Operations Research Summary

Czech Technical University in Prague

Faculty of Electrical Engineering

Department of Electroenergetics

Operation ResearchSummary

Author:Minh-Quan Dang

Draft N o1

November 19, 2015Prague

Abstract

This is the summary of the essay.

Keywords

1

Contents

Abstract 1

Keywords 1

List of Figures 3

List of Tables 4

1 Introduction 11.1 Definition Operation Research . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Problem when conducting OR . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

2 Linear Programming 22.1 Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22.2 Type of LP problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22.3 Assumptions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

2.3.1 Proportionality(linearity) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32.3.2 Additivity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32.3.3 Divisibility . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32.3.4 Non-negativity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42.3.5 Certainty . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

2.4 Solving the LP problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

3 Graphic Solution for LP Problem 73.1 Possible outcome of graphical method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

4 Simplex Method 104.1 Algorithm . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104.2 LP task standard form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104.3 Possible outcome from simplex method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114.4 Solving LP problem with Simplex Method . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

4.4.1 Case of one optimal solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114.4.2 Case of Multiple optimal solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134.4.3 Case of Unbounded feasible region . . . . . . . . . . . . . . . . . . . . . . . . . . . 144.4.4 Case of No feasible solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

4.5 Adapting to other model forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 154.5.1 Equality constrains . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 154.5.2 Functional constraints in ≥ form . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

5 Dual task 165.1 Rule to create dual task . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165.2 Duality theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 175.3 Solving Dual task . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

6 Allocation tasks 186.1 Definition of allocation tasks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 186.2 Transportation task . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 186.3 Solving balanced transportation tasks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

7 Terminology 24

References 25

LIST OF FIGURES LIST OF FIGURES

List of Figures

1 Proportionality Assumption . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32 Graphically expression of constrains . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73 Graphically expression of CF . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74 Graphically interpret when most favorable value exist . . . . . . . . . . . . . . . . . . . . 85 Graphically interpret when no most favorable value exist . . . . . . . . . . . . . . . . . . . 86 Pair of symmetric dual tasks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

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LIST OF TABLES LIST OF TABLES

List of Tables

1 Data for the example problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52 Basic simplex table . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113 Possible outcome from simplex method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114 Initial simplex table of one OS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125 First iteration simplex table of one OS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 126 2nd iteration simplex table of one OS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 127 Initial simplex table of Multiple OS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 138 First iteration simplex table of Multiple OS . . . . . . . . . . . . . . . . . . . . . . . . . . 139 Second iteration simplex table of Multiple OS . . . . . . . . . . . . . . . . . . . . . . . . . 1410 Initial simplex table of Unbounded feasible region . . . . . . . . . . . . . . . . . . . . . . . 1411 Example Balanced Transportation Task . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1912 Starting solution by North West corner method . . . . . . . . . . . . . . . . . . . . . . . . 1913 Starting solution by row minimum method . . . . . . . . . . . . . . . . . . . . . . . . . . 2014 Starting solution by column minimum method . . . . . . . . . . . . . . . . . . . . . . . . 2015 Starting solution by column minimum method . . . . . . . . . . . . . . . . . . . . . . . . 2016 Starting solution by Vogel method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

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1 INTRODUCTION

1 Introduction

1.1 Definition Operation Research

If we define by meaning of terms so the Operations Research(OP) is:

• Operations: The activities in which a organization is involved. The activity could be related tosuch diverse areas as manufacturing, transportation, construction, telecommunications, financialplanning,health care, the military, and public services, to name just a few.

• Research: The systematic investigation into and study of materials and sources in order to establishfacts and reach new conclusions.

Then the basic concept of OR could be expressed as three steps.

• Step 1

Define the task, collecting data, define the criteria function and all variable and constrains.

• Step 2

Base on collected data about criteria function, variable and constrains, we will construct amathematical model that attempts to simulate the real problem behavior.

