Operations Research
First Year Computer
Text Book: Operations Research
8th Edition
Hamdy A. Taha
Course Outline
Part1: Deterministic Models
Chapter 2: Introduction to Linear Programming
2.1 Introduction.
2.2 Construction of the LP Model.
2.3 Graphical LP Solution.
2.3.1 Solution of a Maximization Model.
2.3.2 Solution of a Minimization Model.
2.3.3 Slack, Surplus, and Unrestricted Model.
2.4 Graphical Sensitivity Analysis.
2.4.1 Changes in the Objective Function Coefficient.
2.4.2 Unit Worth of a Resource.
2.5 Computer Solution of Linear Programming Problems.
2.6 Analysis of Selected Linear Programming Model.
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Chapter 3: The Simplex Method
3.1 Introduction.
3.2 Standard Linear Programming Form and Its Solutions.
3.2.1 Standard Linear Programming Form.
3.2.2 Determination of Basic Solutions.
3.2.3 Unrestricted Variables and Basic Solution.
3.3 The Simplex Algorithm.
3.4 Simplex Method Application.
3.5 Special Cases in Simplex Method Application.
3.5.1 Degeneracy.
3.5.2 Alternative Optima.
3.5.3 Unbounded Solution.
3.5.4 Infeasible Solution.
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Chapter 4: Duality.
4.1 Introduction.
4.2 Definition of the Dual Problem.
4.3 Relationship between the Optimal Primal and Dual.
Chapter 5: Transportation Model.
5.1 Definition of the Transportation Model.
5.3 The Transportation Algorithm.
5.3.1 Determination of the Starting Solution.
5.3.2 Iterative Computation of the Algorithm.
5.4 The Assignment Model.
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Chapter 6: Network Models
6.1 Scope of Network Application.
6.2 Network Definitions.
6.3 Minimal Spanning Tree Algorithm.
6.4 Shortest Route Problem.
6.5 Maximal Flow Model.
6.7 CPM and PERT.
6.7.1 Network Representation.
6.7.2 Critical Path Computation.
6.7.3 Construction of the Time Schedule.
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Introduction
Operations Research (OR)
It is a scientific approach to determine the optimum
(best) solution to a decision problem under the restriction
of limited resources. Using the mathematical techniques
to model, analyze, and solve the problem.
Phases of OR
1. Definition of the problem includes:•
The description of the decision variables
(alternatives)
• The determination of the objective of the study
• The specification of the limitations under which
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Dr.Farouk Sha'banthe modeled system operates.
2. Model Construction
Translating the real world problem into
mathematical relationships (the most suitable
model to represent the system, LP, dynamic
programming, integer programming, ………..)
3.Solution of the model
Using well-defined optimization techniques.
An important aspect of model solution is
sensitivity analysis.
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4.Model validity
Testing and evaluation of the model. A common
method for testing a validity of a model is to
compare its performance with some past data
available for the actual system.
5.Implementation of the solution
Implementation of the solution of validated model
involves the translation of the model's results into
instructions issued in understandable form to the
individual
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Basic components of the model
1.Decision Variables
It is the unknown to be determined from the solution
of a model (what does the model seek to determine).
It is one of the specific decisions made by a decision
maker (DM).
2.Objective Function
It is the end result (goal) desired to be achieved by
the system. A common objective is to maximize
profit or minimize cost. It is expressed as a
mathematical function of the system decision
variables.
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3.Constraints
These are the limitations imposed on the
variables to satisfy the restriction of the modeled
system. They must be expressed as mathematical
functions of the system decision variables (D.V.).
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Example 1:
A company manufactures two products A&B. with
profit 4 & 3 units. A&B take 3&2 minutes
respectively to be machined. The total time
available at machining department is 800 hours
(100 days or 20 weeks). A market research
showed that at least 10000 units of A and not
more than 6000 units of B are needed. It is
required to determine the number of units of A&B
to be produced to maximize profit.
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Problem Formulation
Decision variables
X1= number of units produced of A.
X2= number of units produced of B.
Objective Function
Maximize Z= 4 X1 + 3 X2
Constraints
3 X1 + 2 X2
X1
X2
X1, X2
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<= 800x60
>=10000
<=6000
>=0
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Example 2: Feed mix problem
A farmer is interested in feeding his cattle at minimum
cost.
Two feeds are used A&B. Each cow must get at least
400 grams/day of protein, at least 800 grams/day of
carbohydrates, and not more than 100 grams/day of
fat. Given that A contains 10% protein, 80%
carbohydrates and 10% fat while B contains 40%
protein, 60% carbohydrates and no fat. A costs 2
L.E/kg, and B costs 5 L.E/kg. Formulate the problem to
determine the optimum amount of each feed to
minimize cost.
