This document outlines the steps of the simplex method to solve linear programming problems. It begins with putting the problem into standard form and introducing slack variables to transform inequality constraints into equalities. A tableau is then set up to perform row operations. The optimal solution is checked by ensuring all values in the objective row are greater than or equal to zero. If not optimal, a pivot variable is identified and a new tableau is created by optimizing the pivot variable. This process repeats, identifying new pivot variables and creating new tableaus, until an optimal solution is reached where all objective row values are non-negative. Finally, the optimal values of the variables are determined based on their classification as basic or non-basic in the final optimal tableau

1. 1
A Report on Case study of Simplex method
Course name: Operations Research
Course code: F-409
Submitted to:
Shabnaz Amin Auditi
Associate professor
Department of Finance, University of Dhaka.

2. 2
Group profile
SL No. Name ID Remarks
01 Zakia Sultana 22-046
02 Jannatun Naim Taney 22-127
03 Nawshin Tabassum 22-121
04 Najia Akter Jim 22-136
05 Jakia Sultana Jerin 22-161

3. 3
Letter of Transmittal
25 November, 2019
Shabnaz Amin Auditi
Department of Finance,
University of Dhaka
Subject: Submission of a Report on “Case study of Simplex method”.
Dear Madam,
It is our pleasure that we have the opportunity to submit a term paper on “Case study of
Simplex method”. We have completed our report as a part of BBA curriculum. Practical
exposure is very much necessary to adopt with the environment where we have to work.
Term paper/Assignment program helps a student to gather those skills by making a bridge
between the academic and the practical knowledge.
We tried our level best to put meticulous effort for the preparation of this report. Any
shortcomings or flaw may arise unintentionally. We will wholeheartedly welcome any
clarification and suggestion about any view and conception disseminated in our report that
you might have.
Sincerely Yours,
On behalf of the group
Zakia Sultana
ID:22-046 ,Section :A
Signature:

4. 4
Contents
Letter of Transmittal...........................................................................................................................3
Executive Summary............................................................................................................................5
Objective of the report ........................................................................................................................6
Methodology of the report...................................................................................................................6
Limitations of the report......................................................................................................................6
Theoretical Overview..........................................................................................................................7
Simplex Method .............................................................................................................................7
Basic Terminology - Simplex Method Linear Programming...............................................................7
Why use simplex method.................................................................................................................9
Steps in the General Simplex Algorithm ...........................................................................................9
Some technical issues in the simplex model....................................................................................19
Case -1.............................................................................................................................................23
Nutrition Problem .........................................................................................................................23
Solution........................................................................................................................................24
Case -2.............................................................................................................................................30
Production Problem.......................................................................................................................30
Solution........................................................................................................................................31
Conclusion .......................................................................................................................................34
References .......................................................................................................................................35

5. 5
Executive Summary
In mathematical optimization, Dantzig's simplex algorithm (or simplex method) is a
popular algorithm for linear programming. At every iteration, it chooses the variable that can make
the biggest modification toward the minimum solution. From our class room learning, on this
report we have tried to show the real-life implication of simplex method. For this, we have
developed two cases. In our first case, we have tried to show how simplex method can be used in
real life to solve nutrition problem. We have worked on a case of Bangladesh Society for paternal
and Enteral Nutrition about its advice to an individual who is suffering from iron and vitamin B
deficiency. In our second problem, we have tried to use simplex method on production problem.
In this case, We have worked on Otobi Furniture Limited. We have used the simplex method to
eradicate the issues in linear programming.

6. 6
Objective of the report
 To get acquainted with real-life implication of simplex method.
 To use class room learning on real-life.
 To use operations research technique for problem solving.
Methodology of the report
 Data collected from organizations
 Differents websites for learning mathematical tools
Limitations of the report
 Lack of vast amount of data
 Issues with sample and selection

