Refinery Blending Problems by Engr. Adefami Olusegun

1
CHAPTER ONE
INTRODUCTION
1.1 Background
Linear programming is the most often used method for solving optimization problems.
Since the development of the simplex algorithm for solving linear programming by George
Dantzig, Linear programming has been used to solve optimization problems in fields as
diverse as petroleum, banking, education, Product Constraints, Raw Material
The problem of solving a system of linear inequalities dates back at least as far as Fourier,
after whom the method of Fourier-Motzkin elimination is named. Linear programming
arose as a mathematical model developed during the secondworld war to plan expenditures
and returns in order to reduce costs to the army and increase losses to the enemy. It was
kept secret until 1947. Postwar, many industries found its use in their daily planning.
Limitations, Safety Restrictions, Product Specifications and Material and Energy Balances,
and many others. It has a great importance in operation research methods. A useful tool for
solving linear programming problems is WHAT’S BEST! Software. This software
employs powerful optimization algorithms leading to dramatically reduced solution time
of the optimization procedure when the problem contains a large numbers of variables and
constraints.

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1.2 Objective of Study
(1) To develop a software to solve linear programming problems using Visual Studio
2005.
(2) Formulation and Solving Problems in linear programming.
(3) To determine application of linear programming
(4) To compare the program with other software e.g. Excel Solver
1.3 Scope of Work
This study is relevant to process data as well as enterprise data, consisting of commercial
and financial information, are used with the methodologies shown to make decisions in a
timely fashion. This model can be optimized to obtain target levels and price for inter-
refinery transfers, crude oil and product allocations to each refinery, inventory targets,
optimal operating conditions, stream allocations, and blends for each refinery.
1.4 Statement of the problem
(1) Linear programming is applicable only to problems where the constraints
and objective function are linear i.e., where they can be expressed as
equations which represent straight lines. In real life situations, when
constraints or objective functions are not linear, this technique cannot be
used.
(2) Factors such as uncertainty, weather conditions etc. are not taken into
consideration

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(3) There may not be an integer as the solution, e.g., the number of men
required may be a fraction and the nearest integer may not be the optimal
solution i.e., Linear programming technique may give practical valued
answer which is not desirable
(4) Only one single objective is dealt with while in real life situations, problems
come with multi-objectives.
(5) Parameters are assumed to be constants but in reality they may not be so.

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CHAPTER TWO
2 RESEARCH METHODOLOGY
2.1 Introduction
Revised simplex method was programmed into Visual Studio 2005 software
through Visual basic application to solve a Linear Programming problem where
optimal solution is determined. This program enables us to solve different problems
in Refinery Blending and in Process chemical industry. The Computer uses linear
Programming models to optimize blending at lowest cost.
2.2 Visual Basic Tools In the Program
The tools used in the program are:
1. TextBox
2. ComboBox
3. Button
4. ListBox
5. Rich TextBox
TEXTBOX -: We used the textbox to display some text that can be edited.
COMBOBOX -: Combox was used to select the type of optimization (either MAXIMIZE
or MINIMIZE).We also used the ComboBox to select the number of variables the objective
function to optimize.
BUTTON-: We used Button to raises an event when clicked. In the program, three buttons
were used which are “EXECUTE”, “CLEAR” & “EXIT”. Any button pressed responds
based on the Click Event code.

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LISTBOX-: We used it in the program to accept each constrain and display them.
RICH TEXTBOX-: We used rich textbox to display the result of the computation.

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CHAPTER THREE
3.1 Program Design Strategy:
Every computer program has to follow some strategies before they can be effectively
written. The strategies used in designing this program consist of three stages which are:
1. Planning stage;
2. Interface stage; and
3. Coding stage
Planning Stage
The stage involves the use of flowchart in order to determine the flow and order the
program will follow so as to write the program effectively. One other tool that was used is
algorithm.
The flowchart for this program is as shown in figure 3.1 while the algorithms follow
immediately.

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Figure 3.1 Program Flow Chart for Design
3.3.1 The Revised Simplex Program Algorithm.
1. Start
2. Choose the number of variable of the objective function
3. Select to either maximize or minimize the function
4. Supply the objective function
5. Supply each of the constrains
Yes
No
Start
Supply the objective function
and constrains and validate
them
Do the Constrains
validates?
Select the No of
objective function
variable
Do the revised simplex computation
Display the output:
Optimal solution
Stop

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6. Check if the objective function and constrains are correct (validation). If not
correct, go to step 9.
7. Do the revised simplex computation.
8. Display the result.
9. Stop.
The Revised Simplex Algorithm
1. Start with some basic index set B1 = {i(1), i(2), … ,i(m)}.
2. Compute a coefficient of the relative cost coefficient
rj = cj - cBAB
-1
aj
for B1 until some index j with rj > 0 is found.
(i).If all rj ≤ 0, termination phase has been reached.
(ii).Only those j M N1 are candidates
3. Find AB-1
aj = aj and AB
-1
b = b and compute the allowable ratios of the jth column , viz.,
bi/aij for those I with aij > 0.
(i). If there are no allowable ratios, then a termination phase has been reached.
4. Find an index k where the minimum ratio is attained.
5. Add j to the basic index set and delete i(k) to get a new basic index set B2.
6. Go back to step 2.
Interface Design Stage
This stage involves the designing of the forms used in the program. The forms were first
designed effectively before the code for the action is written. It suffice us here to explain
the program components used in the design of the forms

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3.2 Program Component
In this program, components used are two window forms (frmSiimplex and frmRetrieve)
and a module (Rsimplex).
3.2.1 A Brief Description:
The “frmSimplex” is the startup form. It allows the user to enter the objective function and
constrains, validate and execute the function. It also allows the user to save the executed
objective function and retrieve the saved equation. The executed function can also be
printed. All these can be achieved by pressing different buttons (Execute, Save, Validate,
Retrieve and Print buttons). It also has “Exit” button which is used to close the program.
The “frmRetrieve” serves as the form that loads when the “Retrieve” button is being
pressed on the “frmSimplex” form. It provide the user the opportunity of choosing from
the formerly saved functions and execute it and get the answer being displayed on the
“frmSimplex” form. It has “Execute” and “Exit” buttons.
The “RSimplex” module contains public variables, many functions and a structure. The
functions are “Determinant”, “FormLU”, “Revised Simplex
The Screen Shot of the following Buttons where display below Execute, Save, Print,
Validate, Retrieve, Clear and Exit.

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Figure 3.2 Showing “Execute Button”
Figure 3.3 Showing the “Save Button”

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Figure 3.4 showing “Print Button”

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Figure 3.5 Showing “Validate Button”

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Figure 3.6 Showing “Retrieve button” of Revised Simplex
Figure. 3.7 “Clear Button”

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Figure 3.8 showing the “Exist” button display.
3.3.2. Step by Step Procedure Used to Design The Interface:
For any work to be done effective, orders had to be followed. Consequently, there are
procedures that are followed in the course of writing this program. They are as follows:
1. The program was first planned using flowchart. This was to make writing the
program fast and effective with little bugs.
2. After this, the database used was designed . This is necessary to know the structure
of the database and program accordingly to avoid unnecessary wasting of effort in
adjusting both the program and database in the course of writing the codes.
3. The “frmSimplex” form was later created after the paper work concerning the
program interface had been made.

