《流水生产组织》ppt课件

1、 N F N F r e 0 r t S i i ei i i s s k m i ei m i i ssk 11 TaskTime (Mins) DescriptionPredecessors A2Assemble frameNone B1Mount switchA C3.25Assemble motor housingNone D1.2Mount motor housing in frameA, C E0.5Attach bladeD F1Assemble and attach safety grillE G1Attach cordB H1.4TestF, G Task Predecess

2、ors ANone A BA B CNone C DA, C D Task Predecessors ED E FE F GB G HE, G H A C B DEF G H 2 3.25 1 1.2.5 1 1.4 1 Answer: Task C is the cycle time of the line and therefore, the maximum rate of production. TaskTime (Mins) DescriptionPredecessors A2Assemble frameNone B1Mount switchA C3.25Assemble motor

3、housingNone D1.2Mount motor housing in frameA, C E0.5Attach bladeD F1Assemble and attach safety grillE G1Attach cordB H1.4TestE, G Max Production = Production time per day Bottleneck time = 420 mins 3.25 mins / unit =129 units Required Cycle Time, C = Production time per period Required output per p

4、eriod C = 420 mins / day 100 units / day = 4.2 mins / unit Answer: Answer: Theoretical Min. Number of Workstations, N N = Sum of task times (T) Cycle time (C) t t N = 11.35 mins / unit 4.2 mins / unit = 2.702, or 3 t A C B DEF G H 2 3.25 1 1.2.5 1 1.4 1 Station 1Station 2Station 3 TaskFollowersTime

5、(Mins) A62 C43.25 D31.2 B2 1 E20.5 F11 G11 H01.4 A C B DEF G H 2 3.25 1 1.2.5 1 1.4 1 Station 1Station 2Station 3 A (4.2-2=2.2) TaskFollowersTime (Mins) A62 C43.25 D31.2 B2 1 E20.5 F11 G11 H01.4 A C B DEF G H 2 3.25 1 1.2.5 1 1.4 1 A (4.2-2=2.2) B (2.2-1=1.2) TaskFollowersTime (Mins) A62 C43.25 D31.

6、2 B2 1 E20.5 F11 G11 H01.4 Station 1Station 2Station 3 A C B DEF G H 2 3.25 1 1.2.5 1 1.4 1 A (4.2-2=2.2) B (2.2-1=1.2) G (1.2-1= .2) Idle= .2 TaskFollowersTime (Mins) A62 C43.25 D31.2 B2 1 E20.5 F11 G11 H01.4 Station 1Station 2Station 3 A C B DEF G H 2 3.25 1 1.2.5 1 1.4 1 C (4.2-3.25)=.95 TaskFoll

7、owersTime (Mins) A62 C43.25 D31.2 B2 1 E20.5 F11 G11 H01.4 A (4.2-2=2.2) B (2.2-1=1.2) G (1.2-1= .2) Idle= .2 Station 1Station 2Station 3 C (4.2-3.25)=.95 Idle = .95 A C B DEF G H 2 3.25 1 1.2.5 1 1.4 1 TaskFollowersTime (Mins) A62 C43.25 D31.2 B2 1 E20.5 F11 G11 H01.4 A (4.2-2=2.2) B (2.2-1=1.2) G

8、(1.2-1= .2) Idle= .2 Station 1Station 2Station 3 C (4.2-3.25)=.95 Idle = .95 A C B DEF G H 2 3.25 1 1.2.5 1 1.4 1 D (4.2-1.2)=3 TaskFollowersTime (Mins) A62 C43.25 D31.2 B2 1 E20.5 F11 G11 H01.4 A (4.2-2=2.2) B (2.2-1=1.2) G (1.2-1= .2) Idle= .2 Station 1Station 2Station 3 A C B DEF G H 2 3.25 1 1.2

