1.
Lec.N
o(1)
Industrial Engineering (IE)
Dr. Khallel Ibrahim Mahmoud
University of Technology
Electromechanical Engineering Dept.
Introduction
Production and
Productivity
Break Even Analysis(B.E)
2011
2.
Industrial Engineering (IE)
Ibrahim Mahmoud

Dr. Khallel
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.
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Industrial Engineering (IE)
Ibrahim Mahmoud

Dr. Khallel
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.
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Industrial Engineering (IE)
Ibrahim Mahmoud

Dr. Khallel
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.
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Industrial Engineering (IE)
Ibrahim Mahmoud

Dr. Khallel
14) Statistical techniques.
15) Quality control.
16)
17) Decision making theory.
18) Replacement models.
19) Assembly line balancing.
20) MRPJITISOTQM.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
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Industrial Engineering (IE)
Ibrahim Mahmoud

Dr. Khallel
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
input
+

+
c
c

2.3 Types of production systems:
6
++%
+%
%
%
7.
Industrial Engineering (IE)
Ibrahim Mahmoud

Dr. Khallel
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: operatio
i) Make to stock production.
output
input
ii) Make to order production.
system
In make to stock production the products are manufactured
kept as ready stock and supplies to customers as orders are
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.
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Industrial Engineering (IE)
Ibrahim Mahmoud

Dr. Khallel
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, tshirts, etc.
The following fig. shows volume variety relation for different
types of manufacturing systems.
High
Job type
Batch production
Variet
y
Mass
Production
Low
Low
Production volume
High
After this brief introduction to types of production system and
productivity let us discuss how the productivity measured
and improved.
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Industrial Engineering (IE)
Ibrahim Mahmoud

Dr. Khallel
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
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Industrial Engineering (IE)
Ibrahim Mahmoud

Dr. Khallel
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
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Industrial Engineering (IE)
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
Dr. Khallel
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Industrial Engineering (IE)
Ibrahim Mahmoud

Dr. Khallel
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)
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Industrial Engineering (IE)
Ibrahim Mahmoud

Dr. Khallel
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 BreakEven analysis (BE)
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.
Revenue
curve
Cost
Profit
Cost
curve
B.E.
P
Total
v.cos
Total
cost
Loss
F.cos
t
Production
volume
Q
Fig. BreakEven Point (B.E.P)
3.1 Steps in B.E.P
To construct breakeven point, we must know:
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Industrial Engineering (IE)
Ibrahim Mahmoud

Dr. Khallel
 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 breakeven 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 Breakeven 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
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Industrial Engineering (IE)
Ibrahim Mahmoud

Dr. Khallel
P= RTc
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
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16.
Lec.N
o(2)
Industrial Engineering (IE)
Mahmoud

Dr. Khallel Ibrahim
Industrial Engineering
(IE)
Dr. Khallel Ibrahim Mahmoud
University of Technology
Electromechanical Engineering Dept.
Linear Programming
Model
2011
17.
Industrial Engineering (IE)
Mahmoud

Dr. Khallel Ibrahim
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:
12345
Define the objective.
Define the decision variables.
Write the mathematical function for the objective (objective function).
Write a one or two word description of each constraint.
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 lefthand 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
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Industrial Engineering (IE)
Mahmoud

Dr. Khallel Ibrahim
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
4
II
2
Processing time required
for the first product
(Hours)
Processing time required
2
4
for the second product
(Hours)
Production capacity
60
48
available (Hours)
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.
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Industrial Engineering (IE)
Mahmoud

Dr. Khallel Ibrahim
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
20.
Industrial Engineering (IE)
Mahmoud

Dr. Khallel Ibrahim
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:
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Industrial Engineering (IE)
Mahmoud

4X1+2X2 ≤ 60
4X1+2X2 = 60
2X1+4X2 ≤ 48
Dr. Khallel Ibrahim
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
2X1+4X2 = 48
X2=30
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
M
X1
0
X2
0
Z
Z=8(0)+6(0)=0
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Industrial Engineering (IE)
Mahmoud

Dr. Khallel Ibrahim
A
15
0
Z=8(15)+6(0)=120
B
12
16
Z=8(12)+6(6)=132
C
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.
X
4X1+2X2 ≤ 60
C
B
2X1+4X2 ≤ 48
Feasible
solution space
X1
M
A
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
23.
Industrial Engineering (IE)
Mahmoud