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4. The code for the form was written and the bugs (errors) were removed through
testing and debugging the codes.
5. It was later viewed to create allowance for users to have the objective function
saved after the execution. This lead the designing “RSimplex” module and
“frmRetrieve” form.
6. The “RSimplex” module was necessary to store public variables and create public
functions which were used by the two forms.
7. The “frmRetrieve” was then designed and the code was written.
8. The overall program was tested using different examples from textbooks, internet,
and formulated ones.
Coding Stage
This is the stage where the code for the program was written for the form objects designed
in “Interface Design Stage”. These codes are shown below.
CHAPTER FOUR
4 Results and Discussion

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4.1 Introduction
The software is written in Visual Studio 2005 through Visual basic (VB). Code Visual to
Flowchart is a program Flow chart generator for code flowcharting and visualization. It
can perform automated reverse engineering of program code into programming flowcharts,
help programmers to document, visualize and understand source code.
4.2 DATA USED FOR THE PROJECT
Data for the Refinery Feeds and Products were formulated and analyzed in Problem 4.1
and optimum value was determined. Both Problem 4.2 and Problem 4.3 are Production
Planning problems and they were formulated and the raw data was input into visual studio
to obtain optimal solution.
4.3 PROGRAM OUTPUT
4.3.1 Correct Data Input:
Correct data was input in the problems below
Problem 4.1 (Guy, Krik R, Nelson J.J Cole E. 2004)
Looking at the chemical process industries in the 1950’s, the information pertaining to the
expected yields of the fifteen types of crude oils when processed by the refinery is as shown

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in page 67. Note that the product distribution from the refinery is quite different for the
fifteen crude oils. The limitations on the established markets for the various products in
terms of the allowed maximum daily production were given.
To set up the linear programming problem, formulate an objective function and constraints
for the refinery operation.
Solution
Let the variable be: X1 = bbl/day of crude #1
X2 = bbl/day of crude #2
X16 = bbl/day of propane

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X17 = bbl/day of butane
X18 = bbl/day of LPG
X19 = bbl/day of Petroleum ether
X20 = bbl/day of ligroin
X21 = bbl/day of straight-run gasoline
X22 = bbl/day of kerosene
X23 = bbl/day of heating oil
X24 = bbl/day of gas oil
X25 = bbl/day of diesel fuel
X26 = bbl/day of lubricating oil
X27 = bbl/day of waxes
X28 = bbl/day of residual oil
X29 = bbl/day of asphalt
X30 = bbl/day of tar
The linear objective function Z (to be maximized).
Other constraints that exist or are implied in this problem are given below. These can be
formulated as inequality constraints:
Propane: X16  48
Butane: X17  51
LPG: X18  46
Petroleum ether: X19  41
Ligroin: X20  47

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Straight-run gasoline: X21  44
Kerosene: X22  43
Heating oil: X23  43
Gas oil: X24  41
Diesel fuel: X25  49
Lubricating oil: X26  64
Waxes: X27  54
Residual oil: X28  56
Asphalt: X29  59
Tar: X30  45
Maximize Z =
151413121110987654321 54243242 
Subject to

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4523222233
5925242224
56232322344
5422332223233
6425233232223
4932223232
412222222
43222222223
432223322
4422222423
474222222
4122223222
4632215223
513222223223
48222322222
151413121110987654321
151413121110987654321
151413121110987654321
151413121110987654321
151413121110987654321
151413121110987654321
151413121110987654321
151413121110987654321
151413121110987654321
151413121110987654321
151413121110987654321
151413121110987654321
151413121110987654321
151413121110987654321
151413121110987654321















X
X

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Figure 4.1 Formulation of Revised Simplex Method.
Figure 4.2 Optimal Solution Revised Simplex Method Using Visual Studio 2005

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4.3.2 Excel Solver Solution:
Microsoft Excel Provides tool called Solver that handles this problem in a manner that is
transparent to the user. The dynamic display of the iterative search process enables the user
to monitor location of the optimum solution by the search algorithm
Figure 4.3 Showing Solver Parameters

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Figure 4.4 Solver Options
Figure 4.5 9(a) Solver Results

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Figure. 4.6 Solver Results Solution

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Table 4.1 COMPARISONS BETWEEN REVISED SIMPLEX METHOD AND
EXCEL SOLVER
No of
Variables
Revised Simplex
Method
Excel Solver
Solution
Observation
X2 0.59091 0.5909091 X2 in Rsimplex method
solution is the same in Excel
solver Solution
solver Solution
solver Solution
solver Solution
solver Solution
z 85.32955 85.32955 Optimal Solution is the same
in Rsimplex and Excel Solver
Solution.

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4.3.3 Analysis of Problem 4.1
The objective function and constraints for the refinery operation was formulated using
visual studio 2005 and the optimal solution was determined in Problem 4.1 This problem
was compared with Excel Solver Solution. The Optimal Solution for both applications is
the same.
Problem 4.2: Production Planning (Vasant P.M, 2003)
Eleven products are processed through different operations. The times (in minutes)
required per unit of each product, the daily capacity of the operations (in minutes per day)
and the profit per unit sold of each product (in dollars) are as shown below.
The zero times indicate that the product does not require the given operation. It is assumed
that all units produced are sold. Moreover, the given profits per unit are net values that
result after all pertinent expenses are deducted. The goal of the model is to determine the
optimum daily production for the eleven products that maximizes profit.

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Max Z= 1110987654321 2323432 
Subject to
512353
643424222
85234244336
7943346523
7532642323
442342
644422332
77434424233
593233244
582324232
47222332
1110987654321
1110987654321
1110987654321
1110987654321
1110987654321
1110987654321
1110987654321
1110987654321
1110987654321
1110987654321
1110987654321











X
X

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Figure 4.8 Optimal Solution in Revised Simplex Method Using Visual Studio 2005

29
Problem 4.3 (Textile Technology Spinning, 2008).
The Concrete Products Corporation has the capability of producing eleven types of
concrete blocks. The plant manager desires to maximize the profit during the next month.
Max Z= 1110987654321 532423 
Subject to
512353
643424222
85234244336
7943346523
7532642323
442342
644422332
77434424233
593233244
582324232
47222332
1110987654321
1110987654321
1110987654321
1110987654321
1110987654321
1110987654321
1110987654321
1110987654321
1110987654321
1110987654321
1110987654321











X
X

30
Figure 4.10 Optimal Solution Revised Simplex Method Using Visual Studio 2005

31
Problem 4.4 (James H. Gary and Glenn E. handwork, 2001)
Reid Vapor Pressure
The desired RVP of a gasoline is obtained by blending n-butane with C5-38oF (C5-193oC)
naphtha. The amount on n-butane required to give the needed RVP is calculated by:
   iRVPMRVP
N
I
ITt 

1
Where
t total moles blended product
  iRVP Specification RVP for product, psi
i moles of component i
  iRVP RVP of component i, psi or Kpa
Base Stock BPD lb/hr MW mol/hr mol% RVP PVP

32
LSR gasoline 4,000 39,320 86 457 21.0 11.1 2.32
Reformate 6,000 69,900 115 617 28.4 2.8 0.80
Alkylate 3,000 30,690 104 295 13.4 4.6 0.62
FCC gasoline 8,000 87,520 108 810 37.2 4.4 1.64
Total 21,000 2,179 100.0
Blend for a 10 psi RVP (n-butane: MW=58, RVP=52).
Butane requirement:
(2,179)(5.38) + M(52.0) = (2,179 + M)(10)
11,723 + 52.0M = 21,790 + 10.0M
42.0M = 10,067
M = 240 Moles 4nC required
BPD lb/hr MW mol/hr
n-butane 1,640 13,920 58 240
Total 10 psi RVP gasoline = 21,000 + 1,640 = 22,640 BPD
4.3.4. Incorrect Data Input

