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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
Example DiagramFigure 2-11
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
Example DiagramFigure 2-15
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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
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 15R 1.67
E 0.95
26.36842105
Output?1.5/0.6
2 25/0 6
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Operation Standard Minutes
(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?
Operation Standard Minutes
(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