Model Answer
0 min readIntroduction
In operations management, optimizing production processes is crucial for efficiency and cost-effectiveness. A key aspect of this optimization involves identifying and managing bottlenecks – the workstations that limit the overall throughput of the system. Understanding the impact of changes in processing times at different workstations on overall production and Work-In-Process (WIP) is fundamental to effective production planning and control. This answer will analyze the given production process, calculate the number of units produced, and assess the effect of a processing time reduction on WIP.
(i) Units Produced in an 8-Hour Day
To determine the number of units produced in an 8-hour day, we need to identify the bottleneck workstation. The bottleneck is the workstation with the longest processing time per unit. From the given information, the processing times are:
- S1: 10 minutes
- S2: 15 minutes
- S3: 12 minutes
- S4: 8 minutes
Therefore, S2 is the bottleneck workstation with a processing time of 15 minutes per unit.
An 8-hour day is equivalent to 480 minutes (8 * 60). The maximum number of units that can be processed by S2 in an 8-hour day is:
Throughput = Total time available / Processing time per unit = 480 minutes / 15 minutes/unit = 32 units.
Since S2 is the bottleneck, the entire system's throughput is limited by S2. Therefore, 32 units are produced in an 8-hour day.
(ii) Reduction in Work-In-Process (WIP) with Reduced S3 Time
If the processing time in S3 is reduced to 8 minutes, there will be a reduction in Work-In-Process (WIP), but not necessarily a direct proportional reduction in the total WIP. Here's why:
Reducing the processing time at S3 improves the flow of units through the system. However, the bottleneck remains at S2. The system's throughput is still limited by S2's capacity of 32 units per day.
WIP is directly related to cycle time. Cycle time is the time it takes for a unit to move through the entire process. Reducing the processing time at S3 reduces the cycle time, but only to the extent that it doesn't impact the bottleneck. Since S2 remains the bottleneck, the cycle time will still be determined by S2's processing time and the number of units it can process.
The reduction in S3 processing time will primarily reduce the queue length *before* S3. Units will spend less time waiting at S3. However, the queue length *before* S2 will likely remain the same, as S2 continues to be the limiting factor. Therefore, the overall WIP will decrease, but the magnitude of the decrease will be less than the reduction in S3 processing time would suggest if S3 were the bottleneck.
To quantify the reduction in WIP, Little’s Law can be applied:
WIP = Throughput * Cycle Time
Initially, Cycle Time ≈ 15 minutes/unit (dominated by S2). WIP = 32 units * 15 minutes = 480 minutes (expressed as minutes of work in the system).
After the reduction, the cycle time will slightly decrease, but remain close to 15 minutes. The reduction in S3 processing time will reduce the waiting time before S3, but the overall cycle time will still be governed by S2. Therefore, the WIP will decrease, but not significantly.
Conclusion
In conclusion, the production process is limited by the bottleneck workstation, S2, resulting in a throughput of 32 units per 8-hour day. Reducing the processing time at S3 will reduce the Work-In-Process, but the reduction will be limited by the continued presence of S2 as the bottleneck. Effective operations management requires focusing on bottleneck resolution to maximize throughput and minimize WIP. Further analysis, including cost-benefit analysis of bottleneck improvements, would be beneficial.
Answer Length
This is a comprehensive model answer for learning purposes and may exceed the word limit. In the exam, always adhere to the prescribed word count.