operation research setyabudi indartono, ph.d assistant ... · pdf filecontoh soal quiz, uts...

2

Operation Research

2013 Edition

Setyabudi Indartono, Ph.D

Assistant Professor at Faculty of Economy

Yogyakarta State University

Course Module Series

3

Table of Contents Preface ................................................................................................................................ 6

Syllabi .................................................................................................................................. 7

Course Components ......................................................................................................... 7

Class Procedures .............................................................................................................. 8

Assessment ...................................................................................................................... 8

Course Schedule ............................................................................................................... 9

Grading Systems ............................................................................................................ 10

Chapter 1: Introduction to Operations Research Concept .................................................. 11

Definition and origin ...................................................................................................... 11

Essential features of the OR approach ............................................................................ 12

Chapter 2: Introduction to Foundation mathematics and statistics Modeling ..................... 21

LP definition, .................................................................................................................. 21

Quantification of factors ................................................................................................ 26

LP and allocation of resources ........................................................................................ 33

Linearity requirement .................................................................................................... 40

Assignment-1 (Homework) ............................................................................................. 45

chapter 3: Expressing Linier programming problems .......................................................... 46

Limitations or constraints ............................................................................................... 46

Types of Constraints in Linear Programming Problems ................................................... 47

Chapter 4: Inventory Model ............................................................................................... 52

Pentingnya pengendalian persediaan ............................................................................. 53

Keputusan Persediaan .................................................................................................... 54

EOQ, mendifinisikan berapa banyak pemesanan ............................................................ 54

Inventory Cost ................................................................................................................ 55

Menentukan EOQ ........................................................................................................... 56

ROP, Menentukan kapan dilakukan pemesanan ............................................................. 57

EOQ dengan asumsi tanpa penerimaan yang tak tentu ................................................... 57

4

Menentukan annual caarrying cost ................................................................................ 58

Menentukan annual setup cost atau Annual ordering cost ............................................. 58

Model Diskon jumlah ..................................................................................................... 60

Pemakaianan safety stock .............................................................................................. 62

Reorder point dengan biaya ketidaktersediaan yang telah diketahui. ............................. 63

Biaya ketidaktersediaan (stockout). ................................................................................ 63

Carrying cost .................................................................................................................. 64

Safety stok dengan biaya yang tidak diketahui................................................................ 64

ABC Analisys ................................................................................................................... 65

Analysis sensitivitas ........................................................................................................ 65

Assignment-2 (Homework) ............................................................................................. 65

Chapter 5: TRANSPORTAtion model ................................................................................... 66

Pendahuluan .................................................................................................................. 66

Seting up transportation problems ................................................................................. 66

Nortwest corner rule ...................................................................................................... 66

Stepping stone method: mencari biaya terkecil .............................................................. 67

MODI method ................................................................................................................ 72

Vogels approximation metod (VAM) .............................................................................. 73

Supply > demmand = dummy destination ....................................................................... 77

Perhitungan total biaya adalah: ...................................................................................... 77

Supply < demmand = dummy source .............................................................................. 78

Degeneracy in transportation ......................................................................................... 79

Degeneracy during later solution stage. ......................................................................... 79

Pilihan solusi yang lebih dari satu pilihan ........................................................................ 79

Analisis Lokasi fasilitas .................................................................................................... 80

Chapter 6: Assignment Model ............................................................................................ 84

Minimization .................................................................................................................. 84

Dummy Row dan Dummy Colums .................................................................................. 89

MAKSIMALISASI PENUGASAN ......................................................................................... 89

5

Chapter 7: Project Analysis ................................................................................................. 93

Pendahuluan .................................................................................................................. 93

PERT............................................................................................................................... 93

CPM ............................................................................................................................... 97

Diskusi Kasus .....................................................................................................................106

Case #1 CUSTOM VANS INC ...........................................................................................106

Case #2 Haygood Company ...........................................................................................111

Case #3 MANAGEMENT VIDEO PROFESIONAL ...............................................................114

Paper ................................................................................................................................117

Paper-1: Presentation of a New and Beneficial Method Through Problem Solving Timing

of Open Shop by Random Algorithm Gravitational Emulation Local Search ....................117

Paper-2: Inverse Optimization for Linear Fractional Programming .................................118

Paper-3: A multi-objective model for designing a group layout of a dynamic cellular

manufacturing system ...................................................................................................118

Paper-4: Integrating truck arrival management into tactical operation planning at

container terminals .......................................................................................................119

Paper-5: Pharmaceutical Inventory Management Issues in Hospital Supply Chains ........120

Paper-6: Improving a Flexible Manufacturing Scheduling using Genetic Algorithm ........120

Contoh Soal Quiz, UTS dan UAS .........................................................................................122

Quiz ..............................................................................................................................122

UTS ...............................................................................................................................124

UAS ...............................................................................................................................125

Penulis ..............................................................................................................................126

refferences .......................................................................................................................133

6

PREFACE

Operations research helps in solving problems in different environments that

needs decisions. The module covers topics that include: linear programming,

Transportation, Assignment, and CPM/ MSPT techniques. Analytic

techniques and computer packages will be used to solve problems facing

business managers in decision environments.

This module aims to introduce students to use quantitative methods and

techniques for effective decisions–making; model formulation and

applications that are used in solving business decision problems.

The Learning Outcomes of this course included:

1. Knowledge and understanding: Be able to understand the

characteristics of different types of decision-making environments and

the appropriate decision making approaches and tools to be used in

each type.

2. Cognitive skills (thinking and analysis): Be able to build and solve

Transportation Models and Assignment Models.

3. Communication skills (personal and academic): Be able to design new

simple models, like: CPM, MSPT to improve decision –making

and develop critical thinking and objective analysis of decision

problems.

4. Practical and subject specific skills (Transferable Skills): Be able to

implement practical cases, by using TORA, WinQSB

7

SYLLABI

Course Components

1. Operations Research: Applications and Algorithms

Author: Wayne L Winston

2. Introduction to Operations Research (OR)

3. Introduction to Foundation mathematics and statistics

4. Linear Programming (LP), LP and allocation of resources, LP

definition, Linearity requirement

5. Maximization and Minimization problems.

6. Graphical LP Minimization solution, Introduction, Simplex method

definition, formulating the Simplex model.

7. Linear Programming – Simplex Method for Maximizing.

8. Simplex maximizing example for similar limitations, Mixed limitations

9. Example containing mixed constraints, Minimization example for

similar limitations.

10. Sensitivity Analysis: Changes in Objective Function, Changes in RHS,

The Transportation Model

11. Basic Assumptions

12. Solution Methods:

a. Feasible Solution: The Northwest Method, The Lowest Cost

Method;

b. Optimal Solution: The Stepping Stone Method, Modified;

Distribution (MODI) Method.

13. The Assignment Model:

a. Basic Assumptions

b. Solution Methods: Different Combinations Method,

c. Short-Cut Method (Hungarian Method)

8

14. MSPT:- The Dijkestra algorithm, and Floyd’s Algorithm {Shortest Route

Algorithm}

Class Procedures

1. Presence . Students are required to attend lectures at least 80% of

total lectures held. Any violation against this rule may cause the

ineligibility to get a final grade.

2. Class participation . Students are highly expected to contribute ideas,

thoughts, experiences, and arguments to the class discussion.

Although overviews of key points and issues are provided, we require

that students comprehend the materials in details, raise questions and

ideas, and create a “lively” class, meaning that students must read and

prepare readings assigned prior to coming to the class.

3. An experiential approach . Continuously and consistently, students

are assigned to do assignment and report the findings to the lecturer.

Students will be randomly distributed into groups that consist of four

members.

4. Internet exploration . Students are encouraged to harness the

advancement of information and communications technology (ICT) in

exploring knowledge and opportunities. Remember that in current

circumstances, a clever person is not she who can answer all

questions; rather, it is she who knows where to find answers.

Assessment

• Class participation/Discussion 15%

• Presentations 20%

• Mid-term examination (UTS) 30%

• Final examination (UAS) 35%

9

Course Schedule

Class 1. Introduction to Operations Research (OR) Course: Assignment and

Assessment

Class 2. Introduction to Operations Research concept: Definition and origin.

Essential features of the OR approach.

Class 3. Introduction to Foundation mathematics and statistics modeling

a. Quantification of factors and Linearity requirement

b. Expressing linier programming problems: Minimization and

maximization

Class 4. Inventory Model

Class 5. Transportation Model

a. MODI

b. Northwest corner rule

c. Stepping stone method: mencari biaya terkecil

d. VAM

Class 6. Assignment Model

a. Minimization (Hungarian/Floods’ technique)

b. Maximization

Class 7. Project Analysis (PERT – CPM)

Class 8. Mid Exam

Class 9. Case Study #1



Class 12. Quiz

Class 13. Paper Discussion #1 - #2



Class 16. Final Examination

10

Grading Systems

• E~F means at least one of assignment and test found plagiarism

• D (50-59) means that you are able to summarize and order readings

relevant to the topic.

• C (60-69) means that you do this with some greater precision and flair

or more comprehensively and/or accessibly.

• B (70-79) indicates that you have shown evidences of substantial and

well argued independence of thoughts, insightful evaluation, or original

research.

• A (80-100) indicates that you have added significant new values to

existing knowledge or understanding through logic or evidence of

some ingenuity,creativity, or excellence

11

CHAPTER 1: INTRODUCTION TO OPERATIONS RESEARCH

CONCEPT

Definition and origin

Operations research/management science is the development and

application of mathematical techniques and models, for the purpose of

solving problems in a wide array of fields, and producing better quality

decisions. It is applied to Identifies better methods to coordinate financial,

material, equipment, and human resources toward achievement of an

organization’s goals, drawing from mathematics, science, and engineering.

The use of techniques such as statistical inference and decision theory,

mathematical programming, probabilistic models, network and computer

science to solve complex operational and strategic issues. Scientists

describe Operations research in several ways. Winston: “a scientific

approach to decision making, which seeks to determine how best to design

and operate a system, usually under conditions requiring the allocation of

scarce resources.” Kimball & Morse: “a scientific method of providing

executive departments with a quantitative basis for decisions regarding the

operations under their control.

Operation researchers use the technique to solve problems in different

ways, propose alternative solutions to management for consideration, and

responsible for monitoring the implementation of the selected solution, and

working with others to ensure its success. Various issues are analyzed using

the concept of operation research included High-level Strategy, Short-Term

Planning, Intermediate-Term Planning, Forecasting, Resource Allocation,

Performance Measurement, Scheduling, Supply Chain Management, Pricing,

Design of Facilities and Systems, Transportation and Distribution, Analysis of

12

Large Data Bases, Statistical Quality Control, Project Management, Waiting

Lines, Artificial Intelligence, and Risk Assessment.

Essential features of the OR approach

In many large corporations, OR reports directly to the CEO. It is related

to Organizational Structure. Some firms centralize OR in one department and

or within each division. Other firms contract for OR services with an

independent consulting firm and Analysts may also work closely with senior

managers to identify and solve a variety of operational problems.

Presentation to Management process of operation research application

included: Analyst presents management with first recommendations based on

the model results. Additional computer runs of the model may be needed to

consider different assumptions. The O.R. analyst presents the final

recommendation. Once management reaches a selection decision, the O.R.

analyst will usually work with others in the organization to ensure the plan’s

successful implementation.

The Mathematical Model used by scientist Attempts to describe the

system or problem being studied, enables the analyst to assign values to the

various components and to clarify the relationships between them, Values

can be altered in order to examine what may happen to the system or

solution under different circumstances.

Concept of Operations research provides rational basis for decision

making include:

• Solves the type of complex problems that turn up in the modern business

environment

• Builds mathematical and computer models of organizational systems

composed of people, machines, and procedures

13

• Uses analytical and numerical techniques to make predictions and

decisions based on these models

It is believed that concept of Operations research draws upon

engineering, management, mathematics, and closely related to the "decision

sciences" applied mathematics, computer science, economics, industrial

engineering and systems engineering. The using of operational research

concept, Managers describe the symptoms of the problem to the operations

analyst who then formally defines the problem. Then The analyst breaks the

problem down into its major components, and gathers information about each

of them from many sources, both internal and external. The analyst that in

turn selects the most appropriate analytical model or technique. Winston,

Wayne L (1996) draws a basic methodology of Operations Research at The

Seven Steps to a Good OR Analysis. It is included:

1. Identify the Problem or Opportunity: What are the objectives? Is the

proposed problem too narrow? Is it too broad?

2. Understand the System: What data should be collected? How will data

be collected? How do different components of the system interact with

each other?

3. Formulate a Mathematical Model: What kind of model should be used? Is

the model accurate? Is the model too complex?

4. Verify the Model : Do outputs match current observations for current

inputs? Are outputs reasonable? Could the model be erroneous?

5. Select the Best Alternative : What if there are conflicting objectives?

Inherently the most difficult step. This is where software tools will help us!

6. Present the Results of the Analysis: Must communicate results in layman’s

terms and System must be user friendly!

7. Implement and Evaluate: Users must be trained on the new system.

System must be observed over time to ensure it works properly.

14

Various technique used in the application of operation research are

Markov Processes, Econometric Models, Game Theory, Forecasting,

Inventory Control, Assignment Algorithms, Goal Programming, Quality

Control Models, Network Models, Simulation, Linear Programming, Queuing

Models, Dynamic Programming, Nonlinear Programming, Integer

Programming, Neural Networks, Expert Systems, and Decision Analysis.

Sample of the used of the concept of operation researches is to make

a decision on employment. O.R. analysts held 64,000 jobs in 2012 such as in

Telecommunications Company, Aerospace manufacturers, Computer

systems design firms, Engineering, Higher education, Financial institutions,

Insurance carriers, Airline industry, Management service firms, and State and

federal governments. At least a masters degree in OR, computer science,

engineering, math, information systems, or business, coupled with a

bachelors degree in any of the above, or economics or statistics need have

Required Credentials of operation research concept. Concept of operation

researches for employment decision is used to job outlook. It is good,

because firms will strive to improve their productivity, effectiveness, and

competitiveness, or perish! And because firms are sitting on hugh data bases

that need to be analyzed by professionals who have the skills to do so. The

number of professionals needed to support the growing demand, together

with those needed to replace retirees, are expected to exceed the number of

graduates for decades to come For first career position for routine work under

the supervision of experienced analysts, Eventual assignment of more

complex tasks and greater autonomy to design models and solve problems,

Analysts advance by assuming positions as technical specialists and

supervisors. Furthermore, for Long-Term Prospects, The skills acquired by

O.R. analysts are useful for a variety of higher level management jobs,

consequently, experienced analysts can leave the profession to assume non-

15

technical managerial or administrative positions, Analysts with significant

experience may become professors and consultants.

Related occupation that closely need operation research approaches

are Systems analyst, Defense analyst, Database administrator, Computer

scientist, Computer programmer, Engineers, Mathematician, Forecaster,

Materials manager, Quality & Reliability manager, Statistician, Economist,

Market researcher, IT manager, Consultant, Logistician, Technology director,

Financial analyst, Risk analyst, and Project manager. Table 1 shows a result

of operation research concept analysis on Sample Job Titles &

Compensation.

Table 1 Sample Job Titles & Compensation.

Source: http://swz.salary.com/salarywizard - Spring 2009

Title Median Salary Experience

Analyst – Level I $50,913.00 0-3 years

Analyst – Level III $80,472.00 4-7 years

Analyst – Level V $130,772.00 8-10 years

Manager $143,451.00 10+ years

Operation management

Specialist – Level I $32,886.00 0-2 years

Opns Unit Manager $46,154.00 5+ years

Opns Manager $81,527.00 8+ years

Opns Director $141,591.00 10+ years

Top Opns Executive $232,865.00 15+ years

Specialist – Level I $32,886.00 0-2 years

16

Operations research has been applied to a wide variety of situations,

and has had a dramatic impact on the effectiveness of many organizations A

small sampling of the many successful applications include the following.

Burger King uses linear programming to determine how different cuts of

meat should be blended together to produce hamburger patties at minimum

cost while still meeting certain specifications such as fat content, texture,

freshness, and shrinkage. As the cost of different cuts of meat changes, the

firm reevaluates its model to determine whether its recipe should be modified.

Scheduling aircraft crews is a complex problem involving such factors as the

type of aircraft to be flown, the cities of origination and termination for the

flight, the intermediary cities visited by the aircraft, and the length of the flight.

Federal and union rules govern the placement of personnel on the aircraft.

To address these issues, American Airlines has developed an integer linear

programming model that allows the company to quickly determine an optimal

flight schedule for its personnel. The marriage of the microprocessor with the

Global Positioning Satellite System has enabled Sony Corporation to develop

an onboard navigation system capable of giving directions to a car’s driver.

This information is especially valuable during traffic conditions such as rush

hour congestion. The software is based on an operations research model

known as a shortest path network.

Following the California earthquake in January 1994, Interstate 10, a

main freeway serving the Los Angeles area, needed to be rebuilt quickly.

The project prime contractor was given a fairly short 5 month deadline to

reopen the roadway. To encourage the work to be done as quickly as

possible, the contractor was offered a bonus of $500,000.00 for each day by

which it was able to beat the deadline. Using a project scheduling technique

known as the critical path method , the contractor was able to schedule work

crews so as to be able to complete the repair work a month earlier than the

project deadline. As a result, the contractor collected a $15 million dollar

17

bonus. Mrs. Fields operates a nationwide chain of cookie shops specializing

in fresh-baked chocolate chip cookies. The chain has equipped each shop

with a PC-based information system to aid personnel in deciding when

additional cookies should be baked and the amounts that should be

produced. This system relies on the operations research techniques of

demand forecasting and inventory modeling. Lines form as visitors of

Disneyland & Disneyworld await their turn to ride or view the most popular

attractions. Disney incorporates waiting line or queuing models into its

overall design plans for the park. These models mirror customer behavior

and tolerance for waiting in line. As a result, Disney developed an entirely

new “industry” of waiting line entertainment to maintain customer satisfaction

levels and enhance the value and excitement of the ride or attraction. NYC

handles over 20,000 tons of garbage per day. To dispose of this trash, the

city operates 3 incinerators. Refuse is also sent by barge from marine transfer

stations to the Fresh Kills Landfill. To determine future operational plans for

this landfill, the Department of Sanitation undertook an operations research

analysis. The result was development of the BOSS (barge operation systems

simulation) model. This simulation model enabled the department to

determine the number of additional barges that should be purchased to

handle future demands. It also helped plan the dispatching of these barges.

