Bagian 3 PTI

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Management Science Era And Integrated Approach

Sasaran

Memahami proses optimasi dan pendekatan sistemik terintegrasi dalam

menyelesaikan permasalahan

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

MS In Decision Making Process

Data /Information

Real System

IEModel

StandardProblem

SolutionDecision

Action

Management Science Approach

Analysis of Real System

Problem Formulation

Analysis ofModel

ModelBuilding

Implementationof Finding

Management Science Era

ClassicalQuantitative Approach

Operation Research

SimulationModel

Classical Quantitative Approach

• Using Classical Mathematical and Statistical Approach to Solve the Quantitative Problem to

Obtained Optimal Solution Analytically

• Unconstraint problems

Taxonomi Problem

Pasti Probabilistik Tak Tentu ( Determitic ) ( Probabilistic) ( Uncertainty ) S = 0 S 0 S 0 Pola Diketahui Pola Tak Diketahui

:Catatan

Wilson Formula

The first mathematical approach used to solve the inventory problem

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 ?

Formulasi Problem

Bagaimana Menentukan Kebijakan Inventori Optimal ?

• Berapa Ukuran Lot Pemesanan Ekonomis ?( Economic Order Quantity: EOQ )

• Kapan Saat Pemesanan Dilakukan ( Re-Order Point: ROP )

Performance Criteria

Ongkos Inventori Total ( Ot )

Ot = Ob + Op + Os

Ot: Ongkos Inventori Total

Ob: Ongkos Beli

Op: Ongkos Pesan

Os: Ongkos Simpan

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

Hubungan Ot dan Q0

q0* = 2500 q0

Ongkos Pesan(Op)

Ongkos Simpan (Os)

Ongkos Total(OT)

OTOngkos

Ongkos Beli (Ob)100

10.000

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

Posisi Inventori

q0 = 5.000 unit

m = 1/2q0GF

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)

Formulasi Model

Min Ot = Ob + Op + Os

Dimana:Ob = DpOp = AD/Qo

Os = hQo/2

Formulasi Model

Min Ot = Ob + Op + Os

= Dp + AD/Qo + hQo/2

Solusi Model

• Syarat Ot minimal:

Ot/Qo = 0

-AD/Qo + h/2 =0

Qo = {2AD/h}

Solusi Optimal

Qo = {2AD/h}

= { 2. 1000000.10000/2000}

Qo = 3165 unit

Mekanisme Model Wilson

BarangTersedia ?

Transaksi PengeluaranBarang

Barang DiGudang Habis ?

PesanBarang (q0)

PermintaanBarang (User)

PemakaianBarang (User)

Operation Research Approach

• Use Modeling Approach to Solve the Problem to Obtained Optimal Solution

• Constraint problems

How To Get The Solution?Problem

Common Sense?

Model Standard?

New Alternatives Build New Model Simulation

Solution

Use&Choose

Formulate

Problem Solving Approach

• Define Problem

• Generate Alternatives

• Choose Standard Model

• Get The Best Solution

• Make Decision

• Implementation/Action

Operation Research Model

Linear Non Linear -Linear Prog - Queing

-Transportation - Inventory

-Transhipment - Dynamics Prog

- Network - Stochastics Prog.

- Etc - Etc

Representation of System for Special Purposes

Representation Purpose– Model Iconic - Model Descriptive

– Model Analog - Model Predictive

– Model Symbolic - Model Normative

Advantages Using Model

• Minimize destructive experiment

• Minimize complexity of real world

• Minimize negative impacts

• Minimize Cost

Performance of Model

• Valid

• Simple

• Robust

• Adaptive

• Complete

• Controllable

• Communicable

Solution

• Solve the Problem

• Reflect Variable Decision

• Input For Making Decision

Feasible Best Optimal

( Simulasi ) ( Heuristic) ( Analytic)

