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( a ) What is the total processing time ( per unit ) for this manufacturing cell consisting of 3 machines? ( Note that we assume that yield = 1 0 0 % for each process ) . ( b ) What is the capacity of this manufacturing cell per day? In other words, how many units of finished products can be produced by this cell per day? ( c ) Calculate Process Cycle Efficiency for this manufacturing cell based on per unit process time. give all step by step answer and give final summarya T80/s0rs0/30/day | E Forecast Forecasts PRODUCTION ee seco fori entrees pte Weekly Daily P fax bam ies 37200 pisceiwok B "1.0001" Weald Schcte 22005" y-60 pieces Fem] x Week Daly Ship EN / sa | [wn “ess ERCE & Fon [Faas Spe: T i | | ing A i / Y L) 3 staging Sheets 1000 “L” 1050 “L” 500 “L" 5 days 1250 “s™ 2300 “8” 975 “8” |C/T=3 seconds C/T=22 seconds. | C/T=35 seconds C/0=2 hours. C/0=30 minutes. C/0=45 minutes Uptime=90% Uptime=100% Uptime=100% I 25,200 sec. avail. Shift tsi | EPE=1 week" 25,200 sec. avail. 25,200 sec. avail. i 5 day 3.5 days 5.2 days 2.3 days | Lead Time 3 seconds 22 seconds 35 seconds FE Note this is processing time per unit. C/T: cycle time; C/O: changeover time; EPE: every part every

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( a ) What is the total processing time ( per unit ) for this manufacturing cell consisting of 3 machines? ( Note that we assume that yield = 1 0 0 % for each process ) . ( b ) What is the capacity of this manufacturing cell per day? In other words, how many units of finished products can be produced by this cell per day? ( c ) Calculate Process Cycle Efficiency for this manufacturing cell based on per unit process time. give all step by step answer and give final summaryUploaded Imagea T80/s0rs0/30/day | E Forecast Forecasts PRODUCTION ee seco fori entrees pte Weekly Daily P fax bam ies 37200 pisceiwok B "1.0001" Weald Schcte 22005" y-60 pieces Fem] x Week Daly Ship EN / sa | [wn “ess ERCE & Fon [Faas Spe: T i | | ing A i / Y L) 3 staging Sheets 1000 “L” 1050 “L” 500 “L" 5 days 1250 “s™ 2300 “8” 975 “8” |C/T=3 seconds C/T=22 seconds. | C/T=35 seconds C/0=2 hours. C/0=30 minutes. C/0=45 minutes Uptime=90% Uptime=100% Uptime=100% I 25,200 sec. avail. Shift tsi | EPE=1 week" 25,200 sec. avail. 25,200 sec. avail. i 5 day 3.5 days 5.2 days 2.3 days | Lead Time 3 seconds 22 seconds 35 seconds FE Note this is processing time per unit. C/T: cycle time; C/O: changeover time; EPE: every part every

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Asked by: lincy
Created at: 2025-09-29 13:00:22

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# Step-by-Step Solution Let's answer each part step by step using the information from the diagram. --- ## (a) **Total Processing Time (per unit) for 3 Machines** We are asked for the **total processing time per unit** for the three machines in the cell: - **PRESS**: Cycle time (C/T) = 3 sec - **PIERCE & FORM**: C/T = 22 sec - **FINISH GRIND**: C/T = 35 sec **Total processing time per unit:** \[ \text{Total Processing Time} = 3\,\text{sec} + 22\,\text{sec} + 35\,\text{sec} = 60\,\text{sec} \] --- ## (b) **Capacity of the Manufacturing Cell per Day** Let's find out **how many units can be produced per day** (assuming 1 shift, 25,200 sec available per machine per day). ### Step 1: Find the bottleneck (slowest step) - **PRESS**: 3 sec/unit ⇒ \(25,200/3 = 8,400\) units/day - **PIERCE & FORM**: 22 sec/unit ⇒ \(25,200/22 \approx 1,145\) units/day - **FINISH GRIND**: 35 sec/unit ⇒ \(25,200/35 \approx 720\) units/day **Bottleneck:** FINISH GRIND at 35 sec/unit (slowest, limits output) ### Step 2: Calculate daily capacity \[ \text{Capacity per day} = \frac{25,200\,\text{sec}}{35\,\text{sec/unit}} \approx 720\,\text{units/day} \] --- ## (c) **Process Cycle Efficiency (PCE) Based on Per Unit Process Time** **Formula:** \[ \text{Process Cycle Efficiency (PCE)} = \frac{\text{Value-Added Time}}{\text{Total Lead Time}} \] - **Value-Added Time** = Total processing time per unit = **60 sec** (from part a) - **Total Lead Time** = Production Lead Time = **16. days** (from bottom right of diagram) - Convert days to seconds: \(16. \text{ days} \times 24 \text{ hr/day} \times 60 \text{ min/hr} \times 60 \text{ sec/min} = 1,382,400 \text{ sec}\) \[ \text{PCE} = \frac{60}{1,382,400} \approx .0000434 = .00434\% \] --- ## **Summary Table** | Item | Value | |-------------------------------|---------------------------------| | (a) Total Processing Time | **60 seconds per unit** | | (b) Daily Capacity | **720 units per day** | | (c) Process Cycle Efficiency | **.0043%** | --- ## **Final Summary** 1. **Total processing time per unit** in the manufacturing cell (through all three machines) is **60 seconds**. 2. **Cell capacity** is limited by the slowest (bottleneck) machine (Finish Grind), resulting in **720 units per day**. 3. **Process Cycle Efficiency** is very low, at **.0043%**, indicating that most of the production lead time is spent waiting, not actually processing. --- **Let me know if you need further breakdown or explanation for any step!**

Question:

give 2 sentense summary in each question

This is AI generated Response.
Asked by: lincy
Created at: 2025-09-29 13:02:43

Final Answer:

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# Concise Summary of Each Question ## (a) Total Processing Time (per unit) The total processing time per unit for the manufacturing cell is **60 seconds**, which is the sum of the cycle times for all three machines. This indicates that each product takes a total of 60 seconds to complete the manufacturing process. ## (b) Capacity of the Manufacturing Cell per Day The manufacturing cell has a capacity of **720 units per day**, which is determined by the bottleneck machine (Finish Grind). This capacity reflects the maximum number of finished products that can be produced in a single day. ## (c) Process Cycle Efficiency (PCE) The Process Cycle Efficiency (PCE) for the manufacturing cell is approximately **0.0043%**, indicating a very low efficiency. This suggests that most of the production time is spent in non-value-added activities rather than actual processing.

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