During the 1970s, U.S. automobile manufacturers saw a steady decline

in their market share due to competition from Japanese and European

manufacturers. In response, Ford Motor Co. embarked on a “Quality Is Job

One” campaign. Suppliers were held to tighter standards, and new quality

control procedures were developed. As a result of these quality management

activities, the firm was able to reverse its decline in market share and

profitability. Concept of operation research shows successfully applied at

work. Table 2 shows that various big companies have successfully make a

big deal of annual saving.

18

Table 2 Successful companies applied concept of operation research

Company Year Problem Techniques Used Annual Savings

Hewlett Packard 1998 Designing buffers into production line Queuing models $280 million

Taco Bell 1998 Employee scheduling IP, Forecasting, Simulation $13 million

Proctor & Gamble 1997 Redesign production & distribution system Transportation models $200 million

Delta Airlines 1994 Assigning planes to routes Integer Programming $100 million

AT&T 1993 Call center design Queuing models, Simulation $750 million

Yellow Freight Systems, Inc. 1992 Design trucking network

Network models, Forecasting, Simulation

$17.3 million

San Francisco Police Dept. 1989 Patrol Scheduling Linear Programming $11 million

Bethlehem Steel 1989 Design an Ingot Mold Stripper Integer Programming $8 million North American Van Lines 1988 Assigning loads to drivers Network modeling $2.5 million

Citgo Petroleum 1987 Refinery operations & distribution

Linear Programming, Forecasting $70 million

United Airlines 1986 Scheduling reservation personnel LP, Queuing, Forecasting $6 million

Dairyman's Creamery 1985 Optimal production levels Linear Programming $48,000

Phillips Petroleum 1983 Equipment replacement Network modeling $90,000

19

Operations Research Time Line:

1. 1890s: Frederick Taylor applies the scientific approach to improving

operations in production (industrial engineering)

2. 1900s: Henry Gantt develops control charts for minimizing machine job

completion times (project scheduling) Andre Markov studies how

systems change over time.

3. 1910s: Ford Harris develops approaches to determine the optimal

inventory quantity to order (inventory theory) E.K. Erlang develops a

formula for determining the average waiting time for telephone callers

(queuing theory)

4. 1920s: William Shewhart introduces the concept of control charts.

Dodge and Romig develop the technique of acceptance sampling

(quality control)

5. 1930s: John von Neuman and Oscar Morgenstern develop strategies for

evaluating competitive situations (game theory)

6. 1940s: World War II provides the impetus for the application of

Mathematical modeling for solving military problems. George Dantzig

develops the simplex method for solving Problems with a linear

objective and constraints (linear programming)

7. 1950s:

a. Harry Kuhn determines required conditions for optimality for problems

with a nonlinear structure (nonlinear programming)

b. Ralph Gomory develops a solution procedure for problems in which

some variables are required to be integer valued (integer

programming) PERT and CPM are developed (project scheduling)

c. Richard Bellman develops a methodology for solving multistage

decision problems (dynamic programming)

20

8. 1960s: John Little proves a theoretical relationship between the average

length of a waiting line and the average time a customer spends in line

(queuing theory) Specialized simulation languages such as SIMSCRIPT

and GPSS are developed (simulation)

9. 1970s: The microcomputer is developed

10. 1980s: N. Karmarkar develops a new procedure for solving large-scale

linear programming problems (LP). The personal computer is developed

Specialized OR software packages that can run on microcomputers are

developed

11. 1990s: Spreadsheet packages begin to play a major role in modeling

and solving management science models TIMS and ORSA merge to

form the Institute for Operations Research and Management Science

(INFORMS)

As the 21st century begins, there is overwhelming evidence that major

organizations are looking for individuals in all disciplines who have strong

quantitative, computer, and communications skills. Operations research, with

its emphasis on all these areas, as well as its direct application to problems of

optimization and efficiency, is recognized as an important element in a well-

rounded business education.

CHAPTER 2:

MATHEMATICS AND STAT

LP definition,

Linear programming (LP, or linear optimization) is a mathematical

method for determining a way to achieve t

maximum profit or lowest cost) in a given mathematical model for some list of

requirements represented as linear relationships. Linear programming is a

specific case of mathematical programming (mathematical optimization).

More formally, linear programming is a technique for the optimization of a

linear objective function, subject to linear equality and linear inequality

constraints. Its feasible region is a convex polyhedron, which is a set defined

as the intersection of finitely m

linear inequality. Its objective function is a real

on this polyhedron. A linear programming algorithm finds a point in the

polyhedron where this function has the smallest (or la

point exists. Linear programs are problems that can be expressed in

canonical form

where x represents the vector of variables (to be determined), c and b

are vectors of (known) coefficients, A is a (known) matrix of coefficients,

(.)T is the matrix transpose. The expression to be maximized or minimized is

called the objective function (c

CHAPTER 2: INTRODUCTION TO FOUNDATION

MATHEMATICS AND STATISTICS MODELING


method for determining a way to achieve the best outcome (such as




mally, linear programming is a technique for the optimization of a



as the intersection of finitely many half spaces, each of which is defined by a

linear inequality. Its objective function is a real-valued affine function defined


polyhedron where this function has the smallest (or largest) value if such a

Linear programs are problems that can be expressed in

:


are vectors of (known) coefficients, A is a (known) matrix of coefficients,

is the matrix transpose. The expression to be maximized or minimized is

called the objective function (cTx in this case). The inequalities Ax

21

DATION

MODELING


he best outcome (such as




mally, linear programming is a technique for the optimization of a



any half spaces, each of which is defined by a

valued affine function defined


rgest) value if such a

Linear programs are problems that can be expressed in


are vectors of (known) coefficients, A is a (known) matrix of coefficients, and

is the matrix transpose. The expression to be maximized or minimized is

x in this case). The inequalities Ax ≤ b and x ≥

22

0 are the constraints which specify a convex polytope over which the

objective function is to be optimized. In this context, two vectors are

comparable when they have the same dimensions. If every entry in the first is

less-than or equal-to the corresponding entry in the second then we can say

the first vector is less-than or equal-to the second vector.

The mathematical technique LP is used for analysing optimum

decisions subject to certain constraints in the form of linear inequalities.

Mathematically it applies to those problems which require the solution of

maximisation problems subject to a system of linear inequalities stated in

terms of certain variables. If and b, the two variables are the function of c, the

value of c is maximised when any movement from that point results in a

decreased value of c. The value of c is minimised when even a small

movement results in an increased value of c.

An illustration could explain us the model of linear programming. Let us

consider a linear programming problem and solve it by algebraic method. An

important thing that has to be understood is to ascertain the given problem as

linear programming, is to write the objective function and the constraints in

the form of equations or inequalities.

Illustration 1

Presume an industry manufactures two commodities M1 and M2.

Each unit of commodity M1 supplies $30 to profit and each unit of commodity

M2 donates $40 to profits. The manufacture of these commodities requires

inputs X, Y and Z and their available volume are 14, 10 and 4 relatively.

Assumed that the manufacture of one commodity M1 procures 2 units of input

X, 1 unit of input Y and does not require input Z and the manufacture of one

unit of commodity M2 requires 2 units of input X, 2 units of Y and 4 units of Z.

Derivate the above problem into linear programming and solve it with

algebraic method.

23

Solution

First we shall convert the above write-up as linear equations in order to

determine solution.

1. Objective Function –> π = 30M1 + 40M2

2. Input X Constraint –> 2M1 + 2M2 ≤ 14

3. Input Constraint Y –> M1+ 2M2 ≤ 10

4. Input Constraint Z –> M2 ≤ 4

5. Non- Negative Constraint –> M1, M2 ≥ 0

With respect to solve these linear equation problems by algebraic method we

first ascertain possible area and its intense points. This is represented in the

below diagram which could give us a graphical solution to the equation.

Three constraint lines XY, ZI and JL denotes input constraints have been

constructed to obtain the region JKPY as the possible area. There are three

J, K, P and Y corner points or intense points of this possible area.

The intense point J is ascertained by only one constraint of input Z and

according to it 4 units of commodity M2 and no amount of M1 are produced.

Substituting this in profit function we get profits at intense point J.

24

Thus profits at intense point J = 30 x 0 + 40 x 4 = $160

Hence J =160

At intense point Y, 7 units of commodity M1 and no unit of commodity M2 are

manufactured.

Thus π at intense point Y = 30 x 7 + 40 x 0 = $210

Hence Y = 210

The intense point K is ascertained by the intersections constraints of inputs Y

and Z. As will be noted from the possible area, according to the intense point

K, 4 units of commodity M2 are manufactured. To get the quantity

manufactured of commodity M1, we substitute M2 = 4 in the constraint

equation of input Y.

Thus M1 + 2 x 4 = 10

M1 = 10 – 8 = 2

Now with M1 = 2, and M2 = 4, at the intense point K, profits are

π = 30 M1 + 40 M2

= 30 x 2 + 40 x 4 = 60 +160 = 220 Hence K = 220

Let us take the intense point P, which is ascertained by intersection of the

constraints of inputs X and Y. The productivity of two commodities at intense

point be got by solving the constraints equation of input X and Y.

Thus we get,

2M1 + 2M2 = 14 …………….Equation (1)

1M1 + 2M2 = 10 …………….Equation (2)

If we deduct the Equation (2) from Equation (1), we get the following,

M1 = 4

Now we have to substitute the value of M1 in the Equation (2), we obtain

4 + 2M2 = 10

2M2 = 6

M2 = 3

25

Thus at intense point P, 4 units of commodity M1 and 3 units of commodity

M2 are manufactured. With these productivities of M1 and M2 profits are:

π = 30 M1 + 40 M2

= 30 x 4 + 40 x 3

= 120 + 120 = 240; Hence, P = 240

Now after ascertaining the profits earned in the intense points, we are going

to present the tablet containing productivities at intense points with their

respective profits.

Productivities and Revenue Earned at Different Inte nse points of the

Possible Area

Intense Points Productivities of

Commodities M1 and

M2

Revenue earned

Value in $

P M1 = 4; M2 = 3 240

Y M1 = 7; M2 = 0 210

K M1 = 2; M2 = 4 220

J M1 = 0; M2 = 4 160

Intense P represents the optimum or profit maximising productivities of

two commodities, such as 4 units of commodity M1 and 3 units of Commodity

M2. Intense Y represents the optimum or profit maximising productivities of


M2. Intense K represents the optimum or profit maximising productivities of


M2. Intense J represents the optimum or profit maximising productivities of


M2.

26

Quantification of factors

Linear programming can be applied to various fields of study. It is used

in business and economics, but can also be utilized for some engineering

problems. Industries that use linear programming models include

transportation, energy, telecommunications, and manufacturing. It has proved

useful in modeling diverse types of problems in planning, routing, scheduling,

assignment, and design.

Linear programming is a considerable field of optimization for several

reasons. Many practical problems in operations research can be expressed

as linear programming problems. Certain special cases of linear

programming, such as network flow problems and multicommodity flow

problems are considered important enough to have generated much research

on specialized algorithms for their solution. A number of algorithms for other

types of optimization problems work by solving LP problems as sub-problems.

Historically, ideas from linear programming have inspired many of the central

concepts of optimization theory, such as duality, decomposition, and the

importance of convexity and its generalizations. Likewise, linear programming

is heavily used in microeconomics and company management, such as

planning, production, transportation, technology and other issues. Although

the modern management issues are ever-changing, most companies would

like to maximize profits or minimize costs with limited resources. Therefore,

many issues can be characterized as linear programming problems.

Standard form

Standard form is the usual and most intuitive form of describing a linear

programming problem. It consists of the following three parts:

• A linear function to be maximized

e.g.

• Problem constraints

e.g.

• Non-negative variables

e.g.

The problem is usually expressed in

becomes:

Other forms, such as minimization problems, problems with constraints on

alternative forms, as well as problems involving negative

always be rewritten into an equivalent problem in standard form

Example

Suppose that a farmer has a piece of farm land, say

either wheat or barley or some combination of the two. The farmer has a

limited amount of fertilizer,

square kilometer of wheat requires

of insecticide, while every square kilometer of barley requires

fertilizer, and P2 kilograms of insecticide. Let S

per square kilometer, and S

area of land planted with wheat and barley by

profit can be maximized by choosing optimal values for

Problem constraints of the following form

negative variables

The problem is usually expressed in matrix form, and then

becomes:


alternative forms, as well as problems involving negative

always be rewritten into an equivalent problem in standard form

Suppose that a farmer has a piece of farm land, say L km2, to be planted with


limited amount of fertilizer, F kilograms, and insecticide, P

of wheat requires F1 kilograms of fertilizer, and

of insecticide, while every square kilometer of barley requires

kilograms of insecticide. Let S1 be the selling price of wheat

per square kilometer, and S2 be the selling price of barley. If we denote the

area of land planted with wheat and barley by x1 and x2

profit can be maximized by choosing optimal values for

27

, and then


alternative forms, as well as problems involving negative variables can

always be rewritten into an equivalent problem in standard form.

, to be planted with


P kilograms. Every

kilograms of fertilizer, and P1 kilograms

of insecticide, while every square kilometer of barley requires F2 kilograms of

be the selling price of wheat

e the selling price of barley. If we denote the

respectively, then

profit can be maximized by choosing optimal values for x1 and x2. This

problem can be expressed with the following linear programming

the standard form:

Maximize:

Subject

to:

Which in matrix form become

maximize

subject to

Augmented form (slack form)

Linear programming problems must be converted into

being solved by the

negative slack variables

constraints. The problems can then be written in the following

matrix form:

Maximize Z:

x, xs ≥ 0

problem can be expressed with the following linear programming

(maximize the revenue—

"objective function")

(limit on total area)

(limit on fertilizer)

(limit on insecticide)

(cannot plant a negative area).

Which in matrix form becomes:

Augmented form (slack form)

Linear programming problems must be converted into augmented form

being solved by the simplex algorithm. This form introduces non

slack variables to replace inequalities with equalities in the

constraints. The problems can then be written in the following

28

problem can be expressed with the following linear programming problem in

—revenue is the

(cannot plant a negative area).

augmented form before

form introduces non-

to replace inequalities with equalities in the

constraints. The problems can then be written in the following block

where xs are the newly introduced slack variables, and

maximized.

Example

The example above is converted into the following augmented for

Maximize: S1x1 + S

Subject to: x1 + x2 +

F1x1 + F

P1x1 + P

x1, x2, x

where x3, x4, x5 are (non

example the unused area, the amount of unused fertilizer, and the amount of

unused insecticide.

In matrix form this becomes:

Maximize Z:

Duality

Every linear programming problem, referred to as a

converted into a dual problem

are the newly introduced slack variables, and Z is the variable to be

The example above is converted into the following augmented for

S2x2 (objective function)

+ x3 = L (augmented constraint)

F2x2 + x4 = F (augmented constraint)

P2x2 + x5 = P (augmented constraint)

x3, x4, x5 ≥ 0.

are (non-negative) slack variables, representing in this


unused insecticide.

In matrix form this becomes:

Every linear programming problem, referred to as a primal

dual problem, which provides an upper bound to the optimal

29

is the variable to be

The example above is converted into the following augmented form:

lack variables, representing in this


primal problem, can be

, which provides an upper bound to the optimal

30

value of the primal problem. In matrix form, we can express

the primal problem as:

Maximize cTx subject to Ax ≤ b, x ≥ 0;

with the corresponding symmetric dual problem, Minimize bTy subject

to ATy ≥ c, y ≥ 0. An alternative primal formulation is: Maximize cTx subject

to Ax ≤ b; with the corresponding asymmetric dual problem,

Minimize bTy subject to ATy = c, y ≥ 0.

There are two ideas fundamental to duality theory. One is the fact that

(for the symmetric dual) the dual of a dual linear program is the original primal

linear program. Additionally, every feasible solution for a linear program gives

a bound on the optimal value of the objective function of its dual. The weak

duality theorem states that the objective function value of the dual at any

feasible solution is always greater than or equal to the objective function

value of the primal at any feasible solution. The strong duality theorem states

that if the primal has an optimal solution, x*, then the dual also has an optimal

solution, y*, such that cTx*=bTy*.

A linear program can also be unbounded or infeasible. Duality theory

tells us that if the primal is unbounded then the dual is infeasible by the weak

duality theorem. Likewise, if the dual is unbounded, then the primal must be

infeasible. However, it is possible for both the dual and the primal to be

infeasible (See also Farkas' lemma).

Example

Revisit the above example of the farmer who may grow wheat and barley with

the set provision of some L land, F fertilizer and P insecticide. Assume now

that unit prices for each of these means of production (inputs) are set by a

planning board. The planning board's job is to minimize the total cost of

procuring the set amounts of inputs while providing the farmer with a floor on

the unit price of each of his crops (outputs), S1 for wheat and S2 for barley.

This corresponds to the following linear programming problem:

Minimize:

Subject

to:

Minimize:

Subject to:

The primal problem deals with physical quantities. With all inputs

available in limited quantities, and assuming the unit prices of all outputs is

known, what quantities of outputs to produce so as to maximize total

revenue? The dual problem deals with economic values. With floor

guarantees on all output unit prices, and assuming the available quantity of all

inputs is known, what input unit pricing scheme to set so as to minimize total

expenditure?

To each variable i

satisfy in the dual space, both indexed by output type. To each inequality to

satisfy in the primal space corresponds a variable in the dual space, both

indexed by input type.

The coefficients that bound the

used to compute the objective in the dual space, input quantities in this

(minimize the total cost of the

means of production as the


(the farmer must receive no less

than S1 for his wheat)


than S2 for his barley)

(prices cannot be negative).

Which in matrix form becomes:







To each variable in the primal space corresponds an inequality to



indexed by input type.