Dr. Khallel Ibrahim
B/ X1+ X2=200
X1=80
Then X2=20080
=120
Point
A
B
X2
200
120
X1
0
80
Z
3(0)+8(200)=1600
3(80)+8(120)=1200
Min Z=1200
X1=80
X2=120
X
20
A
X1+ X2=200
16
X1≤80
B
12
80
X2≥60
40
X
40
80
12
16
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.
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Industrial Engineering (IE)
Mahmoud

Dr. Khallel Ibrahim
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
8
Region satisfying
3rd constraint
6
Region satisfying
first 2 constraints
4
2
X
2
4
6
8
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 openended.
Let us consider a simple example to illustrate the situation.
Z = 3X1+5X2
Subject to: X1≥ 5
25.
Industrial Engineering (IE)
Mahmoud

Dr. Khallel Ibrahim
X2≤ 10
X1+2 X2≥10
X1, X2 ≥ 0
X
X1≥ 5
15
X2≤10
10
Feasible Region
5
X1+2 X2≥10
X
5
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
26.
Industrial Engineering (IE)
Mahmoud

Dr. Khallel Ibrahim
X1, X2 ≥ 0
X
35
30
2X1+ X2≤30
25
20
Redundan
t
15
10
X1≤25
Feasibl
e
X1+ X2≤20
5
X
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.
27.
Industrial Engineering (IE)
Mahmoud

Dr. Khallel Ibrahim
Ex: Max Z= 3X1+2X2
Subject to: 6X1+4X2≤ 24
X1≤ 3
X1, X2 ≥ 0
X
6
A
Optimum solutions consists of all
combinations of X1,X2 along the AB
5
4
Isoprofit line for
12/line segment AB
3
B
2
1
X
1
2
3
4
5
6
Isoprofit
line for 8
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:
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Industrial Engineering (IE)
Mahmoud

Dr. Khallel Ibrahim
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
4
3
A
2
1
Problems:
X
B
1
2
3
4
5
6
7
8
Q1/ Montana wood products manufacturers two highquality 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.
29.
Industrial Engineering (IE)
Mahmoud

Dr. Khallel Ibrahim
a
To maximize total profit, how many chairs and bookshelf
units should be produced next week?
bHow 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
X1X2≥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
30.
Industrial Engineering (IE)
Mahmoud
Colour
Standard
Economy

3
1
2
Dr. Khallel Ibrahim
1
1
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:
Item
Chopping
board
Knife holder
Time per unit (minutes)
Profit/unit
Cutting
Gluing
$2
1.4
5
$6
0.8
13
Finishing
12
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.
31.
Industrial Engineering (IE)
Mahmoud
ab

Dr. Khallel Ibrahim
Determine the optimal quantities of decision variables.
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
32.
Industrial Engineering (IE)
Mahmoud

Point
A
B
C
Dr. Khallel Ibrahim
X2
100
133.3
100
X1
100
100
150
Z
3600
4299.3
4350
X
350
300
250
20
15
B
100
A
C
50
X
50
Q2: Max Z = 60X1+40X2
Subject to: 2X1+X2≤60
X1≤25
X1, X2 ≥ 0
X2≤35
10
15
20
25
30
33.
Industrial Engineering (IE)
Mahmoud

Dr. Khallel Ibrahim
X1=0 X2=306.6
X2=0
X1=30
Point
M
A
B
C
D
B: 2X1+X2= 60
X2=35
X1
0
0
12.5
25
25
X1= (6035)/2 = 12.5
X2
0
35
35
10
0
Z
0
1400
2150
1900
1500
It is clear that Z equal to 2150 is Maximum at B when X1=12.5 X2=35.
X
60
X1≤25
2X1+X2≤60
50
B
40
A
X2≤35
30
20
C
10
X
D
M
5
10
Q3: Max Z = 10X1+15X2
Subject to: 2X1+X2≤26
2X1+4X2≤ 56
X1X2 ≥5
X1+X2 ≤5
15
20
25
30
34.
Industrial Engineering (IE)
Mahmoud

Dr. Khallel Ibrahim
X1, X2 ≥ 0
X1=5 X2=5
X2
0
5
11
10
0
Point
X1
M
0
A
0
B
6
C
8
D
13
Hence, Z is Maximum (230) at C.
Z
0
75
225
230
130
X
28
2X1+X2≤26
24
X1+X2 ≤5
20
16
B
12
C
8
2X1+4X2≤ 56
A
4
X
D
8
4
M
4
Q4: Min Z = 2X1+4X2
Subject to: X1+X2≤14
3X1+2X2≥30
2X1+X2≤18
8
12
16
18
20
24
28
35.
Industrial Engineering (IE)
Mahmoud

