Scheduling Worksheet 21 - Intermediate-Advanced Level (calendar scheduling) | 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 68 minutes. Batch sizes (in order): 30, 10, 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 (30 units): 33.6 minutes
Batch 2 (10 units): 21.7 minutes
Batch 3 (20 units): 19.4 minutes

4. Total time: 74.6 ≈ 75 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

Hospital OR scheduling with 2 operating rooms (8 hours each): - Emergency: 58 min, Priority 1 - Urgent: 83 min, Priority 2 - Elective A: 73 min, Priority 3 - Elective B: 86 min, Priority 3 - Routine: 121 min, Priority 4 Can all surgeries be completed in one day?

Step-by-step solution:

1. Total surgery time: 421 min = 7.0 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

surgery, operating-room, hospital, healthcare, critical

Question 3

A factory has 3 machines. Each requires 2 day of preventive maintenance every 30 days. If maintenance is staggered, what is the maximum number of machines that can be operational at any time?

Step-by-step solution:

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

Answer: 2 machines

maintenance, preventive, downtime, industrial, reliability

Question 4

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: - Prof. Jones can only speak at 11:00-12:00 - Machine Learning and Blockchain cannot be in the same time slot - Prof. Brown and Prof. Wilson must speak in consecutive time slots - Blockchain must be in Hall A Which speaker presents the Machine Learning session?

Step-by-step solution:

Scheduling Grid Analysis:
1. Fix direct constraints:
- Prof. Jones at 11:00-12:00
- Blockchain in Hall A
2. Apply consecutive constraint: Prof. Brown and Prof. Wilson in consecutive slots
3. Apply conflict constraint: Machine Learning and Blockchain not together

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

Answer: Prof. Wilson presents Machine Learning

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

multi-constraint, professional, CSP, CAT-level

Question 5

**Data Sufficiency Question** Six lectures are scheduled from Monday to Saturday, one per day. **Question:** Which subject is on Thursday? **Statement (1):** Physics is on Wednesday, two days after Chemistry. **Statement (2):** Mathematics is on Friday, immediately after Biology. **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: Physics is on Wednesday, two days after Chemistry.
This gives partial information but not enough to determine the answer uniquely.

Step 2 - Analyze Statement (2) alone: Mathematics is on Friday, immediately after Biology.
This also gives partial information insufficient by itself.

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

Conclusion: Either statement 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 6

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

edge-coloring, match-scheduling, graph-theory, advanced, olympiad

Question 7

A flow shop has 2 machines (M1 → M2). Jobs and processing times (M1, M2): - Job A: (35, 33) - Job B: (15, 33) - Job C: (20, 44) - Job D: (15, 15) - Job E: (43, 27) 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 E → Job B → Job D → Job C
3. Calculate makespan: 197

Answer: 197

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

Question 8

A factory has 3 machines. Each requires 1 day of preventive maintenance every 30 days. If maintenance is staggered, what is the maximum number of machines that can be operational at any time?

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

maintenance, preventive, downtime, industrial, reliability

Question 9

A hospital needs to schedule 5 staff for 7 days (Friday, Sunday, Saturday...). 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: 2 points
- Bob: 2 points
- Carol: 3 points
- David: 8 points
- Emma: 0 points

4. Fairness gap: 8 - 0 = 8

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 10

Events need to be scheduled in rooms. Their time intervals are: - Event A: 3:00 to 11:00 - Event B: 3:00 to 9:00 - Event C: 17:00 to 23:00 - Event D: 15:00 to 23:00 - Event E: 2:00 to 8:00 - Event F: 18:00 to 25:00 - Event G: 5:00 to 10:00 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 3 to 11
Event B: ██████ from 3 to 9
Event C: ██████ from 17 to 23
Event D: ████████ from 15 to 23
Event E: ██████ from 2 to 8
Event F: ███████ from 18 to 25
Event G: █████ from 5 to 10

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

Answer: 4 rooms needed

interval, overlap, room-allocation, graph-theory, optimization

Question 11

A machine can process up to 5 jobs simultaneously as a batch. Each batch takes 57 minutes. If 12 jobs need to be processed, what is the minimum total time required?

Step-by-step solution:

1. Jobs per batch: 5
2. Number of batches: ⌈12 ÷ 5⌉ = 3
3. Total time: 3 × 57 = 171 minutes

Answer: 171 minutes

batch, parallel-machines, capacity, manufacturing, optimization

Question 12

A student has 9 days to prepare for three exams: Physics, Mathematics, Computer Science. The required preparation days are: - Physics: 2 days - Mathematics: 3 days - Computer Science: 1 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:
- Physics: 2 days
- Mathematics: 3 days
- Computer Science: 1 days
- Total: 6 days

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

Answer: The last exam will be on Day 9

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

academic, deadline, preparation, timeline

Question 13

A hospital needs one doctor on-call each day for 30 days. There are 4 doctors: Dr. Brown, Dr. Lee, Dr. Smith, Dr. Jones. If the schedule is as fair as possible, how many days will each doctor be on-call?

Step-by-step solution:

1. Total on-call days: 30
2. Base days per doctor: 30 ÷ 4 = 7 days
3. Remainder: 2 doctor(s) get one extra day

Answer: 7 days, with 2 doctor(s) getting 8 days

on-call, availability, emergency, medical, fairness

Question 14

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 1 - Assemble: 45 min on Line 3 - Package: 15 min on Line 3 **Product Y:** - Cut: 20 min on Line 2 - Assemble: 60 min on Line 1 - Package: 20 min on Line 2 **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: 120 minutes
- Line 2: 80 minutes
- Line 3: 85 minutes

2. Identify bottleneck: The line with maximum load = Line 1
3. Bottleneck load: 120 minutes

Answer: Line 1 (120 minutes)

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

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

Question 15

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 16

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 17

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 18

A machine needs to process 4 jobs. Processing times: - Job B: 61 minutes - Job D: 67 minutes - Job E: 66 minutes - Job C: 77 minutes The machine breaks down at 89 minutes and takes 43 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 D → Job C
2. Simulate processing with breakdown:
- Job B: 0 → 61
- Job E: Starts at 61, breakdown at 89 (28 min completed), repair 43 min, resume 38 min → completes at 170
- Job D: 170 → 237
- Job C: 237 → 314

3. Total makespan: 314 minutes
4. Delay caused by breakdown: 43 minutes

Answer: 314 minutes

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

manufacturing, breakdown, recovery, simulation, reliability

Question 19

In a single-elimination knockout tournament with 8 teams, how many total matches are played to determine the champion?

Step-by-step solution:

1. Single elimination principle: Each match eliminates exactly one team
2. Teams to eliminate: 8 - 1 = 7 teams must be eliminated
3. Matches needed: 7 matches

Answer: 7 matches

tournament, bracket, knockout, sports, competitive-exams

Question 20

A bus route takes 56 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: 56 × 2 = 112 minutes
2. Headway: 12 minutes
3. Buses needed: ⌈112 ÷ 12⌉ = 10

Answer: 10 buses

bus, timetable, frequency, public-transport, optimization

Scheduling - Intermediate-Advanced Level: calendar scheduling INTERMEDIATE-ADVANCED

What you'll learn in this worksheet: