Line Balancing & Worker Machine Process Chart

LINE BALANCING & WORKER AND MACHINE PROCESS CHART

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.

How to Calculate Line Efficiency ? Standard Minutes Operator  to Perform Operation 1 2 3 4 5

0.49 0.31 0.25 0.44 0.54

Operator 4

Operator 3

Operator 5 Operator 1 Operator 2

Line Efficiency ? 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

2.7

5

 S . M .  E  

1 5

 A. M  1

0.54 0.54 0.54 0.54 0.54

 X 100

Opportunities for Improvements ? 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

0.54 0.54 0.54 0.54 0.54

2.03

2.7 Efficiency

75.19%

0.44 0.54

0.25

0.49

0.31

 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

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?

 N    R  

SM    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

15 1.67 0.95 26.36842105

27 operators

 E 

Output? Operation 1 2 3 4 5 6 7 inutes/unit

1.5/0.6 2.25/0.6 Dst.

(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

600/1000

27

Slowest One? Operation 1 2 3 4 5 6 7

(standard Standard Minutes minutes)/(minutes/u Operators nit) 1.5 2.5 3 0.5 2.25 3.75 4 0.5625 1.25 2.083333333 2 0.625 2.5 4.166666667 5 0.5 3 5 5 0.6 2.75 4.583333333 5 0.55 1.75 2.916666667 3 0.583333333

output  

5  60 3

 100 / hour   1000 / day

(2.5/3)*0.6 (3.75/4)*0.6 Dst.

Line Balancing Problem 3.4 mins B E

2.2 mins A

2.7 mins

C 4.1mins D

F

1.7mins

3.3 mins

G 2.6 mins

Question •

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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?

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 A. 73.2% B. 56.7% C. 69.7% D. 79.6% E. 81.2%

Calculate efficiency

2.2 mins

E

A

3.4 mins B

2.7 mins

C 4.1mins D

F

G

1.7mins

3.3 mins

2.6 mins

(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

Number of Workstation 



 (bottleneck)

20 = 4.88 work stations 4.1

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

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

40 secs

What is the minimum # of work stations? Round down. 59 secs 34 secs

84 secs •

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•

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3 2 4 5 6

56 secs

45 secs

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

318/6 x 84= 318/504 = 63%

100

99 secs

3 Stations ?

40 secs

118 secs 59 secs 34 secs

84 secs

318/3x118

56 secs

45 secs

318/354 = 89.8%

Efficency %  

 task times 

101 secs

number of  stationscycle time 

100

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

Worker and Machine Process Chart

Worker and Machine Process Chart Definisi dan Kegunaan 

Menggambarkan hubungan waktu antara working cycle seseorang dengan operating cycle suatu mesin



membandingkan idle mesin dengan idle pekerja

Worker and Machine Process Chart Contoh: man

Machine 1

loading

loading

Machine 2

walking press loading walking

loading idle

press

Worker and Machine Process Chart

lm N1  l w N = jumlah mesin m = total machine running time w = walking time l = loading and unloading time

Worker and Machine Process Chart •

total expected cost  (TEC)

TEC  N 1 

(l   m)( K 1   N 1 K 2 )  N 1

TEC  N 2  (l   w)( K 1   N 2 K 2 ) K1 = operator rate K2 = cost of machine

Worker and Machine Process Chart •

c o n t o h s o al  

proses

Time (minute)

Pick up plate into press dies Lubricate dies in press

0,1 0,3

Press Walk to next press $ worker (K1) = $ 12 / hour $ machine (K2) = $ 10 / hour

1,2 0,1

Worker and Machine Process Chart •

Ditanya: •

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Jumlah mesin yang dibutuhkan TEC tiap kemungkinan jumlah mesin

Jawaban •

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N = (l+m)/(l+w) = (0.4+1.2)/ (0.4+0.1) = 3.2  Sehingga jumlah mesin yang mungkin adalah 3 atau 4 mesin