task D: 3/10 = 0.300
Total U = 1.153
2. RM schedulability bound for 4 tasks: 0.757
3. Conclusion: Utilization 1.153 > 0.757 (RM bound)

Answer: May not be schedulable

Question 6

An airline crew has the following flights: - Flight 101: 08:00 → 10:00 - Flight 102: 10:30 → 12:30 - Flight 103: 13:00 → 15:00 - Flight 104: 15:30 → 17:30 - Flight 105: 18:00 → 20:00 Crew duty time limit is 8 hours. Minimum connection time between flights is 30 minutes. What is the maximum number of flights a crew can operate in a single duty period?
Step-by-step solution:

1. Convert all times to minutes for easier calculation:
- Flight 101: Departs at 8:00, Arrives at 10:00
- Flight 102: Departs at 10:30, Arrives at 12:30
- Flight 103: Departs at 13:00, Arrives at 15:00
- Flight 104: Departs at 15:30, Arrives at 17:30
- Flight 105: Departs at 18:00, Arrives at 20:00

2. Duty time limit: 480 minutes (8 hours)
3. Minimum connection time: 30 minutes

4. Find optimal sequence of flights:
Best sequence found: Flight 103 → Flight 104 → Flight 105
- Take Flight 103: Departs at 13:00
- Connection time: 30 minutes
- Take Flight 104: Departs at 15:30
- Connection time: 30 minutes
- Take Flight 105: Departs at 18:00

Total duty time: 420 minutes (7 hours, 0 minutes)

5. Maximum flights possible: 3

Answer: 3 flights

✓ Duty time check: 7h 0m ≤ 8h (PASSED)

Question 7

A flow shop has 2 machines (M1 → M2). Jobs and processing times (M1, M2): - Job A: (47, 35) - Job B: (33, 25) - Job C: (31, 30) - Job D: (28, 22) Using Johnson's Rule, what is the minimum makespan?
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 A → Job C → Job B → Job D
3. Calculate makespan: 161

Answer: 161

Question 8

Five patients need appointments. Their preferences are: - Patient A: Dr. Patel at 10:00 AM - Patient B: Dr. Kumar at 10:00 AM - Patient C: Dr. Patel at 10:30 AM - Patient D: Dr. Shah at 10:00 AM - Patient E: Dr. Patel at 11:00 AM Each doctor can see one patient per 30-minute slot. If all preferences are honored, how many patients need to be rescheduled?
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.

Question 9

Computer Science courses with prerequisites: - CS201 requires CS202 - CS301 requires CS201 - CS401 requires CS202 - CS102 requires CS201 Which courses can be taken in the first semester?
Step-by-step solution:

1. Identify courses with prerequisites:
- CS201 needs CS202
- CS301 needs CS201
- CS401 needs CS202
- CS102 needs CS201

2. Courses without prerequisites (can take first): CS202

Answer: CS202

Question 10

Eight people attend seminars in four different months (January, March, May, July) on two dates (5th and 15th). Two people attend per month. The constraints are: - V attends in March - T attends on the 15th - Exactly two people attend between S and R - P attends in the same month as U In which month does S attend?
Step-by-step solution:

Timeline Grid Method:
1. Create month-date grid:
Jan 5 | Jan 15 | Mar 5 | Mar 15 | May 5 | May 15 | Jul 5 | Jul 15

2. Apply constraints:
- V in March (Mar 5 or Mar 15)
- T on 15th (any month, date 15)
- Two people between S and R
(If S at position 1, R at position 4)
- P and U in same month

3. Systematic placement:
- Place V at Mar 5 (satisfies March constraint)
- Place T at Mar 15 (satisfies 15th constraint)
- For 'two between' constraint: If S at Jan 5, R at Mar 15
- P and U together: May 5 & May 15

4. Verification:
All constraints satisfied with S in March

Key Strategy: Use grid to visualize all slots, apply direct constraints first, then deduce positions using gap constraints.

Question 11

Four tasks (Task 2, Task 3, Task 1, Task 4) must be scheduled with these constraints: 1. Task 2 must be before Task 3 2. Task 3 must be before Task 1 3. Task 2 must be before Task 1 4. Task 4 must be after Task 1 Which constraint is REDUNDANT (does not add new information beyond the others)?
Step-by-step solution (Redundancy Detection):

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

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

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

Answer: Constraint 3 is redundant

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

Question 12

A job shop has 3 machines. Jobs and their routes: - Job A: M1 → M2 → M3 with times 12, 24, 39 - Job B: M1 → M2 → M3 with times 23, 24, 15 - Job C: M1 → M2 → M3 with times 36, 34, 24 What is a lower bound on the minimum makespan?
Step-by-step solution:

1. Machine load bound: 82
2. Job processing bound: 94
3. Lower bound: 94

Answer: 94

Question 13

**Data Sufficiency Question** A project has 4 phases: Planning, Design, Development, Testing. **Question:** On which day does Development start? **Statement (1):** Testing starts exactly 5 days after Design ends. **Statement (2):** Planning takes 3 days and ends before Design starts. **Options:** A. Statement (1) ALONE is sufficient, but statement (2) alone is NOT sufficient B. Statement (2) ALONE is sufficient, but statement (1) alone is NOT sufficient C. BOTH statements TOGETHER are sufficient, but NEITHER alone is sufficient D. EACH statement ALONE is sufficient E. Statements (1) and (2) TOGETHER are NOT sufficient
Data Sufficiency Reasoning:

Step 1 - Analyze Statement (1) alone: Testing starts exactly 5 days after Design ends.
This gives partial information but not enough to determine the answer uniquely.

Step 2 - Analyze Statement (2) alone: Planning takes 3 days and ends before Design starts.
This also gives partial information insufficient by itself.