33
Incorrect data was input into the cell of visual studio 2005 to test how efficient the program
is compare to excel solver. Fig 4.3.4 showing “you must supply at least one constrain”,
“The constrain must contain ‘<=’ or ‘>=’ or ‘=’, “you must supply number of variable”.
Problem 4.4
Max Z = 321 335 
333
22
31
321


Figure 4.11 Incorrect Data Input Showing (“you must supply at least one constrain”)

34
Figure 4.12 Incorrect Data Input showing The constrain must contain ‘<=’ or ‘>=’ or
‘=’.

35
Figure 4.13 Incorrect Data Input.
4.3.5 Discussion of Results:
The difference between visual studio 2005 software and excel solver is that visual studio
shows incorrect inputs and gives directions of what to do at every point in time why excel
solver does not show any incorrect input data.
4.4 Special Challenges Encountered
The main challenge I encountered was manipulation of matrices especially the inverse
function. I first used LU method but it did not work properly. Later I used Gaussian
elimination method. The other challenge is getting the equation of 10 or more variables
that can be used to test the application because it was not even available on the
internet as at this time. The cost for the research on this project is expensive.

36
4.5 DISCUSSION OF RESULTS
We present a computation method for solving quadratic programming problems which
reduces to the revised simplex for linear programming when the objective function is
linear. The basis matrix which is maintained is symmetric and may vary in size from
iteration to iteration. Comparison on test problems below indicates that the method is more
efficient than the other available methods. We used the active set algorithm in chapter 3
i.e. the revised simplex method Eqs (1) to Eqs (12), to solve the resulting optimization
problem.
Using Visual Studio 2005 software in a design of optimization application is equivalent to
considering the results of many virtual experiments, each of which is run using different
configurations, including geometrical, material or constraints. The design of optimum
codes in visual basic is a very difficult task but the application is friendly when compared
to Excel Solver and Matlab 7.0 softwares in solving large problems.
In the problem formulated and analysed below Excel Solver Software was compared with
Visual Studio software which is the aim and objective of this project and we discovered
the solutions of both problems were the same. The major difference between Excel Solver
Spreadsheet and Visual basic in Visual Studio 2005 is that Visual Basic shows the incorrect
data input while excel solver does not show it.
Finally, we can deduce that application of visual studio 2005 in Linear Programming is
friendly to the end user and it is useful in refinery blending, Production Planning,
Chemical Processing Industry and Financial Distribution.

37
CHAPTER FIVE
5.0 CONCLUSION AND RECOMMENDATON
Based on the results achieved on chapter four, the following conclusions were drawn.
The results generated through Revised Simplex method (data used for the project) was
compared with Excel Solver Solution and results were the same.
Revised Simplex method is the main function for linear programming with the most
flexibility for specifying the methods used, and it is the most efficient for large-scale
problems.
In the Revised Simplex Method there is an improvement in the objective function in
each step as the algorithm converges to the optimum. However, a situation can arise
where there is no improvement in the objective function from an application of the
algorithm.
In solving Problems in Revised simplex method Visual Studio Software application is
better than Excel solver and Mat Lab 7.0 .

38
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allocation hub location problems” Computer and Operation Research, Vol. 32, pp
3093- 3109.
Aspen Technology, Inc. (1996), Ten Canal Park, Cambridge, MA 02141-2200. Tel:
617-949-1000. Fax: 617-949-1030.
Beasley J.E, (2006): “OR – Linear Programming Formulation Notes”, Field of
Operation Research.
Billy E, and others, (2005), “Professional VB 2005” , Wiley Publishing,Inc.
Bazaraa, M. S., and J. J. Jarris, Linear Programming and Network Flows, John
Wiley and Sons, Inc., New York (1977).
Computing in Science and Engineering, (2000), Vol. 2,
James H.Gary and Glenn E. Handwerk, (2001), “Petroleum Refining” Mercel
Dekker, Inc. 270 Madison Avenue, New York, New York 10016.
Cai P. and Cai J.Y, (1997): On the 100% Rule of Sensitivity Analysis in Linear
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Vol. 1276, pp. 460-469.

39
Charnes, A., and W. W. Cooper, Management Models and Industrial Applications
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Delson, J.K and Shahidehpour S.M (1992): “Linear programming applications to
power system economics, planning and operations”, Power System, IEEE
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Edward J.M (1972):” The Application of Linear Programming to a Distribution
System Orientated Toward Service”, International Journal of Physical Distribution
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Gass, S. I., Linear Programming: Methods and Applications, 5th Ed., McGraw-Hill
Book Co., New York (1985).
Greenberg, cites: V. Klee and G.J. Minty. "How Good is the Simplex Algorithm?"
In O. Shisha, editor, Inequalities, III, pp 159–175. Academic Press, New York,
NY, 1972
Kallrath J.F, (2005): “Solving Planning and Design Problems in the Process
Industry Using Mixed Integral and Global Optimization” Annals of Operations
Research, Vol. 140, pp.339-373.
Optmalon Software, (2007): “GIPALS32” - Linear Programming Library

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OR/MS, (1992): “Linear Programming Software” 49 packages, pp 44-45.
Psacharopoulous G. (2006): “Synthetic Approaches in Manpower Planning”, De
Economist”, Vol. 127, pp 287-297.
Ravichandran N, (1989): “Models in Management Science” A Journal of
Management Research, Vol. 4, pp 1
Smith, C. L., R. W. Pike, P. W. Murrill, Formulation and Optimization of
Mathematical Models, International Textbook Co., Scranton, Pa. (1970)
Trkman, Peter & Gradisar, Miro, 2008. "One-dimensional cutting stock
optimization in consecutive time periods," European Journal of Operational
Research, Elsevier, vol. 179(2), pp. 291-301.
Taiwo, F (2004), “Chemical Engineering Operations and Management” Ladoke
Akintola University of Technology” Vo1.3, pp. 1-23(3), Press, Ile-Ife
Tibor F., (1967): “Blast Furnace Production Planning-A Linear Programming
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Vasant P.M, (2003): “Application of Fuzzy Linear Programming in Production
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Wayne L.W, “Operation Research: Application and Algorithms, Wayne L.
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41
William T.M (1960): “Application of Linear Programming To Financial Budgeting
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American Control Conference, Vol. 2, pp 1911-1912

42
APPENDIX
Appendix 1
Definition of terms
BV Matrix for set of basic Variable
b = Right-hand side vector of the original tableau’s constraints
ja = Column of jx in the constraints of the original problem
B = m x m matrix whose jth
column is the column for BVj in the original constraints
jC = Coefficient of jx in the objective function
BVC 1 x m row vector whose jth
elements is the objective function coefficient for BVj
ix = m x 1 column vector with ith
element 1 and all other elements equals to zero

ajB 1
Column for jx in BV tableau

jjBV CaBC `1
Coefficient of jx in row 0 (Price out)
bB 1
= Right-hand side of constraints in BV tableau

iBV xBC 1
Coefficient of slack variable iS in BV in row 0
 
iBV xBC 1
Coefficient of excess variable ie in BV row 0

bBCBV
1
Right-hand side of BV row 0.

