Operations Research First Year Computer

Operations ResearchFirst Year Computer

Text Book: Operations Research8th Edition

Hamdy A. Taha

Dr.Farouk Sha'ban 2

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.

Dr.Farouk Sha'ban 3

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.

Dr.Farouk Sha'ban 4

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.

Dr.Farouk Sha'ban 5

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

the modeled system operates.Dr.Farouk Sha'ban

Dr.Farouk Sha'ban 7

2. Model ConstructionTranslating the real world problem into mathematical relationships (the most suitable model to represent the system, LP, dynamic programming, integer programming, ………..)

3.Solution of the modelUsing well-defined optimization techniques. An important aspect of model solution is sensitivity analysis.

Dr.Farouk Sha'ban 8

4.Model validityTesting 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 solutionImplementation of the solution of validated model involves the translation of the model's results into instructions issued in understandable form to the individual

Dr.Farouk Sha'ban 9

Basic components of the model

1.Decision VariablesIt 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 FunctionIt 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.ConstraintsThese 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.).

Dr.Farouk Sha'ban

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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

Constraints3 X1 + 2 X2 <= 800x60

X1 >=10000 X2 <=6000

X1, X2 >=0Dr.Farouk Sha'ban


Example 2: Feed mix problemA 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.



X1= weight of feed A kg/day/animalX2= 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 <= 0.1

X1, X2 >=0



X1= weight of feed A kg/day/animalX2= 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 <= 0.1

X1, X2 >=0


Example 3: Blending ProblemAn 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.


Problem Formulation

Decision variables

X1= Fraction of ton to be selected from mine number 1X2= Fraction of ton to be selected from mine number 2X3= Fraction of ton to be selected from mine number 3X4= 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 >= 5 90 X1 + 150 X2 + 75 X3 + 175 X4 >= 10 45 X1 + 25 X2 + 20 X3 + 37 X4 >= 30 X1 + X2 + X3 + X4 = 1 X1, X2, X3, X4 >= 0


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 variablesX1= 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 <= 8

X2 <= 10 8(25) X1+ 8(15) X2 >= 1800 200 X1 + 120 X2 >= 1800

X1, X2 >=0Dr.Farouk Sha'ban

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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

Dr.Farouk Sha'ban


Formulate to minimize the trim loss and the number of rolls needed to satisfy the order.


Problem FormulationDecision variables Xj = Number of standard rolls to be cut according to setting j

j = 1, 2, 3, 4, 5, 6Number 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 - 150Y2= X1 + X2+ 2 X5 - 200Y3= 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 FunctionMinimize 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


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


OR Maximize Z= Σn

j=1 CjXj

ConstraintsΣn

j=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. Minimize Z= Σnj=1 CjXj

2. Σnj=1 AijXi >= Bi for some values of i

3. Σnj=1 AijXi = Bi for some values of i

4. Xi unrestricted in sign for some values of i


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).


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).


Graphical Solution

Construction of the LP modelExample 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.


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.


Problem FormulationDecision variables

X1= Tons produced daily of exterior paint.X2= Tons produced daily of interior paint.

Objective FunctionMaximize Z= 5 X1 + 4 X2

Constraints 6 X1 + 4 X2 <= 24 X1 + 2 X2 <= 6 - X1 + X2 <= 1 X2 <= 2 X1, X2 >=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.


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.


.

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=10And 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$.

Dr.Farouk Sha'ban


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.


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 >= 0X1, X2 >= 0



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.


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.


The Simplex Method

IntroductionIt 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 SolutionExample 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

Dr.Farouk Sha'ban


Determination of basic solutionsThe 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.2. 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

)!(!!mnm

n


Example 2:

Maximize Z= 5X1 + 6 X2

S.T 2X1 + 3X2 >= 182X1 +X2 <= 123X1 + 3X2 = 24X1, X2 >=0

Standard LP form

Maximize Z=5X1 + 6X2

S.T 2X1 +3X2 + X3 = 182X1 + X2 + X4 = 123X1 + 3X2 + X5 = 24X1, X2, X3, X4, X5 >=0


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) 36 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) 30 Yes

7) X2, X5 (8, 0, 2, -4, 0) N0

8) X3, X4 (4.5, 3, 0, 0, 1.5) 40.5 Yes

9) X3, X5 (6 . 2 .0 . -2 .5 ) No

10) X4 . X5 (4. 4 . – 2 . 0. 0) NO

solutionspossiblexxmn

aresolutionsbasicpossibleofnumberMaximum

10!2!3!345

!2!3!5

!)35(!3!5

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The basic feasible solutions are the corner points


The Simplex AlgorithmWe 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 = 0S.T. 6 X1 +4 X2 + X3 <= 24

X1 +2 X2 + X4 <= 6 - X1 + X2 + X5 <= 1

X2 + X6 <=2 X1, X2, X3, X4, X5, X6 >=0

The variables X3, X4, X5, X6 are the slacks associated with the four (<=) constraints.


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.

Dr.Farouk Sha'ban


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 Z-row. 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 solution2. 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 Gauss-

Jordan computation.5. Go to step 1


Special Cases in Simplex Method1. Degeneracy2. Alternative optima (Infinity of Solution)3. Unbounded Solution4. 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.