The coefficients that bound the inequalities in the primal space are


31

ze the total cost of the

means of production as the



for his wheat)


for his barley)

(prices cannot be negative).

form becomes:







n the primal space corresponds an inequality to



inequalities in the primal space are


example. The coefficients used to compute the objective in the primal space

bound the inequalities in the dual space, output unit prices in this

Both the primal and the dual problems make use of the same matrix. In the

primal space, this matrix expresses the consumption of physical quantities of

inputs necessary to produce set quantities of outputs. In the dual space, it

expresses the crea

from set input unit prices.

equality and a slack variable, this means each primal variable corresponds to

a dual slack variable, and each dual variable corr

variable. This relation allows us to speak about complementary slackness.

Another example

Sometimes, one may find it more intuitive to obtain the dual program

without looking at the program matrix. Consider the following linear pr

minimize

subject to

We have m + n conditions and all variables are non

define m + n dual variables:

minimiz

e

subject

to


bound the inequalities in the dual space, output unit prices in this




expresses the creation of the economic values associated with the outputs

from set input unit prices. Since each inequality can be replaced by an


a dual slack variable, and each dual variable corresponds to a primal slack


example


without looking at the program matrix. Consider the following linear pr

,

,

,

conditions and all variables are non-negative. We shall

dual variables: yj and s i. We get:

,

32


bound the inequalities in the dual space, output unit prices in this example.




tion of the economic values associated with the outputs

Since each inequality can be replaced by an


esponds to a primal slack



without looking at the program matrix. Consider the following linear program:

negative. We shall

Since this is a minimization pr

that is a lower bound of the primal. In other words, we would like the sum of

all right hand side of the constraints to be the maximal under the condition

that for each primal variable the sum of its

coefficient in the linear function. For example,

If we sum its constraints' coefficients we get

This sum must be at most

maximize

subject to

Note that we assume in our calculations steps that the program is in standard

form. However, any linear program may be transformed to standard form and

it is therefore not a limiting factor.

LP and allocation of resources

Linear programming is a widely used model type that can solve decision

problems with many thousands of variables. Generally, the feasible values of

the decisions are delimited by a set of

,

,

,

Since this is a minimization problem, we would like to obtain a dual program



that for each primal variable the sum of its coefficients do not exceed its

coefficient in the linear function. For example, x1 appears in

If we sum its constraints' coefficients we get a1,1y1 + a1,2y2 +

This sum must be at most c1. As a result we get:

,

,

,



s therefore not a limiting factor.

LP and allocation of resources



the decisions are delimited by a set of constraints that are described by

33

oblem, we would like to obtain a dual program



do not exceed its

n + 1 constraints.

+ ... + a1,nyn + f1s1.





constraints that are described by

34

mathematical functions of the decision variables. The feasible decisions are

compared using an objective function that depends on the decision variables.

For a linear program the objective function and constraints are required to be

linearly related to the variables of the problem. The examples in this section

illustrate that linear programming can be used in a wide variety of practical

situations. We illustrate how a situation can be translated into a mathematical

model, and how the model can be solved to find the optimum solution.

Resource Allocation Problem (will be discussed in detail in the next chapter)

The type of problem most often identified with the application of linear

program is the problem of distributing scarce resources among alternative

activities. The Product Mix problem is a special case. In this example, we

consider a manufacturing facility that produces five different products using

four machines. The scarce resources are the times available on the machines

and the alternative activities are the individual production volumes. The

machine requirements in hours per unit are shown for each product in the

table. With the exception of product 4 that does not require machine 1, each

product must pass through all four machines. The unit profits are also shown

in the table.

The facility has four machines of type 1, five of type 2, three of type 3

and seven of type 4. Each machine operates 40 hours per week. The

problem is to determine the optimum weekly production quantities for the

products. The goal is to maximize total profit. In constructing a model, the first

step is to define the decision variables; the next step is to write the

constraints and objective function in terms of these variables and the problem

data. In the problem statement, phrases like "at least," "no greater than,"

"equal to," and "less than or equal to" imply one or more constraints.

35

Machine data and processing requirements (hrs./unit)

Machine Quantity Product 1 Product 2 Product 3 Product 4 Product 5

M1 4 1.2 1.3 0.7 0.0 0.5

M2 5 0.7 2.2 1.6 0.5 1.0

M3 3 0.9 0.7 1.3 1.0 0.8

M4 7 1.4 2.8 0.5 1.2 0.6

Unit profit, $ —— 18 25 10 12 15

Variable Definitions

Pj : quantity of product j produced, j = 1,...,5

Machine Availability Constraints

The number of hours available on each machine type is 40 times the number

of machines. All the constraints are dimensioned in hours. For machine 1, for

example, we have 40 hrs./machine ¥ 4 machines = 160 hrs.

M1 : 1.2P1 + 1.3P2 + 0.7P3 + 0.0P4 + 0.5P5 < 160

M2 : 0.7P1 + 2.2P2 + 1.6P3 + 0.5P4 + 1.0P5 < 200

M3 : 0.9P1 + 0.7P2 + 1.3P3 + 1.0P4 + 0.8P5 < 120

M4 : 1.4P1 + 2.8P2 + 0.5P3 + 1.2P4 + 0.6P5 < 280

Non negativity

Pj > 0 for j = 1,...,5

Objective Function

The unit profit coefficients are given in the table. Assuming proportionality, the

profit maximization criterion can be written as:

Maximize Z = 18P1+ 25P2 + 10P3 + 12P4 + 15P5

36

Solution

The model constructed with the Math Programming add-in is shown

below. The model has been solved with the Jensen LP add-in. We note

several things about the solution.

1. The solution is not integer. Although practical considerations may demand

that only integer quantities of the products be manufactured, the solution

to a linear programming model is not, in general, integer. To obtain an

optimum integer answer, one must specify in the model that the variables

are to be integer. The resultant model is called an integer programming

model and is much more difficult to solve for larger models. The analyst

should report the optimal solution as shown, and then if necessary, round

the solution to integer values. For this problem, rounding down the

solution to: P1 = 59, P2 = 62, P3 = 0, P4 = 10 and P5 = 15 will result in a

feasible solution, but the solution may not be optimal.

2. The solution is basic. The simplex solution procedure used by the Jensen

LP add-in will always return a basic solution. It will have as many basic

variables as there are constraints. As described elsewhere in this site,

basic variables are allowed to assume values that are not at their upper or

lower bounds. Since there are four constraints in this problem, there are

four basic variables, P1, P2, P4 and P5. Variable P3 and the slack

variables for the constraints are the nonbasic variables.

3. All the machine resources are bottlenecks for the optimum solution with

the hours used exactly equal to the hours available. This is implied by the

fact that the slack variables for the constraints are all zero.

4. This model does not have lower or upper bounds specified for the

variables. This is an option allowed with the Math Programming add-in.

When not specified, lower bounds on variables are zero, and upper

bounds are unlimited.

37

The sensitivity analysis amplifies the solution. The analysis shows the

results of changing one parameter at a time. While a single parameter is

changing, all other problem parameters are held constant. For changes in the

limits of tight constraints, the values of the basic variables must also change

so that the equations defining the solution remain satisfied.

Variable Analysis

• The "reduced cost" column indicates the increase in the objective function

per unit change in the value of the associated variable. The reduced costs

for the basic variables are all zero because the values of these variables

are uniquely determined by the problem parameters and cannot be

changed.

• The reduced cost of P3 indicates that if this variable were increased from

0 to 1 the objective value (or profit) will decrease by $13.53. It is not

surprising that the reduced cost is negative since the optimum value of P3

is zero. When a non basic variable changes, the basic variables change

so that the equations defining the solution remain satisfied. There is no

information from the sensitivity analysis on how the basic variables

38

change or how much P3 can change before the current basis becomes

infeasible. Note that the reduced costs are really derivatives that indicate

the rate of change. For degenerate solutions (where a basic variable is at

one of its bounds) the amount a nonbasic variable may change before a

basis change is required may actually be zero.

• The ranges at the right of the display indicate how far the associated

objective coefficient may change before the current solution values (P1

through P5) must change to maintain optimality. For example, the unit

profit on P1 may assume any value between 13.26 and 24.81. The "---"

used for the lower limit of P3 indicates an indefinite lower bound. Since P3

is zero at the optimum, reducing its unit profit by any amount will make it

even less appropriate to produce that product.

Constraint Analysis

• A shadow price indicates the increase in the objective value per unit

increase of the associated constraint limit. The status of all the constraints

are "Upper" indicating that the upper limits are tight. From the table we

see that increasing the hour limit of 120 for M3 increases the objective

value by the most ($8.96), while increasing the limit for M4 increases the

objective value by the least ($0.36). Again, these quantities are rates of

change. When the solution is degenerate, no change may actually be

possible.

• The ranges at the right of the display indicate how far the limiting value

may change while keeping the same optimum basis. The shadow prices

remain valid within this range. As an example consider M1. For the

solution, there are 160 hours of capacity for this machine. The capacity

may range from 99.35 hours to 173 hours while keeping the same basis

optimal. Changes above 120 cause an increase in profit of $4.82 per unit,

39

while changes below 120 cause a reduction in profit by $4.82 per unit. As

the value of one parameter changes, the other parameters remain

constant and the basic variables change to keep the equations defining

the solution satisfied.

General Resource Allocation Model

It is common to describe a problem class with a general algebraic model

where numeric values are represented by lower case letters usually drawn

from the early part of the alphabet. Variables are given alphabetical

representations generally drawn from the later in the alphabet. Terms are

combined with summation signs. The general resource allocation model is

below. When the parameters are given specific numerical values the result is

an instance of the general model.

40

Linearity requirement

A system is called linear if it has two mathematical properties:

homogeneity and additivity. If you can show that a system has both

properties, then you have proven that the system is linear. Likewise, if you

can show that a system doesn't have one or both properties, you have proven

that it isn't linear. A third property, shift invariance, is not a strict requirement

for linearity, but it is a mandatory property for most DSP techniques. When

you see the term linear system used in DSP, you should assume it includes

shift invariance unless you have reason to believe otherwise. These three

properties form the mathematics of how linear system theory is defined and

used. Later in this chapter we will look at more intuitive ways of understanding

linearity. For now, let's go through these formal mathematical properties.

41

As illustrated in Fig. 5-2, homogeneity means that a change in the input

signal's amplitude results in a corresponding change in the output signal's

amplitude. In mathematical terms, if an input signal ofx[n] results in an output

signal of y[n], an input of kx[n] results in an output of ky[n], for any input signal

and constant, k

A simple resistor provides a good example of both homogenous and

non-homogeneous systems. If the input to the system is the voltage across

the resistor, v(t), and the output from the system is the current through the

resistor, i(t) , the system is homogeneous. Ohm's law guarantees this; if the

voltage is increased or decreased, there will be a corresponding increase or

decrease in the current. Now, consider another system where the input signal

is the voltage across the resistor, v(t), but the output signal is the power being

42

dissipated in the resistor, p(t). Since power is proportional to the square of the

voltage, if the input signal is increased by a factor of two, the output signal is

increase by a factor of four. This system is not homogeneous and therefore

cannot be linear.

The property of additivity is illustrated in Fig. 5-3. Consider a system

where an input of x1[n] produces an output of y1[n]. Further suppose that a

different input, x2[n], produces another output, y2[n]. The system is said to be

additive, if an input of x1[n] + x2[n]results in an output of y1[n] + y2[n], for all

possible input signals. In words, signals added at the input produce signals

that are added at the output.

The important point is that added signals pass through the system without

interacting. As an example, think about a telephone conversation with your

Aunt Edna and Uncle Bernie. Aunt Edna begins a rather lengthy story about

how well her radishes are doing this year. In the background, Uncle Bernie is

43

yelling at the dog for having an accident in his favorite chair. The two voice

signals are added and electronically transmitted through the telephone

network. Since this system is additive, the sound you hear is the sum of the

two voices as they would sound if transmitted individually. You hear Edna and

Bernie, not the creature, Ednabernie.

A good example of a nonadditive circuit is the mixer stage in a radio

transmitter. Two signals are present: an audio signal that contains the voice

or music, and a carrier wave that can propagate through space when applied

to an antenna. The two signals are added and applied to a nonlinearity, such

as a pn junction diode. This results in the signals merging to form a third

signal, a modulated radio wave capable of carrying the information over great

distances.

As shown in Fig. 5-4, shift invariance means that a shift in the input signal

will result in nothing more than an identical shift in the output signal. In more

formal terms, if an input signal of x[n] results in an output of y[n], an input

signal of x[n + s] results in an output of y[n + s], for any input signal and any

constant, s. Pay particular notice to how the mathematics of this shift is

written, it will be used in upcoming chapters. By adding a constant, s, to the

independent variable, n, the waveform can be advanced or retarded in the

horizontal direction. For example, when s = 2, the signal is shifted left by two

samples; when s = -2, the signal is shifted right by two samples.

44

Shift invariance is important because it means the characteristics of

the system do not change with time (or whatever the independent variable

happens to be). If a blip in the input causes a blop in the output, you can be

assured that another blip will cause an identical blop. Most of the systems

you encounter will be shift invariant. This is fortunate, because it is difficult to

deal with systems that change their characteristics while in operation. For

example, imagine that you have designed a digital filter to compensate for the

degrading effects of a telephone transmission line. Your filter makes the

voices sound more natural and easier to understand. Much to your surprise,

along comes winter and you find the characteristics of the telephone line have

changed with temperature. Your compensation filter is now mismatched and

doesn't work especially well. This situation may require a more sophisticated

algorithm that can adapt to changing conditions.

45

Why do homogeneity and additivity play a critical role in linearity, while

shift invariance is something on the side? This is because linearity is a very

broad concept, encompassing much more than just signals and systems. For

example, consider a farmer selling oranges for $2 per crate and apples for $5

per crate. If the farmer sells only oranges, he will receive $20 for 10 crates,

and $40 for 20 crates, making the exchange homogenous. If he sells 20

crates of oranges and 10 crates of apples, the farmer will receive: . This is the

same amount as if the two had been sold individually, making the transaction

additive. Being both homogenous and additive, this sale of goods is a linear

process. However, since there are no signals involved, this is not a system,

and shift invariance has no meaning. Shift invariance can be thought of as an

additional aspect of linearity needed when signals and systems are involved.

Assignment-1 (Homework)

A Manufacture of five products i.e.: P1 to P5. Each unit of commodity P1

supplies $50 to profit, P2 supplies $40, P3 supplies $60, P4 Supplies 20 and

each unit of product P5 donates $40 to profits. The manufacture of these

products requires inputs A, B, C and D and their available volume are 80, 15,

50, and 20 relatively. Assumed that the manufacture of one product P1

procures 5 units of input A, 2 unit of input B, 2 unit of input C, and 2 unit of

input D. Product P2 procures 7 units of input A, 3 unit of input B, 2 unit of

input C, and 3 unit of input D. Product P3 procures 1 units of input A, 0 unit of

input B, 5 unit of input C, and 5 unit of input D. Product P4 procures 0 units of

input A, 0 unit of input B, 9 unit of input C, and 11 unit of input D. and product

P5 procures 15 units of input A, 2 unit of input B, 0 unit of input C, and 5 unit

of input D.

Derivate the above problem into linear programming and solve it with

algebraic method.

46

CHAPTER 3: EXPRESSING LINIER PROGRAMMING

PROBLEMS

Limitations or constraints

Linear programming has turned out to be a highly useful tool of analysis for

the business executives. It is being increasingly made use of in theory of the

firm, in managerial economics, in inter regional trade, in general equilibrium

analysis, in welfare economics and in development planning.

However, there are limitations and they are discussed below.

1. It is complex to determine the particular objective function

2. Even if a particular objective function is laid down, it may not be so easy to

find out various technological, financial and other constraints which may

be operative in pursuing the given objective.

3. Given a Specified objective and a set of constraints it is feasible that the

constraints may not be directly expressible as linear inequalities.

4. Even if the above problems are surmounted, a major problem is one of

estimating relevant values of the various constant co-efficient that enter

into a linear programming mode, i.e. prices etc.

5. This technique is based on the hypothesis of linear relations between

inputs and outputs. This means that inputs and outputs can be added,

multiplied and divided. But the relations between inputs and outputs are

not always clear. In real life, most of the relations are non-linear.

6. This technique presumes perfect competition in product and factor

markets. But perfect competition is not a reality.

47

7. The LP technique is based on the hypothesis of constant returns. In

reality, there are either diminishing or increasing returns which a firm

experiences in production.

8. It is a highly mathematical and complicated technique. The solution of a

problem with linear programming requires the maximisation or

minimisation of a clearly specified variable. The solution of a linear

programming problem is also arrived at with such complicated method as

the simplex method which comprises of a huge number of mathematical

calculations.

9. Mostly, linear programming models present trial and error solutions and it

is difficult to find out really optimal solutions to the various economic

complexities.

Types of Constraints in Linear Programming Problems

This handout describes the most common types on constraints found in linear

programming problems. It should help you in developing your modeling skills.

Due to the wide range of applications, any or all of these constraints might

appear in a given problem. While most constraints fall into one of these

broad classes, exceptions do occur.

For all constraints, make sure the left-hand-side units (e.g., pounds, dollars,

etc.) are consistent with the right-hand-side units. If the units do not agree,

the constraint cannot be correct.

1. Lower and upper bounds on the values of the deci sion variables.

Example: x1 ≥ 10 (lower limit)

x2 ≤ 25 (upper limit)

Note: The standard non-negativity conditions (≥0) are a special case of

lower bounds.

48

2. Limitation constraints (usually ≤ constraints). These are often used to

model limited resources, such as time, units of material, money, etc.

Example: 3x1 + 5x2 + 2x3 ≤ 50

Description: 50 units of the resource are available. Product 1 requires

3 units of the resource, product 2, 5 units, and product 3, 2 units.

3. Requirement constraints (usually ≥ or = constraints). These are

sometimes referred to as covering constraints . They are used to model

a requirement which much be satisfied, such as satisfying the

requirements of a contract, forcing the investment of all money in a

portfolio, or requiring an advertising campaign to reach a certain number

of viewers.

