dr. khallel ibrahim mahmoud · 2018-01-19 · 19) assembly line balancing. 20) mrp-jit-iso-tqm.etc....

Industrial Engineering (IE) Dr. Khallel Ibrahim Mahmoud

University of Technology

Electro-mechanical Engineering Dept.

Introduction Production and Productivity Break Even Analysis(B.E)

Lec.No(1)

2011

Industrial Engineering (IE) - Dr. Khallel Ibrahim Mahmoud

Introduction: 1.1 Concept of Industrial Engineering (IE)

Industrial Engineering is concerned with the design, Improvement, and installation of integrated system of men, material, and machines for the benefit of mankind .It draws upon specialized knowledge and skills in the mathematical and physical sciences together with the principles and methods of engineering analysis and design to specify, predict and evaluate the results to be obtained from such systems.

1.2 IE Objectives

The basic objectives of Industrial Engineering are:

1- Improving operating methods and controlling costs.

2- Reducing these costs through cost reduction programs.

The aim of IE department is to provide specialized services to production departments, such as methods improvement, time study, Job evaluation and merit rating and to head new projects if required.

1.3 IE Activities

Basic activities of industrial engineering as stated to American Institute of industrial engineering as follows:

1- Processes (and methods) selection.

2- Selection and design of tools and equipment.

2

Industrial Engineering (IE) - Dr. Khallel Ibrahim Mahmoud 3- Facilitates planning, plant location, materials handling and storage facilities.

4- System design for planning and control of production inventory, quality, and plant maintenance and distribution.

5- Cost analysis and control.

6- Develop time standards and performance standards.

7- Value engineering and analysis system design and install.

8- Mathematical tools and statistical analysis technia.

9- Performance evaluation.

10- Project feasibility studies.

1.4 IE Approach

Industrial engineering department uses scientific approaching identifying and solving the problems .It collects factual information regarding the problem analysis the problem, prepares alternative solutions taken into account all the internal and external constraints, selects the best solution for implementation.

This stage is called problem identification .It consists of the following steps:

1) Collect all details about the job, using standard recording techniques like charts, diagrams, models and templates.

2) Recorded facts are subjected to critical examination using a series of questions.

3) Find alternative solutions for the problem.

3

Industrial Engineering (IE) - Dr. Khallel Ibrahim Mahmoud 4) Evaluate the alternatives and find the best solution.

Next the industrial engineering department makes recommendation for the implementation of the best alternatives so that the 1.5 Techniques of Industrial Engineering

The main aim of tools and techniques of industrial engineering is to improve the productivity of the organization by optimum utilization of organizations resources: men, materials, and machines. The major tools and techniques used in industrial engineering are:

1) Production planning and control.

2) Inventory control.

3) Job evaluation.

4) Facilitates planning and material handling.

5) System analysis.

6) Linear programming.

7) Simulation.

8) Network analysis (PERT, CPM).

9) Queuing models.

10) Assignment.

11) Sequencing and transportation models.

12) Games theory and dynamic programming.

13) Group technology.

4

Industrial Engineering (IE) - Dr. Khallel Ibrahim Mahmoud 14) Statistical techniques.

15) Quality control.

16)

17) Decision making theory.

18) Replacement models.

19) Assembly line balancing.

20) MRP-JIT-ISO-TQM.etc.

1.6 The six standards phases of IE:-

1- Formulating the problem.

2- Constructing a mathematical model to represent the system under study.

3- Deriving a solution from the model.

4- Testing the model and the solution derived from it.

5- Establishing controls over the solution.

6- Putting the solution to work: Implementation.

2- Productivity The standard of living of industrialized nations depends upon the economic efficiency of all its industrial enterprise great or small.

2.1 Introduction/Definition of productivity

5

Industrial Engineering (IE) - Dr. Khallel Ibrahim Mahmoud Productivity can be defined as the ratio of output in a period of time to the input in the same period time. Productivity can thus be measured as:

Productivity =

In simple terms productivity is the quantitative relationship between what we produce (output) and the resources (inputs) which we used.

2.2 Productivity and production:

Production is the process of converting the raw materials into finished products by performing a set of manufacturing operations in a predetermined sequence. Production refers to absolute output. Thus, if the input increases the output will normally increase in the same proportion. The productivity remains unchanged if however the output increases with the input of the resources, the productivity increases production means the output in terms of money without any regard to the input of resources , which productivity is a human attitude to produce more and more with less input of resources. There are six cases to increase the productivity as shown

output + + c ++% -% input - c - +% --%

2.3 Types of production systems:

6

Industrial Engineering (IE) - Dr. Khallel Ibrahim Mahmoud A production system consists of plant facilitates, equipment and operating methods arranged in a systematic order. This arrangement depends upon the type of product and the strategy that a company employs to serve its customers. There are two major types of production systems:

i) Make to stock production.

ii) Make to order production.

In make to stock production the products are manufactured kept as ready stock and supplies to customers as orders are

operatio

input output

system

Examples of such items are nuts, bolts, bearings, screws, etc.

In make to order production the products are made only

Types of production

The production or manufacturing systems are classified as follows:

a) Job type production.

b) Batch production.

c) Continuous or mass production.

a) Job type production.

It is characterized by high variety, low volume production, producing one or a few products specially designed and produced according to customer specifications like aircraft, ships, special train.

b) Batch production.

7

Industrial Engineering (IE) - Dr. Khallel Ibrahim Mahmoud Here the product similar in design but different in size and capacities are produced in batches of one size and capacity at one time, at regular interval and stocked at warehouses a waiting sales. Examples as pumps, motor etc, of different capacities and types manufactured in batches.

c) Mass (Repetitive) production.

This production is characterized by high volume low variety in this system several standard products are produced in large quantity and stocked in warehouse awaiting dispatch examples of such production, nuts bearings, t-shirts, etc.

The following fig. shows volume variety relation for different types of manufacturing systems.

Low Production volume High

Job type

Batch production

Mass Production

High

Variety

Low

After this brief introduction to types of production system and productivity let us discuss how the productivity measured and improved.

8


2.4 Measurement of productivity

Productivity measures:

There are three major types of productivity measures as listed below:

1- Partial productivity

2- Total factor productivity

It is the ratio of net output to the sum of associate …..

3- Total productivity:

Tangible means measurable for total tangible input =value of human, material, capital, energy and other inputs used.

Man power material

9


P

Machines 5- Factors affecting productivity:

1. Raw material, its nature and quality.

2. Utilization of manpower.

3. Utilization of plant, equipment and machinery.

4. Basic nature of manufacturing processes employed.

5. Efficiency of plant.

6. Volume, capacity, and uniformity of production.

6- The ways in which the productivity can be increased summarized as under:

1) Increase manpower effectiveness at all levels.

2) Method improvement.

3) Improve basic production processes by research and development.

4) Use better production equipment.

5) Improve / simplify product design and reduce variety.

6) Better production planning and control.

2.1 Productivity Improvement Techniques

10

Industrial Engineering (IE) - Dr. Khallel Ibrahim Mahmoud 1) Technology based

a. CAD/CAM/CIMS

b. Robotics.

c. Laser technology.

d. Modern maintenance technology.

e. Energy technology.

f. Flexible manufacturing system (FMS).