• Step 3

With the hypothesis that our model is a sufficiently precise model of the essential features ofthe problem. Then we will use mathematical technique to find the optimal solution for theproblem. The modification and verification is needed

1.2 Problem when conducting OR

After the brief look on the definition of OR several questions is raised.

Step 1

1. What is the task for our research?

2. What is the feature of the task, is that any assumptions?

3. What is the variable which has influence to our task?

4. What is the criteria function for our task?

5. What is the constrains for our task and variable?

Step 2

1. How we can express our task by mathematical model?

2. Which form is convenient for further process?

3. Is our model reflex the real problem precisely?

4. What is level of uncertainty and how we much we can tolerate?

Step 3

1. Which technique we could use to solve the task?

2. What is the advantage and disadvantage of them?

3. Which one we should choose?

4. Is the solution is correct? How we can confirm ?

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2 LINEAR PROGRAMMING

2 Linear Programming

2.1 Definition

” Briefly, the most common type of application involves the general problem of allocating limited re-sources among competing activities in a best possible (i.e., optimal) way. ” [1]

Linear programming uses a mathematical model to describe the problem of concern. The problem ismaximization or minimization of criterion function value and at the same time fulfillment of all relevantconstraints.

• Linear: All the mathematical functions in this model are required to be linear functions.

• Programming: planning

We can understand that Linear programming is an OR which is:

Step 1: Define the Linear programming task include:

• Linear criterion function

• Set of constrains

• Non-negative assumptions

Step 2: we use linear mathematical model to abstract the problem.

Step 3:we try to obtain an optimal result, the best goal among all feasible alternatives.

2.2 Type of LP problems

Linear programming deal with 3 typical problems:

1. Resorce-allocation (Maximum profit from production )

2. Cost-benefit-trade-off (Blending task to minimum effect)

3. Fixed-requirement

2.3 Assumptions

In the Step 1 when we start the research the problem in the real world is usually complicated anddifficult to comprehend all the aspects. For that reason we need to make some assumptions. The functionof assumption are:

1. Simplify the problem.

2. Fill up all unknown information

When using assumptions we must face a fact that this could be or not be correct. Sometime the assumptionis correct in this task but is incorrect in the other task. Then the necessary to consider the assumptionswith these questions:

1. Is that assumption logically acceptable for the task?

2. Is the validity of this assumption depend on task’s boundary(size, length ...)

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2.3 Assumptions 2 LINEAR PROGRAMMING

Any change of the assumptions will lead to LP model update. There are 5 assumption is made for LPtask:

1. Proportionality(linearity)

2. Additivity

3. Divisibility

4. Non-negativity

5. Certainty

2.3.1 Proportionality(linearity)

”The contribution of each activity to the value of the objective function Z is proportional to the levelof the activity xj, as represented by the cjxj term in the objective function. Similarly, the contribution ofeach activity to the left-hand side of each functional constraint is proportional to the level of the activityxj, as represented by the aijxj term in the constraint.”

This assumption means we assume that all the output and input is linear dependence. In other word,we can have as much input material as we want as long as the value lies within the constrains. Thisassumption could not be correct for example when we buy to much goods the price for that goods willchange and for the same amount of money we could but less amount of goods than it should be.

Figure 1: Proportionality Assumption

2.3.2 Additivity

” Every function in a linear programming model (whether the objective function or the function onthe left-hand side of a functional constraint) is the sum of the individual contributions of the respectiveactivities.”

The basic idea of this assumption is we can define the sum by means of all variables are independentto each other. The will be no cannibal effect, that means no complements nor substitutes. In other word,there will no requirement for buying both 2 types of goods or 1 goods can replace for another goods.

2.3.3 Divisibility

”Decision variables in a linear programming model are allowed to have any values, including nonintegervalues, that satisfy the functional and nonnegativity constraints. Thus, these variables are not restricted

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2.4 Solving the LP problem 2 LINEAR PROGRAMMING

to just integer values. Since each decision variable represents the level of some activity, it is being assumedthat the activities can be run at fractional levels.”