Dr. Khallel Ibrahim
X1, X2 ≥ 0
Point
A
B
C
X2
12
10
6
X1
2
4
6
Z
52
48
36
Thus, Z is Minimum at point C when X1 , X2=6 Z=36 .
X
18
16
14
12
3X1+2X2≥30
A
B
10
8
6
X1+X2≤14
C
4
2X1+X2≤18
2
2
4
6
X
8
10
12
14
16
18
36.
Industrial Engineering (IE)
Mahmoud

Dr. Khallel Ibrahim
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
A
B
C
D
X1
0
4
12
24
X2
24
12
4
0
Z
96000
72000
88000
144000
Thus, to Minimize cost, line I should be run for 4 days and line II for 12 days.
X
28
24
20
A
3X1+X2≥24
16
12
8
B
X1+X2≥16
C
4
37.
Industrial Engineering (IE)
Mahmoud

2X1+6X2≥48
D
4
8
12
16
20
Dr. Khallel Ibrahim
X
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
X2
80
70
60
50
40
B
30
C
20
10
D
X
38.
Industrial Engineering (IE)
Mahmoud
A
1

2
Q7: Let board = X1
30
40
50
60
70
Dr. Khallel Ibrahim
80
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 650650=0
(130,0) (0,50) …………… (2)
Finishing: 3X50=150 360150=210 minutes. (30,0) (0,120) ………… (3)
X
140
120
10
80
60
A
B
40
20
C
X
39.
Industrial Engineering (IE)
Mahmoud
20
D

4
60
Q8: The optimum solution X1=4.2 and
Minimum Z=2(4.2)+3(1.6) = 13.2
80
10
X2=1.6
12
14
Dr. Khallel Ibrahim
16
40.
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Industrial Engineering (IE)
Ibrahim Mahmoud

Dr. Khallel
Industrial Engineering
(IE)
Dr. Khallel Ibrahim Mahmoud
University of Technology
Electromechanical Engineering Dept.
Linear Programming
Model
Simplex Method
2011
41.
Industrial Engineering (IE)
Ibrahim Mahmoud

Dr. Khallel
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.
42.
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Ibrahim Mahmoud

Dr. Khallel
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
43.
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Ibrahim Mahmoud

Dr. Khallel
Step 2/ Construct the simplex tableau. The standard form
can be summarized in a compact tableau form as:
Coefficient
Basic
Values of
of OF
variables
basic
variables
variables
in basis
( Cj)
(A)
(q)
0
0
S1
S2
Zj
CjZj
40
80
0
Coefficient of OF
variables
(Cj)
6
X1
2
8
0
6
X1 = 0
Initial solution:
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 :
8
X2
5
4
0
8
0
S1
1
0
0
0
0
S2
0
1
0
0
44.
Industrial Engineering (IE)
Ibrahim Mahmoud

(Cj – Zj) X1 = 60 =6
,
(Cj – Zj) S1 = 00 =0
(Cj – Zj) X2 = 80 =8
,
Dr. Khallel
(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:
45.
Industrial Engineering (IE)
Ibrahim Mahmoud

Dr. Khallel
Type 2: Computation is applied to the remaining row (S2row) 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)
8
X2
8
8
X2
1
0
S2
48
0
1
Zj
64
8
0
0
0
CjZj
6
X1
0
S1
0
S2
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 (CjZj) row ( )
46.
Industrial Engineering (IE)
Ibrahim Mahmoud

Dr. Khallel
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:
(A)
(q)
8
X2
5
6
S2
Zj
CjZj
85
6
X1
0
8
X2
1
1
( Cj)
0
6
8
0
0
0
S1
0
S2
47.
Industrial Engineering (IE)
Ibrahim Mahmoud

Dr. Khallel
Since none of the CjZj 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
X1
X2
Z
Optimum value
15/2 = 7.5
5
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 MMethod (Big M).
The MMethod 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.
48.
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Ibrahim Mahmoud

Dr. Khallel
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
49.
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Ibrahim Mahmoud

