Question:
Consider an assembly line producing an item through a process involving three activities in the following order: cutting, welding, and polishing. There are three workers available: one is assigned only for cutting, one is assigned only for welding, and one is assigned only for polishing.
The activity time per item is given in the table.
Activity name Cutting Welding Polishing Activity time/item (in minutes) 30 40 25
Assuming that the process is running in steady state, the total number of items coming out of the assembly line in six hours is ______ (in integer).
Note: Assume that workers, items, and machines/tools are available as and when needed.
Consider an assembly line producing an item through a process involving three activities in the following order: cutting, welding, and polishing. There are three workers available: one is assigned only for cutting, one is assigned only for welding, and one is assigned only for polishing.
The activity time per item is given in the table.
Assuming that the process is running in steady state, the total number of items coming out of the assembly line in six hours is ______ (in integer).
Note: Assume that workers, items, and machines/tools are available as and when needed.
The activity time per item is given in the table.
| Activity name | Cutting | Welding | Polishing |
|---|---|---|---|
| Activity time/item (in minutes) | 30 | 40 | 25 |
Assuming that the process is running in steady state, the total number of items coming out of the assembly line in six hours is ______ (in integer).
Note: Assume that workers, items, and machines/tools are available as and when needed.
Show Hint
The slowest activity (largest time per item) sets the pace of the whole assembly line.
Updated On: Aug 3, 2026
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Verified By Collegedunia
Correct Answer: 9
Solution and Explanation
Step 1: Understanding the Concept:
Three workers handle cutting, welding and polishing one after another on a single-item assembly line.
Once the line reaches steady state, every station is working continuously, so the line's overall output rate is limited by whichever station is the slowest.
Step 2: Key Formula or Approach:
Instead of working with the cycle time directly, we can convert each station's time into a production rate (items produced per minute), and the slowest rate becomes the rate of the whole line:
\[ \text{Rate of station} = \frac{1}{\text{time per item at that station}} \]
The station with the lowest rate (equivalently the highest time per item) is the bottleneck, and it sets the pace for the entire line.
Step 3: Detailed Explanation:
Cutting rate \( = \frac{1}{30} \) items/min.
Welding rate \( = \frac{1}{40} \) items/min.
Polishing rate \( = \frac{1}{25} \) items/min.
Comparing the three, \( \frac{1}{40} \) is the smallest rate, so welding is the bottleneck station and it governs the line's throughput.
Total time available is 6 hours:
\[ 6 \text{ hours} = 6 \times 60 = 360 \text{ minutes} \]
Multiply the bottleneck rate by the total available time:
\[ \text{Total items} = 360 \times \frac{1}{40} = 9 \text{ items} \]
Final Answer:
The assembly line, running at steady state and paced by the welding station, produces the following number of items in six hours. \[ \boxed{9 \text{ items}} \]
Three workers handle cutting, welding and polishing one after another on a single-item assembly line.
Once the line reaches steady state, every station is working continuously, so the line's overall output rate is limited by whichever station is the slowest.
Step 2: Key Formula or Approach:
Instead of working with the cycle time directly, we can convert each station's time into a production rate (items produced per minute), and the slowest rate becomes the rate of the whole line:
\[ \text{Rate of station} = \frac{1}{\text{time per item at that station}} \]
The station with the lowest rate (equivalently the highest time per item) is the bottleneck, and it sets the pace for the entire line.
Step 3: Detailed Explanation:
Cutting rate \( = \frac{1}{30} \) items/min.
Welding rate \( = \frac{1}{40} \) items/min.
Polishing rate \( = \frac{1}{25} \) items/min.
Comparing the three, \( \frac{1}{40} \) is the smallest rate, so welding is the bottleneck station and it governs the line's throughput.
Total time available is 6 hours:
\[ 6 \text{ hours} = 6 \times 60 = 360 \text{ minutes} \]
Multiply the bottleneck rate by the total available time:
\[ \text{Total items} = 360 \times \frac{1}{40} = 9 \text{ items} \]
Final Answer:
The assembly line, running at steady state and paced by the welding station, produces the following number of items in six hours. \[ \boxed{9 \text{ items}} \]
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