43
Appendix 2
Execute Button Code
rtbDisplay.Text = ""
Dim result As MsgBoxResult
bttnExecute.ForeColor = Color.Red
result = MsgBox("Do you want to execute",
MsgBoxStyle.YesNo, My.Application.Info.Title & "-Execute")
If result = MsgBoxResult.No Then
bttnExecute.ForeColor = Color.Black
Exit Sub
End If
Dim blnValidate As Boolean = False
Try
Validation(False)
Catch ex As Exception
MsgBox(ex.Message, MsgBoxStyle.Critical)
Exit Sub
End Try
If cboContrains.Items.Count > 0 Then
blnValidate = True
Else
MsgBox("You must supply at least one constrain",
MsgBoxStyle.Critical, My.Application.Info.Title)
cboContrains.Focus()
Exit Sub
End If
Me.Enabled = False
Try
If blnValidate = True Then
NC = cboContrains.Items.Count - 1
NV = Convert.ToInt32(cboVariable.Text)
Dim blnOptimum As Boolean = False
Dim dblBConstants(NC) As Double
Dim dblZConstants(NV - 1) As Double
Dim dblXConsArray(NC, NV + NC) As Double
For i As Int32 = 0 To
cboContrains.Items.Count - 1
Dim strConstrain As String =
cboContrains.Items(i)
If strConstrain.IndexOf(">=") <> -1 Then
dblBConstants(i) = -1 *
Microsoft.VisualBasic.Right(strConstrain,
strConstrain.Length - (strConstrain.IndexOf("=") + 1))

44
Else
dblBConstants(i) =
Microsoft.VisualBasic.Right(strConstrain,
strConstrain.Length - (strConstrain.IndexOf("=") + 1))
End If
Next
dblZConstants =
GetMatrixZ(UCase(Trim(txtObjFxn.Text)))
If cboOptimal.SelectedIndex = 0 Then
blnOptimum = True
ElseIf cboOptimal.SelectedIndex = 1 Then
blnOptimum = False
End If
For i As Int32 = 0 To NC
Dim myXConstant() As Double =
GetMatrix(cboContrains.Items(i))
For j As Int32 = 0 To NV - 1
dblXConsArray(i, j) = myXConstant(j)
Next
Next
For i As Int32 = 0 To NC
For j As Int32 = NV To
dblXConsArray.GetUpperBound(1)
Dim myJ As Int32 = j - NV
If i = myJ Then
dblXConsArray(i, j) = 1
Else
dblXConsArray(i, j) = 0
End If
Next
Next
Dim dblAnswer() As Variable =
MatrixHelper.RevisedSimplex(dblZConstants, dblXConsArray,
dblBConstants, blnOptimum)
Call Display(dblZConstants, dblAnswer)
bttnSave.Enabled = True
End If
MsgBox(ex.Message)
Me.Enabled = True
End Try
Me.Enabled = True
End Sub

45
Appendix 3
Save Button Code
Dim blnDuplicate As Boolean = False 'To make sure equation of
the same type does not exist in the database
bttnSave.ForeColor = Color.Red
result = MsgBox("Do you want to Save",
MsgBoxStyle.YesNo, My.Application.Info.Title & "-Save")
bttnSave.ForeColor = Color.Black
Exit Sub
End If
If rtbDisplay.Text = "" Then
MsgBox("Cannot save this equation",
MsgBoxStyle.Critical, My.Application.Info.Title & "-Save")
Exit Sub
End If
Try
Validation(False)
MsgBox(ex.Message & " Cannot save this equation",
MsgBoxStyle.Critical)
Exit Sub
End Try
If myCon.State = ConnectionState.Open Then
myCon.Close()
End If
myCon.Open(strCon)
'Submit equation to database
Dim myData As New ADODB.Recordset
strQry = "SELECT * FROM Equation"
myData.Open(strQry, myCon,
ADODB.CursorTypeEnum.adOpenStatic,
ADODB.LockTypeEnum.adLockOptimistic)
For i As Int32 = 0 To myData.RecordCount - 1
If myData.Fields("strEquation").Value =
txtObjFxn.Text.Trim.ToUpper Then
blnDuplicate = True
End If
myData.MoveNext()
Next
If blnDuplicate = False Then

46
myData.AddNew()
myData.Fields("strEquation").Value =
txtObjFxn.Text.Trim.ToUpper
myData.Fields("strOptimal").Value =
cboOptimal.Text.Trim
myData.Fields("intVarNo").Value =
cboVariable.Text.Trim
myData.Update()
Else
MsgBox("Cannot save this equation. The equation
already exist. ", MsgBoxStyle.Critical,
My.Application.Info.Title)
txtObjFxn.Focus()
Exit Sub
End If
myData.Close()
'Submit the constrains
strQry = "SELECT * FROM Constrains"
For i As Int32 = 0 To cboContrains.Items.Count - 1
myData.AddNew()
myData.Fields("strConstrain").Value =
cboContrains.Items(i).ToString
myData.Update()
Next
myData.Close()
'Submit the answers
strQry = "SELECT * FROM Answer"
For i As Int32 = 0 To DataArray.GetUpperBound(0)
myData.AddNew()

47
myData.Fields("strVariable").Value =
DataArray(i).strVariable
myData.Fields("dblAnswer").Value =
DataArray(i).dblValue
myData.Update()
Next
myData.Close()
myCon.Close()
MsgBox("The equation is saved",
MsgBoxStyle.Information, My.Application.Info.Title)
End Sub
Appendix 4A
Retrieve Button Code
Private Sub bttnRetrieve_Click(ByVal sender As System.Object,
ByVal e As System.EventArgs) Handles bttnRetrieve.Click
frmRetrieve.Show()
End Sub
Appendix 4B
Print Button Code
Private Sub bttnPrint_Click(ByVal sender As System.Object,
ByVal e As System.EventArgs) Handles bttnPrint.Click
Dim PS As New System.Drawing.Printing.PageSettings
PS.Margins.Left = 300
PS.Margins.Right = 175
PS.Margins.Top = 200
PS.Margins.Bottom = 200
PrintDialog1.PrinterSettings =
PrintDocument1.PrinterSettings
If PrintDialog1.ShowDialog() =
Windows.Forms.DialogResult.OK Then
PrintDocument1.DefaultPageSettings = PS
PrintDocument1.PrinterSettings =
PrintDialog1.PrinterSettings
PrintDocument1.Print()
End If
End Sub

48
Appendix 4C
Clear Button Code
bttnClear.ForeColor = Color.Red
result = MsgBox("Do you want to Clear",
MsgBoxStyle.YesNo, My.Application.Info.Title & "-Clear")
bttnClear.ForeColor = Color.Black
Exit Sub
End If
txtObjFxn.Text = ""
txtObjFxn.Focus()
cboOptimal.SelectedIndex = 0
cboVariable.SelectedIndex = -1
cboContrains.Text = ""
cboContrains.Items.Clear()
rtbDisplay.Text = ""
bttnSave.Enabled = False
bttnClear.ForeColor = Color.Black
End Sub
Appendix 4D
Validate Button Code
bttnValidate.ForeColor = Color.Red
result = MsgBox("Do you want to Validate",
MsgBoxStyle.YesNo, My.Application.Info.Title & "-Validate")
bttnValidate.ForeColor = Color.Black
Exit Sub
End If
Try
Validation(True)
MsgBox(ex.Message)
End Try
bttnValidate.ForeColor = Color.Black
End Sub
Appendix 4E