2) Employee based

a. Incentives

b. Promotion

c. Job design

d. Quality circle

3) Material based

a. Material planning and control

b. waste elimination

c. Recycling and reuse of waste materials

d. Purchasing Logistics

4) Process based

a. Method engineering and work simplification

b. Process design

11

Industrial Engineering (IE) - Dr. Khallel Ibrahim Mahmoud c. Human factors engineering

5) Product based

a. Reliability engineering

b. Product mix and promotion

c. Value analysis/value engineering.

6) Management based

a. Management technique

b. Communication

c. Work culture

d. Motivecation

e. Promoting group activity

Questions:

1. Write short notes on:

a) Techniques of Industrial engineering (Ans. 1.5)

b) Productivity measurement models (Ans. 2.4)

2. Define the term productivity. How is it different from production? Give examples using your own numb (Ans. 2.1, 2.2)

3. There are six cases to increase the production. Explain in brief. (Ans. 2.2)

12

Industrial Engineering (IE) - Dr. Khallel Ibrahim Mahmoud 4. Summarize the ways in which the productivity can be increased. (Ans. 2.6)

5. Discuss the factors that affect productivity. (Ans. 2.5)

6. Write short notes on productivity improvement techniques. (Ans. 2.7)

3- Break-Even analysis (B-E)

Break even production …….

At which the production cost equals income from sales. By this value the company sets profit and below this value suffers a loss.

Fig. Break-Even Point (B.E.P)

Cost

Production volume

Revenue curve

Profit

Loss

Total cost

Total v.cos

F.cost

Cost curve

B.E.P

Q

3.1 Steps in B.E.P

To construct break-even point, we must know:

13


- Total fixed expenses at a certain target production. - Total variable cost at the same target production. - Total sales value at the same target production.

We find that the initially zero production rate, fixed cost will remain as it is variable cost will be zero. The company still suffers a loss equal to its fixed cost. As production volume increases this loss decreases but the break-even point…… 3.2 Assumptions

a) All the units remains fixed for any production volume.

b) Variable cost increase is linear.

c) Selling prices will remain constant at all levels.

d) Production and sales quantities are equal.

3.3 Formulation of linear Break-even model

This will define the minimum quantity that should be produced without any loss or profit.

Notations:

Let Q: the quantity sold

b: price (the income per unit)

R: bQ (Revenue or income)

F: fixed cost

v: variable cost per unit

p: profit

Tc: total cost = F + vQ

14

Industrial Engineering (IE) - Dr. Khallel Ibrahim Mahmoud P= R-Tc

Quantity increase in price by making:

- Better product. - Advertisement. - New product limited by market price.

Increase in planned quantity: Increase share of market and increase those products with high profit or increase share of market with the increase price according to quantity of market.

Ref: - M.I.Khan , Industrial Engineering ,2nd Edition ,2008

- Maynard,H.B ,”Industrial Engineering Handbook ,new York ,2004

15





Lec.No(2)

Linear Programming Model

2011


Linear Programming

LP is a mathematical modeling technique designed to optimize the usage of limited resources, such as available materials, labour and machine time.

The LP model includes three basic elements:

- Decision variables that we seek to determine. - Objective (goal) that we aim to optimize. - Constraints that we need to satisfy.

Steps in formulating LP problems:

1- Define the objective. 2- Define the decision variables. 3- Write the mathematical function for the objective (objective function). 4- Write a one- or two- word description of each constraint. 5- Write the right – hand side (RHS) of each constraint, including the unit of

measure. 6- Write ≤, = or ≥ for each constraint. 7- Write all the decision variables on the left-hand side of each constraint. 8- Write the cofficient for each decision variable in each constraint.

Formulation of linear programming model (LP)

The general form of each model will be:

Z= c1x1+ c2x2+……. ckxk

Subject to:

a11x1+ a12x2+……. a1kxk b1

a21x1+ a22x2+……. a2kxk b2

am1x1+ am2x2+……. amkxk b1


where: Cj is a known (cost or profit) coefficient Xj.

Xj is an unknown variable.

aij is a known constant.

bj is a known constant.

≤, = or ≥ for each constraint.

Example: A manager wants to know many units of each product to produce on a daily basis in order to achieve the highest contribution to profit. Production requirement for the products are shown in the following table

Production Departments I II

Processing time required for the first product

(Hours)

4 2

Processing time required for the second product

(Hours)

2 4

Production capacity available (Hours)

60 48

The profit is £8 for each unit of the first product and £6 for each unit of the second product

Solution:

1- Define the objective. The problem is a maximum problem. 2- Define the decision variables. We need to determine the number of units to

be produced. Let: Xi be the number of units of type i (i= 1,2)

Therefore : X1= number of units of the first product.

X2= number of units of the second product.


X1, X2 are the decision variables, when we know their values the problem will be solved.

3- The objective function for the LP is:

Maximize Z = 8X1+6X2

This means the profit Z depend on how many units (X1, X2) are manufactured. Z= c1x1+ c2x2 where c1, c2 are the respective profits for each type of product. c1=£8 c2=£6 and Z = 8X1+6X2 and we should select values of the decision variables X1, X2 that result in the maximum value of Z.

4- There are two constraints :

Maximum production capacity for Dep.I ≤ 60 hours

Maximum production capacity for Dep.II ≤ 48 hours

( Note: because all the constraints in this problem are maximum capacity , all constraints are the ≤ type)

Now we have to write the coefficient for each decision variable in each constraint.

The two constraints can be expressed as:

4X1+2X2 ≤ 60 constraint Dep.I

2X1+4X2 ≤ 48 constraint Dep.II

Consider the first constraint ( Dep.I) what is the coefficient of X1 in this constraint? It is the processing time (Hours) required per unit of X1. In other word, it is the processing time used in manufacturing each unit, first product, or 2 hours. Similarly, the coefficient of X2 in this first constraint is 1 hour.

Therefore the two constraints can be expressed as :

4X1+2X2 ≤ 60

2X1+4X2 ≤ 48

And the no. negatively restriction is all Xj≥ 0 , X1, X2 ≥ 0


Graphical LP solution:

Steps in the graphical method

1- Formulate the objective and constraint functions. 2- Draw a graph with one variable on the horizontal axis and one on the

vertical axis. 3- Plot each of the constraints as if they were lines or equlities. 4- Outline the feasible solution space. 5- Circle the potential solution points .These are the intersections of the

constraints or axes on the inner (minimization) or outer (maximization) perimeter of the feasible solution space.

6- Substitute each of the potential solution point values of the two decision variables into the objective function and solve for Z.

7- Select the solution point that optimizes Z.

Solution of maximization model

To demonstrate the steps of the graphical solution of a maximization problem we use the previous example:

Z = 8X1+6X2

Subject to:

4X1+2X2 ≤ 60

2X1+4X2 ≤ 48

X1 ≥ 0 , X2 ≥ 0

Solution:

1- Plot the constraints (shown in the following figure) change constraints to equalities:


4X1+2X2 ≤ 60 4X1+2X2 = 60

2X1+4X2 ≤ 48 2X1+4X2 = 48

For each constraint ( set X1=0 and solve for X2 , then set X2=0 and solve for X1)

the graph the constraint as if it were an equality

4X1+2X2 = 60 X1=15 X2=0

X1=0 X2=30

2X1+4X2 = 48 X1=24 X2=0

X1=0 X2=12

2- Outline the feasible solution space the values of X1 and X2 at points M,A,B and C are four potential solutions to problem.