7. 7
Theoretical Overview
Simplex Method
The Simplex Method or Simplex Algorithm is used for calculating the optimal solution to the
linear programming problem. In other words, the simplex algorithm is an iterative procedure
carried systematically to determine the optimal solution from the set of feasible solutions.
Basic Terminology - Simplex Method Linear Programming
Slack variable
It is a variable that is added to the left-hand side of a less than or equal to type constraint to convert
the constraint into an equality. In economic terms, slack variables represent left-over or unused
capacity.
Specifically:
X1 + x2 + x3 + .........+ xn ≤ bi can be written as x1 + x2 + x3 + .........+ xn + si = bi
Where i = 1, 2, ..., m
Surplus variable
It is a variable subtracted from the left-hand side of a greater than or equal to type constraint to
convert the constraint into an equality. It is also known as negative slack variable. In economic
terms, surplus variables represent over fulfillment of the requirement.
Specifically:
x1 + x2 + x3 + .........+ xn ≥ bi can be written as x1 + x2 + x3 + .........+ xn - si = bi
Where i = 1, 2, ..., m

8. 8
Artificial variable
It is a nonnegative variable introduced to facilitate the computation of an initial basic feasible
solution. In other words, a variable added to the left-hand side of a greater than or equal to type
constraint to convert the constraint into an equality is called an artificial variable.
Zj
Amount or profit reduced or lost for including any variable in the solution.
Cj
Per unit profit of each variable.
Basic variables
Are variables that are non-negative in terms of the optimal solution.
Constraints
Are a series of equalities and inequalities that are a set of criteria necessary to satisfy whenfinding
the optimal solution.
Inequality
Is an expression that does not have one definite solution and is distinguishable by its ‘greater than’
or ‘less than’ symbols in the place of a traditional equal sign.
Linear program
It is a model used to achieve the best outcome given a maximum or minimum equation with linear
constraints.
Non-basic variables
Are variables that are zero in terms of the optimal solution.
Optimal solution of a maximization linear programming model are the values assigned to the
variables in the objective function to give the largest zeta value. The optimal solution would exist
on the corner points of the graph of the entire model.
Pivot variable
Is used in row operations to identify which variable will become the unit value and is a key factor
in the conversion of the unit value.

9. 9
Simplex method
An approach to solving linear programming models by hand using slack variables, tableaus, and
pivot variables as a means to finding the optimal solution of an optimization problem.
Simplex tableau
Is used to perform row operations on the linear programming model as well as for checking
optimality.
Why use simplex method
The simplex method is used to eradicate the issues in linear programming. The simplex method
uses a systematic strategy to generate and test candidate vertex solutions to a linear program. At
every iteration, it chooses the variable that can make the biggest modification toward the minimum
solution.
Steps in the General Simplex Algorithm
The Simplex method is an approach to solving linear programming models by hand using slack
variables, tableaus, and pivot variables as a means to finding the optimal solution of an
optimization problem. A linear program is a method of achieving the best outcome given a
maximum or minimum equation with linear constraints. Most linear programs can be solved using
an online solver such as MatLab, but the Simplex method is a technique for solving linear programs
by hand. To solve a linear programming model using the Simplex method the following steps are
necessary:
● Standard form
● Introducing slack variables
● Creating the tableau
● Pivot variables
● Creating a new tableau
● Checking for optimality

10. 10
This document breaks down the Simplex method into the above steps and follows the
example linear programming model shown below throughout the entire document to find
the optimal solution.
Step 1: Standard Form
Standard form is the baseline format for all linear programs before solving for the optimal solution
and has three requirements: (1) must be a maximization problem, (2) all linear constraints must be
in a less-than-or-equal-to inequality, (3) all variables are non-negative. These requirements can
always be satisfied by transforming any given linear program using basic algebra and substitution.
Standard form is necessary because it creates an ideal starting point for solving the Simplexmethod
as efficiently as possible as well as other methods of solving optimization problems.
To transform a minimization linear program model into a maximization linear program model,
simply multiply both the left and the right sides of the objective function by -1.
Transforming linear constraints from a greater-than-or-equal-to inequality to a less-than-or-equal-
to inequality can be done similarly as what was done to the objective function. By multiplying by
-1 on both sides, the inequality can be changed to less-than-or-equal-to.
Once the model is in standard form, the slack variables can be added as shown in Step 2 of the
Simplex method.

