ReasoningAbility.com

Scheduling - Beginner Level: schedule logic BEGINNER

This foundation builder 🌟 worksheet contains 20 beginner-level scheduling problems. Worksheet 1 of 30 focuses on schedule logic. Practice schedule logic, time allocation, day scheduling with our step-by-step solutions. Difficulty: foundational concepts and basic patterns. Recommended for entry-level learners.

📝 Worksheet 1 of 30 • 20 questions • ⏱️ Estimated time: 20 minutes • 🎯 Beginner level

What you'll learn in this worksheet:
Your progress through Scheduling
Worksheet 1 of 30 (3% complete)

Question 1

Given these scheduling constraints: - Task C must be before Task A - Task B must be after Task A - Task D must be immediately after Task C Is a valid schedule possible?
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
feasibility, satisfiability, constraint-check, logic, verification

Question 2

A project involves two events, Event A (Meeting) and Event B (Training). The constraints are: - **Event A:** Duration 90 minutes. Must start between 9:00 AM and 11:00 AM. - **Event B:** Duration 60 minutes. Must finish by 3:00 PM. - **Gap:** A minimum of 2 hours is required between the end of Event A and the start of Event B. Assuming all constraints must be met, what is the earliest possible start time for Event B?
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.
time-arithmetic, gap-constraint, optimization, banking-exams

Question 3

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)
airline, crew, flight, aviation, rostering

Question 4

A job shop has 3 machines. Jobs and their routes: - Job A: M2 → M1 → M3 with times 14, 17, 36 - Job B: M1 → M3 → M2 with times 32, 40, 23 - Job C: M3 → M2 → M1 with times 26, 28, 36 What is a lower bound on the minimum makespan?
Step-by-step solution:

1. Machine load bound: 102
2. Job processing bound: 95
3. Lower bound: 102

Answer: 102
job-shop, routing, complex, np-hard, advanced

Question 5

A factory produces Widgets with a 85% learning curve (each doubling of cumulative production reduces time by 15%). First unit takes 97 minutes. Batch sizes (in order): 40, 30, 10 units. What is the TOTAL production time for all batches (in minutes, rounded to nearest minute)?
Step-by-step solution (Learning Curve):

1. Learning curve formula: T_n = T_1 × n^-0.234
where exponent = log(0.85)/log(2) = -0.234

2. Calculate cumulative time using integration:
Cumulative time for N units = T_1 × N^0.765534746362977 / (learning_exponent + 1)

3. Time per batch:
Batch 1 (40 units): 53.4 minutes
Batch 2 (30 units): 38.0 minutes
Batch 3 (10 units): 35.3 minutes

4. Total time: 126.7 ≈ 127 minutes

Key Strategy: Learning curve reduces time with repetition; use cumulative average method for batch calculations.
learning-curve, improvement, batch-processing, industrial, experience

Question 6

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: - Prof. Brown can only speak at 11:00-12:00 - AI Ethics and Cybersecurity cannot be in the same time slot - Dr. Taylor and Prof. Wilson must speak in consecutive time slots - Blockchain must be in Hall C Which speaker presents the Cybersecurity session?
Step-by-step solution:

Scheduling Grid Analysis:
1. Fix direct constraints:
- Prof. Brown at 11:00-12:00
- Blockchain in Hall C
2. Apply consecutive constraint: Dr. Taylor and Prof. Wilson in consecutive slots
3. Apply conflict constraint: AI Ethics and Cybersecurity not together

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

Answer: Dr. Smith presents Cybersecurity

Key Strategy: Use a grid to solve the assignment problem and satisfy all constraints sequentially.
multi-constraint, professional, CSP, CAT-level

Question 7

A project involves two events, Event A (Meeting) and Event B (Training). The constraints are: - **Event A:** Duration 90 minutes. Must start between 9:00 AM and 11:00 AM. - **Event B:** Duration 60 minutes. Must finish by 3:00 PM. - **Gap:** A minimum of 2 hours is required between the end of Event A and the start of Event B. Assuming all constraints must be met, what is the earliest possible start time for Event B?
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.
time-arithmetic, gap-constraint, optimization, banking-exams

