Scheduling Worksheet 13 - Intermediate Level (daily schedule) | 20 Questions

Question 1

A job shop has 3 machines. Jobs and their routes: - Job A: M3 → M1 → M2 with times 37, 10, 19 - Job B: M1 → M2 → M3 with times 27, 34, 40 - Job C: M2 → M1 → M3 with times 34, 24, 18 What is a lower bound on the minimum makespan?

Step-by-step solution:

1. Machine load bound: 95
2. Job processing bound: 101
3. Lower bound: 101

Answer: 101

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

Question 2

A passenger travels from Chicago to Miami via Dallas. The minimum layover at Dallas is **45 minutes**. **Flights Chicago -> Dallas:** - F1-1: Dep 6:30 AM, Arr 8:48 AM - F1-2: Dep 9:30 AM, Arr 11:48 AM - F1-3: Dep 12:30 PM, Arr 2:48 PM **Flights Dallas -> Miami:** - F2-1: Dep 10:00 AM, Arr 1:14 PM - F2-2: Dep 12:00 PM, Arr 3:14 PM - F2-3: Dep 2:00 PM, Arr 5:14 PM What is the minimum total elapsed time for the journey from Chicago to Miami?

1. Timeline Approach & Constraint Application (Minimum Layover: 45 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 404 minutes is achieved by combining F1-1 (Arr: 8:48 AM) and F2-1 (Dep: 10:00 AM, Arr: 1:14 PM).
Total Elapsed Time = Final Arrival Time - Initial Departure Time.

3. Final Answer: The minimum elapsed time is 6 hours and 44 minutes.

transportation, time-interval, optimization, CAT-level

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

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 5

In a round-robin tournament with 6 teams, in Round 3, Team D plays against which team?

Step-by-step solution:

1. Round-robin schedule using circle method
2. Round 3 matches:
- Team F vs Team C
- Team D vs Team B
- Team E vs Team A

3. Team D plays against Team B

Answer: Team B

tournament, sports, combinations, fairness, competitive-exams

Question 6

In a round-robin tournament with 4 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_4
2. Vizing's theorem: χ'(K_n) = n-1 for even n, n for odd n
3. For 4 teams: 3 colors/rounds needed

Answer: 3 rounds

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

Question 7

A project consists of the following tasks: - Task A (Requirements Analysis): 2 days, Depends on: None - Task B (Design): 3 days, Depends on: A - Task C (Database Setup): 2 days, Depends on: A - Task D (Development): 5 days, Depends on: B, C - Task E (Testing): 3 days, Depends on: D What is the minimum number of days required to complete the entire project?

Step-by-step solution:

Critical Path Method (CPM):
1. Identify dependencies and calculate earliest start times:
- Task A: Starts on Day 0, Duration 2 days
Finishes on Day 2
- Task B: Starts on Day 2, Duration 3 days
Finishes on Day 5
- Task C: Starts on Day 2, Duration 2 days
Finishes on Day 4
- Task D: Starts on Day 5, Duration 5 days
Finishes on Day 10
- Task E: Starts on Day 10, Duration 3 days
Finishes on Day 13

2. Task timeline:
Task A: Days 0-2
Task B: Days 2-5 (after A)
Task C: Days 2-4 (after A)
Task D: Days 5-10 (after B and C)
Task E: Days 10-13 (after D)

3. Critical path: A -> B -> D -> E (or A -> C -> D -> E)
4. Total project duration: 13 days

Key Strategy: Calculate earliest start time for each task based on predecessor completion times; the longest path determines total duration.

project-management, dependencies, critical-path, professional

Question 8

Five subjects are scheduled on five different days of the week (Monday to Friday), one subject per day. The following information is given: - Biology is scheduled on Wednesday - Mathematics is scheduled immediately after English - There are exactly two classes between Chemistry and History - English is not on Monday On which day is Mathematics scheduled?

Step-by-step solution:

Table Method:
1. Create a timeline for Monday to Friday
2. Apply direct constraints:
- Biology is on Wednesday (fixed)
- English is not on Monday
3. Apply consecutive constraint:
- Mathematics immediately follows English
- Possible pairs: (Tue-Wed), (Wed-Thu), (Thu-Fri)
- Since Wednesday is occupied, options are (Tue-Wed) or (Thu-Fri)
4. Apply gap constraint:
- Two classes between Chemistry and History
5. Final schedule:
- Monday: Chemistry
- Tuesday: English
- Wednesday: Biology
- Thursday: Mathematics
- Friday: History

Answer: Mathematics is scheduled on Thursday

Key Strategy: Fix direct constraints first, then work with consecutive and gap constraints.

weekly, academic, day-based, constraints

Question 9

A factory produces Widgets with a 80% learning curve (each doubling of cumulative production reduces time by 19%). First unit takes 88 minutes. Batch sizes (in order): 20, 40, 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^-0.322
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 (20 units): 49.5 minutes
Batch 2 (40 units): 27.4 minutes
Batch 3 (10 units): 23.0 minutes

4. Total time: 99.8 ≈ 100 minutes

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

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

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 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 1 **Product Z:** - Cut: 40 min on Line 1 - Assemble: 30 min on Line 2 - 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: 130 minutes
- Line 2: 45 minutes
- Line 3: 110 minutes

2. Identify bottleneck: The line with maximum load = Line 1
3. Bottleneck load: 130 minutes

Answer: Line 1 (130 minutes)

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

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

Question 11

An event runs for 8 hours. Staff needed per hour: - Hour 1: 6 - Hour 2: 5 - Hour 3: 6 - Hour 4: 6 - Hour 5: 7 (PEAK) - Hour 6: 3 - Hour 7: 3 - Hour 8: 4 What is the minimum number of staff needed if staff can work multiple consecutive hours?