Dr. Khallel
X2+ 0S1S2+A1+0A2= 60
X1+ X2+0S1+0S2+0A1+A2= 200
Min Z = 3X1+ 8X2+0S1+0S2+MA1+MA2
( Cj)
(A)
(q)
0
M
S2
A1
A2
Z
CjZj
80
60
200
260M
M
3
8
X1
X2
1
0
0
1
1
1
M
2M
3M 82M
0
S1
1
0
0
0
0
0
S2
0
1
0
M
M
M
A1
0
1
0
M
0
M
A2
0
0
1
M
0
Step 2/ identify the pivot column:
This is done by selecting the none basic variable having the
largest negative value in CjZj .
When all the elements in the CjZj row are positive or zero
the optimal solution is reached .
Step3/ identify the pivot row:
80/0 = 0
,
60/1 = 60 ,
Pivot = 1
A1
200/1 = 200
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
50.
Industrial Engineering (IE)
Ibrahim Mahmoud

Dr. Khallel
0(1X1) = 1
0(1X1) = 1
1(1X0) = 1
( Cj)
(A)
(q)
0
8
M
S1
X2
A2
Zj
80
60
140
480+
140M
8
X2
0
1
0
8
0
S1
1
0
0
0
0
M
M
S2 A1 A2
0
0
0
1
1
0
1
1
1
M8 8M M
3M
CjZj
3
X1
1
0
1
M
0
0
8M 2M8
As CjZj 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
0
51.
Industrial Engineering (IE)
Ibrahim Mahmoud

Dr. Khallel
Step5/ create the new tableau
( Cj)
(A)
(q)
3
8
M
X1
X2
A2
Zj
3
X1
1
0
0
3
80
60
60
720+
60M
0
0
M
M
S1
S2 A1 A2
1
0
0
0
0
1
1
0
1
1
1
1
3M M8 8M M
0
CjZj
8
X2
0
1
0
8
0
M3 8M 2M8
0
As CjZj 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
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 (1X1) = 1
1 – (1X1) = 0
1 (1X1) = 0
S1
52.
Industrial Engineering (IE)
Ibrahim Mahmoud

Dr. Khallel
0 – (1X1) = 1
( Cj)
(A)
(q)
3
8
0
X1
X2
S2
Zj
CjZj
80
120
60
1200
3
X1
1
0
0
3
0
8
X2
0
1
0
8
0
0
S1
1
1
1
5
5
0
S2
0
0
1
0
0
M
A1
0
0
1
0
M
As CjZj is either positive or zero under all column, the
solution is an optimal solution.
X1 = 80 ,
X2 = 120
, S2 = 60
, Z= 1200
M
A2
0
1
1
8
M8
53.
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Mahmoud

Dr. Khallel Ibrahim
Industrial Engineering (IE)
Dr. Khallel Ibrahim Mahmoud
University of Technology
Electromechanical Engineering Dept.
The Dual Model
The assignment model
2011
54.
Industrial Engineering (IE)
Mahmoud

Dr. Khallel Ibrahim
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
55.
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Mahmoud

Dr. Khallel Ibrahim
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 lefthandside 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
56.
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Mahmoud

Dr. Khallel Ibrahim
Subject to: 2y1+8y2 s1+A1 = 6
5y1+ 4y2 s2+ A2 = 8
Cj
A
M
M
A1
A2
Zj
CjZj
Y2
A2
Zj
CjZj
Y2
Y1
Zj
CjZj
80
M
80
40
q
40
80
0
0
M
M
Y1
Y2
S1
S2
A1
A2
6
2
8
1
0
1
0
8
5
4
0
1
0
1
14M
7M
12M
M
M
M
M
407M 8012M
M
M
0
0
¾
¼
1
1/8
0
1/8
0
5
4
0
½
1
1/2
1
60+5M 20+4M
80
1/2M10 M 101/2M
M
204M
0
101/2M
M
1/2M10
0
7/16
0
1
5/32
1/16
5/32
1/16
5/4
1
0
1/8
1/4
1/8
1/4
85
40
80
7.5
5
7.5
5
0
0
7.5
5
M7.5
M5
:( آﺎن آﻤﺎ ﻳﻠﻲprimal) ﺣﻞ اﻟﻨﻤﻮذج اﻷوﻟﻲ
Cj
A
q
8
6
X2
X1
Zj
CjZj
5
7.5
85
6
X1
0
1
6
0
Dual
S1 = 7.5
S2 = 5
Y1 = 5/4
Y2 = 7/16
Ź = 85
Ex1/ Obtain the dual of the following LP:
Max Z =5X1 + 10X2+8X3
St.to: 3X1 + 5X2+2X3≤60
8
X2
1
0
8
0
0
S1
¼
1/8
5/4
5/4
Primal
X1 = 7.5
X2 = 5
S1 = 5/4
S2 = 7/16
Z = 85
0
S2
1/16
5/32
7/16
7/16
57.
Industrial Engineering (IE)
Mahmoud

