Chapter 2 Problem Solving Tools

7/28/2019 Chapter 2 Problem Solving Tools

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PROBLEM-SOLVING TOOLS


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Exploratory Tools Pareto Analysis

Fish Diagrams

Gantt Chart PERT Chart

Job / Worksite Analysis Guide


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Pareto Analysis Items identified and ordered on

common scale in decreasingfrequency, creating a cumulative

distribution 80/20 Rule: 20% of the items account

for 80% of the problems

Allows the company to concentrateresources on the jobs with the mostproblems


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Pareto Analysis

20

80

20

80

Causes Problems


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Pareto Analysis Example Diagram


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Pareto Analysis

80

58

42 42

0

20

40

60

80

100

120

140

160

180

200

220

1 2 3 4

FrekuensiKu

mulatif

0%

10%

20%

30%

40%

50%

60%

70%

80%

90%

100%

ProsentaseKumulatif

Keterangan:

1 = Pergantian sistem penyimpanan status

2 = Pasien lama tidak bawa kartu

3 = Status baru terlalu banyak

4 = Status lama tidak ketemu

Keterangan:

1 = Pergantian sistem penyimpanan status

2 = Pasien lama tidak bawa kartu

3 = Status baru terlalu banyak

4 = Status lama tidak ketemu

3124

19 16

0

20

40

60

80

100

120

140

160

180

200

220

1 2 3 4

FrekuensiKu

mulatif

0%

50%

100%

150%

200%

ProsentaseKumulatif


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Fishbone Diagrams Cause-and-effect diagrams

Identified problem or undesirable resultis the head

Contributing factors are the bones

Typical categories include: Human,

machine, methods, materials,environment, and administrative

Estimates associated probabilities


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Fishbone Diagrams Example Diagram

Figure 2-3


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Gantt Chart

Used for planning of complex projects

Shows expected start and completiontimes, also duration of events

Similarly, major events can be broken intosmaller sub-tasks

Shade the bars to show actual completiontime


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Gantt Chart Example Diagram

Figure 2-4


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PERT Chart Program Evaluation and Review Technique

(PERT) is a planning and control tool

Also known as Network Diagram or Critical

Path

Graphically portrays the optimum way toobtain a desired objective with respects to

time

Optimistic, average, and pessimistic timeestimates utilized


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PERT Chart


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Job / WorksiteAnalysis Guide

Perform a walkthrough observing thearea, worker, task, environment,administrative constraints, etc.

Develop an overall perspective of thesituation

Particularly useful in workstationredesign


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Job / WorksiteAnalysis Guide Example Guide

Figure 2-6


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Recording andAnalysis Tools

Operation Process Chart

Flow Process Chart

Flow Diagram Worker and Machine Process Charts

Gang Process Charts


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Operation ProcessChart

Chronological sequence of alloperations, inspections, timeallowances, materials

Depicts entrance and exit of allcomponents and sub-assemblies andproducts

Provides information on the number ofemployees required time for jobs andinspections


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Basic Symbols

process symbol

Operasi

transportasi

Delay

Inspeksi

Storage


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Operation Process Chart


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OperationProcess Chart

Example DiagramFigure 2-8


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Flow Process Chart More detailed, fit for closer observation

of smaller components or assemblies

Shows all moves (distances) and storagedelays (times) for product movement inplant

Aids in the reduction of hidden costs,

Muda. Can be beneficial for plant layout

suggestions


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ypes o


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ypes oFlow Process Chart

Currently Use

Product/material

(see figure 2-11)

Operative/person

(see figure 2-12)


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Flow ProcessChart


Fl P


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Flow ProcessChart

Example Diagram

Figure 2-12


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Flow Diagram Pictorial representation of the layout

of the plant

Good supplement to the Flow Process Chart


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Flow Diagram

PRODUKSI GUDANG

1 2 3 45


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Flow Diagram Example Diagram Figure 2-13


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Worker and MachineProcess Charts

Used to study, analyze, and improveone workstation

Shows the time relationship betweenworking cycle of the person and theoperating cycle of the machine

Reveals idle time for both machinesand workers

W k d


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Worker andMachineProcess Charts



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Gang Process Chart Example Diagram Figure 2-16


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Quantitative Tools Synchronous Servicing

Random Servicing

Complex Relationships

Line Balancing


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SYNCHRONOUSSERVICING

Assigning more than one machine

to an operator seldom results in ideal case where both the workerand the machine are occupied during the whole cycle.

n = l + m

l

n = Number of machine the operator is assigned

l = Total operator loading and unloading

(servicing) time per machine

m = Totalmachine running time (automatic

power feed).

Synchronous


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SynchronousServicing

Example: assume a total cycle time offour minutes to produce a product, asmeasured from the start of theunloading of the previously completed

product to the end of the machine cycletime. Operator servicing, which includesboth the unloading of the completedproduct and the loading of the rawmaterials is one minute,while the cycletime of the automatic machince cycle isthree minutes.


