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【正文】 p e r i o dC = 4 2 0 m i n s / d a y1 0 0 u n i t s / d a y = 4 . 2 m i n s / u n i tQuestion: Suppose we want to assemble 100 fans per day. What would our cycle time have to be? Answer: Example of Line Balancing: Determine Theoretical Minimum Number of Workstations Question: What is the theoretical minimum number of workstations for this problem? Answer: T h e o r e t i c a l M i n . N u m b e r o f Wo r k s t a t i o n s , NN = S u m o f t a s k t i m e s ( T )C y c l e t i m e ( C )ttN = 1 1 . 3 5 m i n s / u n i t4 . 2 m i n s / u n i t = 2 . 7 0 2 , o r 3tExample of Line Balancing: Rules To Follow for Loading Workstations ? Primary: Assign tasks in order of the largest number of following tasks. ? Secondary (tiebreaking): Assign tasks in order of the longest operating time A C B D E F G H 2 1 .5 1 1 Station 1 Station 2 Station 3 Task Followers Time (Mins) A 6 2 C 4 D 3 B 2 1 E 2 F 1 1 G 1 1 H 0 A C B D E F G H 2 1 .5 1 1 Station 1 Station 2 Station 3 A (=) Task Followers Time (Mins) A 6 2 C 4 D 3 B 2 1 E 2 F 1 1 G 1 1 H 0 A C B D E F G H 2 1 .5 1 1 A (=) B (=) Task Followers Time (Mins) A 6 2 C 4 D 3 B 2 1 E 2 F 1 1 G 1 1 H 0 Station 1 Station 2 Station 3 A C B D E F G H 2 1 .5 1 1 A (=) B (=) G (= .2) Idle= .2 Task Followers Time (Mins) A 6 2 C 4 D 3 B 2 1 E 2 F 1 1 G 1 1 H 0 Station 1 Station 2 Station 3 A C B D E F G H 2 1 .5 1 1 C ()=.95 Task Followers Time (Mins) A 6 2 C 4 D 3 B 2 1 E 2 F 1 1 G 1 1 H 0 A (=) B (=) G (= .2) Idle= .2 Station 1 Station 2 Station 3 C ()=.95 Idle = .95 A C B D E F G H 2 1 .5 1 1 Task Followers Time (Mins) A 6 2 C 4 D 3 B 2 1 E 2 F 1 1 G 1 1 H 0 A (=) B (=) G (= .2) Idle= .2 Station 1 Station 2 Station 3 C ()=.95 Idle = .95 A C B D E F G H 2 1 .5 1 1 D ()=3 Task Followers Time (Mins) A 6 2 C 4 D 3 B 2 1 E 2 F 1 1 G 1 1 H 0 A (=) B (=) G (= .2) Idle= .2 Station 1 Station 2 Station 3 A C B D E F G H 2 1 .5 1 1 C ()=.95 Idle = .95 D ()=3 E ()= Task Followers Time (Mins) A 6 2 C 4 D 3 B 2 1 E 2 F 1 1 G 1 1 H 0 A (=) B (=) G (= .2) Idle= .2 Station 1 Station 2 Station 3 A C B D E F G H 2 1 .5 1 1 C ()=.95 Idle = .95 D ()=3 E ()= F ()= Task Followers Time (Mins) A 6 2 C 4 D 3 B 2 1 E 2 F 1 1 G 1 1 H 0 A (=) B (=) G (= .2) Idle= .2 Station 1 Station 2 Station 3 Which station is the bottleneck? What is the effective cycle time? A C B D E F G H 2 1 .5 1 1 C ()=.95 Idle = .95 D ()=3 E ()= F ()= H ()=.1 Idle = .1 Task Followers Time (Mins) A 6 2 C 4 D 3 B 2 1 E 2 F 1 1 G 1 1 H 0 A (=) B (=) G (= .2) Idle= .2 Station 1 Station 2 Station 3 Example
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