This assumption we will ignore the fact that some goods we must buy as a whole pack, variable canbe taken as fraction.

2.3.4 Non-negativity

We assume the value of variable must not smaller than zero. The reason for this assumption relatedto the task and variable definition. It will be non sense to deal will negative for example table or chair.

2.3.5 Certainty

”The value assigned to each parameter of a linear programming model is assumed to be a known con-stant.”

This assumption is also related to task definition while the task boundary is fixed. Then the inputparameter will not change by time or other factor.

2.4 Solving the LP problem

To understand the process for solving the LP problem we will go though an example and then makethe summary from that. The example is called Prototype Example [1]. Solving process includes:

1. Problem analyze

2. Identifine the needed data

3. Formulation as a LP Problem

4. Finding the optimal solution

Problem analyze:

A company which product glass (Plant 1), windows (Plant 2) and glass doors (Plant 3) decides toproduce new products.

Product 1: An 8-foot glass door with aluminum framingProduct 2: A 4x6 foot double-hung wood-framed window

The Product 1 requires the production capacity in Plant 1 and Plan 3 and Product 2 requires Plant 2and 3. The question is how the company should produce in order to archive the biggest profit .

Definition of the problem

The task is determination of the production rate should be for the two products in order to maximizethe output profit, subject to the restriction imposed by the limited production capacities available in thethree plants. The production rate is defined as the number of batches produced per week. Productionrate can be none or as much as possible.

Identifine the needed data

Base on the task definition we recognize that we need to collect some data.

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1. Number of available production hour per week of each plant.

2. Number of hour required for produce one batch.

3. Profit per batch produced of each new product

Each category of data will be collected from other sources. Such as the manufacturing division , manu-facturing engineer, marketing division, accounting department. The collected data will be gathered intoa table.

Table 1: Data for the example problem

Production Timeper Batch, Hours

Production

Plant 1 2Production Time

Available per Week, Hours1 1 0 42 0 2 123 3 2 18

Profit per batch $ 3,000 $ 5,000

Formulation as a LP Problem

We will assign symbol for each decisiion variable

x1 = number of batches of product 1 produced per weekx2 = number of batches of product 2 produced per week

Z = total profit per week

The problem now can express

Maximize Z = 3x1 + 5x2

subject to the restrictions

x1 ≤ 4

2x2 ≤ 12

3x1 + 2x2 ≤ 18

x1 ≥ 0,

x2 ≥ 0

By word we can say the total profit Z is the sum of profit from each product. Plant 1 can produces1 batch of production 1 per hour and for maximum 4 hours per week. Plant 2 can produces 2 batch ofproduction 2 per hour and for maximum 12 hours per week. Plant 2 can produces 3 batch of production1 and 2 batch of production 2 per hour and for maximum 18 hours per week. The number of batchescannot be negative value.

Now we have enough information to proceed to find the optimal solution.

Finding the optimal solution

The technique used for finding the optimal solution are:

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1. Graphic Solution

2. Simplex Method

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3 GRAPHIC SOLUTION FOR LP PROBLEM

3 Graphic Solution for LP Problem

The basic principle for this method base on the fact that LP with 2 variable can be express graphically.The constrains will some how limit an area of the plain then by moving the CF we can find the optimalsolution. Take a look on a example with 2 constrains

6x + 8y = Max

Subject to

Constrain 1: 30x + 20y ≤ 300

Constrain 2: 5x + 10y ≤ 110

x, y ≥ 0

(a) Constrain 1: 30x+ 20y ≤ 300 (b) Constrain 2: 5x+ 10y ≤ 110

Figure 2: Graphically expression of constrains

As we can see on the chart the constrain will intersect two axis, which are non-negative assumption,then limit an area of the plain. If we have the lower or equal constrain then the area will be the lowerpart under the line. In contrast, the greater constrain will limit the higher part upon the line. All theconstrain and non-negative assumption will limit a feasible area where all constrains is sufficient by allvariable. The next step is looking for the optimal solution.

