Bagian 3 PTI
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Transcript of Bagian 3 PTI
Management Science Era And Integrated Approach
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Sasaran
Memahami proses optimasi dan pendekatan sistemik terintegrasi dalam
menyelesaikan permasalahan
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Management Science
• Due to Limited Resources
•Using Mathematical and Statistical Approach to Solve the Real Problem to
Obtained Solution
• As a Tool for Decision Making
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MS In Decision Making Process
Data /Information
Real System
IEModel
StandardProblem
SolutionDecision
Action
MSOR
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Management Science Approach
Analysis of Real System
Problem Formulation
Analysis ofModel
ModelBuilding
Implementationof Finding
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Management Science Era
ClassicalQuantitative Approach
Operation Research
Model
SimulationModel
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Classical Quantitative Approach
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Classical Quantitative Approach
• Using Classical Mathematical and Statistical Approach to Solve the Quantitative Problem to
Obtained Optimal Solution Analytically
• Unconstraint problems
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Taxonomi Problem
Pasti Probabilistik Tak Tentu ( Determitic ) ( Probabilistic) ( Uncertainty ) S = 0 S 0 S 0 Pola Diketahui Pola Tak Diketahui
1n
)X(XS
:Catatan
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Wilson Formula
The first mathematical approach used to solve the inventory problem
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Contoh Problem Inventory
Diketahui:D = 10.000 unit/tahun
A = Rp.1000.000,-/pesan
p = Rp. 10.000,-/unit
h = 20% dari harga/unit/tahun
Bagaimana kebijakan inventori optimal ?
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Formulasi Problem
Bagaimana Menentukan Kebijakan Inventori Optimal ?
• Berapa Ukuran Lot Pemesanan Ekonomis ?( Economic Order Quantity: EOQ )
• Kapan Saat Pemesanan Dilakukan ( Re-Order Point: ROP )
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Performance Criteria
Ongkos Inventori Total ( Ot )
Ot = Ob + Op + Os
Ot: Ongkos Inventori Total
Ob: Ongkos Beli
Op: Ongkos Pesan
Os: Ongkos Simpan
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Solusi Praktis
111,01,010,0100,0Sepuluh kali belif = 10;qo = 1.000
109,251,258,0100,0Delapan kali belif = 8; qo = 1.250
107,02,05,0100,0Lima kali belif = 5; qo = 2.000
106,52,54,0100,0Empat kali belif = 4; qo= 2.500
107,05,02,0100,0Dua kali belif = 2; qo= 5.000
111,010,01,0100,0Satu kali belif = 1; qo= 10.000
OngkosTotal (OT)
Ongkos SimpanOs = ½q0 × h
Ongkos PesanOp = f × A
Ongkos BeliOb = p × D
Cara dan UkuranPengadaan
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Hubungan Ot dan Q0
q0* = 2500 q0
Ongkos Pesan(Op)
Ongkos Simpan (Os)
Ongkos Total(OT)
OTOngkos
(Rp.)
Ongkos Beli (Ob)100
10.000
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Asumsi Model Wilson
1. Demand Deterministik Dan Barang Datang Secara Uniform
2. Ukuran Lot Pemesanan Tetap Untuk Setiap Kali Pemesanan
3. Barang Yang Dipesan Akan Datang Secara Serentak Pada Saat Pemesanan
4. Harga Barang Konstan Baik Terhadap Lot Maupun Waktu
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Posisi Inventori
q0 = 5.000 unit
C
t
m = 1/2q0GF
A B
E
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Posisi Inventori
SOP = SOH + SOO dimana:
SOP : Posisi inventori (stock on position)
SOH : Inventori tersedia (stock on hand)
SOO : Inventori dalam pesanan (stock on order)
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Formulasi Model
Min Ot = Ob + Op + Os
Dimana:Ob = DpOp = AD/Qo
Os = hQo/2
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Formulasi Model
Min Ot = Ob + Op + Os
= Dp + AD/Qo + hQo/2
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Solusi Model
• Syarat Ot minimal:
Ot/Qo = 0
2
-AD/Qo + h/2 =0
1/2
Qo = {2AD/h}
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Solusi Optimal
1/2
Qo = {2AD/h}
= { 2. 1000000.10000/2000}
Qo = 3165 unit
½
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Mekanisme Model Wilson
BarangTersedia ?