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Problem Formulation
Decision variables
X1= weight of feed A kg/day/animal
X2= weight of feed B kg/day/animal
Objective Function
Minimize Z= 2 X1 + 5 X2
Constraints
Protein
Carbohydrates
Fats
0.1 X1 + 0.4 X2
0.8 X1 + 0.6 X2
0.1 X1
X1,
X2
>=0.4
>=0.8
<= 0.1
>=0
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Problem Formulation
Decision variables
X1= weight of feed A kg/day/animal
X2= weight of feed B kg/day/animal
Objective Function
Minimize Z= 2 X1 + 5 X2
Constraints
Protein
0.1 X1
+ 0.4 X2
>=0.4
Carbohydrates
0.8 X1
+ 0.6 X2
>=0.8
Fats
0.1 X1
X1,
<= 0.1
X2
>=0
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Example 3: Blending Problem
An iron ore from 4 mines will be blended. The analysis has shown
that, in order to obtain suitable tensile properties, minimum requirements
must be met for 3 basic elements A, B, and C. Each of the 4 mines contains
different amounts of the 3 elements (see the table). Formulate to find the
least cost blend for one ton of iron ore.
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Problem Formulation
Decision variables
X1= Fraction of ton to be selected from mine number 1
X2= Fraction of ton to be selected from mine number 2
X3= Fraction of ton to be selected from mine number 3
X4= Fraction of ton to be selected from mine number 4
Objective Function
Minimize Z= 800 X1 + 400 X2 + 600 X3 + 500 X4
Constraints
10 X1 + 3 X2 + 8 X3 +
2X4
90 X1 + 150 X2 + 75 X3 + 175 X4
45 X1 + 25 X2 + 20 X3 + 37 X4
X1 +
X2 +
X3 +
X4
X1, X2, X3, X4
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>= 5
>= 10
>= 30
=1
>= 0
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Example 4: Inspection Problem
A company has 2 grades of inspectors 1&2. It is required that at
least 1800 pieces be inspected per 8 hour/day. Grade 1 inspectors can
check pieces at the rate of 25 per hour with an accuracy of 98%. Grade
2 inspectors can check at the rate of 15 pieces per hour with an
accuracy of 95%. Grade 1 costs 4 L.E/hour, grade 2 costs 3 L.E/hour.
Each time an error is made by an inspector costs the company 2 L.E.
There are 8 grade 1 and 10 grade 2 inspectors available. The company
wants to determine the optimal assignment of inspectors which will
minimize the total cost of inspection/day.
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Problem Formulation
Decision variables
X1= Number of grade 1 inspectors/day.
X2= Number of grade 2 inspectors/day.
Objective Function
Cost of inspection = Cost of error + Inspector salary/day
Cost of grade 1/hour = 4 + (2 X 25 X 0.02) = 5 L.E
Cost of grade 2/hour = 3 + (2 X 15 X 0.05) = 4.5 L.E
Minimize Z= 8 (5 X1 + 4.5 X2)
= 40 X1 + 36 X2
Constraints
X1
X2
8(25) X1+ 8(15) X2
200 X1 + 120 X2
X1,
X2
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<= 8
<= 10
>= 1800
>= 1800
>=0
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Example 5: Trim-loss Problem.
A company produces paper rolls with a standard width of 20
feet. Each special customer orders with different widths are produced
by slitting the standard rolls. Typical orders are summarized in the
following tables.
Possible knife settings
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Formulate to minimize the trim loss and the number of rolls
needed to satisfy the order.
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Problem Formulation
Decision variables
Xj = Number of standard rolls to be cut according to setting j
j = 1, 2, 3, 4, 5, 6
Number of 5 feet rolls produced = 2 X2 + 2 X3 + 4 X4 + X5
Number of 7 feet rolls produced = X1 + X2+ 2 X5
Number of 9 feet rolls produced = X1 + X3+ 2 X6
Let Y1, Y2, Y3 be the number of surplus rolls of the 5, 7, 9 feet rolls thus
Y1= 2 X2 + 2 X3 + 4 X4 + X5 - 150
Y2= X1 + X2+ 2 X5 - 200
Y3= X1 + X3+ 2 X6 - 300
The total trim losses = L (4X1 +3 X2+ X3 + X5 + 2 X6 + 5Y1+ 7Y2+ 9Y3)
Where L is the length of the standard roll.