Note: point B can be determined as follow:

4X1+2X2 = 60

(2X) 2X1+4X2 = 48

4X2+2X2 = 60

4X2+8X2 = 96

6X2=36

X2=36/6 = 6

Then: 4X1+2X2 = 60

4X1+2(6) = 60

4X1 = 48

X1= 48/4 = 12

Points X1 X2 Z M 0 0 Z=8(0)+6(0)=0


A 15 0 Z=8(15)+6(0)=120B 12 16 Z=8(12)+6(6)=132C 0 12 Z=8(0)+6(12)=72

To maximize Z, the optimal solution is point B, where X1=12 and X2=6 and Z=£132 profit.

B

X1

A

X

Feasible solution space

4X1+2X2 ≤ 60

2X1+4X2 ≤ 48

C

M

Minimization case:

Consider the following LP problem:

Min Z = 3X1+8X2

Subject to:

X1≤ 80

X2≥ 60

X1+ X2=200

X1, X2 ≥ 0

Solved problem: X1=80 X2=60

X1=0 X2=200

X1=200 X1=0


B/ X1+ X2=200

X1=80

Then X2=200-80

=120

Point X1 X2 Z A 0 200 3(0)+8(200)=1600 B 80 120 3(80)+8(120)=1200

Min Z=1200

X1=80 X2=120

X2≥60

X

201612

40

40

80

80

12

16

XA

X1≤80

X1+ X2=200

B

20

Special cases in LP

Five special cases and difficulties arise at times when using the graphical approach to solve LP problems:

1) Infeasibility: infeasibility is a condition that arises when there is no solution to LP problem that satisfies all of constraints given.


Graphically, it means that no feasible solution region exists- a situation that might occur if the problem was formulated with conflicting constraints.

Let us consider the following three constraints: X1+2X2 ≥6

2X1+X2 ≥8

X1 ≥7

X

Region satisfying 3rd constraint

Region satisfying first 2 constraints

X8 6 4 2

8

6

4

2

As seen in the figure there is no feasible solution region for this problem because of the presence of conflicting constraints.

2) Unboundedness:

Sometimes a linear program will not have a finite solution. This means that in a maximization problem, for example, one or more solution variables, and the profit, can be made infinitely large without violating any constraints. If we try to solve such a problem graphically, we will note that the feasible region is open-ended.

Let us consider a simple example to illustrate the situation.

Z = 3X1+5X2

Subject to: X1≥ 5


X2≤ 10

X1+2 X2≥10

X1, X2 ≥ 0

X

X2≤10

Feasible Region

X1≥ 5

X1+2 X2≥10

15

10

5

X5 10 15 20

As you see, because this is a maximization problem and the feasible region extends infinitely the right, there is unboundedness or unbounded solution.

3) Redundancy: The presence of redundant constraints occurs in large LP formulations, a redundant constraint is simply on that does not affect the feasible solution region.

Let us kook at the following example:

Max Z= X1+2X2

Subject to: X1+X2≤ 20

2X1+X2≤ 30

X1≤ 25


X1, X2 ≥ 0

X

X1≤25

Redundant

Feasible

X1+ X2≤20

2X1+ X2≤30

X

35

30

25

20

15

10

5

5 10 15 20 25 30

The third constraint, X1≤25 is redundant and unnecessary in the formulation and solution of the problem because it has no effect on the feasible region set.

4) Alternate Optimal Solutions:

An LP problem may on occasion, have two or more alternate optimal solutions. Graphically, this is the case when the objective function’s isoprofit or isocost line runs perfectly parallel to one of the problem’s constraints.


Ex: Max Z= 3X1+2X2

Subject to: 6X1+4X2≤ 24

X1≤ 3

X1, X2 ≥ 0

1

1

2

2

3

3

4

4 5

5

A

Isoprofit line for 12/line segment AB

6

B

Optimum solutions consists of all combinations of X1,X2 along the AB

X

Isoprofit line for 8

X

6

As you see any point along the line between A and B provides an optimal X1 and X2 combination Z=12

5) Degenercy:

Degenercy is a condition that arises when one of the decision variable equal zero. Look to the following example:


Max Z = 3X1+9X2

Subject to: X1+4X2≤ 8

X1+2X2≤ 4

X1, X2 ≥ 0

Solution: X1=8 X2=2

X1=4 X2=2

Then: at point A

X1=0 X2=2 Max Z=18

5

XB

A

4

3

2

1

1 2 3 4 5 6 7 8 Problems:

Q1/ Montana wood products manufacturers two high-quality products, chairs and bookshelf units. Its profit is $15 per chair and $21 per bookshelf unit. Next week’s production will be constrained by two limited resources, labor and wood. The labor available next week is expected to be 920 labor hours, and the amount of wood available is expected to be 2400 board feet. Each chair requires 4 labor hours and 8 board feet of wood. Each bookshelf unit requires 3 labor hours and 12 board feet of wood. Management would like to produce at least 100 units of each product.


a- To maximize total profit, how many chairs and bookshelf units should be produced next week?

b- How much profit will result?

Q2: solve graphically the following linear programming problem:

Max Z = 60X1+40X2

Subject to: 2X1+X2≤ 60

X1≤ 25, X2≤ 35

X1, X2 ≥ 0

Q3: solve graphically:

Max Z = 10X1+15X2

Subject to: 2X1+X2≤ 26

2X1+4X2≤ 56

X1-X2≥-5

X1, X2 ≥ 0

Q4: Consider the following problem and solve graphically:

Minimize Z = 2X1+4X2

Subject to: X1+X2≤ 14

3X1+2X2≥ 30

2X1+X2≤18

X1, X2 ≥ 0

Q5: Tellitell Television company operates two assembly lines, line I and line II. Each line is used to assemble the components of three types of televisions: colour, standard and Economy. The expected daily production on each line is as:

TV Model Line I Line II


Colour 3 1 Standard 1 1 Economy 2 6

The daily running costs for two lines average £6000 for line I and £4000 for line II. It is given that the company must produce at least 24 colour, 16 standard, and 48 economy TV sets for which an order is pending.

You are required to formulate the above problem as (LP) taking the objective function as minimization of total cost. Also determine the number of days that the two lines should be run to meet the requirements.

Q6: Suppose two types of television sets are produced with a profit of 6 units from each television of type II. In addition 2 and 3 units of raw materials are needed to produce one television of type I and II respectively. And 4 and 2 units of time are required to produce one television of type I and II respectively. If 100 units of raw materials and 120 units of time are available. How many units of each type of television should be produced to maximize profit?

Q7: A wood product firm uses available time at the end of each week to make goods for stocks. Currently two products on the list of items are produced for stock: a chopping board and a knife holder. Both items require three operations: cutting, gluing, and finishing.

The manager of the firm has collected the following data on these products:

Time per unit (minutes) Item Profit/unit Cutting Gluing Finishing

Chopping board

$2 1.4 5 12

Knife holder $6 0.8 13 3

The manager has also determine that during each week, 56 minutes are available for cutting, 650 minutes are available for gluing and 360 minutes are available for finishing.


a- Determine the optimal quantities of decision variables. b- Which resources are not completely used by your solution?

How much of each resource is unused?

Q8: solve the following problem using graphical linear programming.

Minimize Z = 2X1+3X2

Subject to: 4X1+2X2≥ 20

2X1+6X2≥ 18

X1+2X2≤12

X1, X2 ≥ 0

Solutions: Lec. No.2

Q1: Let X1: number of chairs to produce next week.

