Scheduling - Intermediate-Advanced Level: shift planning INTERMEDIATE-ADVANCED

Ready to master scheduling? This time-bound test features 20 intermediate-advanced-level challenges. Worksheet 22 of 30 sharpens your shift planning skills. Master schedule logic, time allocation, day scheduling through guided practice. Perfect for advanced developing test preparation.

📝 Worksheet 22 of 30 • 20 questions • ⏱️ Estimated time: 20 minutes • 🎯 Intermediate-advanced level

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Worksheet 22 of 30 (73% complete)

Question 1

Project tasks with uncertain durations (optimistic, likely, pessimistic) in days: - Design: (3, 6, 9) - Development: (2, 3, 4) - Testing: (2, 4, 6) - 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: (3 + 4×6 + 9)/6 = 6.0
- Development: (2 + 4×3 + 4)/6 = 3.0
- Testing: (2 + 4×4 + 6)/6 = 4.0
- Deployment: (3 + 4×6 + 7)/6 = 5.7

2. Total expected duration: 18.7 days

Answer: 18.7 days

Question 2

In a round-robin tournament with 6 teams, in Round 2, Team B plays against which team?
Step-by-step solution:

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

3. Team B plays against Team D

Answer: Team D

Question 3

Real-time tasks with Rate Monotonic Scheduling (shorter period = higher priority): - Task A: Execution 3, Period 10 - Task B: Execution 9, Period 30 - Task C: Execution 2, Period 10 Is the task set schedulable under RM?
Step-by-step solution:

1. Calculate utilization:
- Task A: 3/10 = 0.300
- Task B: 9/30 = 0.300
- Task C: 2/10 = 0.200
Total U = 0.800
2. RM schedulability bound for 3 tasks: 0.780
3. Conclusion: Utilization 0.800 > 0.780 (RM bound)

Answer: May not be schedulable

Question 4

A hospital needs one doctor on-call each day for 30 days. There are 5 doctors: Dr. Brown, Dr. Patel, 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 ÷ 5 = 6 days
3. Remainder: 0 doctor(s) get one extra day

Answer: 6 days each

Question 5

A bus route takes 53 minutes one-way. Peak frequency: 1 bus every 10 minutes. What is the minimum number of buses needed to maintain this frequency in both directions?
Step-by-step solution:

1. Round trip time: 53 × 2 = 106 minutes
2. Headway: 10 minutes
3. Buses needed: ⌈106 ÷ 10⌉ = 11

Answer: 11 buses

Question 6

A traveler needs to go from City A to City D. The transport schedule is: - T1: City A to City B, Departs 06:30, Arrives 8:41 AM - T2: City A to City C, Departs 10:00, Arrives 1:04 PM - T3: City B to City D, Departs 12:30, Arrives 2:23 PM - T4: City B to City C, Departs 15:00, Arrives 4:31 PM - T5: City C to City D, Departs 16:30, Arrives 7:23 PM - T6: City C to City B, Departs 15:30, Arrives 5:24 PM - T7: City E to City D, Departs 16:30, Arrives 6:54 PM - T8: City E to City B, Departs 14:30, Arrives 4:20 PM Minimum connection time is 30 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

2. Best route found:
- T1: City A to City B (06:30 - 8:41 AM)
- Connection time: 229 minutes
- T3: City B to City D (12:30 - 2:23 PM)

Earliest arrival: 2:23 PM

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

Question 7

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.

Question 8

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

Question 9

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

Question 10

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

Table Method:
1. Create a timeline for Monday to Friday
2. Apply direct constraints:
- English is on Wednesday (fixed)
- History is not on Monday
3. Apply consecutive constraint:
- Physics immediately follows History
- 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 Mathematics and Biology
5. Final schedule:
- Monday: Mathematics
- Tuesday: Biology
- Wednesday: English
- Thursday: History
- Friday: Physics

Answer: Physics is scheduled on Friday

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

Question 11

Given these scheduling constraints: - Task B must be before Task C - Task D must be after Task C - Task A must be immediately after Task B 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

Question 12

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.

Question 13

A machine must process four products (P2, P4, P1, P3) to minimize total time. Processing times: P2: 55, P4: 62, P1: 83, P3: 47 Setup times: | From ↓ / To → | P2 | P4 | P1 | P3 | | P2 | 0 | 33 | 27 | 36 | | P4 | 35 | 0 | 22 | 36 | | P1 | 25 | 21 | 0 | 32 | | P3 | 28 | 35 | 18 | 0 | What sequence minimizes total time?
Optimal sequence examined for all 24 permutations.
Minimum time: 321 minutes. Sequence: P3 - P1 - P4 - P2.
Calculation: P3 → P1 → P4 → P2.
Time: PT(P3) + (ST(P3→P1) + PT(P1)) + ...

Question 14

In a round-robin tournament with 6 teams, in Round 2, Team B plays against which team?
Step-by-step solution:

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

3. Team B plays against Team A

Answer: Team A

Question 15

A hospital needs one doctor on-call each day for 30 days. There are 5 doctors: Dr. Brown, Dr. Jones, Dr. Lee, Dr. Smith, Dr. Patel. 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 ÷ 5 = 6 days
3. Remainder: 0 doctor(s) get one extra day

Answer: 6 days each

Question 16

Computer Science courses with prerequisites: - CS102 requires CS301 - CS101 requires CS301 - CS202 requires CS101 - CS401 requires CS301 Which courses can be taken in the first semester?
Step-by-step solution:

1. Identify courses with prerequisites:
- CS102 needs CS301
- CS101 needs CS301
- CS202 needs CS101
- CS401 needs CS301

2. Courses without prerequisites (can take first): CS301

Answer: CS301

Question 17

A traveler needs to go from City A to City E. The transport schedule is: - T1: City A to City B, Departs 09:00, Arrives 12:19 PM - T2: City A to City C, Departs 09:30, Arrives 11:42 AM - T3: City B to City E, Departs 13:00, Arrives 3:54 PM - T4: City B to City C, Departs 10:00, Arrives 12:21 PM - T5: City C to City E, Departs 10:30, Arrives 1:24 PM - T6: City C to City B, Departs 15:30, Arrives 5:08 PM - T7: City D to City E, Departs 10:30, Arrives 1:21 PM - T8: City D to City B, Departs 11:00, Arrives 1:25 PM Minimum connection time is 60 minutes. What is the earliest arrival time at City E?
Step-by-step solution:

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

No feasible route found with 60 min connection requirement.

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

Question 18

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.

Question 19

In a double round-robin tournament with 6 teams, each team plays every other team twice (once home, once away). How many home matches does each team play?
Step-by-step solution:

1. Total matches per team in double round-robin: 2 × (6 - 1) = 10 matches
2. Half are home matches: 10 ÷ 2 = 5 home matches

Answer: 5 home matches

Question 20

Project tasks with uncertain durations (optimistic, likely, pessimistic) in days: - Design: (4, 6, 7) - Development: (2, 4, 5) - Testing: (2, 5, 6) - Deployment: (4, 5, 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 + 4×6 + 7)/6 = 5.8
- Development: (2 + 4×4 + 5)/6 = 3.8
- Testing: (2 + 4×5 + 6)/6 = 4.7
- Deployment: (4 + 4×5 + 7)/6 = 5.2

2. Total expected duration: 19.5 days

Answer: 19.5 days
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