49
Exit Button Code
Private Sub bttnExit_Click(ByVal sender As System.Object,
ByVal e As System.EventArgs) Handles bttnExit.Click
bttnExit.ForeColor = Color.Red
result = MsgBox("Are you sure you want to Exit",
MsgBoxStyle.YesNo, My.Application.Info.Title & "-Exit")
bttnExit.ForeColor = Color.Black
Exit Sub
End If
Me.Close()
End Sub
Appendix 5
Inverse Function Code
Public Function Inverse(ByVal sourceMatrix(,) As Double) As
Double(,)
Dim eachCol As Integer
Dim eachRow As Integer
Dim rowsAndCols As Integer
If (UBound(sourceMatrix, 1) <> UBound(sourceMatrix,
2)) Then
Throw New Exception("Matrix must be square.")
End If
Dim rank As Integer = UBound(sourceMatrix, 1)
' ----- Clone a copy of the matrix (not just a new
reference).
Dim workMatrix(,) As Double = _
CType(sourceMatrix.Clone, Double(,))
' ----- Variables used for backsolving.
Dim destMatrix(rank, rank) As Double
Dim rightHandSide(rank) As Double
Dim solutions(rank) As Double
Dim rowPivots(rank) As Integer
Dim colPivots(rank) As Integer

50
' ----- Use LU decomposition to form a triangular
matrix.
workMatrix = FormLU(workMatrix, rowPivots,
colPivots, rowsAndCols)
' ----- Backsolve the triangular matrix to get the
inverted
' value for each position in the final matrix.
For eachCol = 0 To rank
rightHandSide(eachCol) = 1
BackSolve(workMatrix, rightHandSide, solutions,
rowPivots, colPivots)
For eachRow = 0 To rank
destMatrix(eachRow, eachCol) =
solutions(eachRow)
rightHandSide(eachRow) = 0
Next eachRow
Next eachCol
' ----- Return the inverted matrix result.
Return destMatrix
End Function
Appendix 6
Revised Simplex Code
Public Function RevisedSimplex(ByVal objFunc() As Double,
ByVal constFunc(,) As Double, ByVal dblB() As Double, ByVal
blnOptimal As Boolean) As Variable()
Dim m, t As Int32
Dim zolution(,) As Double
Dim strBasis(NC) As String
Dim intBasis(NC) As Int32
Dim strNBasis(NV - 1) As String
Dim intNBasis(NV - 1) As Int32
Dim Xb(objFunc.GetUpperBound(0), 0) As Double
Dim Xa(objFunc.GetUpperBound(0), 0) As Double
Dim Ratio(NC) As Double
Dim matrixA(constFunc.GetUpperBound(0),
constFunc.GetUpperBound(1)) As Double
Dim objFuncWork(objFunc.GetUpperBound(0)) As Double
Dim dblBWork(dblB.GetUpperBound(0)) As Double
Dim dblBWorkUse(dblB.GetUpperBound(0), 0) As Double
Dim B(matrixA.GetUpperBound(0), NC) As Double
Dim P(NC, 0) As Double
Dim Y(,) As Double

51
Dim Z(intNBasis.GetUpperBound(0)) As Double
Dim intIterationCounter As Int32 = 0
Dim Cb(0, intBasis.GetUpperBound(0)) As Double
Dim invB(matrixA.GetUpperBound(0), NC) As Double
Dim RepeatLoop As Boolean = True
Array.Copy(constFunc, matrixA, constFunc.Length)
Array.Copy(objFunc, objFuncWork, objFunc.Length)
Array.Copy(dblB, dblBWork, dblB.Length)
'Note: based strbasis on the no of varIABLE e.g
variable strbasis(0)="X" & 5
For count As Int32 = 1 To NV
intNBasis(count - 1) = count
Next
For count As Int32 = NV + 1 To
matrixA.GetUpperBound(1) + 1
intBasis(count - (NV + 1)) = count
Next
For j As Int32 = 0 To dblBWorkUse.GetUpperBound(0)
dblBWorkUse(j, 0) = dblBWork(j)
Next
While RepeatLoop
For k As Int32 = 0 To intBasis.GetUpperBound(0)
Try
Cb(0, k) = objFuncWork(intBasis(k) - 1)
Catch
Cb(0, k) = 0
End Try
Next
For i As Int32 = 0 To matrixA.GetUpperBound(0)
For j As Int32 = 0 To
intBasis.GetUpperBound(0)
B(i, j) = matrixA(i, intBasis(j) - 1)
Next
Next
invB = Inverse(B)
Xb = Multiply(invB, dblBWorkUse)
zolution = Multiply(Cb, Xb)
Y = Multiply(Cb, invB)
Dim a As Int32 = 0
Do While a <= intNBasis.GetUpperBound(0)
matrixA.GetUpperBound(0)
P(i, 0) = matrixA(i, intNBasis(a) - 1)

52
Next
Dim Zme(0, 0) As Double
Zme = Multiply(Y, P)
Try
Z(a) = Zme(0, 0) -
objFuncWork(intNBasis(a) - 1)
Catch
Z(a) = Zme(0, 0) - 0
End Try
a += 1
Loop
'to choose to either maximise or minimise
If blnOptimal = True Then 'maximize
m = Minimum(Z, intNBasis)
Else
m = Maximum(Z, intNBasis)
End If
If m < 0 Then
'write out Xb and Z
Dim var(Xb.GetUpperBound(0) + 1) As Variable
For i As Int32 = 0 To Xb.GetUpperBound(0)
var(i).dblValue = Xb(i, 0)
var(i).strVariable = "X" & intBasis(i)
Next
var(Xb.GetUpperBound(0) + 1).strVariable =
"Z"
var(Xb.GetUpperBound(0) + 1).dblValue =
zolution(0, 0)
intCounter = intIterationCounter
Return var
RepeatLoop = False
Exit Function
End If
P(i, 0) = matrixA(i, m) ' Note in case of
error
Next
Xa = Multiply(invB, P)
Dim x As Double = Xb(i, 0) / Xa(i, 0)
If Double.IsInfinity(x) Then
Ratio(i) = 0
Else
Ratio(i) = x
End If
Next

53
If LessEqualZero(Ratio) Then
Next
"Z"
zolution(0, 0)
Return var
RepeatLoop = False
Throw New Exception("The Problem has no
bounded solution")
Exit Function 'write: the problem has no
bounded solution
Else
t = MinRatio(Ratio)
End If
'to avoid double entry of each value
Dim blnDoubleEntry As Boolean = False
For i As Int32 = 0 To intBasis.GetUpperBound(0)
If intBasis(i) = m + 1 Then
blnDoubleEntry = True
End If
Next
If blnDoubleEntry = False Then
If intBasis(i) = t + NV + 1 Then
intBasis(i) = m + 1
End If
Next
intNBasis.GetUpperBound(0)
If intNBasis(i) = m + 1 Then
intNBasis(i) = t + NV + 1
End If
Next
Else
Next