Step-by-step solution:

1. Identify peak demand: 7 staff at hour 5
2. Staff can work multiple hours → schedule around peak
3. Minimum staff needed: 7

Answer: 7 staff

event, staff, venue, hospitality, logistics

Question 12

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 13

A hospital needs one doctor on-call each day for 30 days. There are 4 doctors: Dr. Patel, Dr. Brown, Dr. Jones, Dr. Smith. 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 production line needs to manufacture: - Product A: 3 units (each takes 3 hours) - Product C: 1 units (each takes 2 hours) - Product D: 3 units (each takes 1 hours) - Product B: 3 units (each takes 4 hours) Setup time required when switching products: - P->P: 2 hour What is the minimum total time if production starts with Product C?

Step-by-step solution:

Production Sequencing with Setup Times:
1. Calculate total production time (without setup):
- Product A: 3 x 3 = 9 hours
- Product C: 1 x 2 = 2 hours
- Product D: 3 x 1 = 3 hours
- Product B: 3 x 4 = 12 hours
- Base production time: 26 hours

2. Minimize setup time by batching:
- Optimal sequence: Product C -> Product A -> Product A -> Product A -> Product D -> Product D -> Product D -> Product B -> Product B -> Product B
3. Total with setups:
- Product C: 2 hours
- Setup Product C→Product A: 2 hour + Product A: 3 hours
- Product A: 3 hours (no setup)
- Product A: 3 hours (no setup)
- Setup Product A→Product D: 2 hour + Product D: 1 hours
- Product D: 1 hours (no setup)
- Product D: 1 hours (no setup)
- Setup Product D→Product B: 2 hour + Product B: 4 hours
- Product B: 4 hours (no setup)
- Product B: 4 hours (no setup)

Total: 32 hours

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

manufacturing, setup-time, sequencing, optimization

Question 15

A machine needs to process 4 jobs. Processing times: - Job C: 82 minutes - Job D: 32 minutes - Job E: 44 minutes - Job A: 82 minutes The machine breaks down at 69 minutes and takes 45 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 E → Job C → Job A
2. Simulate processing with breakdown:
- Job D: 0 → 32
- Job E: Starts at 32, breakdown at 69 (37 min completed), repair 45 min, resume 7 min → completes at 121
- Job C: 121 → 203
- Job A: 203 → 285

3. Total makespan: 285 minutes
4. Delay caused by breakdown: 45 minutes

Answer: 285 minutes

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

manufacturing, breakdown, recovery, simulation, reliability

Question 16

Project scheduling with two objectives: minimize time and minimize cost. - Schedule A: 100 days, $50K - Schedule B: 120 days, $40K - Schedule C: 90 days, $60K - Schedule D: 110 days, $45K - Schedule E: 95 days, $55K Which schedules are on the Pareto frontier (not dominated in both objectives)?

Step-by-step solution:

1. Pareto dominance: Schedule X dominates Y if X is better in at least one objective and not worse in others
2. Pareto frontier schedules: Schedule A, Schedule B, Schedule C, Schedule D, Schedule E

Answer: Schedule A, Schedule B, Schedule C, Schedule D, Schedule E

pareto, multi-objective, trade-off, advanced, optimization

Question 17

**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.

data-sufficiency, GMAT, CAT, reasoning, test-prep

Question 18

A traveler needs to go from City A to City D. The transport schedule is: - T1: City A to City B, Departs 08:30, Arrives 11:55 AM - T2: City A to City C, Departs 07:00, Arrives 9:38 AM - T3: City B to City D, Departs 11:00, Arrives 12:57 PM - T4: City B to City C, Departs 15:30, Arrives 6:17 PM - T5: City C to City D, Departs 10:00, Arrives 12:58 PM - T6: City C to City B, Departs 11:30, Arrives 2:04 PM - T7: City E to City D, Departs 14:00, Arrives 4:56 PM - T8: City E to City B, Departs 12:30, Arrives 2:57 PM Minimum connection time is 45 minutes. What is the earliest arrival time at City D?

Step-by-step solution:

Network Path Analysis:
1. Identify all possible routes from City A to City D:
- City A→City B -> City B→City D
- City A→City C -> City C→City D
- City A→City B -> City B→City C -> City C→City D

No feasible route found with 45 min connection requirement.

Key Strategy: Enumerate all possible routes, verify connection times meet minimum requirements.

transportation, connections, optimization, railway-exams

Question 19

Project tasks with uncertain durations (optimistic, likely, pessimistic) in days: - Design: (4, 7, 10) - Development: (2, 3, 5) - Testing: (3, 4, 7) - Deployment: (3, 6, 7) 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: (4 + 4×7 + 10)/6 = 7.0
- Development: (2 + 4×3 + 5)/6 = 3.2
- Testing: (3 + 4×4 + 7)/6 = 4.3
- Deployment: (3 + 4×6 + 7)/6 = 5.7

2. Total expected duration: 20.2 days

Answer: 20.2 days

fuzzy, uncertainty, duration, advanced, real-world

Question 20

**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.