(a) Area divided (b) Moving CF

Figure 3: Graphically expression of CF

The CF will divides the plain into two parts. When the CF line moves to the right that means we arelooking for the maximum value and when it moves to the left then we find the minimum value.

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3.1 Possible outcome of graphical method 3 GRAPHIC SOLUTION FOR LP PROBLEM

3.1 Possible outcome of graphical method

The optimal solution depends on how the CF line intersect with the feasible area. There are 4possibilities could happen:

1. One optimal solution only

2. Multiple optimal solutions

3. Unbounded solution

4. No feasible solution

(a) One optimal solution only (b) Multiple optimal solutions

Figure 4: Graphically interpret when most favorable value exist

One optimal solution only

The optimal solution is one corner point of the feasible region. We determine which corner is optimalsolution by moving the CF line.

Multiple optimal solutions

When the CF line is align to one edge of the feasible region then there will be infinite number ofsolution, each have the same optimal value of objective function.

(a) Unbounded solution (b) No feasible solution

Figure 5: Graphically interpret when no most favorable value exist

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3.1 Possible outcome of graphical method 3 GRAPHIC SOLUTION FOR LP PROBLEM

No feasible solution

This possibility happen when constrains are not consistent and there is no feasible region exist.This outcome occurs only if:

1. It has no feasible solutions

2. The constraints do not prevent improving the value of the objective function (Z) indefinitely in thefavorable direction (positive or negative). Unbounded objective.

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4 SIMPLEX METHOD

4 Simplex Method

4.1 Algorithm

1. Trasform the task into standard form

2. Identify the starting feasible solution (slack variable, artificial base)

3. Evaluate the (zj − cj) value for non-based value and presence of the optimal solution in initialsimplex table.

4. If no optimal solution reached then one new vector is included into base and one original base vectoris excluded to create new base using Max-Min principle.

5. Recalculate of simplex table until all base vector is elementary vectors

6. The optimal solution reached when all (zj − cj) value is non-positive for non-base vectors.

4.2 LP task standard form

Standard form of LP is important for solving especially when using simplex method and dual task.

f(x) = cx = Min

Ax = b

x ≥ 0

If the LP task is not in the standard form we can transform it into standard form.

• If creterion function is looking for Max instead of Min then we will multiple both side with -1

f(x) = Max => −f(x) = Min

• Inequalities in constrains are changed to equalities using slack or surplus variable (method of ar-tificial base).

Using slack variable x3

x1 + x2 ≤ 1 −→ x1 + x2 + x3 = 1

Using artificial base

x1 + x2 = Min −→ x1 + x2 + wu1 + wu2 = Min

x1 − x2 ≥ 1 −→ x1 − x2− x3 + u1 = 1

−x1 + x2 ≥ 2 −→ −x1 + x2 − x4 + u2 = 2

x1, x2, x3, x4 ≥ 0

u1, u2 ≥ 0

The Simplex Tabular

f(x) =∑

cixi = Min∑aixi = bi

x ≥ 0

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4.3 Possible outcome from simplex method 4 SIMPLEX METHOD

Table 2: Basic simplex table

cj c1 c2 c3 c4B cB P0 P1 P2 P3 P4 hP3 c3 b1 a11 a21 a31 a41 b1/ai1P4 c4 b2 a12 a22 a32 a42 b2/ai2zj c3b1 + c4b2 c3a11 + c4a12 c3a21 + c4a22 c3a31 + c4a32 c3a41 + c4a42

zj-cj z0 z1-c1 z2-c2 z3-c3 z4-c4

4.3 Possible outcome from simplex method

1. One optimal solutionThe value (zj − cj) < 0 for all non-base vector Pj That means there will be no improvement incriterion function.