Examples:

a. x1 + x2 + x3 = 10

Total production must equal 10 units, in any combination of

product 1, 2, and 3.

b. 300x1 + 500x2 ≥ 60,000

Advertising campaign must reach 60,000 people, where

medium 1 reaches 300 people per ad, and medium 2, 500

people.

4. Ratio constraints . Also, weighted average and percentage

constraints are very similar. Relationship can be ≤, =, or ≥. These are

used to model situations where the value of one (or more) variable,

compared with the value of another (one or more) variable, must satisfy

some relationship.

Examples:

a. x1/x2 ≥ 2

That is, the ratio of x1 to x2 must be at least 2. This constraint

can occur when a certain product mix ratio is desired between

49

two products. This is the same as the constraint x1≥2x2; that is,

the value of x1 must be at least twice the value of x2. The

constraint can also be written as x1 – 2x2 ≥ 0.

b. x1/(x1+x2+x3) ≤ 0.25

That is, the ratio of x1 to the sum of x1, x2, and x3 must be no

more than 0.25. Another way of thinking of this is that x1 can

comprise at most 25% of the total of x1, x2, and x3. This

situation is very common in blending problems, where "recipes"

may have a certain amount of flexibility. This type of constraint

is also very useful in portfolio investment problems, where one

wants to stay within certain asset allocation guidelines (for

example, at most 25% of total portfolio should be invested in

bonds).

c. (15x1+35x2)/(x1+x2) ≥ 20

This is a constraint on the weighted average of x1 and x2, where

the weights are 15 and 35, respectively. Suppose car 1 gets 15

miles per gallon (mpg) and car 2, 35 mpg. There is an EPA

requirement that the weighted average mpg of all cars sold by a

company be at least 20. This constraint models this

requirement. Note: Avoid the common error of dividing by the

number of variables (2, in this case) instead of the sum of the

variables (x1+x2).

Caution.

Be careful implementing ratio, weighted average, an d percentage

constraints in Excel. The reason is that these constraints can be written

in different forms, some which are linear and some which are nonlinear.

Even though they are algebraically equivalent, the nonlinear form will

50

cause problems with Solver's "Assume Linear Model option. If one

implements a constraint in a nonlinear form, the Solver message "The

conditions for Assume Linear Model are not satisfied" will be received.

Therefore, decide how you're going to implement these constraints before

constructing the spreadsheet model. Taking each example in turn,

a. x1/x2 ≥ 2 is nonlinear because x1 is divided by x2, another decision

variable. The linear forms of this constraint are

x1 ≥ 2x2

and x1 - 2x2 ≥ 0

Either one of these can be implemented in Excel/Solver.

b. x1/(x1+x2+x3) ≤ 0.25 is nonlinear. The linear forms are:

x1 ≤ 0.25*(x1 + x2 + x3) � x1 ≤ 0.25x1 + 0.25x2 + 0.25x3

or 0.75x1 - 0.25x2 - 0.25x3 ≤ 0

c. (15x1+35x2)/(x1+x2) ≥ 20 is nonlinear. The linear forms are:

15x1 + 35x2 ≥ 20*(x1+x2)

or -5x1 + 15x2 ≥ 0

Keep in mind that two main purposes of a spreadsheet model are to

facilitate analysis , and to enhance communication. Choose an

implementation of these constraints which provides the best balance of

these two sometimes conflicting objectives.

5. Balance Constraints. These are used to model processes where the

"inputs" must equal the "outputs." For example, the process of carrying

inventory is modeled with a balance constraint. The process of a

warehouse receiving product from several sources and sending product

out to multiple destinations is another process which must be balanced.

Example: I1 + P2 - S2 = I2

Beginning inventory (end of month 1) plus current (month 2)

production, less month 2 sales, equals ending month 2 inventory.

51

Often (but not always), balance constraints can be implemented directly in

the spreadsheet as formulas, and do not have to be explicitly stated to

Solver.

52

CHAPTER 4: INVENTORY MODEL

Persediaan merupakan aset yang sangat mahal dan penting dalam

sebuah perusahaan yang mewakili sekitar 50%total investasi. Oleh

karenanya pengengalian persediaan merupakan sebuah keputusan

manajerial yang sangat krusial. Pengendalian persediaan ini akan

mempengaruhi pengendalianefektifitas dan efisiensi keuangan. Persediaan

(inventory) merupakan sumberdaya cadangan yang digunakan untuk

memenuhi kebutuhan saat ini maupun waktu yang akan datang. Contoh

inventory misalnya adalah raw material, work in proces dan barang jadi. Level

persediaan untuk barang jadi merupakan fungsi langsung dari adanya

permintaan. Berbagai macam perusahaan memiliki sistem persediaan yang

berbeda. Misalnya persediaan Bank dalam bentuk cash, Rumah sakit dalam

bentuk persediaan darah atau obat misalnya.

Sistem perencanaan dan pengendalian persediaan:

Rencana persediaan

yang harus disediakan

dan bagaimana

mendapatkannya

Perhitungan

permintaan

(Demand)

Pengnedalian

level persediaan

Feedback

53

Pentingnya pengendalian persediaan

1. The Decopupling Function. Jika kita tidak mempersiapkan

persediaan maka akan terjadi keterlambatan (delay) dan in efisiensi

dalam sebuah proses, kaarena proses akan berhenti menunggu

raw material –misalnya- tersedia untuk diproses.

2. Storing Resources. Bahan makanan atau hasil bumi biasanya ada

yang memiliki musim panen tertentu. Padahal kebutuhan atau

permintaan pasar tidak musiman. Oleh karenanya dibutuhkan

persediaan sumberdaya. Sumberdaya itu ity sendiri dapat

terseimpan dalam bentuk proses kerja. Misalnya di sebuah gudang

terdapat 100 mobil dan 1000 roda. Maka persediaan roda sejumlah

100x4 ditambah dengan 1000.

3. Irregular Supply and Demand. Jika permintaan dan persediaan

tidak tetap, maka menyediakan sejumlah barang permintaan

sangatlah penting. Misalnya adanya perbedaan perbedaaan

permintaan ssatu barang di musim hujan yang berbeda dengan

ketika musim kemarau.

4. Quantity Discount. Jika sebuah pemesanan barang dalam jumlah

tertentu akan mendapatkan diskon, maka melakukan pemesanan

barang sejumlah tertentu yang tidak harus sesuai dengan

kebutuhan saat ini harus diperhitungkan dengan baik.

5. Avoiding Stockout and Shortages. Memiliki persediaan untuk

permintaan costumer adalah hal yang sangat mahal. Oleh

karenanya jangan sampai customer kehilangan kepercayaan ketika

kita tidak bisa memberikan kebutuhannya.

54

Keputusan Persediaan

� How much to Order

� When to order

Tujuan model persediaan adalah untu meminimalisasikan biaya persediaan

yang terdiri dari:

1. Cost of item

2. cost of ordering

3. cost of carrying or holding inventory

4. cost of safety stock

5. cost of stockout

EOQ, mendifinisikan berapa banyak pemesanan

Teknik ini di dipublikasikan oleh Ford W. Harris tahun 1915 dan masih

digunakan banyak organisasi saat ini. Teknik ini mudah dalam pemakaiannya

namun harus memiliki asumsu tertentu yaitu:

1. Permintaan diketahui dan konstan

2. The Lead time, yaitu waktu penempatan dan penerimaan order diketahui

dan konstan

3. persediaan dari saat kedatangan dalam satu angkutan dan dalam satu

waktu tertentu.

4. tidak ada diskon

5. biaya variabelnya terdiri dari placing cost, ordering cost dan carrying cost.

6. jika permintaan datang pada waktu yang tepat, maka tidak terjadi

kekosongan persediaan.

55

Inventory Cost

Tujuan model persediaan adalah untuk meminimalisasi biaya persediaan.

Hal ini didapatan pada pemesanan sejumlah order tertentu (optiomal)

yang terjadi saat kurva ccarrying cost sama dengan ordering cost.

Total Cost

time

Inventory

Level

Jumlah

order

Biaya

Ordering

Cost

Carrying

cost

56

Menentukan EOQ

Q* = Ch

CoD..2

Q* = Jumlah optimal pemesanan

D = Demand

Co = Ordering Cost of pieces per order

Ch = Carrying cost per unit per year

Jika carrying cost (Ch) diketahui dalam bentuk prosentase (I) dari harga

barang (P)maka Ch = IP

Contoh:

Sebuah perusahaan manufacture tiap tahun memiliki permintaan sejumlah

1000 unit. Biaya order sebesar $10 per order dan rata-rata caarrying costnya

sebesar $0,50 per tahun. Berapa biaya inventori tiap tahunnya?

Jumlah optimal pemesanan untuk 1000 unit adalah:

Q* = Ch

CoD..2

Q* = 50.0

10.1000.2= 200 unit

Biaya inventory untuk 1000 unit adalah:

TC = CoQ

D+ Ch

Q

2

57

TC = 10200

1000+ 50.0

2

200= $100

Jika Q yang diambil lebih atau kurang darri 200 unit maka Total Costnya akan

lebih besar dari $100.

ROP, Menentukan kapan dilakukan pemesanan

ROP = Demand per day x Leadtime untuk order baru (dalam hari)

ROP = d x L

Contoh .

Sebuah perusahaan komputer memiliki permintaan 8000 chips tiap tahun.

Permintaah hariannya adalah 40 unit. Rata-rata pengiriman order

membutuhkan waktu 3 hari kerja. Maka ROPnya adalah

ROP = d x L = 40 x 3 = 120 unit.

EOQ dengan asumsi tanpa penerimaan yang tak tentu

time

Inventory

Level

t

58

Menentukan annual caarrying cost

= ½ maximum inventory level x Ch

= ½ x Q(1-d/p) x Ch

Q = number of pieces per order or production run

Ch = carrying cost per year

p= daily production rate

d= daily demand rate

t = lenght of production run in day

Q = pt

Menentukan annual setup cost atau Annual ordering c ost

Annual setup cost = CsQp

D

Annual Ordering cost = CoQ

D

D = Annual Demand in units

Qp = Quantity produce in one batch

Cs = Setup cost per setup

Menentukan Optimal Order Quantity dan Production qu antity

Optimal Order Quantity =

−

p

dCh

CoD

1

..2

59

Production quantity =

−

p

dCh

CsD

1

..2

Contoh:

Perusahaan manufacture memproduksi mesin pendingin dalam satu satuan.

Perusahaan memprediksikan menghasilkan 10.000 unit dalam setahun.

Biaya pembuatannya $100 dan carrying cost sebesar 50 sen per unit per

tahun. Hasil yang diperoleh dari proses adalah 80 unit sehari. Selama proses

produksi mampu menghasilkan 60 unit tiap hari. Perusahaan ini

memproduksi 167 hari tiap tahun.

Berapa produksi yang dihasilkan tiap satu satuan? Berapa lama putaran

produksi tiap produknya?

Production quantity =

−

p

dCh

CsD

1

..2

Qp =

−unit

unitunit

80

6015.0$

100.$10000.2

= 4.000 unit

Lama putaran produksi = Q/p = 4000/80 = 50 hari.

Oleh karenanya alat produksi harus di set untuk menghasilkan 50 hari

produksi.

60

Model Diskon jumlah

Rumusan

Total cost = Material cost + ordering cost + carrying cost

TC = DC + CoQ

D + Ch

Q

2

Contoh:

Sebuah toko menjual mainan dengan harga $5. jika pembelian 1000-1999

unit maka akan mendapat diskon sehingga harganya $4.8 per unit. Dan untuk

pembelian lebih dari 2000 harga per unit menjadi $4.75 per unit. Biaya order

$49 per order. Permintaah mainan tiap tahun sebanyak 5000 unit. Carrying

cost adalah 20% harga barang.

Berapa total cost minimum untuk mendapatkan EOQ?

Q1= IP

CoD..2= )5(2.0

49.5000.2=700 mainan per order

Total

Biaya

Diskon 3

Diskon 2 Diskon 1

Q* untuk

Diskon 2

61

Q2= IP

CoD..2= )8.4(2.0


Q3= IP

CoD..2= )75.4(2.0


Penyesuaian dengan diskon. Maka:

Q1 = 700 unit. (tidak ada penyesuaian)

Q2 = 1000 unit. (penyesuaian diskon 1)

Q3 = 2000 unit. (penyesuaian diskon 2)

Annual Material cost (DC)

Dx C1 = 5000 x $5.00 = $ 25,000

Dx C2 = 5000 x $4.80 = $ 24,000

Dx C3 = 5000 x $4.75 = $ 23,750

Annual Ordering Cost = CoQ

D

CoQ

D

1 = 49700

5000=350

CoQ

D

2 = 491000

5000=245

62

CoQ

D

3 = 492000

5000=122.5

Annual Carrying Cost = ChQ

2

ChQ

2 1 = )5$2.0(2

700x = $ 350

ChQ

2 2 = )8.4$2.0(2

700x = $ 48

ChQ

2 3 = )7.4$2.0(2

700x = $ 950

Total Cost = Annual Material cost (DC) + Annual Ordering Cost CoQ

D +

Annual Carrying Cost ChQ

2

TC1= $25,000 + $350.0 + $350 = $25,700,0

TC2= $24,000 + $245.0 + $480 = $24,725,0

TC3= $23,750 + $122.5 + $950 = $24,822.5

Pemakaianan safety stock

Dengan saaafety stock akan menghilangkan ketidaktersediaan barang,

sehingga ada ekstra stok dimiliki.

63

Penganan safety stok terbaik adalah dipergunakan untuk menentukan

reorder poin.

ROP = d x L

Sehingga dengan adanya safeaty stok ini maka

ROP = d x L + SS dimana SS = Safety stok

Reorder point dengan biaya ketidaktersediaan yang t elah diketahui.

• penting diketahui probabilitas demand

• biaya ketersediaan dihitung per unit

• targetnya meminimalisir total cost

contoh:

Carrying cost $5, stockout cost per unit $40. Optomal order per year 6.

Number of units Probability 30 0.2 40 0.2 50 0.3 60 0.2 70 0.1

1.0

Biaya ketidaktersediaan (stockout).

Jika ROP 30 unit.

Pada demand 40 unit

= (40unit-30unit)x$40x6 order per year = $2,400

Pada demand 50 unit


Pada demand 60 unit


Pada demand 70 unit


64

Carrying cost

Jika ROP 30 unit.

Pada demand 40 unit

= (40unit-30unit)x$5 = $50

Pada demand 50 unit

= (50unit-30unit)x$5 = $100

Pada demand 60 unit

= (60unit-30unit)x$5 = $150

Pada demand 70 unit

= (70unit-30unit)x$5 = $200

EMV = �{(probabilitas)i x (alternatives result)i}

4320 = 0.2x0 + 0.2x2400 + 0.3x4800 + 0.2x7200 + 0.1x9600

Safety stok dengan biaya yang tidak diketahui

Untuk menentukan safety stok digunakan servis level dan distribusi normal.

Service level = 1 – probability of a stockout

Contoh:

Sebuah perusahaan diketahui data statistik demand dalam periode tertentu

adalah 350 unit untuk rata-rata demand dengan standar deviasi 10 Berapa

safety stok yang harus di kendalikan?

Probability 0,20 0,20 0,30 0,20 0,10Alternatives 30 40 50 60 70

30 -$ 2.400$ 4.800$ 7.200$ 9.600$ 4.320,00$ 40 50$ -$ 2.400$ 4.800$ 7.200$ 2.410,00$ 50 100$ 50$ -$ 2.400$ 4.800$ 990,00$ 60 150$ 100$ 50$ -$ 2.400$ 305,00$ 70 200$ 150$ 100$ 50$ -$ 110,00$

EMV

65

Jika dipakai kurva normal 5% maka nilai service level, Z pada titik 1-5% =

0.95 adalah 1.65

Z = σSS

=1.65

Maka SS = 1.65 x 10 = 16.5 unit

= 17 unit (pembulatan)

ABC Analisys

Tujuan analysis ABC adalah untuk membedakan perusahaan seluruh jenis

persediaan perusahaan dalam 3 grup, A, B dan C. Kemudian sesuai dengan

masing-masing grup ditentukan level persediaan yang akan di kendalikan

secara umum. Analisis ini untuk membedakan tingkat kepentingan masing-

masing item persediaan yang di kelola.

Misal:

Grup Dolar usage (%) Inventory item (%)

Quantitaitive control used?

A 70 10 Yes B 20 20 In some cases C 10 70 no

Analysis sensitivitas

Perhitungan ini digunakan jika terjadi perubahan variable dalam perhitungan

EOQ

Assignment-2 (Homework)

An Operation Research text book is sold at $5 price. When student buys for

200-299 pieces, he/she will get a discount of $5 each, and buy more than 300

pieces will have $5.5 discount each. The ordering cost is $.5 each ordered.

This book store has to supply 1500 pieces yearly with carrying cost 20% of

prices. Count for minimum cost gaining EOQ?

66

CHAPTER 5: TRANSPORTATION MODEL

Pendahuluan

Metode ini adalah sebuah metode yang dapat memberikan

penyelesaian lebih efisien dalam hal prosedur perhitungan dari pada model

simplex. Perhitungan ini adalah bagian dari network flow problem.

Model transportasi dapat diartikan sebagai distribusi dari sebuah

barang ke tujuan-tujuan tertentu. Tujuan perhitungan ini adalah untuk

penjadwalan pengiriman ke masing-masing tujuan sehingga biaya

transportasi dan produksi dapat diminimalkan.

Sedang model assignment dapat diartikan sebagai penugasan

seseorang pada proyek tertentu, sales ke wilayah tertentu, kontrak ke

penawar tertentu, dan lain sebagainya, dengan tujuan meminimalisir total

cost atau total waktu yang diperlukan dalam penyelesaian tugas. Karakteristik

yang dimiliki oleh model assignment adalah satu orang hanya untuk satu

pekerjaan tertentu, dst.

Seting up transportation problems

Problem transportasi dapat dideskripsikan dengan “bagaimana untuk

memilih rute pengiriman dan jumlah bagian yang dikirim tiap rute” untuk

meminimalisasi biaya total transportasi.