Transaksi PengeluaranBarang
Barang DiGudang Habis ?
PesanBarang (q0)
Tidak
Tidak
Ya
Ya
PermintaanBarang (User)
PemakaianBarang (User)
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Operation Research Approach
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Operation Research Approach
• Use Modeling Approach to Solve the Problem to Obtained Optimal Solution
• Constraint problems
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How To Get The Solution?Problem
Common Sense?
Model Standard?
New Alternatives Build New Model Simulation
Solution
Use&Choose
Formulate
Yes
YesNo
MS
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Problem Solving Approach
• Define Problem
• Generate Alternatives
• Choose Standard Model
• Get The Best Solution
• Make Decision
• Implementation/Action
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Operation Research Model
Linear Non Linear -Linear Prog - Queing
-Transportation - Inventory
-Transhipment - Dynamics Prog
- Network - Stochastics Prog.
- Etc - Etc
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Model
Representation of System for Special Purposes
Representation Purpose– Model Iconic - Model Descriptive
– Model Analog - Model Predictive
– Model Symbolic - Model Normative
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Advantages Using Model
• Minimize destructive experiment
• Minimize complexity of real world
• Minimize negative impacts
• Minimize Cost
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Performance of Model
• Valid
• Simple
• Robust
• Adaptive
• Complete
• Controllable
• Communicable
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Solution
• Solve the Problem
• Reflect Variable Decision
• Input For Making Decision
Feasible Best Optimal
( Simulasi ) ( Heuristic) ( Analytic)
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Performance Criteria
A. Types1. Single Criteria
2. Multi Criteria
B. Level of Management1. Company level
2. Business level
3. Operational level
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Components Model
• Performance Criteria
• Decision Variable
• Constraints
• Parameter
• Logical Relationship
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Model Formulation
Determine the relationship among performance criteria, variables,
parameters and constraints
Objective function : V = f ( Xi, Yi, Ai )
Constraints : f ( Xi, Yi, Ai ) < Bi
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Linear Programming
• Asumsi :– Proprotionality– Additivity– Integrality
• Model Formulation :
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General ModelObjective Function:
Min Z = c1X1 + c2X2 + c3X3 + c4X4 + ………….+ cnXn
Subject to:
1. a11X1 + a12X2 +a13 X3 + a14X4 + ………+ a1nXn <= B1
2. a21X1 + a22X2 +a23 X3 + a24X4 + ………+ a2nXn <= B2
3. a31X1 + a32X2 +a33 X3 + a34X4 + ………+ a3nXn <= B3
. .
. .
m. am1X1 + am2X2 +am3 X3 + am4X4 + ……+ amnXn <= Bm
X1, X2, X3, X4 ……Xn >= 0
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Contoh Linear Programming
PT XYZ produces sport jackets and slacks. The profit on each jacket is $ 10, and for pair of slacks is $ 15. Each jacket
requires 2 m2 of material and 4 manhours of sewing, which each pair of slacks requires 5 m2 of material and 2
manhous of sewing. If there has 50 m2 of material and 36 manhours of work available each week, how many jackets
and pairs of slacks shoul be produced
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Component of Model• Performance Criteria
– Profit/week z• Decision Variables
– Number of jackets produced/week x1
– Number of slack produced/week x2
• Constraints– Material : 50 m2/week– Sewing : 36 manhours/week
• Parameter Jacket Slack
– Profit 10 15 ($/unit)– Usage of material 2 5 (m2/unit) – Sewing requirement 4 2 (manhour/unit)
• Logical Relationship – liner
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Formulation of ModelObjective function : V = f ( Xi, Yi, Ai )
Constraints : f ( Xi, Yi, Ai ) < Bi
Objective function : Z = 10 x1 + 15 x2
Constraints:– Material : 2 x1 + 5 x2 <= 50 – Sewing : 4 x1 + 2 x2 <= 36
x1 , x2 > = 0
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Solution
Feasible Solution
Optimal Solution
Graphis Simplex
Graphical Method
X1
X2
Material
Sewing 4 x1 + 2 x2 <= 36
2 x1 + 5 x2 <= 50
259
10
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FeasibleSolution
OptimalZ= 10 x1 + 15 x2
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Optimal Solution
Produce Jacket X1 : 5 unit/week
Produce Slack X2 : 8 unit/week
Profit Z : $ 170/week
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P Q
1 2 3 8
Transportation Model
. . . . . . .