9、.5 1 1.4 1 C (4.2-3.25)=.95 Idle = .95 D (4.2-1.2)=3 E (3-.5)=2.5 TaskFollowersTime (Mins) A62 C43.25 D31.2 B2 1 E20.5 F11 G11 H01.4 A (4.2-2=2.2) B (2.2-1=1.2) G (1.2-1= .2) Idle= .2 Station 1Station 2Station 3 A C B DEF G H 2 3.25 1 1.2.5 1 1.4 1 C (4.2-3.25)=.95 Idle = .95 D (4.2-1.2)=3 E (3-.5)=

10、2.5 F (2.5-1)=1.5 TaskFollowersTime (Mins) A62 C43.25 D31.2 B2 1 E20.5 F11 G11 H01.4 A (4.2-2=2.2) B (2.2-1=1.2) G (1.2-1= .2) Idle= .2 Station 1Station 2Station 3 Which station is the bottleneck? What is the effective cycle time? A C B DEF G H 2 3.25 1 1.2.5 1 1.4 1 C (4.2-3.25)=.95 Idle = .95 D (4

11、.2-1.2)=3 E (3-.5)=2.5 F (2.5-1)=1.5 H (1.5-1.4)=.1 Idle = .1 TaskFollowersTime (Mins) A62 C43.25 D31.2 B2 1 E20.5 F11 G11 H01.4 A (4.2-2=2.2) B (2.2-1=1.2) G (1.2-1= .2) Idle= .2 Station 1Station 2Station 3 Efficiency = Sum of task times (T) Actual number of workstations (Na) x Cycle time (C) Effic

12、iency = 11.35 mins / unit (3)(4.2mins / unit) =.901 1 2 3 4 5 6 7 8 10 9 11 6 2 5 7 1 2 6 3 5 4 5 Station 1 Minutes per Unit 6 Station 2 7 Station 3 3 Question: Suppose you load work into the three work stations below such that each will take the corresponding number of minutes as shown. What is the

13、 cycle time of this line? Answer: The cycle time of the line is always determined by the work station taking the longest time. In this problem, the cycle time of the line is 7 minutes. There is also going to be idle time at the other two work stations. TaskTime (Mins) DescriptionPredecessors A2Assem

14、ble frameNone B1Mount switchA C3.25Assemble motor housingNone D1.2Mount motor housing in frameA, C E0.5Attach bladeD F1Assemble and attach safety grillE G1Attach cordB H1.4TestF, G Task Predecessors ANone A BA B CNone C DA, C D Task Predecessors ED E FE F GB G HE, G H A C B DEF G H 2 3.25 1 1.2.5 1

15、1.4 1 Answer: Task C is the cycle time of the line and therefore, the maximum rate of production. TaskTime (Mins) DescriptionPredecessors A2Assemble frameNone B1Mount switchA C3.25Assemble motor housingNone D1.2Mount motor housing in frameA, C E0.5Attach bladeD F1Assemble and attach safety grillE G1

16、Attach cordB H1.4TestE, G Max Production = Production time per day Bottleneck time = 420 mins 3.25 mins / unit =129 units Required Cycle Time, C = Production time per period Required output per period C = 420 mins / day 100 units / day = 4.2 mins / unit Answer: Answer: Theoretical Min. Number of Wor

17、kstations, N N = Sum of task times (T) Cycle time (C) t t N = 11.35 mins / unit 4.2 mins / unit = 2.702, or 3 t A C B DEF G H 2 3.25 1 1.2.5 1 1.4 1 Station 1Station 2Station 3 TaskFollowersTime (Mins) A62 C43.25 D31.2 B2 1 E20.5 F11 G11 H01.4 A C B DEF G H 2 3.25 1 1.2.5 1 1.4 1 Station 1Station 2S