Objective Function
Minimize Z= L(4X1 +3 X2+ X3 +
X5 + 2 X6 + 5Y1+ 7Y2+ 9Y3)
Constraints
2 X2+ 2 X3+ 4 X4+ X5
- Y1
= 150
X1 + X2
+2X5
- Y2
= 200
X1 + X3
+ 2 X6
- Y3 = 300
X1, X2, X3, X4, X5, X6
>= 0
Y1, Y2, Y3 >= 0
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General form of a LP problem with m constraints and n
decision variables is:
Maximize Z = C1X1+ C2X2+ …………………. + CnXn
Constraints
A11X1 + A12X2+……………………+ A1nXn <= B1
A21X1 + A22X2+……………………+ A2nXn <= B2
.
.
.
.
Am1X1 + Am2X2+……………………+ AmnXn <= Bm
X1, X2, ……………………………………, Xn >= 0
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OR
Maximize Z= Σnj=1 CjXj
Constraints
Σnj=1 aijXi
<= bi
Xi >=0
i = 1, …., m j = 1, …., n
Where
n = Number of activities.
Xj = Level of activity j
Cj =Contribution of the objective function/unit of activity j
m = Number of resources
Bi = Amount of resource i available.
Aij=Amount of resource i consumed by one unit of activity j
Other forms
1.
2.
3.
4.
Minimize Z= Σnj=1 CjXj
Σnj=1 AijXi >= Bi
for some values of i
Σnj=1 AijXi = Bi
for some values of i
Xi
unrestricted in sign for some values of i
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Terminology of solutions for a LP model:
A Solution
Any specifications of values of X1, X2, ………, Xn is
called a solution.
A Feasible Solution
Is a solution for which all the constraints are satisfied.
An Optimal Solution
Is a feasible solution that has the most favorable value
of the objective function (largest for maximize or
smallest for minimize).
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Note
•If there is exactly one optimal solution it must be a corner
point feasible solution.
•If there are multiple optimal solutions, then at least two of
them must be adjacent corner- point feasible solutions.
Two corner-point feasible solutions are said to be
adjacent if the line segment connecting them lies on the
boundary of the feasible region (one of the constraints).
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Graphical Solution
Construction of the LP model
Example 1:
The Reddy Mikks Company
Reddy Mikks produces both interior and exterior paints
from two raw materials, M1&M2. The following table
provides the basic data of the problem.
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A market survey indicates that the daily demand for interior
paint cannot exceed that of exterior paint by more than 1 ton.
Also, the maximum daily demand of interior paint is 2 ton.
Reddy Mikks wants to determine the optimum (best)
product mix of interior and exterior paints that maximizes the
total daily profit.
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Problem Formulation
Decision variables
X1= Tons produced daily of exterior paint.
X2= Tons produced daily of interior paint.
Objective Function
Maximize Z= 5 X1 + 4 X2
Constraints
6 X1 + 4 X2
X1 + 2 X2
- X1 + X2
X2
X1,
X2
<= 24
<= 6
<= 1
<= 2
>=0
Any solution that satisfies all the constraints of the model is
a feasible solution. For example, X1=3 tons and X2=1 ton is a
feasible solution. We have an infinite number of feasible
solutions, but we are interested in the optimum feasible
solution that yields the maximum total profit.
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Graphical Solution
The graphical solution is valid only for two-variable problem
which is rarely occurred.
The graphical solution includes two basic steps:
1. The determination of the solution space that defines
the feasible solutions that satisfy all the constraints.
2. The determination of the optimum solution from
among all the points in the feasible solution space.
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.
ABCDEF consists of an infinite number of points; we need a
systematic procedure that identifies the optimum solutions. The
optimum solution is associated with a corner point of the
solution space.
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To determine the direction in which the profit function increases
we assign arbitrary increasing values of 10 and 15
5 X1 + 4 X2=10
And 5 X1 + 4 X2=15
The optimum solution is mixture of 3 tons of exterior and 1.5 tons of
interior paints will yield a daily profit of 21000$.
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Example 2: The Diet Problem
Farm uses at least 800lb of special feed daily. The special feed is a
mixture of corn and soybean with the following composition.
The food mixture must contain at least 30% protein and at most
5% fiber. They want to determine the daily minimum cost feed
mix.
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Problem Formulation
Decision variables
X1= lb of corn in the daily mix.
X2= lb of soybean in the daily mix.