X2: number of bookshelves to produce next week.

Max Z = 15X1+21X2

Subject to: 4X1+3X2≤920 …….. (1)

8X1+12X2≤ 2400 …….. (2)

X1≥100 X2≥100

X1, X2 ≥ 0

X1=0 X2=306.6

X2=0 X1=230 …………….. (1)

X1=0 X2=200

X2=0 X1=300 …………… (2)

Z= 4350 with X1=150 X2=100


Point X1 X2 Z A 100 100 3600 B 100 133.3 4299.3 C 150 100 4350

50 10 15 20 25

B

CA

X

30

350

300

250

20

15

100

50 X

Q2: Max Z = 60X1+40X2

Subject to: 2X1+X2≤60

X1≤25 X2≤35

X1, X2 ≥ 0


X1=0 X2=306.6

X2=0 X1=30

Point X1 X2 Z M 0 0 0 A 0 35 1400 B 12.5 35 2150 C 25 10 1900 D 25 0 1500

B: 2X1+X2= 60 X2=35 X1= (60-35)/2 = 12.5

It is clear that Z equal to 2150 is Maximum at B when X1=12.5 X2=35.

X

X2≤35

X1≤25

C

B

2X1+X2≤60

DX

60

50

40

A

30

20

10

M

5 10 15 20 25 30

Q3: Max Z = 10X1+15X2

Subject to: 2X1+X2≤26

2X1+4X2≤ 56

X1-X2 ≥-5 -X1+X2 ≤5


X1, X2 ≥ 0

X1=-5 X2=5

Point X1 X2 Z M 0 0 0 A 0 5 75 B 6 11 225 C 8 10 230 D 13 0 130

Hence, Z is Maximum (230) at C.

4

4

8

8

12

12

16

16 18

20

X

-8 -4 M

28

X

A

24

B

2X1+4X2≤ 56

C

2X1+X2≤26 -X1+X2 ≤5

D

20

24 28

Q4: Min Z = 2X1+4X2

Subject to: X1+X2≤14

3X1+2X2≥30

2X1+X2≤18


X1, X2 ≥ 0

Point X1 X2 Z A 2 12 52 B 4 10 48 C 6 6 36

Thus, Z is Minimum at point C when X1 , X2=6 Z=36 .

2

2

4

4

6

6

8

8 10

10

12

B

12

14

18

16

X

A

C

X1+X2≤14

3X1+2X2≥30

2X1+X2≤18

14

X

16 18


Q5: Let X1= the number of days the line I is run.

X2= the number of days the line II is run.

Min Z = 6000X1+4000X2

Subject to: 3X1+X2≥24

X1+X2≥16 Production Requirement

2X1+6X2≥48

X1, X2 ≥ 0

Point X1 X2 Z A 0 24 96000 B 4 12 72000 C 12 4 88000 D 24 0 144000

Thus, to Minimize cost, line I should be run for 4 days and line II for 12 days.

4

8

12

16

20

B

24

X1+X2≥16

3X1+X2≥24

28

X

C

A


2X1+6X2≥48

D X

4 8 12 16 20 24

Q6: let X1: be the number of units of type I produced.

X2: be the number of units of type II produced.

Max Z = 6X1+4X2

Subject to: 2X1+3X2≤100 (raw material time)

4X1+2X2≤ 120

X1, X2 ≥ 0

The optimum solution at point C (X1=20, X2=30) Hence, 20 units of type I and 20 units of type II should be produced to yield a maximum profit of Z=6(20)+4(20)=200

10

20

30

40

50

B

60

C

80

D

70

X2

X


A

1 2 30 50 40 70 60 80

Q7: Let board = X1 holder= X2

Max = 2X1+6X2

Subject to: 1.4X1+0.8X2≤56 …………… (1)

5X1+13X2≤ 650 ………………. (2)

12X1+3X2≤ 360 ………………. (3)

a) The solution at point A X1=0 X2=50

Z= 2(0) +6(50) = 300

b) Cutting: 56- 0.8(50) = 15 minutes. (40,0) (0,70) ………… (1)

Gluing: 13X50=650 650-650=0 (130,0) (0,50) …………… (2)

Finishing: 3X50=150 360-150=210 minutes. (30,0) (0,120) ………… (3)

X

20

40

60

80

10

B

120

A

140

C

X


D 20 4 60 1080 1412 16

Q8: The optimum solution X1=4.2 and X2=1.6

Minimum Z=2(4.2)+3(1.6) = 13.2





Lec.No(3)

Linear Programming Model

Simplex Method

2011


Simplex Method The linear programming situation with two decision variables is easy tackled by a graphical method of solution. However most practical linear programming problem contain more than two decision variables. Graphic representation is difficult (three dimensional) and requires more time for the determination of an optimal solution. The solution method in such cases is the simplex method. The simplex method is an iterative procedure that consists of moving from one basic feasible solution to another in such a way that the value of the objective function does not decrease (in the maximization problem). This process continue until an optimal solution is reached, if one exist.

The steps required to solve the problem – maximizing are:

Step 1: Define the problem in standard form. This indicates the construction of a linear programming model.

Step 2: convert the inequalities into equation by insert slack variables.

Step 3: Construct a matrix of the coefficients of these equations.

Step 4: This step amends the previous feasible solution so as to improve the profit.

The simplex method will be illustrated with the following example.


Example/ Consider the following LP:

Maximize Z= 6X1+8X2

Subject to:

2X1+5X2≤ 40

8X1+4X2≤ 80

X1,X2 ≥ 0

Step 1/ Conversion of inequalities into equations Add slack variables to remove inequalities

2X1+5X2 +S1 = 40

8X1+4X2 + S2= 80

Where S1,S2 slack variables (surplus) putting X1 =0 and X2 =0 we get

S1 = 40 , S2= 80

Max Z =6X1+8X2 +oS1 + oS2

Subject to:

2X1+5X2 +S1 + oS2= 40

8X1+4X2 + oS1 + S2= 80


Step 2/ Construct the simplex tableau. The standard form can be summarized in a compact tableau form as:

Coefficient of OF

variables in basis

Basic variables

Values of basic

variables

Coefficient of OF variables

(Cj)

( Cj) (A) (q) 6 X1

8 X2

0 S1

0 S2

0 S1 40 2 5 1 0 0 S2 80 8 4 0 1 Zj 0 0 0 0 0 Cj-Zj 6 8 0 0

Initial solution: X1 = 0

X2 = 0

S1 = 40

S2 = 80

Z = 0

Zj:

Zj (q) = 40(0) + 80(0) = 0

Zj (X1) = 2(0) + 8(0) = 0

Zj (X2) = 5(0) + 4(0) = 0

Zj (S1) = 1(0) + 0(0) = 0

Zj (S2) = 0(0) + 1(0) = 0

Cj – Zj :


(Cj – Zj) X1 = 6-0 =6 , (Cj – Zj) S1 = 0-0 =0

(Cj – Zj) X2 = 8-0 =8 , (Cj – Zj) S2 = 0

Step 3/ identify the pivot column

X2 has max positive value (8) Thus the entering variable will be X2 in the new solution

Step 4/ identify the pivot row

, , the smallest value = 8

Thus, S1 , is the leaving variable. Replacing the leaving variable S1 with the entering variable X2 produces the new basic solution (X2, S2) and the pivot element = 5

Types:

1. Pivot row : new pivot row = current pivot row ÷ pivot element

2. All other rows , including Z:

new row = (Current row) – (its pivot column coefficient)X(new pivot row)

Type 1: computation is divide the pivot row (S1 – row) by the pivot element (5) .Thus the new row:


Type 2: Computation is applied to the remaining row (S2- row) as follow:

80 – (4X8) = 48

8 – (4X 2/5) = 32/5

4- (4X1) = 0

0 – (4X1/5) = -4/5

1 – (4X0) = 1

The new tableau corresponding to the new basic solution (X2,S2) thus becomes:

( Cj) (A) (q) 6 X1

8 X2

0 S1

0 S2

8 X2 8

1

0

0 S2 48

0

1

Zj 64

8

0

Cj-Zj 0

0

Observe that the new tableau yields the new basic solution (X2 =8 , S2= 48 ) with the new value of Z = 64 . An examination of the last tableau shows that it is not optimal solution because the variable X1 has a positive coefficient in the (Cj-Zj) row ( )


An increase in X1 is advantageous because it will increase the value of Z. Thus , X1 is the entering variable.