Dr. Khallel Ibrahim
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
58.
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Mahmoud
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+6Y26Y3 8Y4
St. to: 3Y1+4Y2+2Y3≥5
Y1+Y2Y3≤10
Y1Y4≤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

Dr. Khallel Ibrahim
59.
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Dr. Khallel Ibrahim
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
workers
1
2
n
ai
2
M
bj
C11
C21
Cn1
1
C12
C22
Cn2
1
C1m
C2m
Cnm
1
1
1
1
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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=
Subject to:
ij
Xij
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.
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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.
Jobs
M1
25
15
22
A
B
C
Machines
M2
31
20
19
M3
35
24
17
Solution:
A
B
C
M1
10
0
7
M2
12
1
0
M3
18
7
0
A
B
C
M1
0
0
8
M2
1
0
0
M3
7
6
0
The assignment AM1
Min Z = 62
Maximization case:
, BM2 20 ,
A
B
C
M1
0
0
7
M2
2
1
0
M3
8
7
0
A
B
C
M1
0
0
7
M2
2
1
0
M3
8
7
0
CM3
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Use the Hungarian method to solve the same problem for the maximum
productivity.
Job
A
B
C
M1
25
15
22
A
B
C
M3
35
24
0
M1
10
20
13
M1
0
0
0
A
B
C
M2
31
20
19
M2
4
15
16
M2
0
1
9
M3
0
11
18
M3
0
1
15
Cij: No. of units produced per hour
Max Cij= 35
A
B
C
A
B
C
M1
0
10
3
M1
1
0
0
M2
0
11
12
M2
0
0
8
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:
M3
0
11
18
M3
0
0
14
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Use the Hungarian algorithm to solve the assignment problem having the following
time table.
Job
J1
J2
J3
Solution:
J1
J2
J3
Dummy
J4
J1
J2
J3
J4
M1
80
50
50
M3
120
130
230
M4
100
80
150
M1
80
50
50
0
M2
110
160
100
0
M3
120
130
230
0
M4
100
80
150
0
M1
10
0
0
30
M2
10
80
20
0
M3
20
50
150
0
M4
0
0
70
0
M1
0
0
0
40
J1
J2
J3
J4
M2
110
160
100
M2
0
70
10
0
M3
10
40
140
0
Note: M<N
So we added dummy row with
Cij= 0
J1
J2
J3
J4
M2
30
110
50
0
M3
40
80
180
0
M4
20
30
100
0
J1
J2
J3
J4
M4
0
0
70
10
M1
0
0
0
0
M1
0
0
0
20
M2
10
90
30
0
M3
20
60
160
0
M4
0
10
80
0
The optimal assignment:
J1 – M2 110
J2 – M4 80
J3 – M1 50
J4 – M3 0
_________
240
Example 2: Use the Hungarian algorithm to solve the following assignment
problem for the total maximum productivity.
Machines
Task
T1
T2
T3
M1
20
60
80
M2
60
30
100
M3
50
80
90
M4
55
75
80
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T4
T5

