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.

Operator 1 Operator 2

Operator 3

Operator 4

Operator 5

Operator

Standard Minutes

to Perform

Operation

1 0.49

2 0.31

3 0.25

4 0.44

5 0.54

How to Calculate Line Efficiency ?

100

.

..

5

1

5

1 X

MA

MS

E

Operator

Standard Minutes

to Perform

Operation

Wait Time Based on

Slowest Operator

Allowed

Standard

Minutes

1 0.49 0.05 0.54

2 0.31 0.23 0.54

3 0.25 0.29 0.54

4 0.44 0.1 0.54

5 0.54 0 0.54

2.03 2.7

Line Efficiency ?

0.49 0.31

0.25

0.44

0.54

Operator

Standard Minutes

to Perform

Operation

Wait Time Based on

Slowest Operator

Allowed

Standard

Minutes

1 0.49 0.05 0.54

2 0.31 0.23 0.54

3 0.25 0.29 0.54

4 0.44 0.1 0.54

5 0.54 0 0.54

2.03 2.7

Efficiency 75.19%

Opportunities for Improvements ?

Operation Standard Minutes

1 1.5

2 2.25

3 1.25

4 2.5

5 3

6 2.75

7 1.75

Assembly Line

(7 operations)

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?

E

SMRAMRN

27 operators

Operation Standard Minutes

1 1.5

2 2.25

3 1.25

4 2.5

5 3

6 2.75

7 1.75

minutes/unit 0.6

∑SM 15

R 1.67

E 0.95

26.36842105

Operation Standard Minutes

(standard

minutes)/(minutes/u

nit)

Operators

1 1.5 2.5 3

2 2.25 3.75 4

3 1.25 2.083333333 2

4 2.5 4.166666667 5

5 3 5 5

6 2.75 4.583333333 5

7 1.75 2.916666667 3

minutes/unit 0.6 27

Output? 1.5/0.6

2.25/0.6

Dst.

600/1000

Slowest One?

Operation Standard Minutes

(standard

minutes)/(minutes/u

nit)

Operators

1 1.5 2.5 3 0.5

2 2.25 3.75 4 0.5625

3 1.25 2.083333333 2 0.625

4 2.5 4.166666667 5 0.5

5 3 5 5 0.6

6 2.75 4.583333333 5 0.55

7 1.75 2.916666667 3 0.583333333

dayhouroutput /1000/1003

605

(2.5/3)*0.6

(3.75/4)*0.6

Dst.

Line Balancing Problem

A

B

C

4.1mins

D

1.7mins

E

2.7 mins

F

3.3

mins

G

2.6 mins

2.2 mins

3.4 mins

• 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?

Question

Calculate efficiency • A. 73.2%

• B. 56.7%

• C. 69.7%

• D. 79.6%

• E. 81.2%

A

B

C

4.1mins

D

1.7mins

E

2.7 mins

F

3.3

mins

G

2.6 mins

2.2 mins

3.4 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

time cycle

times taskN

(bottleneck)

20

4.1 = 4.88 work stations

Number of Workstation

Line Balancing Solution

A

B

C

4.1

D

1.7

E

2.7

F

3.3

G

2.6

Station 1

Station 2

Station 3

Station 4

2.2

3.4

All under 6 minutes?

(6.0)

(5.6)

(5.8)

4 Stations 20/24=83.3%

Max prod./hour

60/6

10 units/hour

Line Balancing Problem

A

B

C

4.1mins

D

1.7mins

E

2.7 mins

F

3.3

mins

G

2.6 mins

2.2 mins

3.4 mins 5.6

5.0

20/5.6x5 = 20/28 = 71.4%

5 Stations

Max Prod./hour

60/5.6

10.7 units/hour

40 secs

59 secs

84 secs 34 secs

56 secs 45 secs

What is the minimum # of work stations?

Round down.

• 3

• 2

• 4

• 5

• 6

timecycle

task timesN

40+59+84+56+34+45 = 318

318/84 = 3.78 or 3 work stations

What is the efficiency with 6 operators?

100

timecyclestations ofnumber

task times%Efficency

318/6 x 84=

318/504 =

63%

100

timecyclestations ofnumber

task times%Efficency

40 secs

59 secs

84 secs 34 secs

56 secs 45 secs

99 secs

118 secs

3 Stations ?

318/3x118

318/354 = 89.8% 101 secs

40 secs

59 secs

84 secs 34 secs

56 secs 45 secs

4 Stations? 99 secs

84 secs

56 secs

79 secs

318/4 x 99 =

318/396 =

80.3%

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 Machine 2

loading loading

idle

walking press

loading loading

walking press

Worker and Machine Process Chart

wl

mlN1

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)

1

211 ))((1 N

KNKmlTECN

K1 = operator rate

K2 = cost of machine

))(( 2212KNKwlTECN

Worker and Machine Process Chart

• contoh soal

proses Time (minute)

Pick up plate into press dies 0,1

Lubricate dies in press 0,3

Press 1,2

Walk to next press 0,1

$ worker (K1) = $ 12 / hour

$ machine (K2) = $ 10 / hour

Worker and Machine Process Chart

• Ditanya:

• Jumlah mesin yang dibutuhkan

• TEC tiap kemungkinan jumlah mesin

Jawaban

• 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

Jawaban

• TEC3 = (l+m)(K1+n1K2)/n1 = (0.4+1.2)(12+3 x 10)/3/60

= $ 0.3733/ unit

• TEC4 = (l+w)(K1+n2K2) = (0.4+0.1)(12+4x 10)/60

= $ 0.4333 /unit