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Line balancing application

Date post:01-Nov-2014
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Line balancing of production line is very important to get perfect takt time , to monitor value stream line in process and get kanban consumption pattern evenly
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  • 1. LINE BALANCING
  • 2. Definisi Line balancing merupakan suatu metode penugasan sejumlah pekerjaan yang saling berkaitan dalam satu lini produksi sehingga setiap stasiun kerja memiliki waktu yang tidak melebihi waktu siklus dari stasiun kerja tersebut. Line balancing berusaha menyeimbangkan seluruh lintasan yang ada dalam lini perakitan sehingga aliran produksi berjalan lancar.
  • 3. Standard Minutes Operator to Perform Operation 1 2 3 4 5 0.49 0.31 0.25 0.44 0.54 How to Calculate Line Efficiency ? Operator 4 Operator 5 Operator 3 Operator 2 Operator 1
  • 4. Standard Minutes Allowed Wait Time Based on Operator to Perform Standard Slowest Operator Operation Minutes 1 2 3 4 5 0.49 0.31 0.25 0.44 0.54 0.05 0.23 0.29 0.1 0 2.03 5 E= S .M . 1 5 A.M 1 X 100 Line Efficiency ? 0.54 0.54 0.54 0.54 0.54 2.7
  • 5. Standard Minutes Allowed Wait Time Based on Operator to Perform Standard Slowest Operator Operation Minutes 1 2 3 4 5 0.49 0.31 0.25 0.44 0.54 0.05 0.23 0.29 0.1 0 2.03 0.54 0.54 0.54 0.54 0.54 2.7 Efficiency Opportunities for Improvements ? 75.19% 0.44 0.54 0.25 0.31 0.49
  • 6. Assembly Line (7 operations) Operation Standard Minutes 1 1.5 2 2.25 3 1.25 4 2.5 5 3 6 2.75 7 1.75
  • 7. Desired Rate of Production = 1000/day Efficiency = 95% 1 day = 10 hours = 600 minutes Thus R = 1000/600 = 1.67 units/minute Numbers of Operator Needed in Assembly Line?
  • 8. SM N = R AM =R Operation Standard Minutes 1 2 3 4 5 6 7 1.5 2.25 1.25 2.5 3 2.75 1.75 minutes/unit 0.6 SM R E E 15 1.67 0.95 26.36842105 27 operators
  • 9. Operation 1 2 3 4 5 6 7 minutes/unit (standard Standard Minutes minutes)/(minutes/u Operators nit) 1.5 2.5 3 2.25 3.75 4 1.25 2.083333333 2 2.5 4.166666667 5 3 5 5 2.75 4.583333333 5 1.75 2.916666667 3 0.6 27 Output?
  • 10. Slowest One? Operation 1 2 3 4 5 6 7 (standard Standard Minutes minutes)/(minutes/u Operators nit) 1.5 2.5 3 2.25 3.75 4 1.25 2.083333333 2 2.5 4.166666667 5 3 5 5 2.75 4.583333333 5 1.75 2.916666667 3 0.5 0.5625 0.625 0.5 0.6 0.55 0.583333333 5 60 output = = 100 / hour = 1000 / day 3
  • 11. Line Balancing Problem 3.4 mins B E 2.2 mins A 2.7 mins C 4.1mins D F G 1.7mins 3.3 mins 2.6 mins
  • 12. Question 1. What process is the bottleneck? 2. How much is the maximum production per hour? 3. How much the efficiency? 4. How to minimize work stations? 5. How should they be grouped? 6. New efficiency?
  • 13. A. 73.2% B. 56.7% C. 69.7% D. 79.6% E. 81.2% Calculate efficiency 3.4 mins B E 2.2 mins A 2.7 mins C 4.1mins D F G 1.7mins 3.3 mins 2.6 mins
  • 14. (2.2+3.4+4.1+2.7+1.7+3.3+2.6) 4.1x7 20 28.7 69.7% 1-69.7%=30.3% Balance Delay
  • 15. Number of Workstation ( task times) N= cycletime (bottleneck) 20 = 4.88 work stations 4.1
  • 16. 4 Stations 20/24=83.3% Line Balancing Solution (5.6) 3.4 B Station 1 Max prod./hour 60/6 Station 3 10 units/hour E 2.2 A C 2.7 4.1 Station 2 (5.8) All under 6 minutes? (6.0) Station 4 D F G 1.7 3.3 2.6
  • 17. 5 Stations Line Balancing Problem 3.4 mins 20/5.6x5 = 20/28 = 71.4% 5.6 B E 2.2 mins A 2.7 mins C 4.1mins Max Prod./hour 60/5.6 10.7 units/hour D F G 1.7mins 3.3 mins 2.6 mins 5.0
  • 18. 40 secs What is the minimum # of work stations? Round down. 59 secs 3 2 4 5 6 34 secs 84 secs 56 secs 45 secs
  • 19. ( task times) N= cycle time 40+59+84+56+34+45 = 318 318/84 = 3.78 or 3 work stations What is the efficiency with 6 operators? Efficency ( % ) = ( task times) ( number of stations )( cycle time) (100) 318/6 x 84= 318/504 = 63%
  • 20. 99 secs 3 Stations ? 40 secs 118 secs 59 secs 34 secs 84 secs 318/3x118 56 secs 318/354 = 89.8% Efficency ( % ) = ( task times) 45 secs 101 secs ( number of stations )( cycle time) (100)
  • 21. 99 secs 40 secs 4 Stations? 84 secs 79 secs 59 secs 34 secs 84 secs 318/4 x 99 = 318/396 = 80.3% 56 secs 56 secs 45 secs

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