Next, we determine the leaving variables as follow :

8 ÷ = 20 , 48 ÷ = , Thus the pivot element =

S2 is the leaving variable , X1 is the entering variable . and the new pivot row :

48 ÷ = , ÷ = 1 , 0 , ÷ =

new X2 – row : 8- ( X )= 5

- ( X 1) = 0

1- ( X0 ) = 1

- ( X ) =

These Computations produce the following tableau:

( Cj) (A) (q) 6 X1

8 X2

0 S1

0 S2

8 X2 5 0 1

6 S2

1 0

Zj 85 6 8

Cj-Zj 0 0


Since none of the Cj-Zj row coefficient associated with the basic variables is positive , the last tableau is optimal.

The optimum solution can be read from the simplex tableau in the following manner

Decision variables Optimum value X1 15/2 = 7.5 X2 5 Z 85

The M- Method ( Big – M) In the previous example , starting the simplex iterations at a basic feasible . For the LPs in which all the constraints are of the (≤) type, the stacks offer a convenient starting basic feasible solution. A natural question then arises : How can we find a starting basic solution for models that involve (=) and (≥) constraints?

The most common procedure for starting LPs that do not have convenient slacks is to use artificial variables and the closely related method is proposed for effecting this result : the M-Method (Big- M).

The M-Method starts with the LP in the standard form. For any equation (i) that does not have a slack , we augment an artificial variable (Ai) and assign them a penalty in the objective function to force them to zero level at a later iteration of the simplex algorithm.


Given M is a sufficiently large positive value, the variable Ai is penalized in the objective function using -MAi in the case of maximization and +MAi in the case of minimization.

The following example provides the details of the method:

Min Z = 3X1 + 8X2

Subj. to:

X1≤ 80, X2 ≥ 60 , X1+ X2 = 200 , X1, X2 ≥ 0

Step1/ for convenient inequalities into equalities:

X1+ S1= 80

X2- S2 = 60

X1+ X2 = 200

Putting X1, X2 = 0 we get S1= 80 S2 =- 60

Therefore, we introduce artificial variables and the above constraints can be written as:

X1+ S1= 80

X2- S2 + A1 = 60

X1+ X2 + A2 = 200 A1 , A2 Artificial variables

Now addition of this artificial variable destroy the equality required by the L.p model . Therefore, A1 , A2 must not appear in the final solution . To achieve this it is assigned a very large penalty (TM) since Z is to be minimized in the objective function . Therefore we get:

X1+ S1+0S2+0A1+0A2= 80


X2+ 0S1-S2+A1+0A2= 60

X1+ X2+0S1+0S2+0A1+A2= 200

Min Z = 3X1+ 8X2+0S1+0S2+MA1+MA2

( Cj) (A) (q) 3 X1

8 X2

0 S1

0 S2

M A1

M A2

0 S2 80 1 0 1 0 0 0 M A1 60 0 1 0 -1 1 0 A2 200 1 1 0 0 0 1

M Z 260M M 2M 0 -M M M Cj-

Zj 3-M 8-2M 0 M 0 0

Step 2/ identify the pivot column:

This is done by selecting the none basic variable having the largest negative value in Cj-Zj .

When all the elements in the Cj-Zj row are positive or zero the optimal solution is reached .

Step3/ identify the pivot row:

80/0 = 0 , 60/1 = 60 , 200/1 = 200

Pivot = 1 A1 X1

X2/ 60/1 , 0 , 1/1 , 0 , -1/1 , 1, 0

Row S1: will be remain , Row A2 /200-(1X60) = 140

1- (1X0) = 1

1-( 1X1) = 0

0- (1X0) = 0


0-(1X-1) = 1

0-(1X1) = -1

1-(1X0) = 1

( Cj) (A) (q) 3 X1

8 X2

0 S1

0 S2

M A1

M A2

0 S1 80 1 0 1 0 0 0 8 X2 60 0 1 0 -1 1 0 M A2 140 1 0 0 1 -1 1 Zj 480+

140M M 8 0 M-8 8-M M

Cj-Zj

3-M 0 0 8-M 2M-8

0

As Cj-Zj is positive under same column the current basic feasible solution is not optimal and needs to be improved .

Step4/ Identify the pivot row after identify the pivot column

80/1 = 80 , 1 , 0 , 1 , 0 , 0 , 0

Row X2 will remain

Row A2: 140 – (1X80) = 60

1- (1X1) = 0

0 – (1X0) = 0

0 – (1X1) = -1

1 – (1X0) = 1

-1 – (1X0) = -1

1 – (1X0) = 1


Step5/ create the new tableau

( Cj) (A) (q) 3 X1

8 X2

0 S1

0 S2

M A1

M A2

3 X1 80 1 0 1 0 0 0 8 X2 60 0 1 0 -1 1 0 M A2 60 0 0 -1 1 -1 1 Zj 720+

60M 3 8 3-M M-8 8-M M

Cj-Zj

0 0 M-3 8-M 2M-8

0

As Cj-Zj is positive under same column the current solution is not optimal and need to improve.

Step6/ identify the pivot column , and pivot row.

80/0 , 60/-1 , 60/1 = 60 A2 S1

S2: 60 , 0 , 0 , -1 , 1 , -1 , 1

Row X1 : will be remain

Row X2:60- (-1X60) = 120

0- (-1X0) = 0

1- ( -1X0) = 1

0- (-1X-1) = -1

-1 – (-1X1) = 0

1- (-1X-1) = 0


0 – (-1X1) = 1

( Cj) (A) (q) 3 X1

8 X2

0 S1

0 S2

M A1

M A2

3 X1 80 1 0 1 0 0 0 8 X2 120 0 1 -1 0 0 1 0 S2 60 0 0 -1 1 -1 1 Zj 1200 3 8 -5 0 0 8 Cj-

Zj 0 0 5 0 M M-

8

As Cj-Zj is either positive or zero under all column, the solution is an optimal solution.

X1 = 80 , X2 = 120 , S2 = 60 , Z= 1200





Lec.No(4)

The Dual Model The assignment model

2011


The dual Problem

The LP model we develop for a situation is referred to as the primal problem. The dual problem is a close relates mathematical definition that can be derived directly from the primal problem.