Question 8

Given these scheduling constraints: - Task D must be before Task C - Task A must be after Task C - Task B must be immediately after Task D Is a valid schedule possible?
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
feasibility, satisfiability, constraint-check, logic, verification

Question 9

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)
airline, crew, flight, aviation, rostering

Question 10

A student has 8 days to prepare for three exams: Mathematics, Physics, Chemistry. The required preparation days are: - Mathematics: 3 days - Physics: 2 days - Chemistry: 2 days If the student follows the optimal schedule starting today, on which day will the last exam be?
Step-by-step solution:

Timeline Planning Method:
1. Calculate total preparation time needed:
- Mathematics: 3 days
- Physics: 2 days
- Chemistry: 2 days
- Total: 7 days

2. Available days: 8 days
3. Extra buffer days: 1 days
4. Optimal schedule:
- Days 1-3: Prepare for Mathematics
- Day 4: Mathematics exam
- Days 5-6: Prepare for Physics
- Day 7: Physics exam
- Days 8-9: Prepare for Chemistry
- Day 10: Chemistry exam

Answer: The last exam will be on Day 8

Key Strategy: Schedule exams immediately after preparation period ends, accounting for all required prep days.
academic, deadline, preparation, timeline

Question 11

A bus route takes 44 minutes one-way. Peak frequency: 1 bus every 12 minutes. What is the minimum number of buses needed to maintain this frequency in both directions?
Step-by-step solution:

1. Round trip time: 44 × 2 = 88 minutes
2. Headway: 12 minutes
3. Buses needed: ⌈88 ÷ 12⌉ = 8

Answer: 8 buses
bus, timetable, frequency, public-transport, optimization

Question 12

A machine needs to process 4 jobs. Processing times: - Job B: 32 minutes - Job D: 77 minutes - Job E: 36 minutes - Job A: 40 minutes The machine breaks down at 99 minutes and takes 31 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 B → Job E → Job A → Job D
2. Simulate processing with breakdown:
- Job B: 0 → 32
- Job E: 32 → 68
- Job A: Starts at 68, breakdown at 99 (31 min completed), repair 31 min, resume 9 min → completes at 139
- Job D: 139 → 216

3. Total makespan: 216 minutes
4. Delay caused by breakdown: 31 minutes

Answer: 216 minutes

Key Strategy: Simulate the timeline, account for breakdown during active job processing.
manufacturing, breakdown, recovery, simulation, reliability

Question 13

A passenger travels from Miami to Denver via Dallas. The minimum layover at Dallas is **60 minutes**. **Flights Miami -> Dallas:** - F1-1: Dep 6:00 AM, Arr 8:45 AM - F1-2: Dep 9:00 AM, Arr 11:45 AM - F1-3: Dep 12:00 PM, Arr 2:45 PM **Flights Dallas -> Denver:** - F2-1: Dep 10:30 AM, Arr 1:11 PM - F2-2: Dep 12:30 PM, Arr 3:11 PM - F2-3: Dep 2:30 PM, Arr 5:11 PM What is the minimum total elapsed time for the journey from Miami to Denver?
1. Timeline Approach & Constraint Application (Minimum Layover: 60 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 431 minutes is achieved by combining F1-1 (Arr: 8:45 AM) and F2-1 (Dep: 10:30 AM, Arr: 1:11 PM).
Total Elapsed Time = Final Arrival Time - Initial Departure Time.

3. Final Answer: The minimum elapsed time is 7 hours and 11 minutes.
transportation, time-interval, optimization, CAT-level

Question 14

A football league has 6 teams. Each team plays every other team twice (home and away). What is the minimum number of rounds needed if each round has the maximum possible matches?
Step-by-step solution:

1. Total matches in double round-robin: 6 × (6-1) = 30
2. Maximum matches per round: 3
3. Minimum rounds: 30 ÷ 3 = 5 rounds

Answer: 5 rounds
league, fixture, home-away, sports, optimization

Question 15

A passenger travels from Denver to Miami via Atlanta. The minimum layover at Atlanta is **60 minutes**. **Flights Denver -> Atlanta:** - F1-1: Dep 9:30 AM, Arr 12:15 PM - F1-2: Dep 12:30 PM, Arr 3:15 PM - F1-3: Dep 3:30 PM, Arr 6:15 PM **Flights Atlanta -> Miami:** - F2-1: Dep 12:00 PM, Arr 3:36 PM - F2-2: Dep 2:00 PM, Arr 5:36 PM - F2-3: Dep 4:00 PM, Arr 7:36 PM What is the minimum total elapsed time for the journey from Denver to Miami?
1. Timeline Approach & Constraint Application (Minimum Layover: 60 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 486 minutes is achieved by combining F1-1 (Arr: 12:15 PM) and F2-2 (Dep: 2:00 PM, Arr: 5:36 PM).
Total Elapsed Time = Final Arrival Time - Initial Departure Time.