11. 11
Step 2: Determine Slack Variables
Slack variables are additional variables that are introduced into the linear constraints of a linear
program to transform them from inequality constraints to equality constraints. If the model is in
standard form, the slack variables will always have a +1 coefficient. Slack variables are needed
in the constraints to transform them into solvable equalities with one definite answer.
After the slack variables are introduced, the tableau can be set up to check for optimality as
described in Step 3.
Step 3: Setting up the Tableau
A Simplex tableau is used to perform row operations on the linear programming model as well as
to check a solution for optimality. The tableau consists of the coefficient corresponding to the
linear constraint variables and the coefficients of the objective function. In the tableau below, the
bolded top row of the tableau states what each column represents. The following two rows
represent the linear constraint variable coefficients from the linear programming model, and the
last row represents the objective function variable coefficients.
Once the tableau has been completed, the model can be checked for an optimal solution as shown
in Step 4.

12. 12
Step 4: Check Optimality
The optimal solution of a maximization linear programming model are the values assigned to the
variables in the objective function to give the largest zeta value. The optimal solution would exist
on the corner points of the graph of the entire model. To check optimality using the tableau, all
values in the last row must contain values greater than or equal to zero. If a value is less than zero,
it means that variable has not reached its optimal value. As seen in the previous tableau, three
negative values exists in the bottom row indicating that this solution is not optimal. If a tableau is
not optimal, the next step is to identify the pivot variable to base a new tableau on, as described in
Step 5.
Step 5: Identify Pivot Variable
The pivot variable is used in row operations to identify which variable will become the unit value
and is a key factor in the conversion of the unit value. The pivot variable can be identified by
looking at the bottom row of the tableau and the indicator. Assuming that the solution is not
optimal, pick the smallest negative value in the bottom row. One of the values lying in the column
of this value will be the pivot variable. To find the indicator, divide the beta values of the linear
constraints by their corresponding values from the column containing the possible pivot variable.
The intersection of the row with the smallest non-negative indicator and the smallest negative
value in the bottom row will become the pivot variable.
In the example shown below, -10 is the smallest negative in the last row. This will designate the
x2 column to contain the pivot variable. Solving for the indicator gives us a value of
10
3
for the first
constraint, and a value of
8
5
for the second constraint. Due to
8
5
being the smallest non-negative
indicator, the pivot value will be in the second row and have a value of 5.

13. 13
Now that the new pivot variable has been identified, the new tableau can be created in Step 6 to
optimize the variable and find the new possible optimal solution.Step 6: Create the New Tableau
The new tableau will be used to identify a new possible optimal solution. Now that the pivot
variable has been identified in Step 5, row operations can be performed to optimize the pivot
variable while keeping the rest of the tableau equivalent.
I. To optimize the pivot variable, it will need to be transformed into a unit value (value
of 1). To transform the value, multiply the row containing the pivot variable by the
reciprocal of the pivot value. In the example below, the pivot variable is originally 5, so
multiply the entire row by
1
5
.
II. After the unit value has been determined, the other values in the column containing
the unit value will become zero. This is because the x2 in the second constraint is being
optimized, which requires x2 in the other equations to be zero.
III. In order to keep the tableau equivalent, the other variables not contained in the pivot
column or pivot row must be calculated by usingthe new pivot values. For each new value,

14. 14
multiply the negative of the value in the old pivot column by the value in the new pivot
row that corresponds to the value being calculated. Then add this to the old value from the
old tableau to produce the new value for the new tableau. This step can be condensed into
the equation on the next page:
New tableau value = (Negative value in old tableau pivot column) x (value in new tableau pivot
row) + (Old tableau value)
Old Tableau:
New Tableau:
Numerical examples are provided below to help explain this concept a little better.
Numerical examples:
I. To find the s2 value in row 1:
New tableau value = (Negative value in old tableau pivot column) * (value in new tableau
pivot row) + (Old tableau value)
New tableau value = (-3) * (
1
5
) + 0 = -
3
5