2. Multiple optimal solutionThe value (zj − cj) = 0 for non-base vector Pr That means if we continue recalculate the simplextable then we only jump from this corner point to another but CF value is constant.

3. Unbounded solutionFor non base vector Pr we get zr − cr > 0, but vector has no positive items)

4. No feasible solutionvalues of artificial variables are for optimum solution positive

Table 3: Possible outcome from simplex method

zj-cjelementary var non-base var artificial var

One optimal solution 0 <0 -Multiple optimal solution 0 0 and < 0 -Unbounded solution 0 >0 -No feasible solution - - >0

4.4 Solving LP problem with Simplex Method

4.4.1 Case of one optimal solution

Original problem:

3x1 + 4x2 = Max

Subject to

x1 + x2 ≤ 5

x1 + 2x2 ≤ 8

x1, x2 ≥ 0

Standard form

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4.4 Solving LP problem with Simplex Method 4 SIMPLEX METHOD

−3x1 − 4x2 = Min

Subject to

x1 + x2 + x3 = 5

x1 + 2x2 + x4 = 8

x1, x2 ≥ 0

Two slack variable x3, x4 is added to transform inequality into equality relation.

Initial simplex table

Table 4: Initial simplex table of one OS

cj c1 c2 c3 c4B cB P0 P1 P2 P3 P4 hP3 0 5 1 1 1 0 5P4 0 8 1 2 0 1 4 minzj 0 0 0 0 0

zj-cj 0 3 4 max 0 0

The value in the row zj − cj still exists positive element then we must apply principle Max-Min. Thevalue z2 − c2 is max so we will include vector P2 into base. The ratio h of P4 is min so we will excludevector P4 from the base.

First iteration

Table 5: First iteration simplex table of one OS

cj c1 c2 c3 c4B cB P0 P1 P2 P3 P4 hP3 0 1 0.5 0 1 -0.5 2 minP2 -4 4 0.5 1 0 0.5 8zj -16 -2 -4 0 -2

zj-cj -16 1 max 0 0 -2

The value in the row zj − cj still exists positive element then we must apply principle Max-Min.

2nd iteration

Table 6: 2nd iteration simplex table of one OS

cj c1 c2 c3 c4B cB P0 P1 P2 P3 P4 hP1 -3 2 1 0 2 -1P2 -4 3 0 1 -1 1zj -18 -3 -4 -2 -1

zj-cj -18 0 0 -2 -1

After 2 iteration we can see that the value of (zj − cj) are zero for elementary vectors and negativefor non-base vector so we know the optimal solution is reached.

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The optimal solution is Min = -18 for (2,3,0,0), but because the original task is find Max then wemust transform it back. The final optimal solution for Max CF is 18.

4.4.2 Case of Multiple optimal solution

Original problem:

2x1 + x2 = Max

Subject to

4x1 + 2x2 ≤ 8

x1 + x2 ≤ 3

x1, x2 ≥ 0

Standard form

−2x1 − x2 = Min

Subject to

4x1 + 2x2 + x3 = 8

x1 + x2 + x4 = 3

x1, x2 ≥ 0


Table 7: Initial simplex table of Multiple OS

cj 2 1 0 0B cB P0 P1 P2 P3 P4 hP3 0 8 4 2 1 0 2 minP4 0 3 1 1 0 1 3zj 0 0 0 0 0

zj-cj 0 2 max 1 0 0

We will include vector P1 and exclude vector P3 from the base.

First iteration

Table 8: First iteration simplex table of Multiple OS

Cj -2 -1 0 0B Cb P0 P1 P2 P3 P4 hP1 -2 2 1 0.5 0.25 0 4P4 0 1 0 0.5 -0.25 1 2 MinZj -4 -2 -1 -0.5 0

Zj-Cj -4 0 0 -0.5 0

Both zj-cj value of both elementary vector are 0 and negative for non-base vectors, that means theoptimal solution is reached. But not all non-base value has negative value zj-cj so we will have multipleoptimal solution. Then we should choose the other elementary vector to include for the next iteration forconfirmation.

