For All The Problems To Receive Full Scores, You Must Show

For All The Problems To Receive Full Scores You Have To Show Your Co

For all the problems, to receive full scores, you have to show your complete and accurate work. If you only provide the final answers without showing your computations, you will not be assigned a full score (or you will miss the opportunity of receiving partial credit).

Problem 1: Line Balancing (9 points) Employees at ABC Assembly Line work from 6am to 2pm Monday through Friday. Everyone on the assembly line is allowed a 30-minute lunch and two 15-minute breaks. Other cross-trained associates cover lunches and breaks for the associates on the assembly line so the line doesn’t stop.

On your second week at ABC Assembly Line, you record the processes of the assembly line are laid out in the following manner:

• Task 1: no predecessor
• Task 2: predecessor is Task 1
• Task 3: no predecessor
• Task 4: predecessor is Task 2
• Task 5: no predecessor
• Task 6: predecessor is Task 3 and Task 5

The intern the summer before recorded the following observations:

• Task 1 average cycle time = 1.1 min
• Task 2 average cycle time = 0.8 min
• Task 3 average cycle time = 52 sec
• Task 4 average cycle time = 36 sec
• Task 5 average cycle time = 1.4 min
• Task 6 average cycle time = 63 sec

Due to quality issues, a new quality check task (Task 4a) needs to be added after Task 4, taking an average of 27 seconds, and must be completed before Task 6 can begin.

The quality assurance team estimates this process will take 27 seconds on average. Prior to adding the quality check, the line output was 300 units per day. The plant manager wants to maintain this output after adding the quality check.

Paper For Above instruction

This paper addresses the problem of line balancing at ABC Assembly Line, considering the addition of a new quality check task, and aims to optimize task allocation, takt time calculation, and efficiency evaluation. The primary goal is to maintain the current output of 300 units per day despite the additional process.

a) Drawing the Precedence Diagram with Processing Times

The precedence diagram illustrates the sequence and dependencies among tasks, along with their cycle times. The tasks and their times are as follows:

• Task 1: 1.1 minutes
• Task 2: 0.8 minutes
• Task 3: 52 seconds (approximately 0.87 minutes)
• Task 4: 36 seconds (0.6 minutes)
• Task 5: 1.4 minutes
• Task 4a (quality check): 27 seconds (0.45 minutes)
• Task 6: 63 seconds (1.05 minutes)

The precedence relations are as follows:

- Tasks 1, 3, and 5 have no predecessors.

- Task 2 depends on Task 1.

- Task 4 depends on Task 2.

- Task 4a depends on Task 4.

- Task 6 depends on Tasks 3, 5, and 4a.

The diagram visually shows these relationships with process times labeled accordingly, ensuring clarity in dependencies and sequence.

b) Calculating Takt Time

Takt time is the available production time divided by customer demand. The total available work time per day, considering the shift, is 8 hours minus breaks (30 minutes lunch + 2×15 minutes breaks = 1 hour). Thus, total net working hours are 7 hours, or 420 minutes.

The desired daily output is 300 units. Therefore, Takt time = Total available time / Output:

Takt time = 420 minutes / 300 units = 1.4 minutes per unit.

This means each unit should be completed approximately every 1.4 minutes to meet the production target.

c) Task Assignment Using the "Greatest Remaining Time" Heuristic

We first calculate each task's cycle time; the revised cycle times after including Task 4a are as follows:

• Task 1: 1.1 min
• Task 2: 0.8 min
• Task 3: 0.87 min
• Task 4: 0.6 min
• Task 4a: 0.45 min
• Task 5: 1.4 min
• Task 6: 1.05 min

Using 1.4 minutes as the maximum cycle time per station (from Takt time), we assign tasks as follows:

• Work Station 1: Tasks 5 (1.4 min)
• Work Station 2: Tasks 1 (1.1 min) + 2 (0.8 min) = 1.9 min (exceeds 1.4 min), so assign Task 1 alone.
• Work Station 2 (continued): Tasks 2 (0.8 min) + 4a (0.45 min) = 1.25 min, acceptable.
• Work Station 3: Tasks 3 (0.87 min) + 4 (0.6 min) + 6 (1.05 min) (exceeds 1.4), so assign Tasks 3 and 4 separately and adjust accordingly.

More precisely, tasks are assigned by starting with the largest task times and adding smaller ones until reaching or just under the cycle time of 1.4 minutes, respecting dependencies.

Final tentative assignment:

- Work Station 1: Task 5 (1.4 min)

- Work Station 2: Task 1 (1.1 min) and Task 2 (0.8 min), but total exceeds 1.4 min, so assign Task 1 alone, then Task 2 separately.

- Work Station 3: Task 3 (0.87 min) and Task 4a (0.45 min) — total 1.32 min.

- Work Station 4: Task 4 (0.6 min) and Task 6 (1.05 min), but total exceeds 1.4 min, so assign Tasks 4 and 6 separately, with some adjustments.

In conclusion, the task assignments are optimized by prioritizing tasks with the greatest remaining time, while respecting process dependencies.

d) Calculating Efficiency and Idle Time

Total task time assigned across all stations sums to approximately:

• Station 1: 1.4 min
• Station 2: Tasks 1 and 2: 1.1 + 0.8 = 1.9 min
• Station 3: Tasks 3 and 4a: 0.87 + 0.45 = 1.32 min
• Station 4: Tasks 4 and 6: 0.6 + 1.05 = 1.65 min

Sum: 1.4 + 1.9 + 1.32 + 1.65 = 6.27 minutes per cycle.

Efficiency is calculated as:

Efficiency = (Total task time / (Number of stations × Takt time)) × 100%.

Number of stations = 4.

Total available time per cycle = 4 × 1.4 min = 5.6 min.

Efficiency = (6.27 / 5.6) × 100% ≈ 112.05% — which indicates that some tasks exceed the Takt time boundary and require further balancing or splitting.

To compute percentage idle time, we consider the difference between the total available work time and actual task time:

Total system capacity per cycle = 4 × 1.4 = 5.6 min

Actual task time = 6.27 min

Since actual task time exceeds capacity, indicates the line needs adjustment or increased capacity to meet demand.

However, to determine individual station idle times, subtract each station’s assigned task time from the Takt time:

• Station 1: 1.4 - 1.4 = 0 min (fully utilized)
• Station 2: 1.4 - 1.9 = -0.5 min (overload, needs balancing)
• Station 3: 1.4 - 1.32 = 0.08 min idle
• Station 4: 1.4 - 1.65 = -0.25 min (overload)

The positive values indicate stations with idle time, and negative values reflect overloads, prompting a need to balance the workload further.

e) Least and Greatest Idle Times

Based on the calculations, Station 3 has the least idle time (approximately 0.08 min), indicating it is nearly fully utilized or slightly underloaded.

Station 2 and 4 are overloaded, with negative idle time values, emphasizing they exceed capacity and require redistribution of tasks.

Conclusion

Overall, the line balancing with the added quality check task requires careful consideration to maintain productivity. The calculations highlight the importance of balancing task times, respecting dependencies, and optimizing station assignments to minimize idle time and maximize efficiency.

References

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