Aquatica makes underwater camera housings for divers. The process begins with a solid rectangular block of aluminum which is used to make a single housing. A Computer Numerically Controlled (CNC) machine drills into the block to make the metal "frame" of the housing. Each block requires 15 minutes of drilling. The frame is then chemically treated in a series of baths. There is ample capacity in the treating stage. Finally, a worker must manually attach the various buttons and components to the metal housing. This assembly takes 120 minutes per housing per worker and Aquatica has 6 workers trained to do this task. The CNC machine requires a 30 minute setup time before the production of a batch can begin.
Housing type Demand rate (housings per hour)
D7000 0.4
5DS Mark IlI 0.5
AN-5n 0.6
D300 0.7
T2i 0.8
a. If the CNC machine produces 12 housing between setups, what would be its capacity (in housings per hour)?
b. Assuming the process is supply constrained, what is the utilization (as a percentage) of the CNC machine if it operates in batches of 12 housings? (Round your answer to 2 decimal places.)
c. Assuming the process is supply-constrained and operates with batches of 12 housings, what is the average inventory of housings?
d. Aquatica actually makes 5 different housings. As already mentioned, the CNO machine must be setup whenever production switches from one housing to the other. In each case, the setup time is 32 minutes and it takes 17 minutes to drill the block for each type of housing. Demands for the housings are given in the table above. If they wish to minimize inventory while satisfying their demand, how many D7000 housings should they produce in each batch?

(b)Capacity for assembly: = 113/6 = 18.8/minute = 60/18.8 = 3.19/hour as assembly takes 113 minutes per housing per worker and 6 workers are working at the moment Assembly time for batch of 12 housings = 12/3.19 hr = 3.77 hr CNC time for batch = 32+(17×12 )= 236 minutes = 3.93 hr Utilization of CNC: = (17*12)/60/3.766 = 0.902815 = 0

O.90 or 90%

(c)Average inventory = 1/2 ×batch size × (1-(capacity for assembly × 1/4))

= 1/2×12*(1-(3.186/4) = 1.22(Approximately)

(d)Total demand of all the housings = 0.4+0.5+0.6+0.7+0.8 = 3/hour

Setup time per housing = 32/60 hr = 0.53 hr

Setup time for 5 housings will be = 5×0.53 = 2.67 hr

The Processing time = 17/60 hr = 0.2833 hr

Total batch size = (2.67×3)/(1-3×0.2833) = 53.36

Batch size of D700 = 53.36*(0.4/3) = 7.11 = 7(approximately)

tallinn tallinn · Answer 2 · 2022-03-15T13:58:34+01:00

(a) The capacity is =3.05hours

(b) The utilization percentage is = 0.90 or 90%

(c) The Average inventory is =1.22

(d) when the Batch size of D700 is approximately equal to 7

How to Calculate Average inventory?

(a) Now To calculate Capacity is:

Then =( Batch-size) /(set up time + processing time)

After that = 12/ (32+17×12) = 12/236 minutes = (12/236) ×60 hours =

3.05 hours

(b) When the Capacity for assembly is = 113/6 = 18.8/minute = then 60/18.8 is = 3.19/hour as assembly takes 113 minutes per housing per worker and also 6 workers are working at the moment.

After that Assembly time for batch of 12 housings is = 12/3.19 hr = 3.77 hr Then C-N-C time for batch is = 32+(17×12 )= 236 minutes = 3.93 hr After that Utilization of C-N-C is = (17*12)/60/3.766 = 0.902815 = 0

O.90 or 90%

(c) Then the Average inventory is = 1/2 ×batch size × (1-(capacity for assembly × 1/4))

Then = 1/2×12*(1-(3.186/4) = 1.22(Approximately)

(d) Then the Total demand of all the housings is = 0.4+0.5+0.6+0.7+0.8 = 3/hour

After that Setup time per housing is = 32/60 hr = 0.53 hr

Then Setup time for 5 housings will be = 5×0.53 = 2.67 hr

Now The Processing time is = 17/60 hr = 0.2833 hr

After that Total batch size is = (2.67×3)/(1-3×0.2833) = 53.36

Then Batch-size of D700 is = 53.36*(0.4/3) = 7.11 = 7(approximately)

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