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Second iteration

Table 9: Second iteration simplex table of Multiple OS

Cj -2 -1 0 0B Cb P0 P1 P2 P3 P4 hP1 -2 1 1 0 0.5 -1P2 -1 2 0 1 -0.5 2Zj -4 -2 -1 -0.5 0

Zj-Cj -4 0 0 -0.5 0

We see that the zj-cj value is not change from the last iteration.All optimal solution are a weighted average of these two optimal solutions

(x1, x2) = w1(2, 0) + w2(1, 2)

where the weight w1 and w2 (convex combination) are numbers that sastifiy the relationships

w1 + w2 = 1 and w1 ≥ 0, w2 ≥ 0

4.4.3 Case of Unbounded feasible region

Original problem:

x1 − x2 = Max

Subject to

−4x1 + 2x2 ≤ 4

x2 ≤ 4

x1, x2 ≥ 0

Standard form

−x1 + x2 = Min

Subject to

−4x1 + 2x2 + x3 = 4

x2 + x4 = 4

x1, x2 ≥ 0


Table 10: Initial simplex table of Unbounded feasible region

Cj -1 1 0 0B Cb P0 P1 P2 P3 P4 hP3 0 4 -4 2 1 0 -1P4 0 4 0 1 0 1 InfZj 0 0 0 0 0

Zj-Cj 0 1 max -1 0 0

The constrains do not prevent the value of the objective function Z from increasing indefinitely, so thesimplex method would stop with the message that Z is unbounded

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4.5 Adapting to other model forms 4 SIMPLEX METHOD

4.4.4 Case of No feasible solution

Original problem:

x1 + x2 = Min

Subject to

x1 − x2 ≥ 1

−x1 + x2 ≥ 2

x1, x2 ≥ 0

Standard form

x1 + x2 + 0x3 + 0x4 + wV1 + wV2 = 0

Subject to

x1 − x2 − x3 + V1 = 1

−x1 + x2 − x4 + V2 = 2

x1, x2 ≥ 0

4.5 Adapting to other model forms

4.5.1 Equality constrains

All the equality constrains could be cover into a pair of inequality constrains.

Ax = b => Ax ≤ b and Ax ≥ b

. However we do not want to increase the number of variable so we use artificial-variable technique.

1. Introducing a nonnegative artificail variable

Ax + x̃ = b

2. Assign an overwhelming penalty to objective function.

Z = cx−Mx̃

4.5.2 Functional constraints in ≥ form

For this kind of constrains we introduce both a surplus variable and and artificial variable x̃

Ax ≥ b =>> Ax− xi + x̃j = b

Z = cx =>> cx + 0xi + Mx̃j

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5 DUAL TASK

5 Dual task

Dual task is derived from the primary task, the meaning of dual task is process directly with specificprofit per unit of resource. Dual task is a deeper analysis of the task and sometime can be faster solved.

Figure 6: Pair of symmetric dual tasks

5.1 Rule to create dual task

The primary task has dimension m x n is transformed to the dual task with dimension n x mThe constrain in primary task can be transform into dual task:

f(x) = c.x = Max ⇐⇒ g(y) = y.b = Minn∑

j=1

aij .xj = bj ⇐⇒ yi ∈ (−∞,+∞)

n∑j=1

aij .xj ≥ bj ⇐⇒ yj ≥ 0

xj ≥ 0 ⇐⇒m∑i=1

aij .yj ≤ cj

xj ∈ (−∞,+∞) ⇐⇒m∑i=1

aij .yj = cj

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5.2 Duality theorems 5 DUAL TASK

5.2 Duality theorems

1st TheoremIf the optimal solution exists then both tasks has optimal solution and they are equal.