54
"Z"
zolution(0, 0)
Return var
RepeatLoop = False
End If
Array.Sort(intNBasis)
For i As Int32 = 0 To intNBasis.GetUpperBound(0)
strNBasis(i) = "X" & intNBasis(i)
Next
'Array.Sort(intBasis)
strBasis(i) = "X" & intBasis(i)
Next
intIterationCounter += 1
If intIterationCounter = 1000 Then
Dim var(0) As Variable
var(0).strVariable = "No solution is found"
var(0).dblValue = 0
Return var
Exit Function
End If
End While
End Function
Appendix 7
VISUAL BASIC CODE FOR THE COMPUTER AIDED DESIGNED
'Start: Thursday
'Stop:
'Programmer: Adefami olusegun A. (08034988098)
segcyt2@yahoo.com
Public myCon As New ADODB.Connection
Public myData As ADODB.Recordset
Public strCon As String =
"Provider=Microsoft.Jet.OLEDB.4.0;Data Source=" &

55
My.Application.Info.DirectoryPath & "RevisedSimplex.mdb"
';Jet OLEDB:Database Password=adtaxymac"
Public strQry As String
Public intCounter As Int32
Public NV, NC As Int32
Public Solution_Problem As Boolean
Public Function MakeDisplayable(ByVal sourceMatrix(,) As
Double) As String
' ----- Prepare a multi-line string that shows the
contents
' of a matrix, a 2D array.
Dim rows As Integer
Dim cols As Integer
Dim result As New System.Text.StringBuilder
' ----- Process all rows of the matrix, generating
one
' output line per row.
rows = UBound(sourceMatrix, 1) + 1
cols = UBound(sourceMatrix, 2) + 1
For eachRow = 0 To rows - 1
' ----- Process each column of the matrix on a
single
' row, separating values by commas.
If (eachRow > 0) Then result.AppendLine()
For eachCol = 0 To cols - 1
' ----- Add a single matrix element to the
output.
If (eachCol > 0) Then result.Append(",")
result.Append(sourceMatrix(eachRow,
eachCol).ToString)
Next eachCol
Next eachRow
' ----- Finished.
Return result.ToString
End Function
Public Function MakeDisplayable(ByVal sourceArray() As
Double) As String
' ----- Present an array as multiple lines of output.
Dim result As New System.Text.StringBuilder
Dim scanValue As Double

56
For Each scanValue In sourceArray
result.AppendLine(scanValue.ToString)
Next scanValue
Return result.ToString
End Function
Public Function Inverse(ByVal sourceMatrix(,) As Double)
As Double(,)
' ----- Build a new matrix that is the mathematical
inverse
' of the supplied matrix. Multiplying a matrix
and its
' inverse together will give the identity
matrix.
Dim rowsAndCols As Integer
' ----- Determine the size of each dimension of the
matrix.
' Only square matrices can be inverted.
2)) Then
Throw New Exception("Matrix must be square.")
End If
' ----- Clone a copy of the matrix (not just a new
reference).
Dim workMatrix(,) As Double = _
CType(sourceMatrix.Clone, Double(,))
' ----- Variables used for backsolving.
Dim rightHandSide(rank) As Double
Dim solutions(rank) As Double
matrix.
colPivots, rowsAndCols)
' ----- Backsolve the triangular matrix to get the
inverted

57
' value for each position in the final matrix.
rightHandSide(eachCol) = 1
BackSolve(workMatrix, rightHandSide, solutions,
rowPivots, colPivots)
destMatrix(eachRow, eachCol) =
solutions(eachRow)
rightHandSide(eachRow) = 0
Next eachRow
Next eachCol
' ----- Return the inverted matrix result.
Return destMatrix
End Function
Public Function Determinant(ByVal sourceMatrix(,) As
Double) As Double
' ----- Calculate the determinant of a matrix.
Dim result As Double
Dim pivots As Integer
Dim count As Integer
' ----- Only calculate the determinants of square
matrices.
2)) Then
Throw New Exception( _
"Determinant only calculated for square
matrices.")
End If
' ----- Make a copy of the matrix so we can work
inside of it.
Dim workMatrix(rank, rank) As Double
Array.Copy(sourceMatrix, workMatrix,
sourceMatrix.Length)
matrix.
colPivots, count)
' ----- Get the product at each of the pivot points.

58
result = 1
For pivots = 0 To rank
result *= workMatrix(rowPivots(pivots),
colPivots(pivots))
Next pivots
' ----- Determine the sign of the result using
LaPlace's formula.
result = (-1) ^ count * result
Return result
End Function
Private Function FormLU(ByVal sourceMatrix(,) As Double,
_
ByRef rowPivots() As Integer, ByRef colPivots()
As Integer, _
ByRef rowsAndCols As Integer) As Double(,)
' ----- Perform an LU (lower and upper) decomposition
of a matrix,
' a modified form of Gaussian elimination.
Dim pivot As Integer
Dim rowIndex As Integer
Dim colIndex As Integer
Dim bestRow As Integer
Dim bestCol As Integer
Dim rowToPivot As Integer
Dim colToPivot As Integer
Dim maxValue As Double
Dim testValue As Double
Dim oldMax As Double
Const Deps As Double = 0.0000000000000001
' ----- Determine the size of the array.
Dim rowNorm(rank) As Double
ReDim rowPivots(rank)
ReDim colPivots(rank)
' ----- Make a copy of the array so we don't mess it
up.
Array.Copy(sourceMatrix, destMatrix,
sourceMatrix.Length)

59
' ----- Initialize row and column pivot arrays.
rowPivots(eachRow) = eachRow
colPivots(eachRow) = eachRow
rowNorm(eachRow) +=
Math.Abs(destMatrix(eachRow, eachCol))
Next eachCol
If (rowNorm(eachRow) = 0) Then
Throw New Exception("Cannot invert a singular
matrix.")
End If
Next eachRow
' ----- Use Gauss-Jordan elimination on the matrix
rows.
For pivot = 0 To rank - 1
maxValue = 0
For eachRow = pivot To rank
rowIndex = rowPivots(eachRow)
For eachCol = pivot To rank
colIndex = colPivots(eachCol)
testValue =
Math.Abs(destMatrix(rowIndex, colIndex)) _
/ rowNorm(rowIndex)
If (testValue > maxValue) Then
maxValue = testValue
bestRow = eachRow
bestCol = eachCol
End If
Next eachCol
Next eachRow
' ----- Detect a singular, or very nearly
singular, matrix.
If (maxValue = 0) Then
Throw New Exception("Singular matrix used for
LU.")
ElseIf (pivot > 1) Then
If (maxValue < (Deps * oldMax)) Then
Throw New Exception("Non-invertible
matrix used for LU.")
End If
End If
oldMax = maxValue
' ----- Swap row pivot values for the best row.