Objective Function
Minimize Z= 0.3 X1 + 0.9 X2
Constraints
X1 +
X2 >= 800
0.09X1+0.6 X2 >= 0.3(X1+X2) = -0.21X1+0.3 X2 >= 0
0.02X1+0.06 X2 <= 0.05(X1+X2) = 0.03X1-0.01 X2 >= 0
X1, X2 >= 0
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Slack, Surplus, and Unrestricted variables
Slack Variable:
For constraints of the type (<=) the R.H.S normally
represents the limit on the availability of a resources and the
L.H.S represents the usage of this limited resource by the
different activities (variables) of the model.
A slack represents the amount by which the available
amount of the resource exceeds its usage by the activities.
For example (6 X1+4 X2<= 24) is equivalent to
(6 X1+4 X2 + S1= 24) provided that S1>=0.
The slack variable S1=24-6 X1-4 X2 represents the unused
amount of raw material M1.
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Surplus Variable:
It is used in the constraints of type (>=) normally set minimum
specification requirements.
Surplus represents the excess of the L.H.S over the minimum
requirement.
For example (X1+X2 >= 800) is equivalent to (X1+X2 - S1= 800)
provided that S1>=0, this signifies that a surplus amount of feed
over the minimum requirement will be produced.
Unrestricted Variable:
The variable which can be positive or negative.
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The Simplex Method
Introduction
It is a general algebraic method to solve a set of linear
equations.
We use simplex method to get extreme (or corner) point
solution.
We must first convert the model into the standard LP form by
using slack or surplus variables to convert the inequality
constraints into equations.
Our interest in the standard LP form lies in the basic
solutions of the simultaneous linear equations.
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Standard LP Form and its Basic Solution
Example 1:
Express the following LP model in standard form.
Maximize Z= 2X1 + 3 X2 + 5X3
S.T
X1 + X2 - X3
>= -5
-6X1 + 7 X2 - 9X3
<= 4
X1 + X2 + 4X3
= 10
X1,
X2
>= 0
X3 unrestricted = X+3 - X-3
where X+3 , X-3
Standard LP form
Maximize Z= 2X1 + 3 X2 + 5 X+3 - 5 X-3
S.T
-X1 - X2 + 3 X+3 - 3 X-3 + X4
=5
-6X1 + 7X2 - 9 X+3 + 9 X-3
+ X5 = 4
X1 + X2 + 4X+3 - 4 X-3
= 10
X1 ,
X2, X+3,
X-3, X4, X5 >=0
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Determination of basic solutions
The standard LP form includes m simultaneous linear equations in n
unknowns or variables (m<n).
We divide the n variables into 2 sets:
1. (n-m) variables to which we assign zero values which are called
non-basic variables.
1. Remaining m variables whose values are determined by solving the
resulting m equations which is called
basic variables.
The resulting solution is Basic Solution(BS). If all values are satisfying
non- negativity then resulting BS is Feasible Basic Solution(BFS),
otherwise, it is infeasible.
The maximum number of possible basic solution for m equations in n
unknowns is
n!
m!( n  m)!
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Example 2:
Maximize Z= 5X1 + 6 X2
S.T
2X1
+ 3X2
2X1
+X2
3X1
+ 3X2
X1,
X2
>= 18
<= 12
= 24
>=0
Standard LP form
Maximize Z=5X1 + 6X2
S.T
2X1
+3X2
+ X3
2X1
+ X2
3X1
+ 3X2
X1, X2, X3, X4, X5
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+ X4
+ X5
= 18
= 12
= 24
>=0
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Maximum number of possible basic solutions are
 n   5
5!
5! 5 x 4 x3!
      

 10 possible solutions
m
3
3
!
(
5

3
)
!
3
!
2
!
3
!
2
!
   
The basic feasible solutions are the corner points
The Non-Basic Variables
Basic Variables
Z(objective Function)
Corner Pointas
Feasible or not
1) X1, X2
(0, 0, 18, 12, 24)
0
Yes
2) X1, X3
(0, 6, 0, 6, 6)
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Yes
3) X1, X4
(0, 12, -18, 0, -12)
No
4) X1, X5
(0, 8, -6, 4, 0)
No
5) X2, X3
(9, 0, 0, -6, -3)
N0
6) X2, X4
(6, 0, 6, 0, 6)
7) X2, X5
(8, 0, 2, -4, 0)
8) X3, X4
(4.5, 3, 0, 0, 1.5)
9) X3, X5
(6 . 2 .0 . -2 .5 )
No
10) X4 . X5
(4. 4 . – 2 . 0. 0)
NO
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Yes
N0
40.5
Yes
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The Simplex Algorithm
We solve the Reddy Mikks model,
where
X1= Tons produced daily of exterior paint.