Consider a linear programming problem concerned with the maximization of an objective function Z with n decision variables and M constraints (primal). The dual this problem is concerned with minimization of the value of the objective function Ź with M decision variables and n constraints. Thus a maximization problem becomes minimization in the dual and vice verse.

In the mathematical form the primal can be stated as:

Max Z= C1X1+ C2X2+ ………. +CnXn

Subject to:

a11x1 +a12x2+ ……………….. + a1nxn≤ b1

a21x1 +a22x2+ ……………….. + a2nxn≤ b2

am1x1 +am2x2+ ……………….. + amnxn≤ bm

X1≥ 0 X2≥ 0 ………… Xn ≥ 0

The dual of this problem may be expressed in the following form:

Min Ź= b1y1+ b2y2+ ………. +bmym

Subject to:

a11y1 +a12y2+ ……………….. + am1yn≥ c1

a12y1 +a22y2+ ……………….. + am2ym≥ c2


a1my1 +a2ny2+ ……………….. + amnym≥cn

y1≥ 0 y2≥ 0 ………… ym ≥ 0

the variables and constraints of the dual problem can be constructed symmetrically from the primal problem as follows:

1- A dual variable is defined for each of the m primal constraint equations. 2- A dual constraint is defined for each primal of the n primal variables. 3- The left-hand-side coefficients of the dual constraint equal the constraint

(column) coefficient of the associated primal variable. Its right – hand side equals the objective coefficient of the same primal variable.

4- The objective coefficient of the dual equal the right- hand side of the primal constraint equations.

The following examples demonstrate the implementation of these rules.

Ex/ Consider the primal model:

Max Z =6X1 + 8X2

St.to: 2X1+5X2≤40

8X1+4X2≤80

X1 , X2 ≥ 0

The dual of this problem as:

Min Ź = 40Y1+80Y2

St. to: 2Y1+8Y2≥6

5Y1+4Y2≥8

Y1 , Y2 ≥ 0

The solution of the dual problem can be obtained by using simplex minimization procedures.

Min Z = 40 Y1+80 Y2+oS1+oS2+MA1+MA2


Subject to: 2y1+8y2- s1+A1 = 6

5y1+ 4y2- s2+ A2 = 8

40 80 0 0 M M Cj A q Y1 Y2 S1 S2 A1 A2

M A1 6 2 8 -1 0 1 0 M A2 8 5 4 0 -1 0 1 Zj 14M 7M 12M -M -M M M Cj-Zj 40-7M 80-12M M M 0 0

80 Y2 ¾ ¼ 1 -1/8 0 1/8 0 M A2 5 4 0 ½ -1 -1/2 1 Zj 60+5M 20+4M 80 1/2M-10 -M 10-1/2M M Cj-Zj 20-4M 0 10-1/2M M 1/2M-10 0

80 Y2 7/16 0 1 -5/32 1/16 5/32 -1/16 40 Y1 5/4 1 0 1/8 -1/4 -1/8 1/4

Zj 85 40 80 -7.5 -5 7.5 5 Cj-Zj 0 0 7.5 5 M-7.5 M-5

:آان آما يلي) primal(حل النموذج األولي

6 8 0 0 Cj A q X1 X2 S1 S2

8 X2 5 0 1 ¼ -1/16 6 X1 7.5 1 0 -1/8 5/32 Zj 85 6 8 5/4 7/16 Cj-Zj 0 0 -5/4 -7/16

Dual Primal S1 = 7.5 X1 = 7.5 S2 = 5 X2 = 5

Y1 = 5/4 S1 = 5/4 Y2 = 7/16 S2 = 7/16 Ź = 85 Z = 85

Ex1/ Obtain the dual of the following LP:

Max Z =5X1 + 10X2+8X3

St.to: 3X1 + 5X2+2X3≤60


4X1 + 4X2+4X3≤72

2X1 + 4X2+5X3≤100

X1 , X2,X3 ≥ 0

Solution: Min Ź = 60Y1+72Y2+100Y3

St. to: 3Y1+4Y2+2Y3≥5

5Y1+4Y2+4Y3≥10

2Y1+4Y2+5Y3≥8

Y1 , Y2 , Y3 ≥ 0

Ex2/ Consider the following LP and write the associated dual problem:

Max Z =7X1 + 5X2

St.to: 3X1+X2≤48

2X1+X2≤40

X1 , X2 ≥ 0

Min Ź = 48Y1+40Y2

St. to: 3Y1+2Y2≥7

Y1+Y2≥5

Y1 , Y2 ≥ 0

Ex3/ Write the dual for each of the following primal problem:

Max Z =10X1 + 12X2

St.to: X1+X2≥15

X1 = 6

X2 ≤ 8


Solution: X1+X2≥15

X1 ≤ 6 -X1≥ -6

X1≥6

X2 ≤ 8 -X2≥-8

Min Z = 10X1 + 12X2

Sub. To: X1+X2≥15

X1≥6

-X1≥ -6

-X2≥-8

X1 , X2 ≥ 0

Thus, the dual : Max Ź = 15Y1+6Y2-6Y3- 8Y4

St. to: 3Y1+4Y2+2Y3≥5

Y1+Y2-Y3≤10

Y1-Y4≤12

Problems

Write the dual for each of the following problems:

1) Min Z = 10X1 + 16X2

Sub. To: X1+X2=100

X1≤800

X2≥400

2) Min Z = 3X1 + 8X2

Sub. To: X1+X2=200


X1≤80

X2≥60

3) Max Z = 23X1 + 32X2

Sub. To: 10X1+6X2≤2500

5X1+10X2≤2000

X1+2X2≤500

4) Write the primal problem for the following:

Min Ź = 60Y1+72Y2+100Y3

St. to: 3Y1+4Y2+2Y3≥5

5Y1+4Y2+4Y3≥10

2Y1+4Y2+5Y3≥8

Y1 , Y2 , Y3 ≥ 0

The assignment model

Suppose a company has M jobs that must be completed and it has at least n workers who can perform any of the M jobs but possibly in a different amount of time. Which worker should be assigned to each job to minimize the overall time to complete all M jobs, if each worker is assigned to one and only one job. This is the classical assignment problem.

The general assignment model with n workers and M jobs is represented in the following matrix:

jobs 1 2 M bj

1 2 n

C11 C21 Cn1

C12 C22 Cn2

C1m C2m Cnm

1 1 1

workers

ai 1 1 1


The element Cij represents the cost of assigning worker i to job j (i= 1,2,…..,n) (j= 1,1,…..,m).

If we let:

Xij = 0 if worker i is not assigned to job j

1 if worker i is assigned to job j

Cij = efficiency associated with assigning worker i to job j.

Then mathematically the assignment problem can be stated as:

Minimize (maximize): Z= ij Xij

Subject to: for j= 1,2,………, m

for i= 1,2,………, n

Xij = 0 or 1 for al i and j

The following is a step by step algorithm that uses the Hungarian method to solve the general assignment problem.

Step 1: for the original cost matrix, identify each row’s minimum, and subtract it from all the entries of the row.

Step 2: For the matrix resulting from step 1, identify each column’s minimum, and subtract it from all the entries of the column.

Step 3: Draw the minimum number of horizontal and vertical lines in the last reduced matrix that will cover all the zero entries. If the number is equal the columns or rows the feasible assignment can be found, otherwise go to step 4.


Step 4: Select the smallest uncover element, and subtract it from every uncovered element, then add it to every element at the intersection of two lines. And repeat step 3 until the feasible assignment found.

