In semiconductor manufacturing, Deposition is a batch process that costs materials on silicon wafers in a heated oven-like facility. Intal Corp. has a semi-conductor wafer fabrication factory that has on Desposition facility that can take up to 50 wafers in one batch. The setup time for heating the facility to the appropriate temperature is 1 hour. It then takes another 2 hours to complete the Desposition for one batch of wafers. The factory runs a push production and wafers depart from the previous process and arrive at the Desposition facility with an average arrival rate of 10.1 wafers per hour. What is the minimum batch size at Desposition in order to have a stable process without a continually growing job queue? a. 10.1*1/(1-1/10.1) = 11.2, so the minimum batch size is 11 b. 10.1*1/(1-1/10.1) = 11.2, so the minimum batch size is 12 c. 10.1*(1+2) = 30.3, so the minimum batch size is 31 d. the minimum batch size is 50, the maximum that the Desposition facility can take. e. 10.1*(1+2) = 30.3, so the minimum batch size is 30

Current on hand inventory = 1308 units

Withdrawal quantity = 95 units

Net quantity post withdrawal = 1308 – 95 = 1213

Since net quantity post withdrawal ( i.e. 1213 ) > Reorder point 1150, IT IS NOT YET TIME TO REORDER

In order to achieve a stable process ,

it is essential that Flow rate ( wafers / hour ) from previous process = Deposition rate ( Wafers/hour ) .

There will be inventory built up if flow rate from previous process > Deposition rate

Given flow rate from previous process = 10.1 wafers/ hour

Total time for deposition process = Heating time + deposition time = 1 hour + 2 hours = 3 hours

Therefore , in order to achieve a stable process, deposition process also must achieve a rate of 10.1 / hour . Thus batch size of the deposition process = 10.1 hour x 3 hours = 30.3 hours

ANSWER : e) 10.1( 1 + 2 ) = 30.3 , SO THE MINIMUM BATCH SIZE IS 30

 
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