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Data
1 2 3 4 5 6 7 8
A B C D E F G H Supply
P 12 24 21 20 21.5 19 17 20 100
Q 24 15 28 20 18.5 19.5 24 28 45
Dmd 22 14 18 17 15 13 15 20
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Component of Model
• Performance Criteria : • Min. Cost
• Variable Decision : • Number of product to be supplied from plant i
(Si)
• Number of product to be transported from plant i to retailer j( Xij)
• Constraints:• Supply
• Demand
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Model Formulation
Min Z = 12X11 + 24X12 + 21X13 + 20X14 + 21.5X15 + 19X16 +
17X17 + 20X18 + 24X21 + 15X22 + 28X23 + 20X24 +
18.5X25 + 19.5 X26 + 24 X27 + 28X28
Subject to: 1). X11 + X12 + X13 + X14 + X15 + X16 +X17 + X18 <=100
2). X21 + X22 + X23 + X24 + X25 + X26 +X27 + X28 <= 45
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Model Formulation
3). X11 + X21 = 22
4). X12 + X22 = 14
5). X13 + X23 = 18
6). X14 + X24 = 17
7). X15 + X25 = 15
8). X16 + X26 = 13
9). X17 + X27 = 15
10). X18 + X28 = 20
All Variable Non Negative
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Optimal Solution
1 2 3 4 5 6 7 8
A B C D E F G H Supply
P 22 - 18 17 - 13 15 4 89
Q - 14 - - 15 - - 16 45
Dmnd 22 14 18 17 15 13 15 20 134
Minimal Cost = $ 2583.50
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P Q
1 4 5 82 3 6 7
Optimal Solution
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Shortest Route Method
A BE I
C
H
G
F
DJ
Origin
Destination
90
90
8484
60
48
348
150
48126
132
126156
132
66
132
138
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Step Solved Its Closest Total Time n’th Nearest Its Min. Its Last Nodes Conc.un Sl Involved Node Time Con’tion
1 A B 90 B 90 AB* A D 138 A C 348 2 A C 138 C 138 AC A D 348 B C 90+66=156 B E 90+84=174
3 A D 348 B E 90+84=174 E 174 BE* C D 138+156=294 C F 138+90=228
4 A D 348 C D 138+156=294 C F 138+ 90=228 F 228 CF E F 174+132=306 E I 174+84 =258
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Step Solved Its Closest Total Time n’th Nearest Its Min. Its Last Nodes Conc.un Sl Involved Node Time Con’tion
5 A D 348 C D 138156=294 E I 174+84=258 I 258 EI* F G 228132=360 F H 228+60=288 6 A D 348 C D 138+156=294 F G 228+132=360 F H 228+ 60=288 H 288 FH I H 258+132=390 I J 258+126=384 7 A D 348 C D 138+156=294 F G 228+132=360 H G 288+ 48=336 G 336 HG I J 258+126=384
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Step Solved Its Closest Total Time n’th Nearest Its Min. Its Last Nodes Conc.un Sl Involved Node Time
Con’tion
8 A D 348 D 348 AD C D 138+156=294 G D 336+ 48=384 G J 336+150=486 H J 288+126=414 I J 258+126=384
9 G J 336+150=486 H J 288+126=414 I J 258+126=384 J 384 IJ*
Shortest Route : A B E I J ( 384)
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Integrated Approach
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Integrated Approach
Memandang Sesuatu Secara Menyeluruh Dan Komprehensif Tidak Bersifat
Parsial
Systemic Approach
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Contoh
A BE I
C
H
G
F
DJ
Origin
Destination
90
90
8484
60
48
348
150
48126
132
126156
132
66
132
138
Shortest Route ?