15. 15
II. To find the x1 variable in row 3:
New tableau value = (Negative value in old tableau pivot column) * (value in new tableau
pivot row) + (Old tableau value)
New value = (10) * (
1
5
) + -8 = -6
Once the new tableau has been completed, the model can be checked for an optimal solution. Step
7: Check Optimality
As explained in Step 4, the optimal solution of a maximization linear programming model are the
values assigned to the variables in the objective function to give the largest zeta value. Optimality
will need to be checked after each new tableauto see if a new pivot variable needs to be identified.
A solution is considered optimal if all values in the bottom row are greater than or equal to zero.
If all values are greater thanor equal to zero, the solutionis considered optimal and Steps 8 through
11 can be ignored. If negative values exist, the solution is still not optimal and a new pivot point
will need to be determined which is demonstrated in Step 8
Step 8: Identify New Pivot Variable
If the solution has been identified as not optimal, a new pivot variable will need to be determined.
The pivot variable was introduced inStep 5 and is used inrow operations to identifywhichvariable
will become the unit value and is a key factor inthe conversionof the unit value. The pivot variable
can be identified by the intersection of the row with the smallest non-negative indicator and the
smallest negative value in the bottom row.

16. 16
With the new pivot variable identified, the new tableau can be created in Step 9.
Step 9: Create New Tableau
After the new pivot variable has been identified, a new tableau will need to be created. Introduced
in Step 6, the tableau is used to optimize the pivot variable while keeping the rest of the tableau
equivalent.
I. Make the pivot variable 1 by multiplying the row containing the pivot variable by
the reciprocal of the pivot value. In the tableau below, the pivot value was
1
5
, so everything
is multiplied by 5.
II. Next, make the other values in the column of the pivot variable zero. This is done
by taking the negative of the old value in the pivot column and multiplying it by the new
value in the pivot row. That value is then added to the old value that is being replaced.

17. 17
Step 10: Check Optimality
Using the new tableau, check for optimality. Explained in Step 4, an optimal solution appears
when all values in the bottom row are greater than or equal to zero. If all values are greater than
or equal to zero, skip to Step 12 because optimality has been reached. If negative values still exist,
repeat steps 8 and 9 until an optimal solution is obtained.
Step 11: Identify Optimal Values
Once the tableau is proven optimal the optimal values can be identified. These can be found by
distinguishing the basic and non-basic variables. A basic variable canbe classified to have a single
1 value in its column and the rest be all zeros. If a variable does not meet this criteria, it is
considered non-basic. If a variable is non-basic it means the optimal solution of that variable is
zero. If a variable is basic, the row that contains the 1 value will correspond to the beta value. The
beta value will represent the optimal solution for the given variable.
Basic variables: x1, s1, z
Non-basic variables: x2, x3, s2
For the variable x1, the 1 is found in the second row. This shows that the optimal x1 value is found
in the second row of the beta values, which is 8.

18. 18
Variable s1 has a 1 value in the first row, showing the optimal value to be 2 from the beta column.
Due to s1 beinga slack variable, it is not actuallyincluded in the optimal solution since the variable
is not contained in the objective function.
The zeta variable has a 1 in the last row. This shows that the maximum objective value will be 64
from the beta column.
The final solution shows each of the variables having values of:
x1 = 8 s1 = 2
x2 = 0 s2 = 0
x3 = 0 z = 64
The maximum optimal value is 64 and found at (8, 0, 0) of the objective function.
Summary of all steps:
1. Set up the expressions describing the problem constraints. (greater than or equal
inequalities , less than or equal inequalities and equation)
2. Convert the inequalities to equations by adding slack variable to less than or equal
constraints and subtracting surplus variable for greater than or equal constraints.
3. Add artificial variables to all constraints that were originally greater than inequalities or
equations. Give the artificial variables objective coefficients of +M in minimization
problems and –M in maximization problem.
4. Enter the equation in the simplex tableau
5. Calculate the Zj and Cj - Zj values for this solution
6. Determine the entering variable (optimal column) by choosing the one with the highest
positive Cj – Zj value for maximization problem and the most negative Cj – Zj value in
minimization problem.