Step 3 - Combine statements:
Together, they provide enough constraints to solve uniquely.

Conclusion: Both statements together are still insufficient.

Key Strategy: Test each statement independently first, then combine only if neither alone works.

Question 14

A manager has 4 tasks to complete over 8 working hours. The task details are: - Report: Priority High, Duration 3 hours, Deadline 5 hours - Email: Priority Low, Duration 1 hours, Deadline 6 hours - Presentation: Priority High, Duration 2 hours, Deadline 4 hours - Analysis: Priority Medium, Duration 2 hours, Deadline 7 hours If tasks are scheduled based on priority first and deadline second, which task should be completed first?
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.

Question 15

Computer Science courses with prerequisites: - CS101 requires CS401 - CS301 requires CS401 - CS102 requires CS401 - CS201 requires CS401 Which courses can be taken in the first semester?
Step-by-step solution:

1. Identify courses with prerequisites:
- CS101 needs CS401
- CS301 needs CS401
- CS102 needs CS401
- CS201 needs CS401

2. Courses without prerequisites (can take first): CS401

Answer: CS401

Question 16

A machine needs to process 4 jobs. Processing times: - Job A: 56 minutes - Job D: 55 minutes - Job C: 79 minutes - Job E: 86 minutes The machine breaks down at 79 minutes and takes 21 minutes to repair. Jobs are scheduled using Shortest Processing Time (SPT) first rule. What is the total completion time (makespan) after handling the breakdown?
Step-by-step solution (Breakdown Recovery):

1. Original SPT order: Job D → Job A → Job C → Job E
2. Simulate processing with breakdown:
- Job D: 0 → 55
- Job A: Starts at 55, breakdown at 79 (24 min completed), repair 21 min, resume 32 min → completes at 132
- Job C: 132 → 211
- Job E: 211 → 297

3. Total makespan: 297 minutes
4. Delay caused by breakdown: 21 minutes

Answer: 297 minutes

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

Question 17

In a round-robin tournament with 10 teams, each round consists of disjoint matches (no team plays twice in a round). What is the minimum number of rounds needed?
Step-by-step solution:

1. **Model as edge coloring of complete graph K_10
2. Vizing's theorem: χ'(K_n) = n-1 for even n, n for odd n
3. For 10 teams: 9 colors/rounds needed

Answer: 9 rounds

Question 18

Events need to be scheduled in rooms. Their time intervals are: - Event A: 20:00 to 21:00 - Event B: 8:00 to 14:00 - Event C: 1:00 to 2:00 - Event D: 7:00 to 15:00 - Event E: 3:00 to 11:00 - Event F: 1:00 to 6:00 - Event G: 3:00 to 11:00 - Event H: 11:00 to 16:00 What is the minimum number of rooms needed to schedule all events without overlap?
Step-by-step solution (Interval Graph):

1. Plot intervals on timeline:
Event A: █ from 20 to 21
Event B: ██████ from 8 to 14
Event C: █ from 1 to 2
Event D: ████████ from 7 to 15
Event E: ████████ from 3 to 11
Event F: █████ from 1 to 6
Event G: ████████ from 3 to 11
Event H: █████ from 11 to 16

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

Answer: 5 rooms needed

Question 19

A conference needs to schedule 6 sessions across 3 time slots and 3 rooms. Each room can hold one session per slot. The constraints are: - Dr. Smith can only speak at 11:00-12:00 - AI Ethics and Machine Learning cannot be in the same time slot - Dr. Lee and Prof. Brown must speak in consecutive time slots - Cloud Computing must be in Hall A Which speaker presents the Machine Learning session?
Step-by-step solution:

Scheduling Grid Analysis:
1. Fix direct constraints:
- Dr. Smith at 11:00-12:00
- Cloud Computing in Hall A
2. Apply consecutive constraint: Dr. Lee and Prof. Brown in consecutive slots
3. Apply conflict constraint: AI Ethics and Machine Learning not together

4. Final Schedule:
9:00-10:00:
- Hall A: Cloud Computing by Dr. Chen
- Hall B: AI Ethics by Prof. Jones
- Hall C: Blockchain by Dr. Lee
10:00-11:00:
- Hall A: Data Science by Prof. Wilson
- Hall B: Machine Learning by Prof. Brown
- Hall C: (empty)
11:00-12:00:
- Hall A: Robotics by Dr. Smith
- Hall B: (empty)
- Hall C: (empty)

Answer: Prof. Brown presents Machine Learning

Key Strategy: Use a grid to solve the assignment problem and satisfy all constraints sequentially.

Question 20

A JIT manufacturing system has 4 jobs with the following data: | Job | Processing (min) | Due Date (min) | Early Penalty/min | Late Penalty/min | |-----|-----------------|----------------|-------------------|------------------| | Component A | 33 | 68 | 2 | 15 | | Component B | 22 | 82 | 5 | 19 | | Component D | 20 | 78 | 3 | 11 | | Component C | 28 | 57 | 2 | 12 | Using the Earliest Due Date (EDD) sequencing rule, what is the total penalty incurred?
Step-by-step solution (JIT Penalty Calculation):

1. EDD Sequence: Component C → Component A → Component D → Component B
2. Calculate completion times and penalties:
- Component C: completes at 28, due 57, early by 29 min → penalty 58
- Component A: completes at 61, due 68, early by 7 min → penalty 14
- Component D: completes at 81, due 78, late by 3 min → penalty 33
- Component B: completes at 103, due 82, late by 21 min → penalty 399

3. Total penalty: 504

Answer: 504 penalty points

Key Strategy: JIT scheduling minimizes total earliness + tardiness penalties, balancing inventory costs and customer satisfaction.
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