A. Types1. Single Criteria

2. Multi Criteria

B. Level of Management1. Company level

2. Business level

3. Operational level

Components Model

• Performance Criteria

• Decision Variable

• Constraints

• Parameter

• Logical Relationship

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

Linear Programming

• Asumsi :– Proprotionality– Additivity– Integrality

• Model Formulation :

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

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

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

Formulation of ModelObjective function : V = f ( Xi, Yi, Ai )

Objective function : Z = 10 x1 + 15 x2

Constraints:– Material : 2 x1 + 5 x2 <= 50 – Sewing : 4 x1 + 2 x2 <= 36

x1 , x2 > = 0

Solution

Feasible Solution

Optimal Solution

Graphis Simplex

Graphical Method

Material

Sewing 4 x1 + 2 x2 <= 36

2 x1 + 5 x2 <= 50

FeasibleSolution

OptimalZ= 10 x1 + 15 x2

Optimal Solution

Produce Jacket X1 : 5 unit/week

Produce Slack X2 : 8 unit/week

Profit Z : $ 170/week

1 2 3 8

Transportation Model

. . . . . . .

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

Component of Model

• Performance Criteria : • Min. Cost

• Variable Decision : • Number of product to be supplied from plant i

• Number of product to be transported from plant i to retailer j( Xij)

• Constraints:• Supply

• Demand

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

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

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

1 4 5 82 3 6 7

Optimal Solution

Shortest Route Method

A BE I

Origin

Destination

126156

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

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

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)

Integrated Approach

Memandang Sesuatu Secara Menyeluruh Dan Komprehensif Tidak Bersifat

Parsial

Systemic Approach

Contoh

A BE I

Origin

Destination

126156

Shortest Route ?

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)

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

Component of Real System

+Natural System and/or Artificial System

Schematic Representation

MachineMan

Boundary

Environment

Input Out-put

: InteractionFeed-back

MS In Decision Making Process

Data /Information

Real System

IEModel

StandardProblem

SolutionDecision

Action

Characteristic of Integrated Approach

• Problem : Real

• Approach : Systemic

• Model : Valid

• Solution : Feasible

• Decision : Effective

• Action : Implemented

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

Model Building Process

Modeler Real SystemAnalyses of System•Problem Formulation•Components Model

Model Formulation

Solution

Valid ?

Analysis of System

• Formulate the Problem

• Determine Performance Criteria

• Identification of Components Model– Decision Variable – Constraints– Parameter– Logical Relationship

Problem Formulation

Do Not Confuse Among !!!

Symptom

Root Causes

Alternative of Solution

Problem

Any unsatisfactory situation

Symptom• Claims

• Difference: Expectation vs Reality

Root Causes

Problem

Symptom

• Indicator of the Problem

• Problem Could Exist Although Without Any Symptom

• Form: Claims, Differences Expectation and Reality

Symptom

Difference

Expectation Reality

Objective Achievement

Root Causes

• Causes of the symptom

• Use systemic approach to identify

Fish Bone Diagram

Material

Boundary

Machine Man

MethodEnvironment

Fish Bone Diagram

Indicator

Efficiency

Effective + Min Cost

Objective Vs

Achievement/Actual

A. Types1. Single Criteria

2. Multi Criteria

B. Level of Management1. Company level

2. Business level

3. Operational level

Components Model

• Performave Criteria

• Decision Variable

• Constraints

• Parameter

• Logical Relationship

Model Formulation

Determine the relationship among performance criteria, variables,

parameters and constraints

Objective function : V = f ( Xi, Yi, Ai )

Solution

• Value of Decision Variable

• Input For Decision Making

Feasible Best Optimal ( Simulation ) ( Heuristic) ( Analytic)

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 ?

Decision

Problem

Decision

Solution + Judgment

Scientific Art

Criteria

Action

How Decision Could Be Implemented ?

Anticipate Managerial Implication

Prepare Implementation Plan

Action

Bagian 3 PTI

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