Scheduling Worksheet 2 - Beginner Level (time allocation) | 20 Questions

Question 1

Project tasks with uncertain durations (optimistic, likely, pessimistic) in days: - Design: (3, 5, 6) - Development: (4, 7, 8) - Testing: (2, 4, 5) - Deployment: (3, 4, 5) 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: (3 + 4×5 + 6)/6 = 4.8
- Development: (4 + 4×7 + 8)/6 = 6.7
- Testing: (2 + 4×4 + 5)/6 = 3.8
- Deployment: (3 + 4×4 + 5)/6 = 4.0

2. Total expected duration: 19.3 days

Answer: 19.3 days

fuzzy, uncertainty, duration, advanced, real-world

Question 2

7 tasks (A, B, C, D, E, F, G) are to be completed one after the other. The following conditions must be met: - Task F must be performed immediately after Task E. - Task A must be completed before Task G. - Task C is neither the first nor the last task to be completed. - Task D is performed exactly 2 positions after Task B. Which task is scheduled in the third position?

Step-by-step solution (Deductive Logic):

1. Apply Consecutive Constraint: 'F immediately after E' -> (E, F)
2. Apply Before Constraint: 'A before G'
3. Apply Exclusion Constraint: 'C not first or last'
4. Apply Gap Constraints: 'D is 2 after B'

Final Sequence: A → B → C → D → E → F → G

Answer: The task in the third position is C.

Key Strategy: Use fixed pairs and gap constraints to anchor positions.

linear-arrangement, relative-constraints, deduction, CAT-level

Question 3

A machine must process four products (P1, P3, P4, P2) to minimize total time. Processing times: P1: 46, P3: 67, P4: 84, P2: 37 Setup times: | From ↓ / To → | P1 | P3 | P4 | P2 | | P1 | 0 | 16 | 13 | 32 | | P3 | 39 | 0 | 40 | 26 | | P4 | 32 | 39 | 0 | 15 | | P2 | 34 | 19 | 34 | 0 | What sequence minimizes total time?

Optimal sequence examined for all 24 permutations.
Minimum time: 281 minutes. Sequence: P1 - P4 - P2 - P3.
Calculation: P1 → P4 → P2 → P3.
Time: PT(P1) + (ST(P1→P4) + PT(P4)) + ...

professional, sequencing, optimization, TSP-lite

Question 4

A PhD thesis defense requires all 5 committee members to be present. Their availability (slots 1-8): - Prof. D: Slots 7, 5 - Prof. E: Slots 8, 1, 6, 3 - Prof. A: Slots 4, 7, 1, 3 - Prof. B: Slots 8, 4, 2 - Prof. C: Slots 4, 5 What is the earliest slot when all can attend?

Step-by-step solution:

1. Find intersection of availability:
Prof. D: [5, 7]
∩ Prof. E: [1, 3, 6, 8]
∩ Prof. A: [1, 3, 4, 7]
∩ Prof. B: [2, 4, 8]
∩ Prof. C: [4, 5]
= ∅ (No common slots)

Answer: No common slot available

thesis, defense, committee, academic, availability

Question 5

A production line needs to manufacture: - Product A: 3 units (each takes 4 hours) - Product C: 1 units (each takes 2 hours) - Product B: 3 units (each takes 4 hours) - Product D: 3 units (each takes 3 hours) Setup time required when switching products: - P->P: 3 hour What is the minimum total time if production starts with Product A?

Step-by-step solution:

Production Sequencing with Setup Times:
1. Calculate total production time (without setup):
- Product A: 3 x 4 = 12 hours
- Product C: 1 x 2 = 2 hours
- Product B: 3 x 4 = 12 hours
- Product D: 3 x 3 = 9 hours
- Base production time: 35 hours

2. Minimize setup time by batching:
- Optimal sequence: Product A -> Product A -> Product A -> Product C -> Product B -> Product B -> Product B -> Product D -> Product D -> Product D
3. Total with setups:
- Product A: 4 hours
- Product A: 4 hours (no setup)
- Product A: 4 hours (no setup)
- Setup Product A→Product C: 3 hour + Product C: 2 hours
- Setup Product C→Product B: 3 hour + Product B: 4 hours
- Product B: 4 hours (no setup)
- Product B: 4 hours (no setup)
- Setup Product B→Product D: 3 hour + Product D: 3 hours
- Product D: 3 hours (no setup)
- Product D: 3 hours (no setup)

