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A particular automated machine is scheduled to run for an 8 hour shift. However, the shift has two breaks each lasting 1 0 minutes and a 3 0 minute lunch period. Also, on the shift in question, 3 0 minutes has been scheduled for preventative maintenance. This preventative maintenance does not happen during break periods. The machine itself has a rated cycle time of 8 s . During the shift there were 1 2 minutes of downtime due to machine failure and 2 7 minutes due to setting up jobs. At the end of the shift, the machine had produced 2 1 9 0 good parts and 4 0 scrap parts. \ table [ [ Shift Duration, 8 Hours ] , [ Scheduled Breaks / Shift , 2 × 1 0 minute breaks 8 1 × 3 0 minute lunch break ] , [ Scheduled Preventative Maintenance, 3 0 minutes ] , [ Machine Cycle Time, 1 8 8 ] , [ Set - Up A particular automated machine is scheduled to run for an 8 hour shift. However, the shift has two breaks each lasting 1 0 minutes and a 3 0 minute lunch period. Also, on the shift in question, 3 0 minutes has been scheduled for preventative maintenance. This preventative maintenance does not happen during break periods. The machine itself has a rated cycle time of 8 s . During the shift there were 1 2 minutes of downtime due to machine failure and 2 7 minutes due to setting up jobs. At the end of the shift, the machine had produced 2 1 9 0 good parts and 4 0 scrap parts. \ table [ [ Shift Duration, 8 Hours ] , [ Scheduled Breaks / Shift , 2 × 1 0 minute breaks 8 1 × 3 0 minute lunch break ] , [ Scheduled Preventative Maintenance, 3 0 minutes ] , [ Machine Cycle Time, 1 8 8 ] , [ Set - Up. Step by step answer with every step explanation and final answer

Question:

A particular automated machine is scheduled to run for an 8 hour shift. However, the shift has two breaks each lasting 1 0 minutes and a 3 0 minute lunch period. Also, on the shift in question, 3 0 minutes has been scheduled for preventative maintenance. This preventative maintenance does not happen during break periods. The machine itself has a rated cycle time of 8 s . During the shift there were 1 2 minutes of downtime due to machine failure and 2 7 minutes due to setting up jobs. At the end of the shift, the machine had produced 2 1 9 0 good parts and 4 0 scrap parts. \ table [ [ Shift Duration, 8 Hours ] , [ Scheduled Breaks / Shift , 2 × 1 0 minute breaks 8 1 × 3 0 minute lunch break ] , [ Scheduled Preventative Maintenance, 3 0 minutes ] , [ Machine Cycle Time, 1 8 8 ] , [ Set - Up A particular automated machine is scheduled to run for an 8 hour shift. However, the shift has two breaks each lasting 1 0 minutes and a 3 0 minute lunch period. Also, on the shift in question, 3 0 minutes has been scheduled for preventative maintenance. This preventative maintenance does not happen during break periods. The machine itself has a rated cycle time of 8 s . During the shift there were 1 2 minutes of downtime due to machine failure and 2 7 minutes due to setting up jobs. At the end of the shift, the machine had produced 2 1 9 0 good parts and 4 0 scrap parts. \ table [ [ Shift Duration, 8 Hours ] , [ Scheduled Breaks / Shift , 2 × 1 0 minute breaks 8 1 × 3 0 minute lunch break ] , [ Scheduled Preventative Maintenance, 3 0 minutes ] , [ Machine Cycle Time, 1 8 8 ] , [ Set - Up. Step by step answer with every step explanation and final answer

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Asked by: Test User 100021
Created at: 2025-05-03 12:09:05

Final Answer:

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Let's analyze this problem step by step. --- ### **Step 1: Convert shift duration to minutes** - Shift duration = 8 hours - 1 hour = 60 minutes - Total shift time in minutes: \[ 8 \text{ hours} \times 60 \text{ min/hour} = 480 \text{ minutes} \] --- ### **Step 2: Calculate total scheduled break time** - Two breaks of 10 minutes each: \[ 2 \times 10 = 20 \text{ minutes} \] - Lunch period: 30 minutes - Total scheduled break time: \[ 20 + 30 = 50 \text{ minutes} \] --- ### **Step 3: Calculate scheduled preventative maintenance time** - Scheduled for 30 minutes - Note: Preventative maintenance does **not** occur during break periods, so no need to subtract breaks from this. - Total scheduled downtime for maintenance: 30 minutes --- ### **Step 4: Calculate total downtime** - Machine failure downtime: 12 minutes - Setup downtime: 27 minutes - Total unscheduled downtime: \[ 12 + 27 = 39 \text{ minutes} \] --- ### **Step 5: Calculate total non-productive time** - Sum of all non-productive times: \[ \text{Breaks} + \text{Maintenance} + \text{Downtime} = 50 + 30 + 39 = 119 \text{ minutes} \] --- ### **Step 6: Calculate net operating time** - Total shift time: 480 minutes - Total non-productive time: 119 minutes - Operating time (actual production time): \[ 480 - 119 = 361 \text{ minutes} \] --- ### **Step 7: Calculate total parts produced** - Good parts: 2190 - Scrap parts: 40 - Total parts produced: \[ 2190 + 40 = 2230 \text{ parts} \] --- ### **Step 8: Calculate total cycle time and verify** - Machine rated cycle time: 8 seconds per part - Convert to minutes: \[ 8 \text{ seconds} \div 60 = \frac{8}{60} = 0.1333 \text{ minutes} \] - Theoretical maximum parts in the actual operating time (assuming 100% efficiency): \[ \frac{\text{Operating time}}{\text{Cycle time}} = \frac{361}{0.1333} \approx 2707.5 \text{ parts} \] --- ### **Step 9: Calculate production efficiency** - Actual parts produced: 2190 - Theoretical maximum: approximately 2707.5 - Efficiency: \[ \frac{2190}{2707.5} \times 100 \approx 80.85\% \] --- ### **Final Answer:** **a) Total non-productive time (breaks, maintenance, and downtime):** \[ \boxed{119 \text{ minutes}} \] **b) Operating time (actual production time):** \[ \boxed{361 \text{ minutes}} \] **c) Total parts produced:** \[ \boxed{2190 \text{ good parts} + 40 \text{ scrap parts} = 2230 \text{ parts}} \] **d) Production efficiency based on cycle time:** \[ \boxed{\approx 80.85\%} \] --- If you need any further analysis or clarification, feel free to ask!

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