For example, suppose that three jobs must be assigned to three machines, each machine must be assigned to only one job, and each job must be assigned to only one machine.

The costs are shown below. Solve the problem by using the Hungarian algorithm.

Machines Jobs M1 M2 M3

A 25 31 35 B 15 20 24 C 22 19 17

Solution:

M1 M2 M3 M1 M2 M3 A 10 12 18 A 0 2 8 B 0 1 7 B 0 1 7 C 7 0 0

C 7 0 0

M1 M2 M3 M1 M2 M3 A 0 1 7 A 0 2 8 B 0 0 6 B 0 1 7 C 8 0 0

C 7 0 0

The assignment A-M1 , B-M2 20 , C-M3

Min Z = 62

Maximization case:


Use the Hungarian method to solve the same problem for the maximum productivity.

Job M1 M2 M3 A 25 31 35 B 15 20 24 C 22 19 0

M1 M2 M3 M1 M2 M3 A 10 4 0 A 0 0 0 B 20 15 11 B 10 11 11 C 13 16 18

C 3 12 18 M1 M2 M3 M1 M2 M3

A 0 0 0 A 1 0 0 B 0 1 1 B 0 0 0 C 0 9 15

C 0 8 14

Cij: No. of units produced per hour

Max Cij= 35

The final assignment :( two solution)

First Second

A M3 A M2

B M2 20 B M3 24

C M1 C M1

Unbalanced cases

If M≠N When M > N added dummy column with Cij = 0

Max or Min When M < N added dummy row with Cij = 0

Example 1:


Use the Hungarian algorithm to solve the assignment problem having the following time table.

Solution:

Job M1 M2 M3 M4 J1 80 110 120 100 J2 50 160 130 80 J3 50 100 230 150

M1 M2 M3 M4 M1 M2 M3 M4 J1 80 110 120 100 J1 0 30 40 20 J2 50 160 130 80 J2 0 110 80 30 J3 50 100 230 150 J3 0 50 180 100

Dummy J4

0 0 0 0

J4 0 0 0 0

M1 M2 M3 M4 M1 M2 M3 M4 J1 10 10 20 0 J1 0 10 20 0 J2 0 80 50 0 J2 0 90 60 10 J3 0 20 150 70 J3 0 30 160 80

J4 30 0 0 0 J4 20 0 0 0

M1 M2 M3 M4 J1 0 0 10

0

J2 0 70 40 0 J3 0 10 140 70 J4 40 0 0 10

The optimal assignment: J1 – M2 110 J2 – M4 80 J3 – M1 50 J4 – M3 0 _________ 240

Note: M<N So we added dummy row with Cij= 0

Example 2: Use the Hungarian algorithm to solve the following assignment problem for the total maximum productivity.

Machines Task M1 M2 M3 M4 T1 20 60 50 55 T2 60 30 80 75 T3 80 100 90 80


T4 65 80 75 70 T5 70 65 60 65

Example 3/ suppose four people can each perform any one of four different jobs but possibly in different amount of time. The following table gives the corresponding time to perform the various jobs. Which person should be assigned to each job to minimize the total time to perform all four jobs?

Time to perform jobs Person 1 2 3 4

A 2 10 9 7 B 15 4 14 8 C 13 14 16 11 D 4 15 13 9

Solution:

Ex 2: first solution, Cij

320

____

T5 – M1 70

T4 – M3 75

T3 – M2 100

Second solution, Cij

T1 – M5 0

T2 – M4 75

T1 – M5 0

T2 – M3 80

T3 – M2 100

T4 – M4 70

T5 – M1 70

____

320





Transportation Model

Lec.No(5)

2011


Transportation Model

Introduction

The transportation model is a special class of the linear programming problem. It deals with the solution in which an item is shipped from sources (e.g. factories) to destinations (e.g. warehouses). The objective is to determine the amounts shipped from each source to each destination that minimize the total shipping cost while satisfying both the supply limits and the demand requirements.

The Mathematical Model

The general problem can be presented by the network.

- There are M sources. S1,S2…………..Sm - And n destinations. D1,D2…………..Dn - The arcs linking the sources and destinations represent the routes between

the sources and the destinations. - The transportations cost per unit = Cij and the amount shipped = Xij - The amount of supply at source i is ai and the amount of demand at

destination j is bj.

The objective of the model is to determine the unknown Xij that will minimize the total transportation cost, while satisfying all the supply and demand restrictions.

a1 a2 ai Units of supply am

S1 S2 Si S Sources

D D Dj

xmx13

cmc13

c12

x12

x11

D

c11

Destination

b1 b2 bj bn


Objective Function:

Min Z: Cij Xij Units of demand

Subject to:

j = 1,2,………………n

i = 1,2,………………m

1) Determination of the starting solution. The transportation model is always balanced ( = ) i.e. the sum of supply= the sum of the demand. Thus, the model has Mth-1 basis variables.

The special structure of the transportation problem allows securing a nonartificial starting basic solution using one of three methods:

a- North-west- corner method. (N.W.C)

b- Least-cost method. (L.C.M) c- Vogal method. A) N.W.C. method:

The method starts at the North West – corner cell of the tableau (variable X11).

Step 1: Allocate as much as possible to the select cell, and adjust the associated amounts of supply and demand by subtracting the allocated amount.


Step 2: Cross out the row or column with zero supply or demand to indicate that no further assignments can be made in that row or column. If both the row and column net to zero simultaneously, cross out one only, and leave a zero supply (demand) in the uncrossed-out row (column).

Step3: If exactly one row or column is left uncrossed-out, stop. Otherwise, move to the cell to the right if a column has just been crossed or the one below if a row has been crossed out. Go to step1.

Example1:

In the following transportation problem use the north west- corner method (N.W.C) to find the starting solution.

S/D D1 D2 D3 Supply S1 5 3 10 150 S2 3 9 8 70 S3 11 10 7 80 S4 6 13 6 40

Demand 160 120 60 340/340 Solution:

S/D D1 D2 D3 ai 5 3 10S1

150

150

3 9 8 S2

10

60

70

11 10 7 S3

60

20

80

6 13 6 S4

40

40

bj 160 120 60 340/340 Z= 5X150 + 3X10 + 9X60 + 10X60 +7X20 + 6X40

= 2300 M+N-1=6

B) Least-Cost method:


The Least-cost method finds a better starting solution by concentrating on the cheapest routes. Instead of starting with the North West cell, we start by assigning as much as possible to cell with the smallest unit cost (ties are broken arbitrary). We then cross out the satisfied row or column, and adjust the amounts of supply and demand accordingly. If both a row and a column are satisfied simultaneously, only one is crossed out. Next, we always look for the uncrossed-out cell with the smallest unit cost and repeat the process until we are left at the end with exactly one uncrossed out row or column.

Example 2: the least -cost is applied to example (1)

S/D D1 D2 D3 ai 5 3 10S1

30

120

150

3 9 8 S2

70

70

11 10 7 S3

20

60

80

6 13 6 S4

40

40

bj 160 120 60 340/340 Z= 5X30 + 3X120 + 3X70 + 11X20 + 7X60 + 6X40

= 1600

C) Vogal Method

Vogal method is an improved of version of the least cost method that generally produces better starting solutions.

Step 1: For each row (each column) determine a penalty measure by subtracting the smallest unit cost element in the row (column) from the next smallest unit cost element in the same row (column).