X2= Tons produced daily of interior paint.
Exterior and interior paints are produced from two types of raw
materials M1 and M2
Maximize Z- 5 X1 - 4 X2 + 0 X3 + 0 X4 + 0 X5 +0 X6
S.T.
6 X1 +4 X2 + X3
X1 +2 X2
+ X4
- X1 + X2
+ X5
X2
+ X6
X1, X2, X3, X4, X5, X6
= 0
<= 24
<= 6
<= 1
<=2
>=0
The variables X3, X4, X5, X6 are the slacks associated with the four
(<=) constraints.
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The Starting Basic Feasible Solution is
Is the starting solution optimal? No, since the coefficients of X1
and X2 are still negative so they can increase the profit of Z.
We choose X1 with more negative (-5) i.e.; X1 is the entering
variable.
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Since none of the Z-row coefficients associated with the non-basic
variables X3, X4 is negative the last table is optimal.
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The rules for selecting the entering and leaving variables are referred to
as the optimality and feasibility conditions
(a) Optimality conditions:
The entering variable in maximization (minimization) problem is the
non-basic variable having the most negative (positive) coefficient in the Zrow. Ties are broken arbitrarily.
The optimum is reached at the iteration where all the Z-row
coefficients of the non-basic variables are non-negative (non-positive)
(b) Feasibility Condition:
for both the maximum and minimum problems the leaving variables is
the basic variables associated with the smallest non-negative ratio. Ties
are broken arbitrarily.
The steps of the simplex method are:
1. Determine a starting basic feasible solution
2. Select an entering variable using the optimality condition. Stop if
there is no entering variable.
3. Select a leaving variable using the feasibility condition.
4. Determine the new basic solution by using the appropriate GaussJordan computation.
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5. Go to step 1
Special Cases in Simplex Method
1.
2.
3.
4.
Degeneracy
Alternative optima (Infinity of Solution)
Unbounded Solution
Non-existing or Infeasible Solution
1) Degeneracy
one or more basic variable(s) has zero value.
If you find more than one leaving variable (i.e.;
we have two or more variables having the same
ratio in the R.H.S) so one or more of the basic
variable(s)will be equal zero in the next iteration.
This condition indicates that the model has at
least one redundant constraint.
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Example 1:
Maximize Z= 3 X1 + 9X2
X1 +4 X2 <= 8
X1 +2 X2 <= 4
X1,
X2 >=0
Standard LP form
Maximize Z=3X1 + 9X2
S.T
X1
+4X2
+ X3
X1 + 2X2
X1, X2, X3, X4
+ X4
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=8
=4
>=0
48
X1=0, X2 =2, X3 =0, X4=0, Z=18
Is it possible to stop at the second iteration (when
degeneracy first appears) even though it is not optimum?
The answer is NO, because the
solution may be temporarily degenerate.
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2) Alternative optima (Infinity ofSolution):
When the objective function is parallel to a binding
constraint. So the objective function (Z) will have the
same optimal value at more than one solution point.
Example 2:
Maximize Z= 2 X1 + 4 X2
X1 + 2X2
X1 + X2
X1,
X2
<= 5
<= 4
>=0
Standard LP form
Maximize Z=2X1 + 4X2
S.T
X1 +2X2 + X3
=8
X1 + X2
+ X4 = 4
X1, X2, X3, X4
>=0
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First Optimum Solution:
X1*=0, X2*=5/2, X3*=0, X4*=3/2, Z*=10
Second Optimum Solution: X1*=3, X2*=1, X3*=0, X4*=0, Z*=10
In the second table: X1=0 , X2= 5/2
In the third table:
X1= 3 , X2= 1
The general form:
X1 = α (0) + (1-α) (3) = 3-3α
X2 = α (5/2) + (1-α) (1) = 1+3/2α
α=0→1
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51
3) Unbounded Solution:
In which the objective function (Z)
can be increased indefinitely without
violating any of the constraints i.e.;
the solution space is unbounded in at
least one direction.
Example 3:
Maximize Z= 2 X1+ X2
S.T
X1- X2 <= 10
2X1
<= 40
X1, X2
>=0
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Standard LP form
Maximize Z=2X1 +X2
S.T
X1 -X2 + X3
= 10
2X1
+ X4= 40
X1, X2, X3, X4
>=0
Both X1, X2 are candidates for entering the solution. But all the constraint
coefficient under X2 are negative or zero, meaning that X2 can be
increased indefinitely without violating any of the constraints. Any
increase in X2 will increase Z so it can be increased ∞ as X2 → ∞. Thus the
problem has no bounded solution.