2nd TheoremIf given constraint of LP task is fulfilled as sharp inequality (i.e. is not effective) for optimum solution

then corresponding constraint in dual task is fulfilled as the equality

5.3 Solving Dual task

Dual task will be solved by the same method as primary task, which is:

1. Graphical method

2. Simplex method

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6 ALLOCATION TASKS

6 Allocation tasks

6.1 Definition of allocation tasks

This task deals with limited and interchangeable sources to individual customers in optimal way.For example we must produce a certain amount bj of products with i machines, each machine has

specific cost cij usd/h, productivity of each machine is kij product/hour and maximum hour of productionis ai hours. We can rewrite the task as follow:

The total cost is sum product of all costs:

m∑i=1

n∑j=1

cij .xij = Min

The maximum production hours constrain:

n∑j=1

xij ≤ ai

Amount of product is productivity x hours:

m∑i=1

kij .xij = bj

xij ≥ 0

6.2 Transportation task

This is a special case of allocation task when we apply two assumption:

1. coefficient kij = 1, that means we assume the same effectiveness of all sources.

2. Balance supplier and demand, ai = bj

Thank to assumptions we can have some remarks:

1. The problem has been remarkably simplified

2. Balanced transportation task always has optimal solution

3. We need only m+n-1 of linear independent vector.

6.3 Solving balanced transportation tasks

To solve this problem we must follow 2 steps:

1. Find the starting basic solutionThere are 5 methods we can use to find the starting basic solution.

(a) North West corner method

(b) Row minimum method

(c) Column minimum method

(d) Smallest minimum method

(e) Vogel method

2. Find the optimum solution by MODI

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6.3 Solving balanced transportation tasks 6 ALLOCATION TASKS

We will see an example for BTT

Table 11: Example Balanced Transportation Task

Productiviy/Plant O1 O2 O3 a (Capacity)D1 /2 /5 /3 10D2 /5 /4 /8 7D3 /1 /3 /6 7

b (Demand) 6 12 6

The value assigned in each DiOj cell is the cost of production.

Starting solution by North West corner method

Procedure:

• upper left corner of the matrix is assigned with maximum possible capacity of transportation

• recalculation of capacities and demands

• exclusion of row or column with fully utilized capacity or satisfied demand

• continuation till the full allocation

Table 12: Starting solution by North West corner method

Productiviy/Plant O1 O2 O3 a (Capacity)D1 6/2 4/5 0/3 10D2 0/5 7/4 0/8 7D3 0/1 1/3 6/6 7

b (Demand) 6 12 6

We start at cell D1O1 then fill max demand of O1 is 6. The other cells in O1 must be 0. Capacity on D1still has 4 left so we fill 4 into cell D1O2. The capacity of D1 is reached then the other cell of D1 must be0. The demand of O2 still has 8 left so we will fill 7 into cell D2O2 because capacity of D2 is 7. The rest 1will be filled to cell D3O2. The capacity of D3 has 6 left as well as 6 needed at O3 which filled to cell D3O3.

Starting solution by row minimum method

Procedure:

• 1st row: maximum allocation in cell with minimum transportation cost



• continuation in 1st row in sub matrix


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Table 13: Starting solution by row minimum method


b (Demand) 6 12 6

We start at row D1, the minimum cost is cell D1O1 then we give 6 here and 4 to the D1O3 with cost3. As a consequence, the rest cells of O1 is 0. For the row D2 the cell D2O2 has cost 4 then we fill 7 herethen the rest cell of D2 is 0. The row D3 we have cell D3O2 with cost 3 then we give here 5 as the restof demand of O2 and the final cell D3O3 is 2.

Starting solution by column minimum method

Table 14: Starting solution by column minimum method


b (Demand) 6 12 6

This method is similar to row minimum method only we looking for the minimum cost on each columninstead of row.