Nortwest corner rule

Contoh:

Biaya transportasi dan kapasitas

67

Distribusi barang

Jumlah biaya

D-A 100 unit x $5 = $ 500

E-A 200 unit x $8 = $1,600

E-B 100 unit x $4 = $ 400

F-B 100 unit x $7 = $ 700

F-C 200 unit x $5 = $1,000

Total $4,200

Stepping stone method: mencari biaya terkecil

Jumlah rute dilalui = jumlah kolom + jumlah baris – 1

Contoh diatas jumlah rute dilalui � 5 = 3 + 3+ 1

Jika jumlah rute kurang dari jumlah rute yang dilalui maka solusinya

dinamakan dengan degenerate.

A (300) B (200) C (200)D (100) $5 $4 $3E (300) $8 $4 $3F (300) $9 $7 $5

Gudang tujuan (Kapasitas)Pabrik (Kapasitas)

A (300) B (200) C (200)D (100) 100E (300) 200 100F (300) 100 200

Pabrik (Kapasitas)

Gudang tujuan (Kapasitas)

68

Menguji hasil untuk peningkatan yang memungkinkan.

Langkah:

1. pilih kotak/jalur yang tidak digunakan (DB-DC-EC-FA) untuk

dievaluasi

2. dengan dimulai dari jalur ini, telusuri jalur dengan jalur tertutup

melewati jalur yang sebenarnya/terpakai.

3. Di jalur yang tidak terpakai, berilah tanda plus. Kemudian jalur

selanjutnya tanda minus dan seterusnya sesuai dengan jalur yang di

kalkulasikan.

4. hitung improvement index dengan menambahkan unit cost sesuai

jalur dengan tanda plus atau minus.

5. Ulangi tahap 1-4 untuk tiap jalur kosong yang ada. Jika dihasilkan

nilai sama atau lebih dari nol, maka solusi optimalnya dapat diketahui.

Namun jika ada yang kurang dari nol maka memungkinkan untuk

meningkatkan hasil sebelumnya dan mengurangi total shipping cost.

Contoh:

+ DB-DA+EA-EB = +4-5+8-4 = +$3

A(300) B(200) C(200)D(100) 100E(300) 200 100F(300) 100 200

Pabrik (Kapasitas)

Gudang tujuan (Kapsitas)

69

+EC-EB+FB-FC = +3-4+7-5 = +$1

+DC-DA+EA-EB+FB-FC = +3-5+8-4+7-5 = +$4

+FA-FB+EB-EA = +9-7+4-8 = -$2

Dengan adanya nilai improvement index kurang dari nol ini, maka cost saving

mungkin akan bisa didapat dari FA. Dalam kasus ini indek negatif terdapat

dalam satu rute, jika terdapat lebih dari satu indek maka diambil nilai indek

negatif terbesar.

A(300) B(200) C(200)D(100) 100E(300) 200 100F(300) 100 200

Pabrik (Kapasitas)


A(300) B(200) C(200)D(100) 100E(300) 200 100F(300) 100 200

Pabrik (Kapasitas)


A(300) B(200) C(200)D(100) 100E(300) 200 100F(300) 100 200

Pabrik (Kapasitas)


70

Langkah selanjutnya adalah menentukan jumlah unit maksimum yang akan

melalui rute baru ini (nilai indek minimal terbesar)

Untuk itu ditentukan terlebih dahulu cell FA dengan tanda plus, dst. Dalam

kasus diambil nilai pengiriman terkecil, karena kita menginginkan pengirian

dalam jumlah besar oleh karena itu cell FA dengan nilai -100 dihilangkan dan

ditambahkan (yang memungkinkan) ke cell EB. Sehingga hasilnya didapat:

maka indek yang terjadi

D ke B = Indek DB = +4-5+8-4=+$3

D ke C = Indek DC = +3-5+9-5=+$2

E ke C = Indek EC = +3-8+9-5= -$1

F ke B = Indek FB = +7-4+8-9=+$2

Sehingga dapat dilakukan improvement pada jalur EC. Jalur EC diberikan

100 unit. Sehingga FA mendapat tambahan 100 unit. Dan terjadi

pengurangan 100 unit di FC.

A(300) B(200) C(200)D(100) 100E(300) 100 200F(300) 100 200

Pabrik (Kapasitas)


A(300) B(200) C(200)D(100) $5 $4 $3E(300) $8 $4 $3F(300) $9 $7 $5

Pabrik (Kapasitas)


71

maka indek yang terjadi

D ke B (jalur DB-DA-FA-FC-EC-EB-DB)

Indek DB = +4-5+9-5+3-4=+$2

D ke C (jalur DC-DA-FA-FC-FC)

Indek DC = +3-5+9-5=+$2

E ke A (jalur EA-FA-FC-EC-EA)

Indek EA = +8-9+5-3=+$1

F ke B (jalur FB-FC-EC-EB-FB)

Indek FB = +7-5+3-4=+$1

Sampai langkah ini didapat seluruh indek lebih besar dari nol, sehingga posisi

jalur ini sudah merupakan hasil yang optimal.

Total Cost yang didapat.

Rute DA 100 unit x $5 = $ 500

A(300) B(200) C(200)D(100) 100E(300) 200 100F(300) 200 100

Pabrik (Kapasitas)


A(300) B(200) C(200)D(100) $5 $4 $3E(300) $8 $4 $3F(300) $9 $7 $5

Pabrik (Kapasitas)


72

Rute EB 200 unit x $4 = $ 800

Rute EC 100 unit x $3 = $ 300

Rute FA 200 unit x $9 = $ 1,800

Rute FC 100 unit x $5 = $ 500

Total $ 3,900

MODI method

Langkah:

Jika R adalah row atau baris dan K adalah kolom dan C adalah biaya yang

terjadi di jalur tersebut, maka:

1. Ri + Kj = Cij, dimana hanya dihitung pada jalur yang terpakai

2. kemudian anggap R1 = 0

3. Hitung sistem rumusan pada semua nilai R dan K

4. hitung indek pada tiap jalur tidak terpakai dengan rumusan I(ij) = C(ij)-Ri-

Kj

5. Pilih indek negatif terbesar, dan teruskan dengn perhitungan seperti

rumusan metode stepping stone.

Contoh:

Distribusi barang

A(300) B(200) C(200)D(100) 100E(300) 200 100F(300) 100 200

Pabrik (Kapasitas)


73

maka:

R1+K1 = 5

R2+K1 = 8

R2+K2 = 4

R3+K2 = 7

R3+K3 = 5

Jika R1=0, maka K1=5, R2=3, K2=1,R3=6,K3=-1.

Kemudian indek yang didapat pada jalur kosong:

Jalur DB (R1K2) = C12-R1-K2 = $4-$0-$1=+$3

Jalur DC (R1K3) = C13-R1-K3 = $3-$0-$1=+$2

Jalur EC (R2K3) = C23-R2-K3 = $3-$3-$1=+$2

Jalur FA (R3K1) = C31-R3-K1 = $9-$6-$5 =-$2

Hasil ini sama dengan perhitungan dengan metode pendekatan stepping

stone.

Vogels approximation metod (VAM)

Metode Vogels approximation metod (VAM) merupakan metode yang tidak

sesimpel nortwest corner namun dapat memberikan solusi yang optimal.

A(300) B(200) C(200)D(100) $5 $4 $3E(300) $8 $4 $3F(300) $9 $7 $5

Pabrik (Kapasitas)


74

Metode ini dapat memberikan gambaran biaya tiap alternati rute. Langkah

perhitungan VAM :

1. tentukan perbedaan antara biaya pengiriman terendah. Perbedaan ini

menggambarakan perbedaan antara biaya distribusi pada ruter terbaik

dalam kolom atau baris dengan rute terbaik keduanya. Misalnya dari

tabel dibawah diketahui untuk baris E, 2 biaya terendah adalah $3 dan

$4, sehingga memilki perbedaan $1. Kolom A, 2 biaya terendah adalah

$8 dan $5, sehingga memilki perbedaan $3.

2. identiikasikan baris atau kolom dengan peluang biaya terbesar, dalam

tabel diatas maka kolom A memiliki perbedaan terbesar, yaitu 3

3. berilah tanda dengan unit, untuk kolom atau baris termurah. Misal

kolom A memiliki biaya termurah di baris B, $5, diberikan 100 unit

sesuai dengan kapasitas pabrik D.

4. Beri tanda X pada baris yang kolom pada baris yang sudah terisi.

3 0 0

A(300) B(200) C(200)D(100) $5 $4 $3 1E(300) $8 $4 $3 1F(300) $9 $7 $5 2

Pabrik (Kapasitas)


A(300) B(200) C(200)D(100) 100 x xE(300)F(300)

Pabrik (Kapasitas)


75

5. Menghitung kembali perbedaan biaya dengan pertimbangan hasil

eliminasi kolom atau baris terisi

6. hitung kembali dari langkah ke-2.

a. Seperti dalam kasus ini,maka peluang biaya terbesar di kolom B

(3). Masukkan berapa unit yang akan dikirim pada baris yang

memungkinkan yaitu pada baris dengan biaya termurah (baris E,

$4, lebih kecil dari baris F, $7), yaitu di baris E, dengan kapasitas

maksimal (kolom B(200), baris E(300)), yaitu 200 unit,yaitu

kapasitas maksimal gudang B.

b. Tentukan kembali perbedaan biaya yang terjadi (Baris E, $5) .

kemudian pada kolom termurah, yaitu kolom C, $3.

1 3 2

A(300) B(200) C(200)D(100) 100 x xE(300) $8 $4 $3 1F(300) $9 $7 $5 2

Pabrik (Kapasitas)


1 3 2

A(300) B(200) C(200)D(100) 100 X X 1E(300) 8 200 3 5F(300) 9 X 5 4

Kapasitas Pabrik


A(300) B(200) C(200)D(100) 100 X XE(300) x 200 100F(300) 200 X 100

Kapasitas Pabrik


76

masukan kapasitas maksimal pada kolom dan baris ini yaitu 100

unit, dari 300unit kapasitas pabrik E-200 unit yang tersalur ke

gudang B.

c. Dari tabel diatas maka akan diketahui sell FA (300-100) dan FC

(20-100) serta FB (300-200-100)

Sehingga dapat diketahui biaya penugasan VAM sebesar :

- 100 unit x $5 = $500

- 200 unit x $4 = $800

- 100 unit x $3 = $300

- 200 unit x $9 = $1,800

- 100 unit x $5 = $500

- Total $ 3,900

A(300) B(200) C(200)D(100) 100 X X

E(300) 8 200 100

F(300) 9 X 5

Kapasitas Pabrik


A(300) B(200) C(200)D(100) 5 4 3E(300) 8 4 3F(300) 9 7 5

Gudang tujuan (Kapasitas)Kapasitas Pabrik

A(300) B(200) C(200) DummyD(250) $5 $4 $3 $0E(300) $8 $4 $3 $0F(300) $9 7 $5 $0

Kapasitas Pabrik


77

unbalance transportation problems

Hal ini terjadi jika permintaan tidak sama dengan supply.

- Supply > demmand = dummy destination (warehouse/surplus)

- Supply < demmand = dummy source (factory)

Kasus ini akan mengakibatkan koeisien biaya pengiriman akan nol.

Supply > demmand = dummy destination

Contoh.

Jika kapasitas pabrik D menjadi 250 unit, sehingga total supply menjadi 850

unit. Sedangkan kapasitas gudang tetap, 700 unit.

Untuk menseimbangkan permasalahan ini maka dibuat dummy column,

dengan kapasitas 850 unit – 700 unit = 150 unit.

Perhitungan total biaya adalah:

250 unit x $ 5 = $1,250

50 unit x $ 8 = $ 400

200 unit x $ 4 = $800

50 unit x $ 3 = $150

150 unit x $ 5 = $750

A(300) B(200) C(200)D(250) $5 $4 $3E(300) $8 $4 $3F(300) $9 7 $5

Gudang tujuan (Kapasitas)Kapasitas Pabrik

A(300) B(200) C(200) DummyD(250) 250E(300) 50 200 50F(300) 150 150

Kapasitas Pabrik


78

150 unit x $ 0 = $0

Total $ 3,350

Supply < demmand = dummy source

Untuk mengantisipasi hal ini dibutuhkan dummy plant.

Contoh:

Jika terjadi jumlah permintaan (500 unit) lebih dari supply (400 unit) maka

dibutuhkan dummy plant dengan kapasitas 50 unit.

sehingga total costnya sebesar:

200 unit x $ 6 = $1,200

50 unit x $ 10 = $ 500

100 unit x $5 = $500

25 unit x $ 8 = $200

75 unit x $ 6 = $450

50 unit x $ 0 = $0

Total $ 2,850

A(250) B(100) C(150)D(200) $6 $4 $9E(175) $10 $5 $8F(75) $12 $7 $6

Dummy (50) $0 $0 $0

Kapasitas Pabrik


A(250) B(100) C(150)D(200) 200E(175) 50 100 25F(75) 75

Dummy (50) 50

Kapasitas Pabrik


79

Degeneracy in transportation

problem ini muncul jika rute<ΣK+ΣR-1. Untuk perhitungannya maka kita harus

meletakkan angka nol pada sel yang tidak terpakai dalam jalur, sehingga

seolah-olah jalur tersebut dipakai/dilalui.

Contoh kasus:

Dalam tabel terlihat, rute<ΣK+ΣR-1

4<3=3-1

dari tabel maka sel DB atau EA dapat dianggap sebagai jalur. Kemudian

dihitung seperti hitungan sebelumnya.

Degeneracy during later solution stage.

Permasalahan transportasi dapat terjadi penurunan jika eliminasi dua rute

yang dilalui ditambahkan pada jalur yang tidak terpakai. Seperti dalam soal 2

jalur yang diberi tanda minus pada satu jalur tertutup (sel DB dab DE)

memiliki jumlah unit yang sama (nol)

Pilihan solusi yang lebih dari satu pilihan

Permasalahan transportasi memmungkinkan memberikan beberapa solusi

dalam arti bahwa jalur transportasi yang didapatkan bisa lebih daru satu

alternatif untuk satu total biaya yang sama.

A(100) B(100) C(100)D(100) 100E(120) 100 20F(80) 80

Kapasitas Pabrik


80

Analisis Lokasi fasilitas

Model transportasi bisa digunakan untuk membantu menentukan lokasi

gudang atau pabrik baru yang akan dibangun untuk memenuhi kebutuhan

perusahaan.

Contoh:

Sebuah perusahaan memiliki tiga lokasi pabrik dan 4 lokasi gudang.

A B C D10000 12000 15000 9000

E 15000 25 55 40 60 48F 6000 35 30 50 40 50G 14000 36 45 26 66 52

Lokasi pabrik alternatifH 60 38 65 27 53I 35 30 41 50 49

Production Cost

Gudang Tujuan (Kapasitas)Pabrik

(Kapasitas)

jawaban

Pabrik

(Kapasitas)

Gudang Tujuan (Kapasitas)

A B C D

10000 12000 15000 9000

E 15000 73 103 88 108

F 6000 85 80 100 90

G 14000 88 97 78 118

Lokasi pabrik

alternatif

H 113 91 118 80

I 84 79 90 99

81

11 10 10

Pabrik

(Kapasitas)


A B C D

10000 12000 15000 9000

E 15000 10000 103 88 108

F 6000 x 80 100 90 10

G 14000 x 97 78 118 19

H x 91 118 80 11

11 10

Pabrik

(Kapasitas)


A B C D

10000 12000 15000 9000

E 15000 10000 103 88 108

F 6000 x 80 100 90 10

G 14000 x x 14000 x

H x 91 118 80 11

Pabrik

(Kapasitas)


A B C D

10000 12000 15000 9000

E 15000 10000 103 88 108

F 6000 x 6000 x x

G 14000 x x 14000 x

H x 91 118 80 11

82

Pabrik

(Kapasitas)


A B C D

10000 12000 15000 9000

E 15000 10000 103 1000 108 5

F 6000 x 6000 x x

G 14000 x x 14000 x

H x 91 x 80 11

Pabrik

(Kapasitas)


A B C D

10000 12000 15000 9000

E 15000 10000 4000 1000 x

F 6000 x 6000 x x

G 14000 x x 14000 x

H x 2000 x 9000

Alternatif ke 2

Pabrik

(Kapasitas)


A B C D

10000 12000 15000 9000

E 15000 10000 1000 4000

F 6000 6000

G 14000 14000

H 2000 9000

83

Total Cost dari kedua alternatif adalah

Alternatif 1 $ 3.704.000

Alternatif 2 $ 3.741.000

Sehingga dipilih lokasi alternatif pertama untuk pembuatan pabrik baru.

84

CHAPTER 6: ASSIGNMENT MODEL

Minimization

Permasalahan penugasan ini biasanya ditunjukkan dalam sebuah matrik

dimana kolom yang ada menggambarkan tujuan dari sebuah penunjukan dan

baris yang ada menunjukkan penunjukkan subyek yang mendapatkan

penugasan tersebut.

Contoh: Ada tiga buah proyek yang akan dikerjakan oleh tiga orang

Biaya Proyek


Typing Editing Printing

Pet

ugas

A $11 $14 $6 B $8 $10 $11 C $9 $12 $7

Alternatif Biaya yang muncul denganberbagai alternatif penugasan:



Typing Editing Printing Typing Editing Printing A B C $11 $10 $7 $28 A C B $11 $12 $11 $34 B A C $8 $14 $7 $29 B C A $8 $12 $6 $26 C A B $9 $14 $11 $34 C B A $9 $10 $6 $25

Dalam contoh diatas kemungkinanyang muncul adalah 6 kemungkinan. Atau

6 kemungkinan ini diadapat dari tiga penugasan, yaitu 3! (tiga faktorial),

85

3!=3x2x1. bisa dibayangkan jika matrik yang terjadi ada 10 baris – 10 kolom,

maka kemungkinan yang muncul adalah 10!=10x9x8x7x6x5x4x3x2x1 atau

akan terjadi kemungkinan sebanyak 3.628.800 kemungkinan!