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wl

mlN1

N = number of machines

m = total machine running time

w = walking time

l = loading and unloading time


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Total Expected Cost

1

211 ))((1

N

KNKmlTECN

K1 = operator rateK2 = cost of machine

))(( 2212 KNKwlTECN


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Alternate Approach:

R

KNKTECN)( 211

1

))(( 2212 KNKwlTECN

11

Nxml

R

wl

R

1


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contoh so al

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


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Ditanya:

Jumlah mesin yang dibutuhkan

TEC tiap kemungkinan jumlah mesin


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Jawaban:

N = (l+m)/(l+w) = (0.4+1.2)/ (0.4+0.1) = 3.2

Sehingga jumlah mesin yang mungkinadalah 3 atau 4 mesin


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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

Synchronous Servicing


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Synchronous Servicingin Design Tools

Worker 1

Machine 1Machine 2

Machine 3

Machine 4

Answer:n = l + m = 1 + 3 = 4 machines

l 1

Worker Working

Loading/unloading

Machine running

Legend:

Idle Time


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RANDOM SERVICING

Helps to determine the number ofmachines to assign to an operatorwhen it is not known exactly when each

machine needs to be serviced or forhow long

The binomial expansion give a useful

approximation of the machine downprobability


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Assuming that each machine is down atrandom times during the day and that theprobability of the down time isp and theprobability of runtime is q = (1-p). Each term

of the binomial expansion can be expressedas a probability of m (out of n) machinesdown:

P(m of n) = n !

m! (n-m)!

pm qn-m


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Total Expected Cost (TEC):

K1 = hourly rate of the operator

K2 = hourly rate of the machine

N = number of machines assignedR = rate of production, pieces from N

machines per hour

R

NKKTEC

)( 21

COMPLEX


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COMPLEXRELATIONSHIPS

Here the servicing time is relativelyconstant, but the machines areserviced randomly


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For seven or more machines, WrightsFormula can be used:

I = interference, expressed as apercentage of the mean servicing time

X = ratio of mean machine running time tomean machine servicing time

N = number of machine units assigned toone operator

NXNNXI

12150

2


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LINE BALANCING

Helps to determine the ideal number ofworkers to be assigned to a productionline

Computer software is available toeliminate the calculations

100

.

..

X

MA

MS

E

E = EficiencySM = Standard minute per operation

AM = Allowed standard minutes per operation


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DefinisiLine balancingmerupakan suatu metode

penugasan sejumlah pekerjaan yangsaling berkaitan dalam satu lini produksisehingga setiap stasiun kerja memiliki

waktu yang tidak melebihi waktu siklusdari stasiun kerja tersebut.

Line balancingberusaha menyeimbangkanseluruh lintasan yang ada dalam lini

perakitan sehingga aliran produksiberjalan lancar.

How to Calculate Line Efficiency ?


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Operator 1Operator 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 ?

Line Efficiency ?


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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 ?

Opportunities for Improvements ?


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0.490.31

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 ?


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Operation Standard Minutes

1 1.5

2 2.25

3 1.25

4 2.5

5 3

6 2.757 1.75

Assembly Line

(7 operations)


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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 Neededin Assembly Line?


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E

SMRAMRN

27 operators


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 15R 1.67

E 0.95

26.36842105

Output?1.5/0.6

2 25/0 6


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(standard

minutes)/(minutes/u

nit)

Operators

1 1.5 2.5 32 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

inutes/unit 0.6 27

Output? 2.25/0.6Dst.

600/1000


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Slowest One?


(standard

minutes)/(minutes/unit)

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 0.66 2.75 4.583333333 5 0.55

7 1.75 2.916666667 3 0.583333333

dayhouroutput /960/9625.1

602

(2.5/3)*0.6

(3.75/4)*0.6Dst.


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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


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1. What process is the bottleneck?

2. How much is the maximumproduction per hour?

3. How much the efficiency?

4. How to minimize work stations?

5. How should they be grouped? 6. New efficiency?

Question


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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


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(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


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t imeyc le t imesa skN

(bottleneck)20

4.1

= 4.88 work stations

Number of Workstation

4 Stations 20/24=83 3%


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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./hour60/610 units/hour

5 Stations


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Line Balancing Problem

A

B

C

4.1mins

D

1.7mins

E2.7 mins

F

3.3

mins

G

2.6 mins

2.2 mins

3.4 mins5.6

5.0

20/5.6x5 = 20/28 = 71.4%

5 Stations

Max Prod./hour60/5.6

10.7 units/hour


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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

timecyclestationsofnumber

task times%Efficency

318/6 x 84=

318/504 =

63%

99 secs 3 Stations ?


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100timecyclestationsofnumber

task times%Efficency

40 secs

59 secs

84 secs34 secs

56 secs 45 secs

118 secs

318/3x118

318/354 = 89.8%101 secs

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Chapter 2 Problem Solving Tools