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53
4) Non-existing or Infeasible
Solution:
If the constraints are not
satisfied simultaneously, the
model has no feasible solution.
This situation can never occur
if all the constraints are of the
type <= (assuming nonnegative R.H.S constraints)
because the slacks provide a
feasible solution.
Example 4:
Maximize Z= 3 X1 +2 X2
S.T
2 X1 + X2 <= 2
3 X1 +4 X2 >= 12
X1,
X2 >=0
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Transportation Model and Its Variants
Definition of the transportation model
The transportation model is a special class of the LP
problem. It deals with the situation in which a commodity
is shipped from source (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.
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55
Example 1:
There are three plants A, B, C and two destinations 1and 2, the
distance between them is as follows
The mile costs 8 cents find the cost/unit round to $
We define
X11 as the amount shipped from A to 1
X12 as the amount shipped from A to 2
X21 as the amount shipped from B to 1
X22 as the amount shipped from B to 2
X31 as the amount shipped from C to 1
X32 as the amount shipped from C to 2
The plants A, B, and C produces 1000, 1500, and1200 cars respectively.
The destination 1&2 demands are 2300 and 1400 cars.
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The LP model is
Minimize Z=80 X11+215 X12 +100 X21 +108 X22 +102 X31 +68 X32
S.T
X11 + X12
= 1000 ……….…… (A)
X21 + X22
= 1500 ……………. (B)
X31 + X32
= 1200 ……………. (C)
X11 + X21+ X31 = 2300 ……………. (1)
X12 + X22+ X32 = 1400 …………..... (2)
Xij >=0 i=1, 2, 3
j=1, 2
These constraints are all equations because the total supply from the three
plants = 1000 + 1500 + 1200 = 3700 cars, equals the total demands at the two
destinations = 2300 + 1400 = 3700 cars.
The LP model can be solved by the simplex method. However, the special
structure of the constraints allows us to solve the problem more conveniently
using the transportation tableau.
When the total supply does not equal the total demand, the
transportation model is said to be unbalanced. We must add a
dummy source or destination.
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Example 2:
In example (1) suppose that plant B
capacity is 1300 cars (instead of 1500). The
total supply =3500cars and the total demand
= 3700cars. So the total supply < the total
demand. Therefore, we have to add dummy
source with capacity 200 cars.
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Example 3:
In example (1) suppose that the demand of
the destination 1 is 1900 cars (instead of 2300). The
total supply =3700cars and the total
demand = 3300cars. So the total supply > the total
demand. Therefore, we have to add dummy
destination with 400 cars.
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The Transportation Algorithm
Step (1): Determine an Initial Basic Feasible Solution (IBFS) by
using one of the following methods.
1.The North-West Corner Method OR,
2.The Least-Cost Method OR,
3.Vogel Approximation Method.
A general T.M. with m sources and n destinations has m + n
constraints, one for each source and each destination. However,
because the T.M. is always balanced, one of these equation is
redundant. Thus, the model has m + n - 1 independent
constraint equations, which means that the starting basic
solution consists of m + n – 1 basic variables.
Step (2): Improve the IBFS
•Determine an entering variable.
•Determine a leaving variable.
•Obtain a new BFS
•If the solution is optimal stop. Otherwise; repeat step (2).
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Steps of North-West Method
1. Allocate as much as possible to a selected
cell X11
2. Cross out the row or column with zero
supply or demand (if both are satisfied at
the same time, only one should be crossed
out).
3. If exactly one row or column is left
uncrossed out stop. Otherwise go to
step(1)
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Example 3:
Solve the following
transportation problem.
Solution
Step (1):
finding the IBFS
•Using the North-West
Corner Method
Transportation Cost = 5(10) +10(2) +5(7) +15(9) +5(20) +10(18) =520
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The Least-Cost Method:
This method gives a better starting solution because
it takes the cost into consideration
Transportation Cost = 15(2) +15(9) +0(7) +10(20) +5(4) +5(18) =475
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Steps of Least-Cost Method
1. Assign as much as possible to the variable with
the smallest unit cost cell.
2. Cross out the row or column with zero supply of
demand (if both are satisfied at the same time,
only one should be crossed out).
3. If exactly one row or column is left uncrossed
out stop. Otherwise go to step(1)
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The Vogel Approximation Method:
This method provides a better starting solution (optimal or
closed to the optimal solution)
Steps of Vogel Approximation Method
1. In each row and column, subtract the lowest two costs from
each other.
2. Determine the row or column with the largest difference.
3. Allocate as much as possible to the variable with the least
cost in the selected row or column. Any row or column with
zero supply or demand would not be used in computing the
following step.