Untuk mempermudah perhitungn dan mendapatkan kemungkinan penugasan

yang menghasilkan biaya yang paling rendah, dengan metode hungarian

(Flood’s technique)

Langkah-langkah perhitungan:

1. Menyusun tabel biaya dari permasalahan yang disajikan

2. mencari biaya peluang

a. kurangkan nilai terendah untuk tiap baris

b. kurangkan nilai terendah untuk tiap kolom

3. menguji tabel biaya untuk melihat penugasan yang paling optimal

dengan menggambarkan garis minimum yang memungkinkan dalam

kolom atau baris yang di kover nilai nol

4. jika hasilnya optimal (jumlah garis = jumlah baris atau jumlah kolom).

Solusi yang optimal terletak pada nilai nol

a. uji tiap barus dan kolom untuk nilai nol dan berilah penugasan

pada persilangan tersebut

b. eliminasi kolom dan baris pada nilai nol yang lain, buat

penugasan selanjutnya

5. jika hasil tidak optimal. Revisi peluan biaya pada langkah kedua

a. kurangkan nilai terendah untuk yang tidak terkover oleh garis

dari tempat tersebut untuk setiap bilai yang tidak terkover

b. tambahkan nomor ini pada tiap persimpangan dari dua garis

86

contoh:

Biaya Proyek



Pet

ugas

A $11 $14 $6 B $8 $10 $11 C $9 $12 $7

Langkah 1 pengurangan dalam setiap baris, dengan nilai terendah pada tiap

baris

Biaya Proyek



Pet

ugas

A $11 $14 $6 B $8 $10 $11 C $9 $12 $7

Hasil pengurangan



Pet

ugas

A $5 $8 $0 B $0 $2 $3 C $2 $5 $0

Kemudian dikurangkan dengan nilai terendah masing-masing kolom



Pet

ugas

A $5 $6 $0 B $0 $0 $3 C $2 $3 $0

87

Dari nilai nol diatas diartikan sebagai:

- C ditugaskan menangani proyek 3 (C3=0)

- B ditugaskan menangani proyek 1 (B1=0) atau ditugaskan

menangani proyek 2 (B2=0)

- A ditugaskan menangani proyek 3 (A3=0)

Dari perhitungan ini kita belum bisa memberikan penugasan kepada ketiga

orang tersebut. Untuk itu dari tabel diatas dibuat garis melalui seluruh nilai nol

yang ada dalam tabel. Jika jumlah garis yang terjadi sama dengan jumlah

baris atau jumlah kolom, maka hal ini manunjukkan penugasan yang paling

optimal (menghasilkan biaya yang paling rendah).

Sehingga dari permasalah diatas garis pemotong yang terjasi adalah:

Ternyata garis yang memotong nilai nol hanya berjumlah 2 garis (kurang dari

jumlah baris atau jumlah kolom), sehingga komposisi seperti ini belum

menghasilkan penugasan yang optimal. Sehingga perlu dilakukan revisi

tabel, dengan cara mengurangkan semua nilai yang terkover pada tabel

terakhir (A1, A2, C1, C2) dengan nilai terkecil dari nilai yang tidak terkover

tersebut (nilai 2 di C1) dan menambahkan nilai terkecil tersebut pada nilai

yang tidak terkover (nilai 3 di B3).

88



Pet

ugas

A $3 $4 $0 B $0 $0 $5 C $0 $1 $0

Kemudian dibuat garis yang memotong nilai nol yang ada:

Garis yang memotong ini jumlahnya ada 3 garis (sama dengan jumlah baris

atau jumlah kolom). Posisi seperti ini menunjukkan bahwa penugasan yang

terjadi (nilai nol menunjukkan adanya kemungkinan penugasan) akan

menghasilkan biaya terendah (penugasan optimal), yaitu:

• A ditugaskan menangani proyek 3 (A3=0)

• B ditugaskan menangani proyek 1 (B1=0) atau ditugaskan menangani

proyek 2 (B2=0)

• C ditugaskan menangani proyek 3 (C3=0) atau ditugaskan menangani

proyek 1 (C1=0)

Sehingga kesimpulannya adalah A menangani proyek 3, C menangani

proyek 1 dan B menangani proyek 2, sehingga biaya yang dikeluarkan:

Biaya Proyek



Pet

ugas

A $11 $14 $6 B $8 $10 $11 C $9 $12 $7

89

A3 = $6, B2 = $10, dan C1 = $9 sehingga jumlahnya adalah $25

Dummy Row dan Dummy Colums

Hal ini terjadi jika jumlah kolom tidak sama dengan jumlah baris. Hal ini dapat

dijembatani dengan menambahkan kolom baru atau menambahkan baris

baru untuk menyamakan jumlah kolom dan jumlah baris dalam matrik yang

ada.

MAKSIMALISASI PENUGASAN

Kasus yang terjadi dilapangan tidak hanya meminimalisir biaya, namun juga

kasus memaksumalkan keuntungan ateu meningkatkan efisiensi. Oleh

karenanya perhitungan ini diperlukan untuk menentukan hasil yang

dimaksudkan.

Tahap perhitungan dimulai dengan menentukan nilai tertinggi dari seluruh

nilai dalam tabel. Nilai ini digunakan untuk mencari selisih dengan seluruh

nilai yang ada dalam tabel tersebut.

Contohnya.

Jika beberapa proyek dikerjakan oleh beberapa petugas akan mendatangkan

keuntungan tertentu. Maka untuk mendapatkan total keuntungan terbesar,

langkah yang dilakukan adalah menentukan nilai tertinggi dalam tabel

tersebut, kemudian mengurangkannya dengan nilai-nilai yang ada di setiap

sel tabel tersebut.

Proyek

A B C D

Pet

ugas

1 $20 $60 $50 $55 2 $60 $30 $80 $75 3 $80 $100 $90 $80 4 $65 $80 $75 $70

90

Proyek

A B C D

Pet

ugas

1 $20 $60 $50 $55 2 $60 $30 $80 $75 3 $80 $100 $90 $80 4 $65 $80 $75 $70

Proyek

A B C D

Pet

ugas

1 $80 $40 $50 $45 2 $40 $70 $20 $25 3 $20 $0 $10 $20 4 $35 $20 $25 $30

Kemudian setiap baris diambil nilai yang paling kecil, yang kemudian

digunakan untuk mengurangi semua nilai dalam baris tersebut.

Proyek

A B C D

Pet

ugas

1 $80 $40 $50 $45 2 $40 $70 $20 $25 3 $20 $0 $10 $20

4 $35 $20 $25 $30

Proyek

A B C D

Pet

ugas

1 $40 $0 $10 $5 2 $20 $50 $0 $5 3 $20 $0 $10 $20 4 $15 $0 $5 $10

Selanjutnya dibuat garis yang memotong nilai nol (jika jumlah garis yang

terjadi sama dengan jumlah baris atau jumlah kolom, akan menunjukkan

perhitungan yang optimal/menghasilkan penugasan maksimal)

91

Jika jumlah garis tidak sama dengan jumlah baris atau jumlah kolom,

dilakukan proses eliminasi dengan cara mengurangkan nilai terendah pada

masing-masing kolom dengan nilai terendah pada masing-masing kolom

tersebut.

Proyek

A B C D

Pet

ugas

1 $40 $0 $10 $5 2 $20 $50 $0 $5 3 $20 $0 $10 $20

4 $15 $0 $5 $10

Proyek

A B C D

Pet

ugas

1 $25 $0 $10 $0 2 $5 $50 $0 $0 3 $5 $0 $10 $15 4 $0 $0 $5 $5

Kemudian diuji lagi dengan membuat garis yang memotong nilai nol dalam

tabel tersebut. Jumlah garis yang sama dengan jumlah baris atau jumlah

kolom yang terjadi menunjukkan adanya penugasan yang paling optimal

untuk mendapatkan nilai maksimal dari perhitungan yang dilakukan.

92

Sehingga penugasan yang terjadi adalah

Proyek

A B C D

Pet

ugas

1 $25 $0 $10 $0 2 2 $5 $50 $0 $0 2 3 $5 $0 $10 $15 1 4 $0 $0 $5 $5 2

1 3 1 2

Petugas 1 mengerjakan proyek D $ 55

Petugas 2 mengerjakan proyek C $ 80

Petugas 3 mengerjakan proyek B $ 100

Petugas 4 mengerjakan proyek A $65

Sehinga total keuntungan yang didapat sebesar $ 300

93

CHAPTER 7: PROJECT ANALYSIS

Pendahuluan

The Program Evaluation and review Technique (PERT) dan Critical Path

Method (CPM) adalah teknik analisis yang paling populer dalam membantu

manajer untuk merencanakan, penjadwalan dan memonitor serta

mengendalikan proyek-proyek besar dan komplek.

Kerangka PERT dan CPM

� mendifinisikan proyek dan aktivitas-aktivitas yang signifikan dalam

sebuah proyek

� mengembangkan hubungan antar aktivitas, menetukan aktivitas mana

yang mendahului dan aktivitas yang didahului

� menggambarkan hubungan jaringan antar aktivitas

� menentukan biaya dan waktu untuk tiap aktivitas

� Menghitung waktu paling panjang pada suatu jalur dalam jaringan

aktivitas tersebut yang disebut critical path

� menggunakan jaringan untuk membantu perencanaan, penjadwalan

monitoring dan pengendalian proyek.

PERT

Metode ini dapat menjawab pertanyaan-pertanyaan:

1. kapan sebuah proyek diselesaikan

2. apa saja aktivitas kritis dalam proyek, yang harus

diperhatikan/didahulukan

3. aktivitas mana yang tidak kritis dalam proyek yang dapat ditunda

4. kemungkinan penyelesaian proyek dalam satuan waktu tertentu

94

5. melihat kapan sebuah proyek dapat dikatakan sesuai jadwal,

mendahului atau terlambat

6. biaya yang terjadi jika ada keterlambatan, percapatan sebuah

penyelesaian proyek

7. apakan sumber daya yang dimiliki mencukupi penyelesaian proyek

tepat waktu

8. jika menginginkan percepatan penyelesaian proyek, apa yang harus

dilakukan.

Contoh.

Aktivitas Aktivitas yang mendahului A - B - C A D B E C F C G D-E H F-G

Waktu Aktivitas:

- Waktu optimis (a), terjadi jika keseluruhan aktivias berjalan

sebaik mungkin

- Waktu Pesimis (b) terjadi jika menghadapi kondisi yang tidak

mendukung

95

- Waktu realistik (m), waktu yang realistis dalam penyelesaian

proyek, dengan memperhatikan berbagai kemungkinan yang

terjadi.

Waktu yang diharapkan memiliki rumusan:

T = 6

4 bma ++

Dengan variance 2

6

−ab

Contoh menentukan Critical path method (CPM)

Aktivitas Waktu optimis (a)

Waktu Realistis (m)

Waktu Pesimis(b)

T =

6

4 bma ++

variance 2

6

−ab

A 1 2 3 2 4/36 B 2 3 4 3 4/36 C 1 2 3 2 4/36 D 2 4 6 4 16/36 E 1 4 7 4 36/36 F 1 2 9 3 64/36 G 3 4 11 5 64/36 H 1 2 3 2 4/36

96

Earliest Start (ES) dan Earliest finish (EF)

latest Start (LS) dan Latest finish (LF)

97

CPM

Aktivitas ES EF LS LF LS-ES CP A 0 2 0 2 0 Y B 0 3 1 4 1 N C 2 4 2 4 0 Y D 3 7 4 8 1 N E 4 8 4 8 0 Y F 4 7 10 13 6 N G 8 13 8 13 0 Y H 13 15 13 15 0 Y

Project Variance

Aktivitas variance 2

6

−ab

A 4/36 C 4/36 E 36/36 G 64/36 H 4/36 112/36 3.11

98

Project standar deviation

11.3 = 1.76 pekan

Jika proyek ini memiliki batas penyelesaian selama 16 pekan, maka

kemungkinan dapat terselesaikan

Z = 75.1

1516−=0.57

Sehingga kemungkinan penyelesaiannya adalah (Z = 0.57) = 71.6%

PERT/COST

Aktivitas ES EF LS LF LS-ES CP Budget Cost/week A 0 2 0 2 0 Y 22.000 11.000 B 0 3 1 4 1 N 30.000 10.000 C 2 4 2 4 0 Y 26.000 13.000 D 3 7 4 8 1 N 48.000 12.000 E 4 8 4 8 0 Y 56.000 14.000 F 4 7 10 13 6 N 30.000 10.000 G 8 13 8 13 0 Y 80.000 16.000 H 13 15 13 15 0 Y 16.000 8.000

99

Anggaran Berdasarkan Earliest Start

Aktivitas 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

A 11 11 B 10 10 10 C 13 13 D 12 12 12 12 E 14 14 14 14 F 10 10 10 G 16 16 16 16 16 H 8 8

21 21 23 25 36 36 36 14 16 16 16 16 16 8 8 komulatif 21 42 65 90 126 162 198 212 228 244 260 276 292 300 308

Anggaran Berdasarkan Latest Start

Aktivitas 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

A 11 11 B 10 10 10 C 13 13 D 12 12 12 12 E 14 14 14 14 F 10 10 10 G 16 16 16 16 16 H 8 8

21 21 23 25 26 26 26 14 16 16 26 26 26 8 8 komulatif 21 42 65 90 116 142 168 182 198 214 240 266 292 300 308

100

Kurve S

Pengendalian Anggaran

Aktivitas Budget % Completed

Value Actual Slack

A 22.000 100 22.000 20.000 -2.000 B 30.000 100 30.000 36.000 6.000 C 26.000 100 26.000 26.000 0 D 48.000 10 4.800 6.000 1.200 E 56.000 20 11.200 20.000 8.800 F 30.000 20 6.000 4.000 -2.000 G 80.000 0 0 0 0 H 16.000 0 0 0 0

Contoh soal

0

50

100

150

200

250

300

350

1 3 5 7 9 11 13 15

AnggaranBerdasarkanES

AnggaranBerdasarkanES

101

Jika diketahui sebuah proyek dengan waktu dan biaya seperti dibawah,

carilah CP-nya dan buatlah kurve S-nya

Akt Mod T $ A 3 900 B 5 100 C B 4 200 D A 3 300 E A 2 800 F E 1 200 G D 6 300 H C 3 300 I D 2 100 J I-H 1 200 K C 4 800 L F-G 2 400 M K 1 200 N J-L-M 7 700

Jawaban.

Mencari jalur kritis

Akt Mod T $ $/T ES EF LS LF LS-ES CP A 3 900 300 0 3 0 3 0 Y B 5 100 20 0 5 0 5 0 Y C B 4 200 50 5 9 5 9 0 Y D A 3 300 100 3 6 3 6 0 Y E A 2 800 400 3 5 9 11 6 N F E 1 200 200 5 6 11 12 6 N G D 6 300 50 6 12 6 12 0 Y H C 3 300 100 9 12 10 13 1 N I D 2 100 50 6 8 11 13 5 N J I-H 1 200 200 12 13 13 14 1 N K C 4 800 200 9 13 9 13 0 Y L F-G 2 400 200 12 14 12 14 0 Y M K 1 200 200 13 14 13 14 0 Y N J-L-M 7 700 100 14 21 14 21 0 Y

CP �� A-B-C-D-G-K-L-M-N

102

Berdasarkan ES

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 Akt Mod T A 3 300 300 300

B 5 20 20 20 20 20

C B 4 50 50 50 50

D A 3 100 100 100

E A 2 400 400

F E 1 200

G D 6 50 50 50 50 50 50

H C 3 100 100 100

I D 2 50 50

J I-H 1 200

K C 4 200 200 200 200

L F-G 2 200 200

M K 1 200

N J-L-M 7 100 100 100 100 100 100 100

320 320 320 520 520 350 150 150 100 350 350 350 600 400 100 100 100 100 100 100 100

320 640 960 1480 2000 2350 2500 2650 2750 3100 3450 3800 4400 4800 4900 5000 5100 5200 5300 5400 5500

103

Berdasarkan LS

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 Akt Mod T A 3 300 300 300

B 5 20 20 20 20 20

C B 4 50 50 50 50

D A 3 100 100 100

E A 2 400 400

F E 1 200

G D 6 50 50 50 50 50 50

H C 3 100 100 100

I D 2 50 50

J I-H 1 200

K C 4 200 200 200 200

L F-G 2 200 200

M K 1 200

N J-L-M 7 100 100 100 100 100 100 100

320 320 320 120 120 150 100 100 100 650 750 600 550 600 100 100 100 100 100 100 100

320 640 960 1080 1200 1350 1450 1550 1650 2300 3050 3650 4200 4800 4900 5000 5100 5200 5300 5400 5500

104

0

1000

2000

3000

4000

5000

6000

1 3 5 7 9 11 13 15 17 19 21

Earliest

Latest

105

Soal Latihan

Buatlah kurve S dari proyek dibawah dan tentukan jalur kritisnya

Aktivitas Pendahulu Waktu Biaya A 10 20 B 12 30 C 14 28 D A 5 15 E A 6 12 F B 4 8 G B 3 6 H C 6 12 I C 7 14 J C 9 27 K D 7 21 L E-F 5 20 M G-H 8 36 N G-H 3 15 O G-H 5 5 P J 8 64 Q K-L-M 6 36 R O-I 9 27 S Q 4 12 T R-U 2 12 U P 6 18 V S-N-T 8 24 W R-U 10 20 X W 9 27 Y V-X 5 20 Z Y 10 20

106

DISKUSI KASUS

Case #1 CUSTOM VANS INC

Perusahaan Custom Van adalah perusahaan yang bergerak dibidang

perombakan standar kendaraan model van kebentuk camper. Dengan

kesesuaian pada jumlah pekerjaan dan perombakan yang dilakukan,

perombakan ini membutukan biaya kurang dari $1000 atau lebih dari $5000.

Pada empat tahun terakhir, Tony Rizzo mampu mengembangkan usahanya

di Gary Indiana hingga mamapu mengembangkannya dengan memiliki outlet

di Chicago,Milwaukee, Minneapolis dan Detroit.