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 = 8

X1 + 2X2 + X4 = 4 X1, X2, X3, X4 >=0


X1=0, X2 =2, X3 =0, X4=0, Z=18Is 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.


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 <= 5 X1 + X2 <= 4

X1, X2 >=0

Standard LP form

Maximize Z=2X1 + 4X2

S.T X1 +2X2 + X3 = 8 X1 + X2 + X4 = 4X1, X2, X3, X4 >=0


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*=10In the second table: X1=0 , X2= 5/2In 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


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 formMaximize Z=2X1 +X2

S.T X1 -X2 + X3 = 102X1 + X4= 40X1, 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.

Dr.Farouk Sha'ban


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 non-negative R.H.S constraints) because the slacks provide a feasible solution.

Example 4:Maximize Z= 3 X1 +2 X2

S.T 2 X1 + X2 <= 23 X1 +4 X2 >= 12X1, X2 >=0

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Transportation Model and Its Variants

Definition of the transportation modelThe 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.

Dr.Farouk Sha'ban


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 1X12 as the amount shipped from A to 2X21 as the amount shipped from B to 1X22 as the amount shipped from B to 2X31 as the amount shipped from C to 1X32 as the amount shipped from C to 2The 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 isMinimize 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, 2These 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.

Dr.Farouk Sha'ban


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.


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.


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).


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)


Example 3: Solve the following transportation problem.

SolutionStep (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


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


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)


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=5X12=15X23=15X14=0, X34=5, X24=10

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.

Dr.Farouk Sha'ban


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 variableUi + Vj = Cij for each basic Xij •For non-basic variableC'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.


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 variableUi + Vj = Cij for each basic Xij

For non-basic variableC’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 iterationGet the entering variable with maximum C'ij = C'14 = 4

The Leaving variable is X24 =10X14 = 10

and the new solution is:

Dr.Farouk Sha'ban


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


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


The Linear Programming Formulation of the Problem

Decision variablesXij job i is assigned to machine j

Objective FunctionMinimize Z= Σn

j=1 Σni=1 Cij Xij

ConstraintsΣn

j=1 Xij =1 i= 1, 2… nΣn

i=1 Xij =1 j= 1, 2… n Xij =1 if the job i is assigned to the machine j

Xij =0 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.


The Algorithm Steps1. 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.


Example 1:

Solve the following assignment problem.

Solution:The optimal assignment is X12, X21, X33

Job1 is assigned to machine2Job2 is assigned to machine1Job3 is assigned to machine3The cost = 10 + 9 + 8 = 27$

Jobs Person


Example 2: Solve the following assignment problem.

Solution:The optimal assignment is X15, X22, X33,

X44, X51

Job1 is assigned to machine5Job2 is assigned to machine2Job3 is assigned to machine3Job4 is assigned to machine4Job5 is assigned to machine1The cost = 3 + 4 + 2 + 5 + 3 = 17$

77

6.Network Models6.1Introduction

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). Program Evaluation and Review Technique (PERT).

Dr.Farouk Sha'ban

78

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.Dr.Farouk Sha'ban


Basic Terminology1. 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 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.

2. A dummy activity is represented by dashed line arrow or solid arrow with zero time estimate.

3. The arrow must be oriented from left to right.

4. The network must have one initial node and one ending node.

5. Any two events may not be connected by more than one activity.

6. Event numbers must not be duplicated in the network.

7. Each activity is represented by one and only one arrow. No single activity can be represented twice in the network.


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.


Solution by Linear Programming Formulation

Each node represents a variable. Each arc represents a constraint.

Example 1:

Consider the following Network problem.Decision variablesLet ti represents the time at which event (i) occurs i= 1, 2, 3, 4, 5.t4: time at which jobs (C) & (D) are completedt5: time at which the project is completedTime of completion of the whole project = t5 – t1

Objective Function Minimize Z= t5 – t1

Constraints t2 – t1 ≥ 3 , t3 – t1 ≥ 1 , t3 – t2 ≥ 0 t4 – t3 ≥ 2 , t4 – t2 ≥ 4 , t5 – t4 ≥ 5 ti ≥ 0 i= 1, 2, 3, 4, 5This problem can be solved by simplex method Z* is the minimum time

required to complete the project.


Solution by Network Analysis

CPM Calculations:

Let Dij = Duration (completion time) of activity ijESi = The earliest start time at which event (i) occursLFi = The latest finish time at which event (i) occurs without

delaying the project beyond its earliest time.

ESi ………. ForwardESi = max [ESi+ Dij] for all (i, j)ESi = max [(2+8), (1+6), (3+5)]


= max [10, 7, 8] =10

Start from the initial node and move to the terminal node for initial node ES1= 0LFi ………. BackwardESi = min [LFj - Dij] for all (i, j)LFi = min [(10-3), (9-5), (13-4)] = min [7, 4, 9] =10

Start from the terminal node and move backward to the initial node for terminal node LFn = ESn


Example1:

Determine the critical path for the project network given below (all the durations are in days)

Solution

The critical path is defined by 1-2-4-5-6The duration of the project = 25 days

Documents

Operations Research First Year Computer