19. 19
7. Determine the row to be replaced by dividing quantity column values by their
corresponding optimal column values and choosing the smallest nonnegative quotient (that
is, only compute the ratios for rows whose elements in the optimal column are greater than
zero)
8. Compute the values for the replacing row (by dividing each number in the replaced row by
the intersectional element of the replaced row)
9. Compute the values for the remaining rows. Element in new row = [ element in old row –
(intersectional element of old row*corresponding element of replacing row)
10. Calculate Zj and Cj – Zj values for this solution.
11. If there is a positive Cj – Zj value in maximization problem , or a negative Cj – Zj in a
minimization problem ,return to step 6
12. If there is no positive Cj – Zj value remaining in a maximization problem or no negative
Cj – Zj in a minimization problem, the optimal solution has been obtained.
Some technical issues in the simplex model
Some problems encountered in solving linear programs with simplex method these are described
below:
Infeasibility
If, when we reach the final solution, one or more artificial variables are still positive, then there is
no feasible solution to the problem.
Degeneracy
A condition resulting from a tie in the ratios determining the replaced row, which produces a basic
variable with zero value. An example of degeneracy is given below:

20. 20
First table
Cj 5 8 0 0 0
Product
mix
Quantity X1 X2 S1 S2 S3
0 S1 24 4 6 1 0 0
0 S2 18 2 1 0 1 0
0 S3 36 3 9 0 0 1
Zj 0 0 0 0 0 0
Cj - Zj 5 8 0 0 0
Second table (after X2 replaced S3 in this solution)
Cj 5 8 0 0 0
Product
mix
Quantity X1 X2 S1 S2 S3
0 S1 0 2 0 1 0 -2/3
0 S2 14 5/3 0 0 1 -1/9
8 X1 4 1/3 1 0 0 1/9
Zj 32 8/3 8 0 0 8/9
Cj - ZJ 7/3 0 0 0 -8/9

21. 21
Third table (after X1 replaced S1 in this solution)
Cj 5 8 0 0 0
Product
mix
Quantity X1 X2 S1 S2 S3
5 X1 0 1 0 1/2 0 -1/3
0 S2 14 0 0 -5/6 1 1/9
8 X1 4 0 1 -1/6 0 2/9
Zj 32 5 8 7/6 0 1/9
Cj - ZJ 0 0 -7/6 0 -1/9
Unboundedness
If there is no nonnegative ratio or if all the ratios are in the form of (16/0) then we say solution is
unbounded because it is not possible to determine exit variable.
Multiple/ Alternative optima
When non basic (that is variable is not in solution) has a zero entry in Cj – Zj row of an optimal
table then bringing that variable into solution we will produce a solution which is identical to the
previous solution. An example of multiple solution is given below:
Initial Table
Cj 8 4 0 0
Product
mix
Quantity T C Sa Sf
8 T 12 1 0 1/3 -1/6
4 C 6 0 1 -1/6 1/3
Zj 120 8 4 2 0
Cj - Zj 0 0 -2 0

22. 22
Second Table (C will be replaced by Sf)
Cj 8 4 0 0
Product
mix
Quantity T C Sa Sf
8 T 15 1 1/2 1/4 0
0 Sf 18 0 3 -1/2 1
Zj 120 8 4 2 0
Cj - Zj 0 0 -2 0

23. 23
Case -1
Nutrition Problem
Bangladesh Society for paternal and Enteral Nutrition advises an individual who is suffering from
iron and vitamin B deficiency to take at least 4800 milligrams (mg) of iron, 4200 mg of vitamin
B1, and 3000 mg of vitamin B2 over a period of time.
Two vitamin pills are suitable, brand-A and brand-B.
Each brand-A pill costs 12 cents and contains maximum 80 mg of iron, 20 mg of vitamin B1, and
10 mg of vitamin B2.
Each brand-B pill costs 16 cents and contains highest 20 mg of iron and 30 mg each of vitamins
B1 and B2.
What combination of pills should the individual purchase in order to meet the minimum iron and
vitamin requirements at the lowest cost?

24. 24
Solution
Let’s first tabulate the given information:
Brand-A Brand-B Minimum
Requirement
Cost 12 16
Iron 80 20 4800
Vitamin B1 20 30 4200
Vitamin B2 10 30 3000
Let,
x = Be the number of brand-A pills to be purchased
y = The number of brand-B pills to be purchased.
The cost C (in cents) is given by
To Min 12x+16y
and it is the objective function to be minimized.
The amount of iron contained in x brand-A pills and y brand-B pills is given by 80x + 20y mg,
and this must be greater than or equal to 4800 mg.
This translates into the inequality,
80x+ 20y >- 4800