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Solusi• Parsial
– Methoda : Logik ambil jarak terpendek pada setiap node
– Hasil : A-B-C-F-H-G-J (494)
• Integrated– Methoda : Dinamik programmning– Hasil : A-B-E-I-J ( 384)
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Systemic Aspect• Structural Aspect
– Man, Machine, Material
• Functional Aspect– Man-man, Man-Material, Man-Machine, Feed-back
• Boundary• Environment
– Stakeholder and Societal
• Objective– Unitary, pluralist, coercive
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Component of Real System
Man
+Natural System and/or Artificial System
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Schematic Representation
MachineMan
Mat
Boundary
Environment
Input Out-put
: InteractionFeed-back
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MS In Decision Making Process
Data /Information
Real System
IEModel
StandardProblem
SolutionDecision
Action
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Characteristic of Integrated Approach
• Problem : Real
• Approach : Systemic
• Model : Valid
• Solution : Feasible
• Decision : Effective
• Action : Implemented
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What IE Has To Do?
1. Problem Identification
2. Generate Alternatives
3. Know the Standard Models
4. Decide Performance Criteria
5. Choose the Best Solution
6. Make Decision
7. Anticipate Managerial Implication
8. Action
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Model Building Process
Modeler Real SystemAnalyses of System•Problem Formulation•Components Model
Model Formulation
Solution
Valid ?
Stop
No
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Analysis of System
• Formulate the Problem
• Determine Performance Criteria
• Identification of Components Model– Decision Variable – Constraints– Parameter– Logical Relationship
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Problem Formulation
Do Not Confuse Among !!!
Symptom
Root Causes
Alternative of Solution
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Problem
Any unsatisfactory situation
Symptom• Claims
• Difference: Expectation vs Reality
Root Causes
Problem
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Symptom
• Indicator of the Problem
• Problem Could Exist Although Without Any Symptom
• Form: Claims, Differences Expectation and Reality
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Symptom
Difference
Expectation Reality
Objective Achievement
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Root Causes
• Causes of the symptom
• Use systemic approach to identify
Fish Bone Diagram
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Material
Boundary
Machine Man
MethodEnvironment
Fish Bone Diagram
Indicator
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Performance Criteria
Efficiency
Effective + Min Cost
Objective Vs
Achievement/Actual
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Performance Criteria
A. Types1. Single Criteria
2. Multi Criteria
B. Level of Management1. Company level
2. Business level
3. Operational level
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Components Model
• Performave Criteria
• Decision Variable
• Constraints
• Parameter
• Logical Relationship
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Model Formulation
Determine the relationship among performance criteria, variables,
parameters and constraints
Objective function : V = f ( Xi, Yi, Ai )
Constraints : f ( Xi, Yi, Ai ) < Bi
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Solution
• Value of Decision Variable
• Input For Decision Making
Feasible Best Optimal ( Simulation ) ( Heuristic) ( Analytic)
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Validation
• Logical Validation (Verification)Is the Model Logic and Rational ?
• Historical ValidationIs the Model Fit With the Past Performance ?
• Result ValidationIs the Model Fit With the Future Performance ?
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Decision
Problem
Decision
Solution + Judgment
Scientific Art
Criteria
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Action
How Decision Could Be Implemented ?
Anticipate Managerial Implication
And
Prepare Implementation Plan
Action