Step 2: Identify the row or column with the largest penalty. Break ties arbitrarily. Allocate as much as possible to the variable with the least unit cost in the selected


row or column. Adjust the supply and demand and cross out the satisfied row or column.

Step 3: a) If exactly one row or column with supply or demand remains uncrossed out, stop.

b) If one row or column with positive supply (demand) remains uncrossed out, determine the basic variables in the row (column) by the least-cost method, stop.

c) If all the uncrossed out rows and columns have (remaining) zero supply and demand, determine the zero basic variables by the least-cost method.

d) Otherwise, go to step 1.

Example 3: Solve the transportation model of example (1) by using vogal method.

S/D D1 D2 D3 ai V1 V2 V3 V4 V5 V6

5 3 10S1

30

120

150 2 5

3 9 8 S2

70

70 5 5 5

11 10 7 S3

20

60

80 3 4 4 4 4

6 13 6 S4

40

40 0 0 0 0 0

bj 160 120 60 340/340V1 2 6 1 V2 2 1 V3 3 1 V4 5 V5

Z= 5X30 + 3X120 + 3X70 + 11X20 + 7X60 + 6X40

= 1600


This solution happens to have the same objective value (1600) as in the Least-cost method (L.C). usually, however vogal method is expected to produce better starting solutions for the transportation method.

Stepping Stone Method (S.S)

After determining the starting ( using any of the three methods) we use one of the following method :

1. Stepping stone method. 2. Modified distribution method.

So to solve a transportation problem, we first find an initial solution ( values of Xij) and then improve the initial solution by reducing the cost through successive iterations until the minimum cost solution is found.

Example:

S/D D1 D2 ai 6 3 S1

70

8 7 S2

100

bj 90 80 170/170 By using the N.W.C

S/D D1 D2 ai 6 3 S1

70

70

8 7 S2

20

80

100

bj 90 80 170/170

Z= 60X70 + 8X20 + 7X80

80

+ 6 -70

S1D = 1140


S/D D1 D2 ai 6 3 S1

70

70

8 7 S2

90

10

100

bj 90 80 170/170 Z= 1140 – 2X70 = 1000

3X70 + 8X90 + 7X10 = 1000

Unbalanced transportation problems

( ≠ )

Demand less than supply or demand greater than supply

( < ) ( > )

From/To A B C Dummy Factory capacity

ai 5 4 3 0 D

250

250

8 4 3 0 E

50

200

50

300

9 7 5 0 F

150

150

300

Requirement bj

300 200 200 150 700\850

-78


Z = 3350

From/To A B C Plant supply

ai 6 4 9Plant W

200

200

10 5 8 Plant X

50

100

15

175

12 7 6Plant Y

75

75

0 0 0Dummy plant

50

50

Requirement bj

250 100 150 500\450

Z= 2850

Problems:

Q1/ Consider the following transportation problem. Find the starting solution by using the following methods:

1- N.W.C method. 2- L.C method. 3- Vogal method.

From/To C1 C2 C3 C4 Supply ai V1 2 2 2 1 3 V2 10 8 5 4 7 V3 7 6 6 8 5

Demand bj 4 3 4 4 15\15 Solutions:

1- N.W.C = 93


2- L.C = 79 3- Vogal = 68

Q2 / Use the north west-corner method to find the starting solution (the initial solution) for the following problem:

Warehouse From/To D1 D2 D3 D4 Supply ai

S1 0 4 0 2 5 S2 1 2 5 6 10 S3 5 3 7 9 15

Factory

Demand bj 20 10 15 15 60\30 Solution: Z= 65

Q3/ determine the initial solution for the following problem(use the three methods).

Markets From/To 1 2 3 4 Supply ai

A 10 8 6 4 2000 B 14 17 5 2 1300 C 18 7 11 9 1700

Warehouse

Demand bj 1000 2000 500 1500 5000\5000 Solutions:

1- N.W.C Z= 52200 2- L.C Z= 30700 3- Vogal Z= 30700

Q4/ Use the northwest corner method to find the starting solution, then determine the optimum solution by using (S.S) method for the following problem.

S/D D1 D2 D3 ai S1 5 3 10 150 S2 3 9 8 70 S3 11 10 7 80 S4 6 13 6 40 bj 160 120 60 340\340


Z0 = 2300

Z1 = 2300 – 8X60 = 1820

Z2 = 1820 – 5X40 = 1620

Z3 = 1620 – 1X20 = 1600

Solutions for transportation method:

Q1/ N.W.C method

From\To C1 C2 C3 C4 ai 2 2 2 1 V1

3

3

10 8 5 4 V2

1

3

3

7

7 6 6 8 V3

1

4

5

bj 4 3 4 4 15\15

4+3-1 = 6

Z= 2X3 + 1X10 + ……. +8X4 = 93

L.C.method

From\To C1 C2 C3 C4 ai 2 2 2 1 V1

3

3


10 8 5 4 V2

2

4

1

7

7 6 6 8 V3

2

3

5

bj 4 3 4 4 15\15

4+3-1 =6

Z = 1X3 + 1X4 + …… + 6X3 = 79

Vogal method

From\To C1 C2 C3 C4 ai U1 U2 U3 U4 U52 2 2 1V1

3

3 1

10 8 5 4V2

3

4

7 1 1 3

7 6 6 8V3

1

3

1

5 0 0 0 0 1

bj 4 3 4 4 15\15

U1 5 4 3 3 U2 3 2 1 4

U3 3 2 1 U4

Z = 2X3 + 5X3 +…… + 6X1 = 68


Q2: ∑ ai ≠ ∑ bj

From \To D1 D2 D3 D4 ai S1 0 4 0 2 5 S2 1 2 5 6 10 S3 5 3 7 9 15 bj 20 10 15 15 60\30

N.W.C method ∑ bj - ∑ ai = 30

From\To D1 D2 D3 D4 ai 0 4 0 2 S1

5

5

1 2

5 6 S2

10

10

5 3 7 9 S3

5

10

15

0 0 0 0 Dummy S4

0

15

15

30

bj 20 10 15 15 60\60

Z = 0X5 + 1X10 +…….. + 0X15 = 56


Q3:

L.C.M

Market From\To 1 2 2 4 ai

10 8 6 4 A

1000

300

500

200

2000

14 17 5 2 B

1300

1300

18 7 11 9 C

1700

1700

warehouse

bj 1000 2000 500 1500 5000\5000

Z= 10X1000 + ……. + 7X1700 = 30700

N.W.C

Market From\To 1 2 2 4 ai

10 8 6 4 A

1000

1000

2000

14 17 5 2 B

1000

300

1300

18 7 11 9 C

200

1500

1700

warehouse

bj 1000 2000 500 1500 5000\5000

Z = 10X1000 + …………….. + 9X1500 = 52200


Vogal method

From\To 1 2 3 4 ai V1 V2 V3 V4 V510 8 6 4A

1000

300

500

200

2000 2 2 2 2

14 17 5 2B

1300

1300 3 3

18 7 11 9C

1700

1700 2 2 2 4

bj 4 3 4 4 15\15

V1 4 1 1 2

V2 1 1

V3 1 5 5

V4 1 5

V5 1

Z = 10X100 + ………. + 7X1700 = 30700

dr. khallel ibrahim mahmoud · 2018-01-19 · 19) assembly line balancing. 20) mrp-jit-iso-tqm.etc....

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