25. 25
The amount of vitamin B1 contained in x brand-A pills and y brand-B pills is given by 20x + 30y
mg, and this must be greater or equal to 4200 mg.
This translates into the inequality
20x+30y >- 4200
The amount of vitamin B2 contained in x brand-A pills and y brand-B pills is given by 10x + 30y
mg, and this must be greater or equal to 3000 mg.
This translates into the inequality
10x+30y >- 3000
In short, we want to minimize the objective function
To Min 12x+16y
subject to the system of inequalities
80x+ 20y >- 4800
20x+30y >- 4200
10x+30y >- 3000
X >- 0
Y >- 0

26. 26
Initial Table
cj 12 16 0 0 0 M M M
PM Q x y S1 S2 S3 A1 A2 A3 Ratio
M A1 4800 80 20 -1 0 0 1 0 0 60
M A2 4200 20 30 0 -1 0 0 1 0 210
M A3 3000 10 30 0 0 -1 0 0 1 300
ZJ 110M 80M -M -M -M M M M
Cj- zj 12-
110M
16-
80M
M M M -M -M -M
Since this is a minimization problem, our objective is to reduce the cost.
By examining the numbers in the Cj-Zj row of initial table, we can see the highest cost reduction
can be done by variable ‘x’ which has the biggest negative number in Cj-Zj row and lowest
quantity is on the row of A1 variable.
So, ‘A1’ will be replaced by ‘x’ and as there is still negative number exists, now we proceed on
the second solution.
Second Table
cj 12 16 0 0 0 M M M
PM Q x y S1 S2 S3 A1 A2 A3 Ratio
12 X 60 1 .25 -.0125 0 0 .0125 0 0 240
M A2 3000 0 25 .25 -1 0 -.25 1 0 120
M A3 2400 0 27.5 .125 0 -1 -.125 0 1 87.27
720+
5400M
12 3+
52.5M
-.15+
.375M
-M -M .15-
.375M
M M
0 13-
52.5M
0.15-
.375M
M M -.15+
0.625M
0 0

27. 27
Now, in the second solution we can see further reduction can be done. As the number in the Cj-Zj
row shows biggest negative number in the column of variable ‘y’ and quantity column shows
lowest number in A3 variable.
So, ‘A3’ will be replaced by ‘y’ and as there is still negative number exists, now we proceed on
the third solution.
Third Table
cj 12 16 0 0 0 M M M
PM Q x y S1 S2 S3 A1 A2 A3 Ratio
12 X 38 1 0 -.0136 0 .009 .0136 0 -.009 4242
M A2 818 0 0 .138 -1 .9 -.138 1 -.9 909
16 Y 87 0 1 .0045 0 -.036 0.0045 0 0.036 N/A
ZJ 12 16 -.09+
.138M
-M -.468
+.9M
0.16-
0.14M
M 0.47-
0.9M
CJ-ZJ 0 0 O.09-
.138M
M .468
-.9M
-0.16+
0.14M
0 0.1M-
0.47
In the third solution, we can see that the number in the Cj-Zj row still shows negative number and
biggest negative number in the column of variable ‘S3’ and quantity column shows lowest number
in A2 variable.
So, ‘A2’ will be replaced by ‘S3’ and as there is still negative number exists, now we proceed on
the further solution.

28. 28
Fourth Table
cj 12 16 0 0 0 M M M
PM Q x y S1 S2 S3 A1 A2 A3 Ratio
12 X 30 1 0 -0.02 0.0099 0 0.015 -0.0999 0
0 S3 909 0 0 0.15 -1.11 1 -0.15 1.11 -1
16 Y 120 0 1 0.0099 -0.04 0 0.0009 0.040 0
ZJ 2280 12 16 -1.64 -0.52 0 0.19 -0.56 0
CJ-
ZJ
00 0 1.64 0.52 0 M-.19 M+0.56 M
In the fourth solution, we can see there is no more negative number exists. So further cost reduction
cannot be done.
So this is our final solution which means to minimize the cost, 30 quantity of ‘x’, 909 quantity of
‘S3’ and 120 quantity of ‘y’ should be combined. And then the cost will be 2280.
1
1
1
1
1
2
3
4

29. 29
Economic explanation
 The quantity column elements(1,2,3,4)
1) 30 units of brand-A pills to be purchased. Because we can see that 30 was the largest
quantity which could be made without increasing additional cost.
2) 909 units will be unused if we purchase 30 units of Brand A pills and 120 units of
Brand B pills.
3) 120 units of brand-B pills to be purchased. Because we can see that 120 was the
largest quantity which could be made without increasing additional cost.
4) The 2280 in Zj cell represents the total cost from the variables in the product mix:
12x+16y = 12*30+16*120= 2280.

