Scheduling Worksheet 26 - Advanced Level (weekly schedule) | 20 Questions

Question 1

A factory produces Widgets with a 80% learning curve (each doubling of cumulative production reduces time by 19%). First unit takes 65 minutes. Batch sizes (in order): 10, 30, 20 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^-0.322
where exponent = log(0.8)/log(2) = -0.322

2. Calculate cumulative time using integration:
Cumulative time for N units = T_1 × N^0.6780719051126377 / (learning_exponent + 1)

3. Time per batch:
Batch 1 (10 units): 45.7 minutes
Batch 2 (30 units): 23.8 minutes
Batch 3 (20 units): 18.5 minutes

4. Total time: 87.9 ≈ 88 minutes

Key Strategy: Learning curve reduces time with repetition; use cumulative average method for batch calculations.

learning-curve, improvement, batch-processing, industrial, experience

Question 2

A delivery company has vehicles with capacity 18 units. Customer demands: - C1: 3 units - C2: 4 units - C3: 4 units - C4: 7 units - C5: 7 units What is the minimum number of vehicles needed to serve all customers?

Step-by-step solution:

1. Total demand: 25
2. Vehicle capacity: 18
3. Minimum vehicles: ⌈25 ÷ 18⌉ = 2

Answer: 2 vehicles

vehicle-routing, logistics, delivery, optimization, VRP

Question 3

A hospital needs to schedule 5 staff for 7 days (Monday, Sunday, Wednesday...). Each day has 3 shifts: Morning, Evening, Night. Undesirable shifts (higher weight = more undesirable): - Weekend Night: weight 3 - Weekend Evening: weight 2 - Any Night: weight 1 After creating a fair schedule, what is the fairness gap (difference between max and min undesirable weights assigned to any staff)?

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: 6 points
- Carol: 1 points
- David: 0 points
- Emma: 4 points
- Frank: 4 points

4. Fairness gap: 6 - 0 = 6

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

fairness, shift-scheduling, workforce, equity, human-resources

Question 4

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 5

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.

appointment, conflict-resolution, multi-person

Question 6

A flow shop has 2 machines (M1 → M2). Jobs and processing times (M1, M2): - Job A: (41, 30) - Job B: (46, 41) - Job C: (32, 32) - Job D: (50, 13) - Job E: (10, 12) 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 B → Job A → Job D → Job E → Job C
3. Calculate makespan: 211

Answer: 211

flow-shop, multi-stage, makespan, johnsons-rule, manufacturing

Question 7

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 | 37 | 109 | 2 | 17 | | Component D | 52 | 115 | 3 | 20 | | Component C | 42 | 82 | 2 | 12 | | Component B | 36 | 113 | 3 | 11 | 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 B → Component D
2. Calculate completion times and penalties:
- Component C: completes at 42, due 82, early by 40 min → penalty 80
- Component A: completes at 79, due 109, early by 30 min → penalty 60
- Component B: completes at 115, due 113, late by 2 min → penalty 22
- Component D: completes at 167, due 115, late by 52 min → penalty 1040

3. Total penalty: 1202

Answer: 1202 penalty points

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

JIT, earliness, tardiness, penalty, inventory, lean-manufacturing

Question 8

A school needs to schedule 6 courses. The following courses have overlapping students and cannot be scheduled at the same time: - Physics conflicts with CS - Physics conflicts with Math - Physics conflicts with History - CS conflicts with English - CS conflicts with Math - CS conflicts with Biology - English conflicts with Math - History conflicts with Math What is the minimum number of time slots needed to schedule all courses without conflicts?