Starting solution by column minimum method

Procedure:

• maximum allocation in cheapest cell



• continuation cheapest cell in sub matrix


Table 15: Starting solution by column minimum method


b (Demand) 6 12 6

We start at the cell with minimum cost D3O1 and fill it with 6. The rest of O2 will be 0 and wecontinue with either cell D1O3 or D3O2. We choose the cell D1O3 and fill it will 6 as demand of O3.Then cell D1O2 will be 4 to full fill capacity D1. Cell D3O2 is minimum cost now and we fill it with 1 to

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archive the capacity of D3. The last cell D2O2 is 7 consequently.

Starting solution by Vogel method

Procedure:

• calculation of the difference between best and second best option for each row and column

• risk reduction = assignment of maximum transportation to the cheapest cell in row/column withhighest difference (i.e. risk)


• continuation cheapest cell in sub matrix


Table 16: Starting solution by Vogel method

Productiviy/Plant O1 O2 O3 a (Capacity) diff 1a diff 2a diff 3a diff 4aD1 4/2 0/5 6/3 10 1 3D2 0/5 7/4 0/8 7 1 1 1 4D3 2/1 5/3 0/6 7 2 2 2 3

b (Demand) 6 12 6

diff 1b 1 1 3diff 2b 1 1diff 3b 4 1diff 4 1

First we calculate the diff 1a and diff 1b by subtract the second smallest cost for the minimum cost inthe respected row or column. Then the biggest value of diff 1b O3 so we fill 6 into cell D1O3 with lowestcost on the column O3. Now we can fill all the cell on the column O3. Next we calculate diff 2a,b by thesame process without participant of O3. The biggest is diff 2a D1 then we will fill 4 into D1O1 with mincost of the row. We exclude row D1 from calculation of diff 3a,b. The max value is on diff 3b O1, so wefill 2 into D3O1 and exclude column O1. The last diff 4a,b is calculated and the last two cells are fill withthe same manner.

MODI Method

This method include 2 steps:

1. Find the initial improvement of inital solution

2. Find close loop and iterate until finding optimal solution.

Step 1

First we will set one of ui or vj to zeros. The other value of ui and vj of occupied cells will bedetermined from this setting by formular:

cij = ui + vj

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Then value dij = cij − ui − vj of the non-occupied cells is calculated:

Then value dij = ui + vj − cij of the non-occupied cells is calculated.

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The cell with dij < 0 could be improved to reach the optimal solution.

Step 2

We identify a close loop to change transportation schedule. This loop starts at the non-occupied celland goes though occupied cells to returns to the starting point. At each corner, the cells will be assignedminus or plus sign alternatively with starting point is plus.

We will reschedule the transport by minus value of transportation at minus-cells and add more valueto plus-cells. The value is chosen as the minimum of all minus-cells to avoid negative value.

We recalculate the value of ui and vj as well as dij of all occupied and non-occupied cells. If some dijstill smaller than 0 then we repeat step 2 until all dij is positive.

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7 TERMINOLOGY

7 Terminology

1. Criterion function: c1x1 + c2x2 + ... + cnxn

function being maximize or minimize.

2. Functional constrains: ai1x1 + ai2x2 + ... + ainxn

3. No negativity constrains: xj ≥ 0

4. Feasible solution: is a solution for which all the constraints are satisfied.

5. Infeasible solution is a solution for which at least one constraint is violated.

6. Additivity The additivity assumption of linear programming holds if every function in the modelis the sum of the individual contributions of the respective activities.

7. corner-point feasible (CPF) solution is a solution that lies at a corner of the feasible region.

8. slope-intercept form: For the geometric representation of a linear programming problem withtwo decision variables, the slope-intercept form of a line algebraically displays both the slope of theline and the intercept of this line with the vertical axis.

x2 = −3

5x1 +

1

5Z2

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REFERENCES REFERENCES

References

[1] Frederick S. Hillier and Gerald J. Lieberman. Introduction to Operation Research. McGraw-Hill Edu-cation, 2 Penn Plaza, New York,2015

[2] J. Knpek and Martin Dobi Lecture of course Operations Research. CVUT, Prague, 2015

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