60
If (rowPivots(pivot) <> rowPivots(bestRow)) Then
rowsAndCols += 1
Swap(rowPivots(pivot), rowPivots(bestRow))
End If
' ----- Swap column pivot values for the best
column.
If (colPivots(pivot) <> colPivots(bestCol)) Then
rowsAndCols += 1
Swap(colPivots(pivot), colPivots(bestCol))
End If
' ----- Work with the current pivot points.
rowToPivot = rowPivots(pivot)
colToPivot = colPivots(pivot)
' ----- Modify the remaining rows from the pivot
points.
For eachRow = (pivot + 1) To rank
rowIndex = rowPivots(eachRow)
destMatrix(rowIndex, colToPivot) = _
-destMatrix(rowIndex, colToPivot) / _
destMatrix(rowToPivot, colToPivot)
For eachCol = (pivot + 1) To rank
colIndex = colPivots(eachCol)
destMatrix(rowIndex, colIndex) += _
destMatrix(rowIndex, colToPivot) * _
destMatrix(rowToPivot, colIndex)
Next eachCol
Next eachRow
Next pivot
' ----- Detect a non-invertible matrix.
If (destMatrix(rowPivots(rank), colPivots(rank)) =
0) Then
Throw New Exception("Non-invertible matrix used
for LU.")
ElseIf (Math.Abs(destMatrix(rowPivots(rank),
colPivots(rank))) / _
rowNorm(rowPivots(rank))) < (Deps * oldMax)
Then
Throw New Exception("Non-invertible matrix used
for LU.")
End If
' ----- Success. Return the LU triangular matrix.
Return destMatrix

61
End Function
Private Function Calculate_Inverse(ByVal MatrixA(,) As
Double) As Double(,)
Dim multiplier_1 As Double
Dim elem1 As Double
Dim temporary_1 As Double
Dim line_1 As Double
Dim k As Int32
Dim m As Int32
Dim N As Int32
Dim System_DIM As Int32 = MatrixA.GetUpperBound(0) +
1
Dim MAX_DIM As Int32 = MatrixA.GetUpperBound(0)
Dim Operations_Matrix(MAX_DIM, (2 * MAX_DIM) + 1) As
Double 'Matrix where the calculations are done
Dim Inverse_Matrix(MAX_DIM, MAX_DIM) As Double
'Matrix with the Inverse of [A]
'Uses Gauss elimination method in order to calculate
the inverse matrix [A]-1
'Method: Puts matrix [A] at the left and the singular
matrix [I] at the right:
'[ a11 a12 a13 | 1 0 0 ]
'[ a21 a22 a23 | 0 1 0 ]
'[ a31 a32 a33 | 0 0 1 ]
'Then using line operations, we try to build the
singular matrix [I] at the left.
'After we have finished, the inverse matrix [A]-1
(bij) will be at the right:
'[ 1 0 0 | b11 b12 b13 ]
'[ 0 1 0 | b21 b22 b23 ]
'[ 0 0 1 | b31 b32 b33 ]
'On Error GoTo errhandler 'In case the inverse cannot
be found (Determinant = 0)
Solution_Problem = False
'Assign values from matrix [A] at the left
For N = 0 To System_DIM - 1
For m = 0 To System_DIM - 1
Operations_Matrix(m, N) = MatrixA(m, N)
Next
Next
'Assign values from singular matrix [I] at the right
For m = 0 To System_DIM - 1

62
If N = m Then
Operations_Matrix(m, N + System_DIM) = 1
Else
Operations_Matrix(m, N + System_DIM) = 0
End If
Next
Next
'Build the Singular matrix [I] at the left
For k = 0 To System_DIM - 1
'Bring a non-zero element first by changes lines
if necessary
If Operations_Matrix(k, k) = 0 Then
For N = k To System_DIM - 1
If Operations_Matrix(N, k) <> 0 Then
line_1 = N : Exit For
End If 'Finds line_1 with non-zero
element
Next N
'Change line k with line_1
For m = k To (System_DIM - 1) * 2
temporary_1 = Operations_Matrix(k, m)
Operations_Matrix(k, m) =
Operations_Matrix(line_1, m)
Operations_Matrix(line_1, m) =
temporary_1
Next m
End If
elem1 = Operations_Matrix(k, k)
For N = k To 2 * (System_DIM - 1)
Operations_Matrix(k, N) =
Operations_Matrix(k, N) / elem1
Next N
'For other lines, make a zero element by using:
'Ai1=Aij-A11*(Aij/A11)
'and change all the line using the same formula
for other elements
If N = k And N = MAX_DIM Then Exit For
'Finished
If N = k And N < MAX_DIM Then N = N + 1 'Do
not change that element (already equals to 1), go for next
one
If Operations_Matrix(N, k) <> 0 Then 'if it
is zero, stays as it is

63
'
multiplier_1 = Operations_Matrix(N, k) /
Operations_Matrix(k, k)
For m = k To 2 * (System_DIM - 1)
Operations_Matrix(N, m) =
Operations_Matrix(N, m) - Operations_Matrix(k, m) *
multiplier_1
Next m
End If
Next N
Next k
'Assign the right part to the Inverse_Matrix
For k = 0 To System_DIM - 1
Inverse_Matrix(N, k) = Operations_Matrix(N,
System_DIM + k)
Next k
Next N
Return Inverse_Matrix
Exit Function
errhandler:
Throw New Exception("An error occured during the
calculation process. Determinant of Matrix [A] is probably
equal to zero.")
Solution_Problem = True
End Function
Private Sub Swap(ByRef firstValue As Integer, ByRef
secondValue As Integer)
' ----- Reverse the values of two reference integers.
Dim holdValue As Integer
holdValue = firstValue
firstValue = secondValue
secondValue = holdValue
End Sub
Private Sub BackSolve(ByVal sourceMatrix(,) As Double, _
ByVal rightHandSide() As Double, ByVal
solutions() As Double, _
ByRef rowPivots() As Integer, ByRef colPivots()
As Integer)
' ----- Solve an upper-right-triangle matrix.
Dim pivot As Integer
Dim rowToPivot As Integer
Dim colToPivot As Integer

64
' ----- Work through all pivot points. This section
builds
' the "B" in the AX=B formula.
For pivot = 0 To (rank - 1)
colToPivot = colPivots(pivot)
For eachRow = (pivot + 1) To rank
rowToPivot = rowPivots(eachRow)
rightHandSide(rowToPivot) += _
sourceMatrix(rowToPivot, colToPivot) _
* rightHandSide(rowPivots(pivot))
Next eachRow
Next pivot
' ----- Now solve for each X using the general formula
' x(i) = (b(i) - summation(a(i,j)x(j)))/a(i,i)
For eachRow = rank To 0 Step -1
colToPivot = colPivots(eachRow)
rowToPivot = rowPivots(eachRow)
solutions(colToPivot) =
rightHandSide(rowToPivot)
For eachCol = (eachRow + 1) To rank
solutions(colToPivot) -= _
sourceMatrix(rowToPivot,
colPivots(eachCol)) _
* solutions(colPivots(eachCol))
Next eachCol
solutions(colToPivot) /=
sourceMatrix(rowToPivot, colToPivot)
Next eachRow
End Sub
Public Function RevisedSimplex(ByVal objFunc() As Double,
ByVal constFunc(,) As Double, ByVal dblB() As Double, ByVal
blnOptimal As Boolean) As Variable()
Dim m, t As Int32
Dim zolution(,) As Double
Dim strBasis(NC) As String
Dim intBasis(NC) As Int32
Dim strNBasis(NV - 1) As String
Dim intNBasis(NV - 1) As Int32
Dim Xb(objFunc.GetUpperBound(0), 0) As Double
Dim Xa(objFunc.GetUpperBound(0), 0) As Double
Dim Ratio(NC) As Double
Dim matrixA(constFunc.GetUpperBound(0),
constFunc.GetUpperBound(1)) As Double