Step-by-step solution (Graph Coloring):

1. Model as graph coloring problem:
- Vertices = Courses
- Edges = Conflicts (courses that cannot be together)
2. Apply greedy coloring algorithm:
- Physics: Slot 1
- CS: Slot 2
- English: Slot 1
- History: Slot 2
- Math: Slot 3
- Biology: Slot 1

3. Colors/slots used: 3

Answer: Minimum 3 time slots

Key Strategy: The chromatic number of the conflict graph gives the minimum slots needed.

timetable, conflict-resolution, graph-theory, chromatic-number, advanced

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 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 2 - Package: 15 min on Line 2 **Product Y:** - Cut: 20 min on Line 1 - Assemble: 60 min on Line 2 - Package: 20 min on Line 2 **Product Z:** - Cut: 40 min on Line 3 - Assemble: 30 min on Line 2 - 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: 20 minutes
- Line 2: 200 minutes
- Line 3: 65 minutes

2. Identify bottleneck: The line with maximum load = Line 2
3. Bottleneck load: 200 minutes

Answer: Line 2 (200 minutes)

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

nested, bottleneck, resource-allocation, multi-line, factory

Question 11

A project has 6 tasks with the following requirements: - T1: Duration 5 days, Requires 1 resources - T2: Duration 4 days, Requires 2 resources - T3: Duration 2 days, Requires 1 resources - T4: Duration 5 days, Requires 2 resources - T5: Duration 4 days, Requires 3 resources - T6: Duration 3 days, Requires 2 resources Dependencies: - T1 must be completed before T2 - T2 must be completed before T3 Maximum 4 resources are available at any time. What is the minimum project completion time?

Step-by-step solution:

Resource-Constrained Scheduling:
1. List all tasks with resource requirements:
- T1: 5 days, 1 resources
- T2: 4 days, 2 resources
- T3: 2 days, 1 resources
- T4: 5 days, 2 resources
- T5: 4 days, 3 resources
- T6: 3 days, 2 resources

2. Available resources: 4 per day

3. Dependencies considered

4. Optimal schedule:
Days 0-4: T5 (3 resources)
Days 0-5: T1 (1 resources)
Days 4-9: T4 (2 resources)
Days 4-7: T6 (2 resources)

5. Total time: 7 days

Key Strategy: Identify task combinations that fit within resource limits.

resource-constraint, parallel-execution, project-management, olympiad

Question 12

Arrange the following activities in chronological order: Office Work, Evening Walk, Lunch Break, Dinner

Step-by-step solution:

Timeline Approach:
1. Convert all times to 24-hour format for easy comparison
- Office Work: 9:00 AM
- Evening Walk: 6:00 PM
- Lunch Break: 1:00 PM
- Dinner: 8:00 PM

2. Arrange in chronological order:
1. Lunch Break at 1:00 PM
2. Evening Walk at 6:00 PM
3. Dinner at 8:00 PM
4. Office Work at 9:00 AM

Final Schedule: Lunch Break -> Evening Walk -> Dinner -> Office Work

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

daily, time-based, chronological

Question 13

A conference needs to schedule 5 sessions across 3 time slots and 3 rooms. Each room can hold one session per slot. The constraints are: - Dr. Taylor can only speak at 9:00-10:00 - Cybersecurity and Data Science cannot be in the same time slot - Prof. Garcia and Dr. Taylor must speak in consecutive time slots - Blockchain must be in Hall B Which speaker presents the Cybersecurity session?

Step-by-step solution:

Scheduling Grid Analysis:
1. Fix direct constraints:
- Dr. Taylor at 9:00-10:00
- Blockchain in Hall B
2. Apply consecutive constraint: Prof. Garcia and Dr. Taylor in consecutive slots
3. Apply conflict constraint: Cybersecurity and Data Science not together

4. Final Schedule:
9:00-10:00:
- Hall A: Data Science by Prof. Jones
- Hall B: Blockchain by Dr. Taylor
- Hall C: Cybersecurity by Prof. Wilson
10:00-11:00:
- Hall A: Robotics by Prof. Garcia
- Hall B: IoT by Dr. Smith
- Hall C: (empty)
11:00-12:00:
- Hall A: (empty)
- Hall B: (empty)
- Hall C: (empty)

Answer: Prof. Wilson presents Cybersecurity

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

multi-constraint, professional, CSP, CAT-level

Question 14

A flow shop has 2 machines (M1 → M2). Jobs and processing times (M1, M2): - Job A: (16, 42) - Job B: (46, 47) - Job C: (47, 33) - Job D: (34, 36) - Job E: (17, 23) 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 C → Job A → Job E → Job D → Job B
3. Calculate makespan: 228

Answer: 228

flow-shop, multi-stage, makespan, johnsons-rule, manufacturing

Question 15

A school has 5 exams in 3 time slots. Each time slot needs 2 invigilators. A teacher can invigilate at most one exam per time slot. What is the minimum number of teachers required?

