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Raw Material Inventory TASK-A (3 sec) TASK-B (17 sec) TASK-C (13 sec) TASK-D (7 sec) Finished Good Inventory Use the above in

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Answer #1
Task no. Time in secs per unit
A 3
B 17
C 13
D 7

a. The task B takes 17 secs per unit. This time is the maximum for any task. This is going to limit the output of the entire line.

∴   task B is the bottleneck operation.

b. Capacity in units per hour =3600/17=211.76 =211 no.s per hour.

If the allowance of 30 minutes in a shift of 9 hrs (i.e 5.55 %) is taken into account, the achievable capacity is :

211.76/(1+0.055)=200.72 or 200 units /hour of the 9 hour shift.

c. Order received for one year=539,000

Time available in the year=9 hrs /day*6 days /week*50 weeks=2700 hours.

Possible output in one year=2700*200==540,000 (theoreticall 540317)

This figure is more than the order.

the system can handle this order by working at an overall equipment effectiness of

(539000/540000)*100=99.8 %. (This achievement is unheard of)

d. A fifth station, obviously has to be added only for task B which is the bottleneck at present. Adding a station for any other task will not reduce the bottleneck task. The additional station will not increase the capacity and will be superfluous (unnnecessary cost)

e. The time in seconds for task B now becomes 17/2 =8.5 secs.

Task no. Time in secs per unit
A 3
B (Two stations) 8.5
C 13
D 7

The bottleneck task, now is C-with a time of 13 seconds.

The capacity/hour, now is=3600/13=276.9 or 277 units/hour.

This gets reduced to the extent of the 30 minute allowanmce per shift and =276.9/(1+0.055)=262.46

or say 262 units/hour.

Capacity per year=262*9*6*50=707400.

The order can now be completeed by working at an OEE of (539000/707400)*100=76.19 or 76 %

This OEE is well achievable.

**

NOTE: Student may also note that there is too large an unbalance in the line and this calls for a reorganization of the entire process by detailed method study.

** **

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