4. If exactly one row or column remains uncrossed out stop.
Otherwise, go to step (1).
5. If only one row or column with positive supply or demand
remains uncrossed out determine the basic variable by the
least-cost method.
6. If all uncrossed out rows and columns have assigned zero
supply and demand, determine the zero basic variable by
the least-cost method.
Note: In all the three methods, if you have a tie you can break
the tie arbitrarily.
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X31=5
X12=15
X23=15
X14=0,
X34=5,
X24=10
Dr.Farouk Sha'ban
Transportation Cost = 15(2) + 15(9) +0(4)
+10(20) +5(4) +5(18) = 475
It is expected to produce better starting
solution than least-cost method.
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Improving the IBFS
Determine an entering variable using method of multipliers (U, V method),
In this method associate the multipliers with row I and column j of the
transportation tableau.
•For each current basic variable
Ui + Vj = Cij for each basic Xij
•For non-basic variable
C'ij = Ui + Vj - Cij for each non-basic Xij
Choose the non-basic variable with the largest positive coefficient (C'ij) to be
the entering variable. The solution is optimal if and only if all C'ij are nonpositive.
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Improving the IBFS
Determine an entering variable using method of multipliers
(U,V) method, In this method associate the multipliers with row I
and column j of the transportation tableau.
For each current basic variable
Ui + Vj = Cij for each basic Xij
For non-basic variable
C’ij = Ui + Vj - Cij for each non-basic Xij
Choose the non-basic variable with the largest positive
coefficient (C’ij) to be the entering variable. The solution is
optimal if and only if all C’ij are non-positive.
Consider the same example:
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The entering variable must have the maximum C'ij which is X31 = 9
To determine the leaving variable make a closed loop of basic variable
starting with maximum C'ij of non-basic variable keeping:
1. Supply and demand requirements remain satisfied.
2. No negative shipments are allowed through any of the routes.
Therefore, θ=5 and X31 will be basic with value 5 and X22 is a
basic variable but with value 0 and X11 will be non-basic (leaving variable)
The cost of the IBFS by Northwest corner method was 520$
The first iteration cost is 475$ (less which is better)
In the second iteration
Get the entering variable with maximum C'ij = C'14 = 4
The Leaving variable is X24 =10
X14 = 10
and the new solution is:
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All C'ij is negative for all non-basic variables, so the solution is optimal which is:
X12 =5, X14 =10, X22 =10, X23 =15, X31 =5, X34 =5
With optimum cost
= 5*2 +10*11 + 10*7 + 15*9 +5*4 + 5*18
= 10+110+70+135+20+90
= 435
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The Assignment Model
There are n machines M1, M2, M3…, Mn. And n
different jobs J1, J2, …., Jn.
The jobs are to be assigned to these machines.
The machine cost for each job depends on the
machine to which it is assigned.
Each machine can work only on one job.
The problem is to assign the jobs to the machines to
minimize the total cost of machining.
Number of possible assignment = n!
If n=5, then 5! = 120 ways
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The Linear Programming Formulation of the Problem
Decision variables
Xij job i is assigned to machine j
Objective Function
Minimize Z= Σnj=1 Σni=1 Cij Xij
Constraints
Σnj=1 Xij =1
Σni=1 Xij =1
Xij =1
Xij =0
i= 1, 2… n
j= 1, 2… n
if the job i is assigned to the machine j
other wise
The Assignment Algorithm (Hungarian Algorithm)
It is the most efficient way or solving the assignment problem.
It is based on the idea that if a constant is added or
subtracted from any row or column the optimal assignment is not
affected.
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The Algorithm Steps
1. Subtract the smallest value in each column from all the
values in that column.
2. Subtract the smallest value in each row from all the
values in that row.
3. Note: step 1 & 2 will produce a cost matrix containing at
least one zero element in each row or column.
4. Determine a feasible assignment using the cells with
zero cost. If this is possible stop, as optimal solution is
found. This step is done as follows:
a) Determine the number of lines required to cover all
zeros.
b) If the number of lines is equal to n, stop (This is the
optimal assignment).
c) If the number of lines is less than n then go to (4).
5. Subtract the smallest value not covered with lines from
all uncovered values and add it to the value at the
intersection of the lines then go to (3).
Note: Obtain the minimum cost from the original table.
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Example 1:
Solve the following
assignment problem.
Jobs
Person
Solution:
The optimal assignment is X12, X21, X33
Job1 is assigned to machine2
Job2 is assigned to machine1
Job3 is assigned to machine3
The cost = 10 + 9 + 8 = 27$
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Example 2:
Solve the following assignment problem.