65
Dim objFuncWork(objFunc.GetUpperBound(0)) As Double
Dim dblBWork(dblB.GetUpperBound(0)) As Double
Dim dblBWorkUse(dblB.GetUpperBound(0), 0) As Double
Dim B(matrixA.GetUpperBound(0), NC) As Double
Dim P(NC, 0) As Double
Dim Y(,) As Double
Dim Z(intNBasis.GetUpperBound(0)) As Double
Dim intIterationCounter As Int32 = 0
Dim Cb(0, intBasis.GetUpperBound(0)) As Double
Dim invB(matrixA.GetUpperBound(0), NC) As Double
Dim RepeatLoop As Boolean = True
Array.Copy(constFunc, matrixA, constFunc.Length)
Array.Copy(objFunc, objFuncWork, objFunc.Length)
Array.Copy(dblB, dblBWork, dblB.Length)
'Note: based strbasis on the no of varIABLE e.g
variable strbasis(0)="X" & 5
For count As Int32 = 1 To NV
intNBasis(count - 1) = count
Next
For count As Int32 = NV + 1 To
matrixA.GetUpperBound(1) + 1
intBasis(count - (NV + 1)) = count
Next
For j As Int32 = 0 To dblBWorkUse.GetUpperBound(0)
dblBWorkUse(j, 0) = dblBWork(j)
Next
While RepeatLoop
For k As Int32 = 0 To intBasis.GetUpperBound(0)
Try
Cb(0, k) = objFuncWork(intBasis(k) - 1)
Catch
Cb(0, k) = 0
End Try
Next
For j As Int32 = 0 To
B(i, j) = matrixA(i, intBasis(j) - 1)
Next
Next
invB = Inverse(B)
zolution = Multiply(Cb, Xb)

66
Y = Multiply(Cb, invB)
Dim a As Int32 = 0
Do While a <= intNBasis.GetUpperBound(0)
matrixA.GetUpperBound(0)
P(i, 0) = matrixA(i, intNBasis(a) - 1)
Next
Dim Zme(0, 0) As Double
Zme = Multiply(Y, P)
Try
Z(a) = Zme(0, 0) -
objFuncWork(intNBasis(a) - 1)
Catch
Z(a) = Zme(0, 0) - 0
End Try
a += 1
Loop
'to choose to either maximise or minimise
If blnOptimal = True Then 'maximize
m = Minimum(Z, intNBasis)
Else
m = Maximum(Z, intNBasis)
End If
If m < 0 Then
'write out Xb and Z
Next
"Z"
zolution(0, 0)
Return var
RepeatLoop = False
Exit Function
End If
P(i, 0) = matrixA(i, m) ' Note in case of
error
Next
Xa = Multiply(invB, P)
Dim x As Double = Xb(i, 0) / Xa(i, 0)

67
If Double.IsInfinity(x) Then
Ratio(i) = 0
Else
Ratio(i) = x
End If
Next
If LessEqualZero(Ratio) Then
Next
"Z"
zolution(0, 0)
Return var
RepeatLoop = False
Throw New Exception("The Problem has no
bounded solution")
Exit Function 'write: the problem has no
bounded solution
Else
t = MinRatio(Ratio)
End If
'to avoid double entry of each value
Dim blnDoubleEntry As Boolean = False
If intBasis(i) = m + 1 Then
blnDoubleEntry = True
End If
Next
If blnDoubleEntry = False Then
If intBasis(i) = t + NV + 1 Then
intBasis(i) = m + 1
End If
Next
intNBasis.GetUpperBound(0)
If intNBasis(i) = m + 1 Then
intNBasis(i) = t + NV + 1
End If
Next

68
Else
Next
"Z"
zolution(0, 0)
Return var
RepeatLoop = False
End If
Array.Sort(intNBasis)
For i As Int32 = 0 To intNBasis.GetUpperBound(0)
strNBasis(i) = "X" & intNBasis(i)
Next
'Array.Sort(intBasis)
strBasis(i) = "X" & intBasis(i)
Next
intIterationCounter += 1
If intIterationCounter = 1000 Then
Dim var(0) As Variable
var(0).strVariable = "No solution is found"
var(0).dblValue = 0
Return var
Exit Function
End If
End While
End Function
Public Function Multiply(ByVal A1(,) As Double, ByVal
B1(,) As Double) As Double(,)
Dim m As Int32 = A1.GetUpperBound(0)
Dim k As Int32 = A1.GetUpperBound(1)
Dim n As Int32 = B1.GetUpperBound(1)
Dim C(m, n) As Double
'Multiply
For i As Int32 = 0 To m
For j As Int32 = 0 To n
C(i, j) = 0 'Initialise to zero
For p As Int32 = 0 To k
C(i, j) += A1(i, p) * B1(p, j)
Next

69
Next
Next
Return C
End Function
Public Function Minimum(ByVal minA() As Double, ByVal
minValues() As Int32) As Int32
Dim minimal As Double
Dim blnFound As Boolean = False
Dim minB(minA.GetUpperBound(0)) As Double
Dim myMinLoc As Int32 = -1
For min As Int32 = 0 To minA.GetUpperBound(0)
If minA(min) < 0 Then
myMinLoc = -1
blnFound = True
End If
Next
If Not (blnFound) Then
Return myMinLoc
Exit Function
End If
minB(min) = -1 * minA(min)
Else
minB(min) = 0
End If
Next
minimal = minB(0)
myMinLoc = 0
If minimal < minB(min) Then
minimal = minB(min)
myMinLoc = min
End If
Next
If minimal = 0 Then
myMinLoc = -1
Else
myMinLoc = minValues(myMinLoc) - 1
End If
Return myMinLoc
End Function

70
Public Function Maximum(ByVal maxA() As Double, ByVal
maxValues() As Int32) As Int32
Dim maximal As Double
Dim maxB(maxA.GetUpperBound(0)) As Double
Dim myMaxLoc As Int32 = -1
For min As Int32 = 0 To maxA.GetUpperBound(0)
If maxA(min) > 0 Then
myMaxLoc = -1
blnFound = True
End If
Next
If Not (blnFound) Then
Return myMaxLoc
Exit Function
End If
For max As Int32 = 0 To maxA.GetUpperBound(0)
If maxA(max) < 0 Then
maxB(max) = 0
Else
maxB(max) = maxA(max)
End If
Next
maximal = maxB(0)
myMaxLoc = 0
For min As Int32 = 1 To maxA.GetUpperBound(0)
If maximal < maxB(min) Then
maximal = maxB(min)
myMaxLoc = min
End If
Next
If maximal = 0 Then
myMaxLoc = -1
Else
myMaxLoc = maxValues(myMaxLoc) - 1
End If
Return myMaxLoc
End Function
Public Function MinRatio(ByVal minA() As Double) As Int32
Dim minimal As Double
Dim minB(minA.GetUpperBound(0)) As Double
Dim myMinLoc As Int32 = -1
If minA(min) > 0 Then

71
blnFound = True
End If
Next
If blnFound = False Then
Return myMinLoc
Exit Function
End If
minB(min) = 0
Else
minB(min) = minA(min)
End If
Next
If minB(0) <> 0 Then
minimal = minB(0)
myMinLoc = 0
Else
minimal = 1000000
End If
If minimal > minB(min) And minB(min) <> 0 Then
minimal = minB(min)
myMinLoc = min
Else
minimal = minimal
myMinLoc = myMinLoc
End If
Next
Return myMinLoc
End Function
Public Function LessEqualZero(ByVal Zero() As Double) As
Boolean
Dim blnFound As Boolean = True
For myCount As Int32 = 0 To Zero.GetUpperBound(0)
If Zero(myCount) > 0 Then
blnFound = False
End If
Next
End Function
Structure Variable
Dim strVariable As String

72
Dim dblValue As Double
End Structure
End Module

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