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

exam, invigilation, proctor, academic, constraints

Question 16

A machine must process four products (P4, P2, P1, P3) to minimize total time. Processing times: P4: 89, P2: 55, P1: 51, P3: 76 Setup times: | From ↓ / To → | P4 | P2 | P1 | P3 | | P4 | 0 | 38 | 20 | 38 | | P2 | 34 | 0 | 11 | 22 | | P1 | 10 | 23 | 0 | 27 | | P3 | 20 | 29 | 19 | 0 | What sequence minimizes total time?

Optimal sequence examined for all 24 permutations.
Minimum time: 321 minutes. Sequence: P3 - P2 - P1 - P4.
Calculation: P3 → P2 → P1 → P4.
Time: PT(P3) + (ST(P3→P2) + PT(P2)) + ...

professional, sequencing, optimization, TSP-lite

Question 17

In a double round-robin tournament (each pair plays twice), 6 teams compete. How many total matches will be played?

Step-by-step solution:

1. Single round-robin formula: n(n-1)/2
2. For 6 teams: 6 × 5 / 2 = 15 matches
3. Double round-robin: 15 × 2 = 30 matches

Answer: 30 matches

tournament, sports, combinations, fairness, competitive-exams

Question 18

A hospital needs to schedule 5 staff for 7 days (Tuesday, Monday, Wednesday...). Each day has 3 shifts: Morning, Evening, Night. Undesirable shifts (higher weight = more undesirable): - Weekend Night: weight 3 - Weekend Evening: weight 2 - Any Night: weight 1 After creating a fair schedule, what is the fairness gap (difference between max and min undesirable weights assigned to any staff)?

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:
- Alice: 1 points
- Bob: 6 points
- Carol: 0 points
- Emma: 5 points
- Frank: 3 points

4. Fairness gap: 6 - 0 = 6

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

fairness, shift-scheduling, workforce, equity, human-resources

Question 19

Arrange the following activities in chronological order: Lunch Break, Breakfast, Office Work, Evening Walk

Step-by-step solution:

Timeline Approach:
1. Convert all times to 24-hour format for easy comparison
- Lunch Break: 1:00 PM
- Breakfast: 7:30 AM
- Office Work: 9:00 AM
- Evening Walk: 6:00 PM

2. Arrange in chronological order:
1. Lunch Break at 1:00 PM
2. Evening Walk at 6:00 PM
3. Breakfast at 7:30 AM
4. Office Work at 9:00 AM

Final Schedule: Lunch Break -> Evening Walk -> Breakfast -> Office Work

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

daily, time-based, chronological

Question 20

A school needs to schedule 6 courses. The following courses have overlapping students and cannot be scheduled at the same time: - CS conflicts with English - CS conflicts with Biology - Math conflicts with English - Math conflicts with Biology - Biology conflicts with Chemistry - Biology conflicts with English - Biology conflicts with History - Chemistry conflicts with English - Chemistry conflicts with History What is the minimum number of time slots needed to schedule all courses without conflicts?

Step-by-step solution (Graph Coloring):

1. Model as graph coloring problem:
- Vertices = Courses
- Edges = Conflicts (courses that cannot be together)
2. Apply greedy coloring algorithm:
- CS: Slot 1
- Math: Slot 1
- Biology: Slot 2
- Chemistry: Slot 1
- History: Slot 3
- English: Slot 4

3. Colors/slots used: 4

Answer: Minimum 4 time slots

Key Strategy: The chromatic number of the conflict graph gives the minimum slots needed.

timetable, conflict-resolution, graph-theory, chromatic-number, advanced

Scheduling - Advanced Level: weekly schedule ADVANCED

What you'll learn in this worksheet:

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Question 11

Question 12

Question 13

Question 14

Question 15

Question 16

Question 17

Question 18

Question 19

Question 20