a. What is the cell\'s curment inventory le b. What is the takt time for this ma

Question

a. What is the cell's curment inventory le b. What is the takt time for this manufa c. What is the production lead tme at 302 PART 2 MANAGING PROCESSES d How many days of work in process inventory is held between Finish Grind and Shipping a. What is the new value steam's proxduction lead time t. What is the new value stream's processing time The manager at Ormonde, Inc. collected the value stream mapping data from the plant's most probilematic manu- facturing cell that fabricates parts for washing machines This data is shown in Table 85 Using this data, calculate the current state performance of the cell and answer the following questions d. What is the total processing time of ing cel e What is the capacity of this neua TABLE 8.5 OPERATIONS DATA FOR ORMONDE, INC. Overall Process Atributes Process Step Average demand: 550/day Batch size: 20 Number of shifts per day: 3 Availability: 8 hours per shift with a 45-minute lunch break uting Cycle time- 120 seconds Setup time-3 minutes Up 100% Operators1 Process Step 2 Bending ycle tme 250 seconds Setup time 5 minutes Operators 2 -500 units (Before Bending Cycle tme 140 seconds Setup time- none Up the 100% Operators = 1 WP-200 urits Before Punting WP-1000 units (Aher Punting Process Step 3 Punching Customer Shipments One shipment of 2,750 units each we Information Flow l communications with customer are electronic There is a weekly arder release to Cutting All material is pushed

Explanation / Answer

Answer A:- Inventory level = 400 + 500 + 200 + 1,000 = 2,100 units

Answer B:- (8 * 60 – 45) * 3 = 1,305

Takt time = 1,305 / 550 = 2.3727

Answer C:-400 / 550 = 0.7273 day

500 / 550 = 0.9091 day

200 / 550 = 0.3636 day

1000 / 550 = 1.8182 days

Lead time = 0.7273 + 0.9191 + 0.3636 + 1.8182 = 3.8282 days

Answer D:- Process time = 120 + 250 + 140 = 510 seconds

Answer E:- Punching is the bottleneck.

140 + 0 / 20 seconds = 2.3333 minutes

Capacity = 1,305 / 2.3333 = 559.29 units

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