Kesuksesan usahan Tony ini sangat dipengaruhi oleh inovasi yang

dikembangkannya dalam merombak bengkel van yang masih kecil menjadi

usaha yang lebih besar dan professional di Midwest. Tony kelihatannya

memiliki kemampuan khusus untuk merancang dan mengembangkan gaya

dan model sehingga mampu meningkatkan permintaan dari para pemilik

mobil van. Sebagai contoh, shower Rific mampi berkembang, setelah

dirombak oleh tony 6 bulan setelah usaha tni dimulai. Sebuah shower kecil

yang didesain dan dikembangkan untuk dapat ditempatkan di berbagai posisi

di dalam van. Shower Rifik dibuat dari fiberglass dan terdiri dari rak handuk,

tampat sapodansabun, dan pintu plastik yang unik. Tiap pembuatan Shower

ini membutuhkan 2 galon fiberglass dan 3 jam kerja.

Hampir semua proses produksi produk ini dikerakan gary di warehouse

yang sama dimana perombakan van ditemukan. Pabrik dari produk ini

menghasilkan 300 shower rific dalam satu bulan, Namun kapasitasnya tidak

pernah mencukupi. Bengkel pembuatan peromabak van di semua lokasi

mendapatkan protes karena tidak mencukupi produk shower ini dan

jugakarena minneapolis terlalu jauh dari gay dibanding lokasi lainnya. Tony

selalu cenderung mengirim shower rific ini ke leokasi lainnya sebelum ke

107

minneapolis. Hal ini membuat marah manager yang ada di minneapolis.

Seleha dilakukan diskusi, tony memutuskan untuk membangun pabrik baru di

Fort Wayne Indiana. Di tempat ini akan diproduksi sejumlah 150 shower tiap

bulan.

Pabrik di fort wayne ini tenyata masih belum mapu memenuhi

permintaan untiuk produk ini. Dan tony juga mengatui bahwa untuk tahun

depan permintaan ini akansemakin meningkat tajam. Setalah berkonsultasi

dengan konsultan dan bank, tony dia harus segara membangun dua pabrik

lainnya secepat mungkin. Tiap pabrik harus memiliki kapasitas yang sama

dengan yang ada di fort wayne. Ada beberapa alternatif tempat untuk

membangun dua pabriknya yaitu, detroit, michigan, rockord, illinois atau

madison, wisconsin. Tony memahami bahwa pemilihan dan penempatan

lokasi yang terbaik adalah keputusan sulit. Biaya transportasi dan

permintaan untuk likasi yang berbeda akam mempengaruhi keputusan ini.

Toko di chicago di manage oleh Bill Burch. Disini tony menemptkan

salah satu outletnya yang kemudian diikuti dengan lokasi lain berikutnya.

Pabrik gary ini mensuply 200 shower tiap bulannya. Walaupun Bill

mengetahui bahwa permintaaan di chicago sebanyak 300 unit. Biaya

transportasi dari gary ke chicago $10, dan biaya transport dari fort wayn dua

kali jumlah tersebut. Bill selalu meminta kepada tony untukk mendapatkan

tambahan 50 unit dari forth wayne. Tambahan 2 pabrik yang dibangun akan

mampu suplai pada bill dengan tambahan 100 unit yng dibutuhkan bill. Dan

biaya trnsportasinya akantergantung dari lokasi yang akan dipilihnya. Jika

dari detroit biay transportasinya $30, dari rockford $5 dan dari madison $10.

Wilma Jackson manajer di Milwaukee juga kesal dengn kekurangn

suplai showerini. Permintaanya adalah 100 unit. Sedang suplai yang

diterimanya selama ini hanya separuh dari kebutuhanya dari pabrik di fort

wayne. Wilma heran kenapa tony tidak mengiriminya 100 unit dari gary

walau biaya transportasinya hanya $20 dari Gary. Sedang biaya transpotasi

108

dari fortwayne $30.Wilma berharap tony akan memilih madison sebagai salah

satu lokasi produksinya. Sehingga dia akan dapat memperoleh produk ini

sesuai kebutuhannya dengan biaya transportasi hanya $5 per unit. Jika buka

di Madison, pabrik baru di rockford akan mampu menyediakan seluruh

kebutuhannya, namun biaya transportasinya akan menjadi duakali lipat

dibanding dari madison. Karena biaya tranportasi per unit dari detoid $40,

Wilma membuat spekulasi kalau-kalau detroid akan menjedi salah satu pabrik

baru yang akan dibangun. Dia tidak akan mengambil dari selain dari Detroit.

Custom Van Inc. Di minneapolis yang di pimpin oleh manager Tom

Poanski, dia membutuhkan 100 shower dari parbrik gary. Permintaan yang

ada 150 unit. Tom mendapatkan biaya transportasi tertinggi dibanding

dengan lokasi lainnya. Biaya transportasi dari Gari sebesa $40 per unit. Hal

ini lebih tinggi $10 jika shower dikirim dari lokasi Fort Wayne. Tom berharap

detroit bukan satu-satunya pabrik baru yang akan dibangun, karena akan

mengakibatkan biaya trsportasi menjadi $60 per unit. Rockford dan Madison

akan memberikan biaya $30 dan $25 ke minneapolis.

Posisi Toko di detroit hampir mirip dengn yang berada di Milwaukee,

hanya saja permintaanya hanya separoh untuk tiap pekannya.Detroit tidak

enerima 10 unit shower ini secara langsung dari Pabrik di fort Wayne, Biaya

transportasinya hanya $15 per unit dari Fort Wayne, sedangkan biaya $25

dari gary. Dick Lopz manaer Custom Vans Inc di detroit, memperhitungkan

kemungkinan jika menempatkan sebuah pabrik di detroit dengan pioritas

yang tinggi. Pabrik yang akan ditempatkan di tengah kota dan biaya

transportasinya hanya akan memakan $5 per unit. Dia akan medapatkan

shower sejupah 150 unit dari pabrik baru di detroit dan 50 unit lainnya dari

Fort Wayne. Jika Detroit tidak enjadi pilihan, maka dua lokasi lainnya tidak

bisa dihindari. Rockford akan memakan biayatransportasi sebesar $35 dan

Madison $40.

109

Tony empertimbangkan dilema pilihan lokasi dua pabrik baru beberapa

pekan sebelum membuat keputusan dengan mengadakan pertemuan semua

manajer dri semua tokonya. Keputusan ini akan sngat rumit, namun jelas

tujuannya yaitu untuk meminimalisir biaya. Pertemuan akan di adakan di

Gary dan semua manajer hadi kecuali Wilma.

Tony: Terimakaih atas kehadiran saudara-saudara pada rapat kali ini.

Sebagaimana kita ketahui berasma bahwa saya telah memutuskan untuk

membuka dua pabrik baru di rockford, madisun atau detroit. Dua lokasi yang

akan kita pilih akan emberikan pengaruh pada praktek pengiriman kita, dan

harapan saya hal ini akan dapat memenuhi kebutuhan pengiriman dan

permintaan untuk tiap toko saudara-saudara.Saya mengetahui bahwa anda

harus menjual lebih banyak, saya ingin anda mengetahui bersma bahwa

saya mohon maaf atas segala situasi ini.

Dick: Tony, saya telah memaparkan kondisi ini dengan beberapa

alasan, dan saya sangat baerharap bahwa salah satu pabriknya akan anda

tempatkan di detroit. Sebagai mana anda ketahui, saya hanya mendapatkan

separoh dari kebutuhan kami. Saudara saya, Leon, sangat terrtarik untuk ikut

menjalankan pabrik ini dan saya tahu dia akan melamar untuk pekerjaan ini.

Tom: Dick, saya tahu Leon akan mendapatkan pekerjaan yang baik ini,

dan saya tahu betapa berat yang dirasakannya sejak di keluarkan dari

industri Auto. Namun kita harus memperhatikan biaya total bukan

permasalahan personal. Saya percaya jika pabrik baru akan di tempatkan di

madison dan di rockford. Saya berpikir panjang jika penempatan pabrik dn

beberapa toko yang direncanakan di tempat ini akan sangat signifikan

mengurangi biaya transportasi.

Dic: Hal itu mungkin benar, namun ada beberapa alasan lainnya. Detroit

adalah salah satu suplier terbear untuk fiberglass, dan saya telah melakukan

pengecekan biayanya. Satu pabrik baru di detroit hanya mebutuhkan $2 per

110

galon untuk fiberglass lebih murah dibandingkan dengan pabrik yang

diusulkan di lokasi lain.

Tom: di madison, kami memiliki biaya pekerja yang sangat menarik. Hal

ini karena banyak pelajar dari Univeritas madison. Mereka adalah pekerja

keras dan mereka hanya dibayar $1 perjam lebih murah dari lokasi lain. Ini

alasan saya.

Bill: Santai saja saudara-saudara, jelas disini kita tidak akan mampu

memuaskan semuanya dari kita dlam memutuskan lokasi pabrik-pabrik baru

kita. Namun kita harus memilih dengan voting yang terbaik dua lokasi yang

akan kita jadikan tempat pembangunan pabrik baru kita.

Tony: saya kira voting bukan hal terbaik bagi ita. Wilma tidak hadir saat

ini, dan kita harus melihat semua faktor secara bersama-sama dengan

semua kondisinya.

DISKUSI:

Dimana lokasi pabrik itu akan ditempatkan?

111

Case #2 Haygood Company

George dan harry haygood adalah sebuah kontraktor bangunan yang

mengkhususnkan pada pembangunan rumah tinggal, gudang dan bisnis

kecil, yang kurang dari 20.000 kaki persegi untuk luasan lantai. Baik george

maupun harry memulai dari program pelatihan asosiasi pengusaha kayu awal

tahun 1990an dan selama mengikuti pelatihan untuk menjadi tenaga ahli

hingga tahun 1996. Sebelum memulai bisnisnya sendiri, merke bekerja pada

kontraktor-kontrakot di wilayah detroit.

Akhirnya heygood bersaudara berhasil memenangkan beberapa tender

poyek perumahn. Dalampenyelesian kontrak, beberapa aspek kontruksi

seperti pemasangan jaringan listrik, saluran pembuangan, pengecatan,

konblok dan lainnya di sub kontarkkan. George danharry hanya menangani

pekerjaan kayunya saja. Namun mereka juga yang membuat perencanaan

dan penjadwalan kerja untuk seluruh operasional pembangunan, dan

masalah keuangan serta mensupervisi semua aktivitas pembangunan

tersebut.

Dengan moto “ Waktu adalah uang” Haygood mencoba untuk

melakukan efisiensi dengan mengendalikan keuangan. Oleh karenanya

jangan sampai ada penundaan pekerjaan dalam proyek ini. Untuk

menganalisisi heygood menggunakan metode PERT. Yant pertama di

jabarkan seluruh aktivitas dalam proyek tersebut. Kemudian dihitung

kebutuhan waktu untuk penyelesaian masing-masing aktivitas tersebut,

hingga konsekuensi pembiayaannya. Setelah diketahui Earliest dan Latest

time untuk keseluruhan aktivita, haygood dapat mengalokasikan sumberdaya

yang dimilikinya untuk penyelesaian proyek.

Berikut aktivitas yang ada padaproyek tersebut:

1. Merancang Keuangan

2. Mencari subkontraktor

112

3. memasang pondasi

4. pemasangan plumbing

5. pemasangan rangka

6. pemasangan atap

7. pemasangan jaringan listrik

8. memasang pintu dan jendela

9. pemasangan jaringan pemanas dan pendingin ruangan

10. pemasangan panel dan plesteran dinding

11. pemasangan kabinet

12. pemasangan konblok

13. Pemasangan strimin luar

14. pemsangan strimin dalam

15. pengecatan

16. Pemasangan lantai

A B C

D

E

F

G

H

I

J

K

L

M

O

N

P Q

113

Data proyek:

Aktivitas optimis realistis Pesimis AB 4 5 6 BC 2 5 8 CD 5 7 9 CE 4 5 6 D 2 4 6

FG 3 5 9 FH 4 5 6 FI 3 4 7 FJ 5 7 9 JK 10 11 12 KL 4 6 8 KM 7 8 9 MN 4 5 10 LO 5 7 9 OP 5 6 7 PQ 2 3 4

Diskusi:

1. Jalur Kritis/ Critical path? Lama waktu proyek yang melalui jalur kritis?

2. Hitung jumlah waktu yang dapat ditunda untuk aktivitas yang ada

tanpa mempengaruhi keterlambatan proyek

3. Berapa keumnginan yang terjadi jika proyek dimulai tanggal 1 agustus

dan selesai tanggal 30 september? Dengan catatan jadwal

penyelesaian 60 hari.

114

Case #3 MANAGEMENT VIDEO PROFESIONAL

Semenjak awal diperkenalkan sebuah system home video untuk televisi,

stave Goodman sudah memimpikan sebuah peluang usahanya untuk sisete

video untuk aktovitas professional. Selama beberapa tahun terakhir stave

melihat beberapa film lama favoritnyadi video ruhamnya dan merancang

pengebangan untuk system video rumahnya. Dia mencoba untuk

menggunakan stasion televisi, agen periklanan dan dan beberapa kelompok

dan perorangan yang menginginkan system vided terbaik.

Dasar dari system ini termasuk ruang pengendali yang lengkap, dua

system videotape yang terpisahkan, videodisk dan satu set televisi

professional yang berkualitas. Semua peralatan ini adalah alat yang saling

terintegrasikan. Sebagai tambahan, dasar dari system ini hadir dengan

tambahan system remote control. System ini dapat mengoperasikan baik

sviseo system, video disk dan system tv dengan mudah. Remote contro

dapat bekerja dengan mengirimkan sinyal infra merah ke kotak pengendali

yang ada dalam system pengenaliannya.

Hal yang unik dalam system video ini adalah terdapat pada kotak

pengendalinya. Kotak pengendali ii terdapat microposesorcanggih yang

memiliki kemampuan menkoordinasikan pemakaianan dan ungsi peraalatan

lain yang terdeteksi.

Sistem video professional stave ini memiliki beberapa keunggulan

disbanding system yang ada pada umumnya. untuk memulainya seperti

adanya efek khusus yang dapat dikendalikan secara mudah. Image pada

videodisk, salah satu sistem video tersebut, dan sistem televisi dapat dengan

mudah ditempatkan pada sistem video lainnya. Sebagai tambahan hal ini

memungkinkan untuk menghubungkan dengan kotak pengendali pada PC

atau Mechintos. Hal ini membuat kemungkinan adanya pengembangan grafis

yang lebih menarik dalam microkomputer dan dapat untuk mentranfer secara

115

langsung ke system video. Begitu juga memungkinakn untuk

mencantelkansistem stereo ke dalam kotak pengendali untuk

mengintegrasikan sistem kualitas suara yang lebih teinggike dalam sistem

dan merekamnya ke dalam salat satu video sistemnya.

Kedua sistem video tersebut juga memiliki fleksibilitas yang luarbiasa

untuk proses editing. Beberapa item khusus telah ditempatkan pada reote

kontrolnya. Hal ini memungkinkan untuk merekam sebuah program pada

salah satu sistem video yang pertama dan mengeditnya dengan sistem video

tape yang lain untuk memberikan tambahan atau mengurangi begian-bagian

tertentu. Salah satu feature terbaik pada sustem yang dikembangkan staveini

adalah harganya. System dasar dan termasuk kotak pengendali, dua sistem

video, video disk, dan system televisi hanya dijual dengan harga $1.995.

Stave menemukan produk untuk sistem televisi, kotak pengendali, dan

sistem videodisk di Amerika. Karena sistem videotapelebih terkenal, stave

memiliki banyak pilihan. Setelah mengadakan penelitian, stave menemukan

dua suplier. Kedua suplier ini adalah perusahaan jepang. Tishiki sebagai

perusahaan baru di luar tokyo jepang. Seperti suplier lainnya, toshiki

memberikan diskon . Untuk pembelian kurang dari 2000 unit, permintaan

produk untuk stave akan diberikan harga $250 pervideo system. Sedangkan

harga $230 akan diberikan untuk pembelian antara 2000 hingga 8000 unit.

Dan untuk pembelian antara 8000 sampai 20.000 unit akan dikenakan biaya

$210 per unit untuk sistem video ini.

Suplier jepang lainnya, Kony. Walaupun asalnya Kony berawal di

jepang, dan diluar tokyo, namun memiliki fasilitas dan kantor di seluruh dunia.

Salah satunya berada di sekitar 100 mil utara atlanta, georgia. Seperti

Toshiki, Diskont yang ditawarkan oleh Kony akan diberikan untuk sejumlah

pembelian. Untuk sejumhalh kurang dari 1000 unit akan diberikan harga

$250 per unit. Untuk julah 1000 hingga 5000 akan diberikan harga $240.

Dan untuk pengadaan lebih dari 5000 akan diberikan harga $220.

116

Karena perusahaan Kony terletak di Amerika, biaya pemesanan dan

waktu pengiriman lebih menarik dari Toshiki. Perkiraan biaya pemesanan

untuk Kony adalah $40, dan pengirimanhanya memakanwaktu 2 pekan.

Sedangkan untuk toshiki biaya pemesanan dan wktu pengirimanakan lebih

tinggi dibnding dengan kony. Belum lagi tambahan biaya $90 untuk

pengadaan dari jepang untuk tiap ordernya. Dan waktu pengiriman selama 3

bulan. Perkiran stave untuk biaya pengiriman akan mencapai lebih 30%.

Untuk tahun pertama stave memutuskan untuk menjual hanya unit dasar

dari system produk yang diciptakan. Yaitu kotak pengendali, televisi, video

disk dan dua sistem videotae. Perminttaan dari sistem keseluruhan relati

konstn selama 6 bulan terakhir. Misalnya untuk permintaan bulan juni

penjualanan mencapai 7979 unit, juli 8070 unit, agustus 7950 unit dan

september 8070unit. Perkembangan ini akan relatif tetap untuk beberapa

bulan mendatang.

Diskusi:

1. Produk apa yang diunggulkan stave?

2. Apa saja masalah yang dihadapi stave untuk memasarkan produk?

3. Berapa ROP untuk Kony dan toshiki?

4. Jika anda menjadi stave, perusahaan mana yang akan anda pilih

sebagai mitraa suplier anda?