Total: 44 hours

Key Strategy: Batch identical products together to minimize setup changes.

manufacturing, setup-time, sequencing, optimization

Question 6

Four employees need to be scheduled for three shifts over three days. The constraints are: - Each employee works exactly one shift per day - No employee works the same shift two days in a row - Alice works Morning shift on Monday - Bob cannot work Night shift - Charlie works Evening shift on Tuesday Who works the Evening shift on Wednesday?

Step-by-step solution:

Table Method with Constraint Elimination:
1. Create a 3D table: Days x Shifts x Employees

2. Apply direct constraints:
- Monday Morning: Alice (fixed)
- Tuesday Evening: Charlie (fixed)
- Bob: Never Night shift (all days)

3. Apply rotation constraint:
- Alice (Morning Mon) cannot be Morning Tue
- Charlie (Evening Tue) cannot be Evening Wed

4. Fill Monday:
- Morning: Alice
- Evening: Charlie (can work evening)
- Night: Diana (Bob can't do night)

5. Fill Tuesday:
- Morning: Bob (Alice can't repeat, Charlie is evening)
- Evening: Charlie (fixed)
- Night: Diana (Bob can't)

6. Fill Wednesday:
- Charlie can't be Evening (was Evening Tue)
- Alice can be Evening (was Morning Mon, okay to shift)
- Answer: Alice works Evening on Wednesday

Key Strategy: Apply fixed constraints first, then use rotation rules to eliminate impossible assignments systematically.

shift-rotation, multi-person, constraints, professional

Question 7

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 8

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 3 - Assemble: 45 min on Line 1 - Package: 15 min on Line 2 **Product Y:** - Cut: 20 min on Line 3 - Assemble: 60 min on Line 3 - Package: 20 min on Line 2 **Product Z:** - Cut: 40 min on Line 1 - Assemble: 30 min on Line 3 - Package: 25 min on Line 1 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: 110 minutes
- Line 2: 35 minutes
- Line 3: 140 minutes

2. Identify bottleneck: The line with maximum load = Line 3
3. Bottleneck load: 140 minutes

Answer: Line 3 (140 minutes)

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

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

Question 9

Round Robin scheduling with time quantum = 2: - P1: Burst time 9 - P2: Burst time 15 - P3: Burst time 7 What is the average completion time?

Step-by-step solution:

1. Round Robin simulation:
2. Completion times:
- P3: 23
- P1: 24
- P2: 31

3. Average: 78 ÷ 3 = 26.0

Answer: 26.0

preemptive, interruptible, round-robin, OS, computer-science

Question 10

A job shop has 3 machines. Jobs and their routes: - Job A: M3 → M1 → M2 with times 21, 19, 31 - Job B: M2 → M3 → M1 with times 26, 37, 39 - Job C: M2 → M3 → M1 with times 24, 33, 19 What is a lower bound on the minimum makespan?

Step-by-step solution:

1. Machine load bound: 91
2. Job processing bound: 102
3. Lower bound: 102

Answer: 102

job-shop, routing, complex, np-hard, advanced

Question 11

A hospital needs one doctor on-call each day for 30 days. There are 4 doctors: Dr. Patel, Dr. Brown, Dr. Lee, 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 12

A school needs to schedule 6 courses. The following courses have overlapping students and cannot be scheduled at the same time: - Physics conflicts with Math - Physics conflicts with History - Physics conflicts with English - Physics conflicts with Biology - Math conflicts with CS - Math conflicts with English - History conflicts with CS - History conflicts with English - History conflicts with Biology - CS conflicts with English 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
- Math: Slot 2
- History: Slot 2
- CS: Slot 1
- English: Slot 3
- Biology: 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