65
70
80
65
Dr. Khallel Ibrahim
75
60
70
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?
1
2
15
13
4
Person
A
B
C
D
Time to perform jobs
2
3
10
9
4
14
14
16
15
13
Cij
Second solution,
4
7
8
11
9
Solution:
Ex 2: first solution,
Cij
T1 – M5
0
T1 – M5
0
T2 – M3
80
T2 – M4
75
T3 – M2
100
T3 – M2
100
T4 – M4
70
T4 – M3
75
T5 – M1
70
T5 – M1
70
____
____
320
320
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Industrial Engineering (IE)
Dr. Khallel Ibrahim Mahmoud
University of Technology
Electromechanical Engineering Dept.
Transportation Model
2011
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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 Units of supply ai
S1
S2
am
Si
S
Sources
c11
x11
cm
c12
c13
x12
xm
x13
D
D
Dj
D
b1
b2
bj
bn
Destination
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Objective Function:
Min Z:
Units of demand
Cij Xij
Subject to:
j = 1,2,………………n
i = 1,2,………………m
1)
Determination of the starting
=
) i.e. the
solution. The transportation model is always balanced (
sum of supply= the sum of the demand. Thus, the model has Mth1 basis
variables.
The special structure of the transportation problem allows securing a nonartificial
starting basic solution using one of three methods:
a
Northwest corner method.
(N.W.C)
bcA)
Leastcost method. (L.C.M)
Vogal method.
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.
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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 uncrossedout row (column).
Step3: If exactly one row or column is left uncrossedout, 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
S1
S2
S3
S4
Demand
Solution:
D2
3
9
10
13
120
D1
5
3
11
6
160
S/D
S1
11
S4
6
10
ai
150
9
8
70
7
80
D1
D3
150
10
60
10
60
13
20
6
bj
160
120
Z= 5X150 + 3X10 + 9X60 + 10X60 +7X20 + 6X40
= 2300
B)
Supply
150
70
80
40
340/340
3
3
S3
D2
5
S2
D3
10
8
7
6
60
40
40
60
M+N1=6
LeastCost method:
340/340
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The Leastcost 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 uncrossedout 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
S1
S3
11
S4
6
8
70
10
3
10
ai
150
9
5
S2
D2
7
80
D1
D3
3
30
120
70
20
60
13
6
40
bj
160
120
Z= 5X30 + 3X120 + 3X70 + 11X20 + 7X60 + 6X40
40
60
340/340
= 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
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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 leastcost method, stop.
c) If all the uncrossed out rows and columns have (remaining) zero supply and
demand, determine the zero basic variables by the leastcost method.
d) Otherwise, go to step 1.
Example 3: Solve the transportation model of example (1) by using vogal method.
S/D
S1
D1
5
S2
3
S3
11
S4
6
D2
10
ai
150
V1 V2 V3 V4 V5 V6
2
5
9
8
70
5
5
5
10
7
80
3
4
4
4
4
40
0
0
0
0
0
3
30
D3
120
70
20
60
13
6
40
bj
160
120
60
340/340
V1
2
6
1
V2
2
1
V3
3
1
V4
5
V5
Z= 5X30 + 3X120 + 3X70 + 11X20 + 7X60 + 6X40
= 1600
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This solution happens to have the same objective value (1600) as in the Leastcost
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.
2.
Stepping stone method.
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
S1
D2
6
3
ai
70
S2
8
7
100
D1
bj
90
By using the N.W.C
S/D
S1
3
bj
ai
70
7
8
170/170
D2
6
S2
80
100
D1
70
20
80
90
80
Z= 60X70 + 8X20 + 7X80
170/170
+
6
= 1140
S1D
70
80
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8
S/D
S1
D1
6
S2
8
Dr. Khallel Ibrahim
7
D2
ai
70
3
70
7
100
90
10
bj
90
Z= 1140 – 2X70 = 1000
80
170/170
3X70 + 8X90 + 7X10 = 1000
Unbalanced transportation problems
(
≠
)
Demand less than supply
(
<
demand greater than supply
)
From/To
D
or
(
A
B
>
4
3
0
Factory
capacity
ai
250
4
5
C
)
Dummy
3
0
300
0
300
250
E
8
50
F
Requirement
bj
9
200
7
300
50
5
200
150
200
150
150
700850
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Z = 3350
From/To
Plant W
A
B
4
9
Plant
supply
ai
200
5
6
C
8
175
200
Plant X
10
50
Plant Y
12
100
7
15
6
75
75
Dummy
plant
Requirement
bj
0
0
250
0
100
50
50
150
500450
Z= 2850
Problems:
Q1/ Consider the following transportation problem. Find the starting solution by
using the following methods:
123From/To
V1
V2
V3
Demand bj
Solutions:
1
N.W.C method.
L.C method.
Vogal method.
C1
2
10
7
4
C2
2
8
6
3
C3
2
5
6
4
C4
1
4
8
4
N.W.C = 93
Supply ai
3
7
5
1515
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Dr. Khallel Ibrahim
L.C = 79
Vogal = 68
Q2 / Use the north westcorner method to find the starting solution (the initial
solution) for the following problem:
Factory
From/To
S1
S2
S3
Demand bj
Solution: Z= 65
D1
0
1
5
20
Warehouse
D2
4
2
3
10
D3
0
5
7
15
D4
2
6
9
15
Supply ai
5
10
15
6030
Q3/ determine the initial solution for the following problem(use the three
methods).
Warehouse
From/To
A
B
C
Demand bj
1
10
14
18
1000
Markets
2
8
17
7
2000
3
6
5
11
500
4
4
2
9
1500
Supply ai
2000
1300
1700
50005000
Solutions:
123
N.W.C Z= 52200
L.C Z= 30700
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
S1
S2
S3
S4
bj
D1
5
3
11
6
160
D2
3
9
10
13
120
D3
10
8
7
6
60
ai
150
70
80
40
340340
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