18、tation 3 A (4.2-2=2.2) TaskFollowersTime (Mins) A62 C43.25 D31.2 B2 1 E20.5 F11 G11 H01.4 A C B DEF G H 2 3.25 1 1.2.5 1 1.4 1 A (4.2-2=2.2) B (2.2-1=1.2) TaskFollowersTime (Mins) A62 C43.25 D31.2 B2 1 E20.5 F11 G11 H01.4 Station 1Station 2Station 3 A C B DEF G H 2 3.25 1 1.2.5 1 1.4 1 A (4.2-2=2.2)

19、 B (2.2-1=1.2) G (1.2-1= .2) Idle= .2 TaskFollowersTime (Mins) A62 C43.25 D31.2 B2 1 E20.5 F11 G11 H01.4 Station 1Station 2Station 3 A C B DEF G H 2 3.25 1 1.2.5 1 1.4 1 C (4.2-3.25)=.95 TaskFollowersTime (Mins) A62 C43.25 D31.2 B2 1 E20.5 F11 G11 H01.4 A (4.2-2=2.2) B (2.2-1=1.2) G (1.2-1= .2) Idle

20、= .2 Station 1Station 2Station 3 C (4.2-3.25)=.95 Idle = .95 A C B DEF G H 2 3.25 1 1.2.5 1 1.4 1 TaskFollowersTime (Mins) A62 C43.25 D31.2 B2 1 E20.5 F11 G11 H01.4 A (4.2-2=2.2) B (2.2-1=1.2) G (1.2-1= .2) Idle= .2 Station 1Station 2Station 3 C (4.2-3.25)=.95 Idle = .95 A C B DEF G H 2 3.25 1 1.2.5

21、 1 1.4 1 D (4.2-1.2)=3 TaskFollowersTime (Mins) A62 C43.25 D31.2 B2 1 E20.5 F11 G11 H01.4 A (4.2-2=2.2) B (2.2-1=1.2) G (1.2-1= .2) Idle= .2 Station 1Station 2Station 3 A C B DEF G H 2 3.25 1 1.2.5 1 1.4 1 C (4.2-3.25)=.95 Idle = .95 D (4.2-1.2)=3 E (3-.5)=2.5 TaskFollowersTime (Mins) A62 C43.25 D31

22、.2 B2 1 E20.5 F11 G11 H01.4 A (4.2-2=2.2) B (2.2-1=1.2) G (1.2-1= .2) Idle= .2 Station 1Station 2Station 3 A C B DEF G H 2 3.25 1 1.2.5 1 1.4 1 C (4.2-3.25)=.95 Idle = .95 D (4.2-1.2)=3 E (3-.5)=2.5 F (2.5-1)=1.5 TaskFollowersTime (Mins) A62 C43.25 D31.2 B2 1 E20.5 F11 G11 H01.4 A (4.2-2=2.2) B (2.2

23、-1=1.2) G (1.2-1= .2) Idle= .2 Station 1Station 2Station 3 Which station is the bottleneck? What is the effective cycle time? A C B DEF G H 2 3.25 1 1.2.5 1 1.4 1 C (4.2-3.25)=.95 Idle = .95 D (4.2-1.2)=3 E (3-.5)=2.5 F (2.5-1)=1.5 H (1.5-1.4)=.1 Idle = .1 TaskFollowersTime (Mins) A62 C43.25 D31.2 B

24、2 1 E20.5 F11 G11 H01.4 A (4.2-2=2.2) B (2.2-1=1.2) G (1.2-1= .2) Idle= .2 Station 1Station 2Station 3 Efficiency = Sum of task times (T) Actual number of workstations (Na) x Cycle time (C) Efficiency = 11.35 mins / unit (3)(4.2mins / unit) =.901 1 r tT iS i x x ki 全部工步 1、2、5 1、2、6 3、5 4、55、7 3、8 7 710 9、10 11 3、94 4、8 10、11 59 5 10 568 5 57 58 66 510 510 5 9 7 6 6 6 6 10 64 65 m i i ei f gs bp 1 )1 ( n i ii CC Cn i ii BB Bn i ii AA A tQ tQ tQ tQ tQ tQ 111 ； ； eCCeBBeAA FF