Question 13

A student has 8 days to prepare for three exams: Physics, Chemistry, Mathematics. The required preparation days are: - Physics: 2 days - Chemistry: 2 days - Mathematics: 3 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
- Chemistry: 2 days
- Mathematics: 3 days
- Total: 7 days

2. Available days: 8 days
3. Extra buffer days: 1 days
4. Optimal schedule:
- Days 1-2: Prepare for Physics
- Day 3: Physics exam
- Days 4-5: Prepare for Chemistry
- Day 6: Chemistry exam
- Days 7-9: Prepare for Mathematics
- Day 10: Mathematics 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 14

A JIT manufacturing system has 4 jobs with the following data: | Job | Processing (min) | Due Date (min) | Early Penalty/min | Late Penalty/min | |-----|-----------------|----------------|-------------------|------------------| | Component D | 31 | 77 | 2 | 18 | | Component B | 56 | 88 | 1 | 12 | | Component A | 22 | 49 | 1 | 10 | | Component C | 60 | 75 | 3 | 10 | 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 A → Component C → Component D → Component B
2. Calculate completion times and penalties:
- Component A: completes at 22, due 49, early by 27 min → penalty 27
- Component C: completes at 82, due 75, late by 7 min → penalty 70
- Component D: completes at 113, due 77, late by 36 min → penalty 648
- Component B: completes at 169, due 88, late by 81 min → penalty 972

3. Total penalty: 1717

Answer: 1717 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 15

Arrange the following activities in chronological order: Evening Walk, Dinner, Office Work, Morning Yoga

Step-by-step solution:

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

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

Final Schedule: Morning Yoga -> 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 16

In a round-robin tournament with 6 teams, in Round 5, Team E plays against which team?

Step-by-step solution:

1. Round-robin schedule using circle method
2. Round 5 matches:
- Team F vs Team A
- Team C vs Team B
- Team E vs Team D

3. Team E plays against Team D

Answer: Team D

tournament, sports, combinations, fairness, competitive-exams

Question 17

A JIT manufacturing system has 4 jobs with the following data: | Job | Processing (min) | Due Date (min) | Early Penalty/min | Late Penalty/min | |-----|-----------------|----------------|-------------------|------------------| | Component D | 51 | 74 | 1 | 20 | | Component A | 48 | 91 | 3 | 10 | | Component C | 23 | 79 | 1 | 11 | | Component B | 32 | 72 | 5 | 17 | 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 B → Component D → Component C → Component A
2. Calculate completion times and penalties:
- Component B: completes at 32, due 72, early by 40 min → penalty 200
- Component D: completes at 83, due 74, late by 9 min → penalty 180
- Component C: completes at 106, due 79, late by 27 min → penalty 297
- Component A: completes at 154, due 91, late by 63 min → penalty 630

3. Total penalty: 1307

Answer: 1307 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 18

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. Taylor can only speak at 9:00-10:00 - AI Ethics and Data Science cannot be in the same time slot - Prof. Jones and Dr. Lee must speak in consecutive time slots - IoT must be in Hall B Which speaker presents the AI Ethics session?

Step-by-step solution:

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

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

Answer: Dr. Taylor presents AI Ethics

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

multi-constraint, professional, CSP, CAT-level

Question 19

A production line needs to manufacture: - Product B: 1 units (each takes 3 hours) - Product A: 3 units (each takes 2 hours) - Product C: 1 units (each takes 3 hours) Setup time required when switching products: - P->P: 3 hour What is the minimum total time if production starts with Product B?

Step-by-step solution:

Production Sequencing with Setup Times:
1. Calculate total production time (without setup):
- Product B: 1 x 3 = 3 hours
- Product A: 3 x 2 = 6 hours
- Product C: 1 x 3 = 3 hours
- Base production time: 12 hours

2. Minimize setup time by batching:
- Optimal sequence: Product B -> Product A -> Product A -> Product A -> Product C
3. Total with setups:
- Product B: 3 hours
- Setup Product B→Product A: 3 hour + Product A: 2 hours
- Product A: 2 hours (no setup)
- Product A: 2 hours (no setup)
- Setup Product A→Product C: 3 hour + Product C: 3 hours

Total: 18 hours

Key Strategy: Batch identical products together to minimize setup changes.