5. Stave memiliki beberaa strategi. Strategi pertama stave adalah akan

menjual koponenya secara terpisah. Strategi kedua adalah akan

memodifikasi kotak pengendali mengikuti sistem videotape untuk

digunakan sebaik sistem video yang dihailkan stave. Secara umum,

apa pengaruh dari penambahan sistem ini pada ROP dan

pengendalian persediaan bagi stave?

117

PAPER

Paper-1: Presentation of a New and Beneficial Metho d Through Problem

Solving Timing of Open Shop by Random Algorithm Gra vitational

Emulation Local Search

Abstract

One of the most important problems of timing in engineering and industry is

timing of open shop. The problem of timing of the open shop induces big and

complicated solve space. So, this problem is a kind of NP-Complete. In timing of

the open shop, there some works, that each work has several operation.

Each operation should do in machine whit should do in the same machine the

aim of timing of the open shop is to catch a suitable timing for doing all of the

operation, how that enough time to minimize to make-span. In problem solve

of timing of the open shop. Until now different algorithm were presented. In

this article, a new algorithm that is called TIME_GELS is presented which

used of a random. Algorithm Gravitational Emulation Local Search (GELS) for

following problem solving. This algorithm is basic of the random local search

use of two of the four main parameter of speed and the power of gravity in

physics. A suggestive algorithm compared with Genetic Algorithm and result

is show that a proposed algorithm has a better efficient and finding the

answer very soon.

Keywords: Timing; Open Shop; Genetic Algorithm; Velocity; Newton law;

Gravitational force

118

Paper-2: Inverse Optimization for Linear Fractional Programming

Abstract:

In this paper, we have proposed an inverse optimization model for linear

ractional programming (LFP) problem. In our proposed model,

the parameters of the objective function are adjusted as little as

ossible (under L1 norm), so that the given feasible solution become optimal.

We formulate the inverse linear fractional programming (ILFP) problem as

a linear programming problem having large number of variables,

which can be solved by many existing methods or

optimization software such as: TORA, EXCELSOLVER etc. The method has

been illustrated by a numerical example also.

Keywords: Inverse optimization, Linear Fractional Programming,

Complementary slackness

Paper-3: A multi-objective model for designing a gr oup layout of a

dynamic cellular manufacturing system

Abstract

This paper presents a multi-objective mixed-integer nonlinear programming

model to esign a group layout of a cellular manufacturing system in a dynamic

environment, in which the number of cells to be formed is variable. Cell

formation (CF) and group layout (GL) are concurrently made in a dynamic

environment by the integrated model, which incorporates with an extensive

coverage of important manufacturing features used in the design of CMSs.

Additionally, there are some features that make the presented model different

from the previous studies. These features include the following: (1) the

variable number of cells, (2) the integrated CF and GL decisions in a dynamic

environment by a multi-objective mathematical model, and (3) two conflicting

119

objectives that minimize the total costs (i.e., costs of intra and inter-cell

material handling, machine relocation, purchasing new machines, machine

overhead, machine processing, and forming cells) and minimize the

imbalance of workload among cells. Furthermore, the presented model

considers some limitations, such as machine capability, machine capacity,

part demands satisfaction, cell size, material flow conservation, and location

assignment. Four numerical examples are solved by the GAMS software to

illustrate the promising results obtained by the incorporated features.

Keywords: Dynamic cellular manufacturing systems, Multi-objective model, Cell formation, Group layout

Paper-4: Integrating truck arrival management into tactical operation

planning at container terminals

Abstract

Truck arrival management (TAM) has been recognized as an effective olution to alleviate the gate congestion at container terminals. To further utilize TAM in improving the overall terminal performance, this study integrates TAM with the other terminal operations at a tactical level. An integrated planning model and a sequential planning model are presented to coordinate the major terminal planning activities, including quayside berth allocation, yard storage space allocation and TAM. A heuristic-based genetic algorithm is developed to solve the models. A range of numerical examinations are performed to compare two planning models. The result shows that: the integrated model can improve the terminal performance significantly from the sequential model alone, particularly when the gate capacity and the yard capacity are relatively low; whereas the sequential model is more efficient than the integrated model in terms of computational time.

Keywords: container terminal; integrated planning; truck arrival management; berth allocation; storage space allocation

120

Paper-5: Pharmaceutical Inventory Management Issues in Hospital

Supply Chains

Abstract

The primary focus of the healthcare sector is to provide patients with the best quality of care. While the healthcare cost is keep on growing, effective healthcare supply chain should be achieved to reduce some unnecessary costs. To address this issue, this study aims to examine inventory management practice in one of Indonesian public hospital and focus on the role of inventory to drive hospital supply chain performance. Three major issues regarding inventory management practice has been identified such as overstock, unjustified forecasting technique and lack of IT support. Proposed (s,Q) policy using continuous review can reduce by 50% total inventory value on hand of oncology medication. Among several forecasting technique that’s presented, Holt’s model appears to be the best adapted for oncology medication. Future study is needed to simulate the outlook condition using proposed policy. By implementing a new inventory policy that cope all the constraints and problems will help hospital to manage its pharmacy inventory in effective and efficient way.

Keywords: Inventory Management, Oncology Medication, Public Hospital, Indonesia

Paper-6: Improving a Flexible Manufacturing Schedul ing using Genetic

Algorithm

Abstract

A Flexible Manufacturing System (FMS) is designed to produce a variety of products, utilizing a set of resources like work stations, robots etc., interlinked by certain means of transport. The prime characteristic of an FMS is that the overall system is under the computer control to realize these essential improvements in a firm; it imposes many challenging problems for planning, scheduling, monitoring and control of manufacturing system. These problems have a fundamental implication on the overall performance of a FMS, and

121

influence the responsiveness of the system to satisfy the changing customer needs. In this study, dispatching rules are used to solve the scheduling problem. Further, the multiple dispatching rule based heuristic is proposed to search the optimal sequence of operations. Genetic Algorithm (GA) is used as a random search optimization technique in the proposed heuristic. Finally, the sequence determined with the proposed heuristic is utilized to develop based intelligent controller.

Keywords: Job shop scheduling, genetic algorithm, priority rule, flexible manufacturing system, heuristics

122

CONTOH SOAL QUIZ, UTS DAN UAS

Quiz

Take Home, dikumpulkan sehari setelah soal ini dibagikan

Dikerjakan dengan kertas bergaris, dengan huruf balok.

soal transportasi

TUJUAN Pasar A Pasar B Pasar C Kapasitas 600 750 650 Gudang 1 (500) Rp. 1.250,- Rp. 1.750,- Rp. 1.000,- Gudang 2 (500) Rp. 2.250,- Rp. 1.500,- Rp. 1.750,- Gudang 3 (1000) Rp. 2.500,- Rp. 1.500,- Rp. 2.500,-

Tentukan distribusi yang menghasilkan biaya terendah dan berapa biaya yang terjadi.

Soal Inventori

Tentukan

- ROP dan EOQ

January 1500 20$ 1,275$ 3February 1600 20$ 1,360$ 4March 1700 20$ 1,445$ 5April 1500 20$ 1,275$ 3May 1600 20$ 1,360$ 5June 1600 20$ 1,360$ 4July 1700 20$ 1,445$ 4August 1500 20$ 1,275$ 5September 1500 20$ 1,275$ 3October 1400 20$ 1,190$ 4November 1400 20$ 1,190$ 5December 1500 20$ 1,275$ 3

sending (day)Data Order

Product Sold

Ordering Cost

Carrying Cost

123

jika ,Stock out cost perunit sebesar 20+X; X adalah digit terakhir NIM anda Berapa nilai EMV terkecil dan pada posisi alternatif order berapa

Pembelian Discount<3500 no Disc3500-5000 30%>5000 45%

124

UTS

Dikerjakan dengan kertas bergaris, dengan huruf balok

Waktu mengerjakan 90 menit, OPEN BOOK.

Kerjakan Sendiri-sendiri

Tentukan Critical path dan S Kurve-nya

aktivitas pendahulu Waktu Biaya (Rp.) A - 3 6 B - 5 10 C - 3 9 D A 6 18 E B 5 15 F C 8 16 G D 6 12 H F 9 27 I G,E,H 8 32

125

UAS

Kerjakan dua soal dari soal-soal berikut:

� Berapa biaya ternendah untuk penyelesaian kasus 2: Custom Vanc

Inc.

� Buatlah kurve S dari kasus 4: Hay Good Company

� Berapa persen efektifitasn yang terjadi dari perubahan layout pada

Kasus 3: New England Casting

� Pada formulasi loket seperti apa yang menghasilkan biaa antrian

terendah antrian pada kasus 6: Management Family

126

PENULIS

Setyabudi Indartono, Ph.D

Lahir di Purwokwero, 20 juli 1972. Menyelesaikan studi SD hingga SMA di Banjarnegara Jawa Tengah. Kemudian melanjutkan studi S1 di Teknis Sipil Universitas Gadjahmada dan S2 di Magister Manajemen di Universitas yang sama, Sedang S3 di dapatkan dari National Central University Taiwan. Pernah bekerja di PT Freeport Indonesia sebagai senior fasilitator/trainer, kemudian Direktur Umum dan Keuangan Rumah Sakit PKU Muhammadiyyah Bantul. Menjadi Direktur Cabang LMT Trustco sejak 1998. Kemudian menjadi Staf Pengajar/Dosen Manajeman di Universitas Negeri Yogyakarta.

Beberapa buku dan modul yang pernah ditulis adalah:

1. Steel Structure Design of PT FI apartments with Staad III Software (1995),

2. Construction Management of PT FI (1997), 3. Justice Party direct Selling (2000), 4. Management Behavior : Mentoring as Solution (2000), Business

Research Method: Memory Research (2000), 5. Yogyakarta Islamic Hospital: Managing Performance (2000), 6. Yayasan Bina Sehat: Organization Change and Developmet as a

priority need (2000), 7. Human Resource Management: Sociaty central health Bantul

Yogyakarta (2000), 8. Organization Design of Region Directorate of Justice Party of

Yogyakarta (2000), 9. PT KPI Tembagapura Compensation applications (2000), 10. SWOT (2003), 11. Advance SWOT (2003), 12. Modul TFT Trustco (2004), 13. Leadership (2005), 14. Training For Beginer (2005), 15. Smart Trainer (2005), 16. Strategic trainer (2005), 17. Marketing Advance (UNY, 2005),

127

18. Lembaga Keuangan (UNY, 2005), 19. Bahan Ajar Perkuliahan Manajemen Konflik (2013) 20. Bahan Ajar Perkuliahan Strategic Human Resourches Management

(2012) 21. Bahan Ajar Perkuliahan Metode Penelitian Bisnis (2012) 22. Bahan Ajar Perkuliahan Metodologi Riset SDM (2011) 23. Bahan Ajar Perkuliahan Manajemen Perubahan (2011) 24. Bahan Ajar Perkuliahan Perilaku Organisasi (2010) 25. Bahan Ajar Perkuliahan Teknik Proyeksi Bisnis (2010) 26. Panduan praktikum Perkuliahan Operation Research (UNY, 2009), 27. Bahan Ajar Perkuliahan Pengantar Manajemen (2009)

Journal Publication 1. Indartono & chen, 2008, Glocalization of Personal Ethical Threshold,

Journal of Education, Vol. 1. No. 1, pg. 39 2. Indartono & chen, 2008, Perception of direct and indirect compensations

fulfillment on hazardous work environment The relationship with age, tenure, employee’s rank and work status, Jurnal Siasat Bisnis, Vol. 12 No.1, pg. 13

3. Indartono, Chou & chen, 2008, The Knowledge Characteriscs Work Design Analysis of Job Fit Influence on Role Performance, Journal of Human Capital, Vol 1 No 1 pg. 81

4. Indartono, 2008, Pengaruh personal job fit terhadap hubungan desain kerja dan kinerja pengajar, Jurnal Humaniora, Vol. 13 no. 2, pg. 33

5. Indartono et al, 2009, The knowledge characteristics work design: Analysis of job fit influence on role performance, Usahawan, No. 01 vol. 38, pg. 33

6. Indartono & chen, 2009, Articulating strategic human resources management: Concept perspective to practice of managing human resources, Journal of Human Capital, Vol 1 No 3., pg.227

7. Indartono , 2009, Contribution of different organizational politics perceptions: Study on interaction among perception organizational politics, performance and trust on the role of compensation, Integritas Jurnal Manajemen Bisnis, Vol 2 no 1., pg 13

8. Indartono, 2009, Mediation effect of trust on the relationship between perception of organization politics and commitment, Jurnal Administrasi Bisnis, Vol. 5 no. 2., pg.160

9. Indartono, 2009, Different effect of Task Characteristics requirement on Job satisfaction: Gender analysis of teacher occupation on WDQ, Jurnal Ekonomika Madani,Vol 1, no. 2., pg.20

10. Indartono, Setyabudi and Vivian Chen, Chun-Hsi , 2010, Moderation of Gender on the relationship between task characteristics and performance,

128

International Journal of Organizational Innovation (IJOI), Vol. 2, no 4, Pg. 195-223

11. Indartono, Setyabudi; Chiou, Hawjeng; Vivian Chen, Chun-Hsi, 2010 The Joint Moderating Impact of Personal Job Fit and Servant Leadership on the Relationship between the Task Characteristics of Job Design and Performance, Interdisciplinary Journal of Contemporary Research in Business,Vol 2, No 8, pg 42-61

12. Indartono, Setyabudi and Vivian Chen, Chun-Hsi , 2011, Moderating Effects of Tenure and Gender on the Relationship between Perception of Organizational Politics and Commitment and Trust, South Asian Journal of Management, Vol18, no1. Pg.7-36

13. Vivian Chen, Chun-Hsi and Indartono, Setyabudi, 2011, Study of commitment antecedents: The dynamic point of view, Journal of Business Ethics, Vol. 103, No.4 , Pg.529-541 (IF2010: 1.125)

14. Indartono, Setyabudi, 2011, The Effect of E-Learning on Character Building: Proposition for Organizational Behavior Course, Jurnal Pendidikan Karakter Vol 1, No. 1, pp.59-73, LPPMP UNY

15. Indartono, Setyabudi; Nafiuddin, Yajid; Sakti K., Lingga; and Praja R. Ega, 2012, Different Perception of Gender on Workplace Spirituality: Case on School Environment, Online Journal of Education Research, Volume 1, Issue 4, Pages: 73-79 Conference proceeding

1. Indartono, Setyabudi, 2010, from statisc to dynamic perspective of behavior: case of organizational commitment”, proceeding “the First Annual Indonesia Scholars Conference in Taiwan: improving nation competitiveness by strengthening and accelerating independent reseearch”, Vol. 1 no. 1, Tainan Taiwan

2. Indartono, Setyabudi, 2009, Measuring the behavior of individual and group performance: Hierarchical linier modeling approach”, proceeding “Doctoral Program National Qolloquium” Gadjahmada University Indonesia

3. Indartono, Setyabudi, 2011, “Effect of Servant Leadership on Knowledge characteristics”, proceeding “the Second Annual Indonesia Scholars Conference in Taiwan: Becoming “Asian Tiger” through modern agriculture-based Industry : revitalization and modernization of education, technology, economy, and investment climate in agricultural sector, Vol. 2. no. 1, Taichung Taiwan

4. Indartono, Setyabudi, 2011, Acceptance and Tolerance Limit Phenomena: an Empirical Approach, International Sustainability Forum on Islamic Economic and Business, Universitas Lambung Mangkurat Indonesia

5. Indartono, Setyabudi, 2012, Reformatting Knowledge and Science Theory Building: Transcendental Point of View, proceeding “the Third Annual

129

Indonesia Scholars Conference in Taiwan: Acceleration and Development of Information and Communication Technology Research basd on Demand: Improving Sustainable Synergy of Academycs, Industry, and Government, Vol. 1.No. 1, Hsinchu Taiwan

6. Indartono, Setyabudi, 2012, Desain Kerja untuk Staf pengajar untuk mencapai Kesesuaian dan Kepuasan Kerja, Konvensi Nasional Pendidikan Indonesia VII 2012, Universitas Negeri Yogyakarta

Membership and Activities 1. Member of Forum Dosen Ekonomi dan Bisnis Islam (FORDEBI) 2011-now 2. Secreatry of board, Indonesia Committee for Science and Technology

Transfer in Taiwan (IC3T), 2010-now 3. Member of Editorial Board of International Journal of Commerce &

Accounting Research (IJCAR), 2011-now 4. Member of Editorial Board of Journal of Arts Science & Commerce

Research (RW-JASCR) , 2011-now 5. Member of Editorial Board of Asian Journal of Business Ethics (AJBE) ,

2012-now 6. Member of Editorial Board of International Journal of Organizational

Analysis (IJOA) , 2012-now 7. Coordinator of Development Division of Economic Faculty, Yogyakarta

State University, 2011-now 8. Member of Research Devision of Economic Faculty, Yogyakarta State

University, 2011-now

Tinggal dengan seorang Istri, dr. Yayuk Soraya, AAK, dan tiga anak laki-lakinya, Aiman Hilmi Asaduddin (1999), Rofiq Wafi’ Muhammad (2001), dan Muhammad Kaisan Haedar (2004) di Jl Arwana No 7 Minomartani. [email protected]

130

REFFERENCES

Winston, Wayne L., Operations Research: Applications and Algorithms, 3rd

Edition, Duxbury Press, 1994, p. 2.

http://en.wikipedia.org/wiki/Linear_programming copy at September, 7th

2013

http://www.me.utexas.edu/~jensen/ORMM/models/unit/linear/subunits/resour

ce_allocation/ copy at September, 7th 2013

http://www.dspguide.com/ch5/2.htm copy at September, 7th 2013

http://www.tutorsonnet.com/homework_help/micro_economics/product_pricin

g/linear_programming_and_its_limitations_assignment_help_online_tuto

ring.htm copy at September, 7th 2013

operation research setyabudi indartono, ph.d assistant ... · pdf filecontoh soal quiz, uts...

Documents