Solution:
The optimal assignment is X15, X22, X33, X44, X51
Job1 is assigned to machine5
Job2 is assigned to machine2
Job3 is assigned to machine3
Job4 is assigned to machine4
Job5 is assigned to machine1
The cost = 3 + 4 + 2 + 5 + 3 = 17$
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6.Network Models
6.1Introduction
There is a multitude of operations research situations that can be
modeled and solved as networks (nodes connected with branches)
We have 5 network algorithms such as:
1. Minimal spanning tree.
2. Shortest route algorithm.
3. Minimum-cost capacitated network algorithm.
4. Critical Path Method (CPM).
5. Program Evaluation and Review Technique (PERT).
6.7Project planning and control with PERT-CPM
The successful management of large-scale projects requires
careful planning, scheduling and control of numerous interrelated
activities especially when these activities have to be performed in a
specified technological sequence, to aid in these tasks special
procedures based on the use of network techniques were developed. The
most important of these procedures are:
 Critical Path Method (CPM).
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 Program Evaluation and Review Technique (PERT).
Applications of PERT & CPM
1. Construction projects ( buildings, highways,
houses, bridges)
2. Preparations of bids and proposals for large
projects.
3. Maintenance planning of oil refineries, ship
repair and other large operations.
4. Manufacture and assembly of large items as
aeroplans and computers.
5. Development of new weapons system.
6. Simple projects such as cleaning and painting.
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Basic Terminology
1. A Project:
A project defines a combination of interrelated activities that
must be done in a certain order before the entire task can be
completed.
2. An Activity:
It is an effort required to complete a plan of the project. Most
activities can not start until some others are completed.
3. Real Activity:
Consumes resources and time.
4. Waiting Activity:
Consumes time.
5. Dummy Activity:
Consumes no time and no resources. It represents the
dependence of one activity upon another.
6. An Event:
Represents a point in time. It signifies the beginning or ending
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of some activities.
Network Representation
1. An activity is represented by a line or an arrow (usually with time
estimate written along it). The arrow begins and ends with an
event which is represented by a node.
1. A dummy activity is represented by dashed line arrow or solid
arrow with zero time estimate.
2. The arrow must be oriented from left to right.
3. The network must have one initial node and one ending node.
4. Any two events may not be connected by more than one activity.
5. Event numbers must not be duplicated in the network.
6. Each activity is represented by one and only one arrow. No
single activity can be represented twice in the network.
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Numbering a network
1. An initial event is one which has arrows
coming out of it only. Number it (1).
2. Delete all arrows emerging from event (1).
This will create at least one more initial
event, number these new events(s) as (2,
3…).
3. Delete all emerging arrows from these
numbered events which will create new
initial events and continue until the last
event which has no arrows emerging from it.
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Solution by Linear Programming Formulation
Each node represents a variable. Each arc represents a constraint.
Example 1:
Consider the following Network problem.
Decision variables
Let ti represents the time at which event (i) occurs i= 1, 2, 3, 4, 5.
t4: time at which jobs (C) & (D) are completed
t5: time at which the project is completed
Time of completion of the whole project = t5 – t1
Minimize Z= t5 – t1
t2 – t1 ≥ 3 , t3 – t 1 ≥ 1 , t3 – t2 ≥ 0
t4 – t3 ≥ 2 , t4 – t2 ≥ 4 , t5 – t4 ≥ 5
ti ≥ 0
i= 1, 2, 3, 4, 5
This problem can be solved by simplex method Z* is the minimum time
required to complete the project.
Objective Function
Constraints
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Solution by Network Analysis
CPM Calculations:
Let
Dij = Duration (completion time) of activity ij
ESi = The earliest start time at which event (i) occurs
LFi = The latest finish time at which event (i) occurs without
delaying the project beyond its earliest time.
ESi ………. Forward
ESi
= max [ESi+ Dij] for all (i, j)
ESi
= max [(2+8), (1+6), (3+5)]
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Start from the terminal node and move backward to the initial node for
terminal node LFn = ESn
= max [10, 7, 8] =10
Start from the initial node and move to the terminal node for initial
node ES1= 0
LFi ………. Backward
ESi = min [LFj - Dij] for all (i, j)
LFi = min [(10-3), (9-5), (13-4)]
= min [7, 4, 9] =10
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Example1:
Determine the critical path for the
project network given below (all
the durations are in days)
Solution
The critical path is defined by
1-2-4-5-6
The duration of the project =
25 days
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