30. 30
Case -2
Production Problem
Otobi Furniture Limited produces space saving equipment. Initially it is launching two products.
One is Bed which can be used as Sofa as well. Another is Almirah which can be used as dining
too. Each bed can be sold at 12000 taka and Almirah will be sold at 16000 taka.
The total production is completed through 2 processes two separate department perform 2 different
tasks. These two departments are –
1) Production department: they have 120 hours weekly available
2) Finishing department: they have 96 hours weekly available
 Each Almirah takes 8 hours at production and 4 hours for finishing.
 Each Bed takes 4 hours at production and 8 hours for finishing.
What combination of sales of Almirah and bed will maximize the profit of the company?

31. 31
Solution
Amirah –A
Bed -B
Here the objective is to maximize =16000A+12000B
Subject to, 8A+4B < 120(Production)
4A+8B < 96(Finishing)
To being equal to solve the problem, adding slack on production and finishing we get following
equation –
8A+4B +Sp+0Sf =120
4A+8B+0Sp+Sf =96
Initial table
Cj 16000 12000 0 0
Product
mix
Hours Almirah Bed Sp Sf Ratio
0 Sp 120 8 4 1 0 15(lowest)
0 Sf 96 4 8 0 1 24
Zj 0 0 0 0 0
Cj-Zj 16000 12000 0 0
Since this is a maximization problem, our objective is to maximize the profit.
By examining the numbers in the Cj-Zj row of initial table, we can see the highest profit
maximization can be done by variable ‘A’ which has the biggest positive number in Cj-Zj row and
lowest quantity is on the row of Sp variable.
So, ‘Sp’ will be replaced by ‘A’ and as there is still positive number exists, now we proceed on
the second solution.

32. 32
Second Table
Cj 16000 12000 0 0
Product
mix
Hours Almirah Bed Sp Sf Ratio
16000 B 15 1 .50 .125 0 30
0 Sf 36 0 6 -.50 1 6(lowest)
Zj 240000 16000 8000 2000 0
Cj-Zj 0 4000 -2000 0
Now, in the second solution we can see further increase can be done. As the number in the Cj-Zj
row shows positive number in the column of variable ‘B’ and quantity column shows lowest
number in Sf variable.
So, ‘Sf’ will be replaced by ‘B’ and as there is still positive number exists, now we proceed onthe
third solution.

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Third Table
Cj 16000 12000 0 0
Product
mix
Hours Almirah Bed Sp Sf Ratio
16000 B 12 1 0 .1667 .083333
12000 A 6 0 1 -.083333 .1667
Zj 264000 16000 12000 1667 667
Cj-Zj 0 0 -1667 -667
No further improvement is possible as all the values are 0 and negative. So this is our final solution
which means to maximize the profit, 6 units of ‘A’, 12 units of B should be produced. And then
the profit will be maximized there at 264,000
Economic Explanation
 The quantity column elements(1,2,3)
1) We found that 12 was the largest number which can be made without violating the
time restriction. So 12 units of B should be sold.
2) 6 units of A should be sold to maximize the profit.
3) The amount 264,000 in the Zj row represents the total profit from the variables in
the product mix: 12000A+!6000B= 12000*A+16000*B= 264,000.
1
2
3

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Conclusion
The Simplex method is an approach for determining the optimal value of a linear program by hand.
The method produces an optimal solution to satisfy the given constraints and produce a maximum
zeta value. To use the Simplex method, a given linear programming model needs to be in standard
form, where slack variables can then be introduced. Using the tabular method and pivot variables,
an optimal solution can be reached.

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References
1. https://en.wikipedia.org/wiki/Simplex_algorithm
2. https://www.otobi.com/
3. https://www.nutritioncare.org/