3. Final Answer: The minimum elapsed time is 8 hours and 6 minutes.
transportation, time-interval, optimization, CAT-level

Question 16

**Data Sufficiency Question** Five friends A, B, C, D, E are standing in a queue. **Question:** Who is standing at the front of the queue? **Statement (1):** A is standing immediately before C. **Statement (2):** There are exactly two people between B and E. **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: A is standing immediately before C.
This gives partial information but not enough to determine the answer uniquely.

Step 2 - Analyze Statement (2) alone: There are exactly two people between B and E.
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 sufficient but neither alone is sufficient.

Key Strategy: Test each statement independently first, then combine only if neither alone works.
data-sufficiency, GMAT, CAT, reasoning, test-prep

Question 17

A clinic operates for 4 hours with 15-minute appointment slots. If 10 patients need appointments, how many can be accommodated?
Step-by-step solution:

1. Total slots available: (4 × 60) ÷ 15 = 16
2. Patients: 10
3. All patients can be scheduled

Answer: All 10 patients can be scheduled
clinic, appointment, doctor, healthcare, wait-time

Question 18

Real-time tasks with Rate Monotonic Scheduling (shorter period = higher priority): - Task A: Execution 5, Period 40 - Task B: Execution 5, Period 30 - Task C: Execution 3, Period 10 - Task D: Execution 4, Period 30 Is the task set schedulable under RM?
Step-by-step solution:

1. Calculate utilization:
- Task A: 5/40 = 0.125
- Task B: 5/30 = 0.167
- Task C: 3/10 = 0.300
- Task D: 4/30 = 0.133
Total U = 0.725
2. RM schedulability bound for 4 tasks: 0.757
3. Conclusion: Utilization 0.725 ≤ 0.757 (RM bound)

Answer: Schedulable
real-time, deadline, rate-monotonic, embedded, OS

Question 19

A factory produces Widgets with a 90% learning curve (each doubling of cumulative production reduces time by 9%). First unit takes 120 minutes. Batch sizes (in order): 20, 30, 40 units. What is the TOTAL production time for all batches (in minutes, rounded to nearest minute)?
Step-by-step solution (Learning Curve):

1. Learning curve formula: T_n = T_1 × n^-0.152
where exponent = log(0.9)/log(2) = -0.152

2. Calculate cumulative time using integration:
Cumulative time for N units = T_1 × N^0.84799690655495 / (learning_exponent + 1)

3. Time per batch:
Batch 1 (20 units): 89.7 minutes
Batch 2 (30 units): 70.3 minutes
Batch 3 (40 units): 63.1 minutes

4. Total time: 223.1 ≈ 223 minutes

Key Strategy: Learning curve reduces time with repetition; use cumulative average method for batch calculations.
learning-curve, improvement, batch-processing, industrial, experience

Question 20

Project tasks with uncertain durations (optimistic, likely, pessimistic) in days: - Design: (2, 3, 6) - Development: (4, 6, 8) - Testing: (4, 6, 7) - Deployment: (4, 6, 8) Using the PERT formula (O + 4M + P)/6, what is the expected total project duration?
Step-by-step solution (PERT):

1. Calculate expected duration for each task:
- Design: (2 + 4×3 + 6)/6 = 3.3
- Development: (4 + 4×6 + 8)/6 = 6.0
- Testing: (4 + 4×6 + 7)/6 = 5.8
- Deployment: (4 + 4×6 + 8)/6 = 6.0

2. Total expected duration: 21.2 days

Answer: 21.2 days
fuzzy, uncertainty, duration, advanced, real-world
Previous Worksheet Next Worksheet