1. The document describes solving a linear programming problem using the simplex method. It provides the standard form of the LP problem with the objective function and constraints. 2. It shows the initial basic feasible solution and constructs the initial tableau. It then performs two iterations of the simplex method by choosing incoming and outgoing variables. 3. After two iterations, it finds that the optimal solution is x1=15 and x2=12, with maximum objective value Z=120. It provides the final simplex table.

1. Solve given LP problem using simplex method and find maximum value of objective function.
Write all
steps.
max : = 4.71 + 6x2 s.t. -x1 + 2 11 12s 27 21 52 90 i 1.2
Solution
the LPP problem according to Simplex method is given as follows:
Let, S1, S2, and S3 are slack variables three constraints. The standard form of given LPP is given
as follows:
Max Z = 4X1 + 5X2 + 0S1 + 0S2 + 0S3
Subject To:
-X1 + X2 + S1 + 0S2 + 0S3 = 11
X1 + X2 + 0S1 + S2 + 0S3 = 27
2X1 + 5X2 + 0S1 + 0S2 + S3 = 90
X1, X2, S1, S2, S3 >= 0
2)
To form an initial basic solution, consider the structural variables as zero, in other words X1 =
X2 = 0. Thus initial basic variables remains are S1, S2, and S3.
Z = 0, S1 = 11, S2 = 27, and S2 = 90
3)
The Initial Tableau is constructed as follows:
Cj
4
5
0
0
0
B (Quantity)
Cb
Xb
x1
x2

2. S1
S2
S3
0
S1
-1
1
1
0
0
11
0
S2
1
1
0
1
0
27
0
S3
2
5
0
0
1
90
Iteration 1:
Cj
4
5
0
0
0
B (Quantity)
RR

3. Cb
Xb
x1
x2
S1
S2
S3
0
S1
-1
1
1
0
0
11
11/1 = 11
0
S2
1
1
0
1
0
27
27/1 = 27
0
S3
2
5
0
0
1
90
90/5 = 18
Zj
0

4. 0
0
0
0
0
j = Cj – Zj
4
5
0
0
0
0
zj=cBj x xj
j= (cj-- zj)
As the j is positive the optimal solution is not obtained.
X2 is having highest j, thus incoming variable is X2. It realizes highest profit for next iteration.
Pivot Column = X2, pivot elements are 1, 1, and 5.
Determine replacement ratio by dividing B value by respective pivot column elements.
The replacement ratio for basis variable S1 is lowest positive, this variable is outgoing variable.
S1 is completely utilized if 11 units of X2 are to be produced.
Pivot row = S1
Key element (KE) = 1
The elements of next variables are obtained as follows:
Incoming Variable value for X1 is obtained as:
Other basis variable is obtained as:
1st row/KE
Old element – (Key column element x I.V. row element)
Cj
4
5
0
0
0
B (Quantity)
RR
Cb

5. Xb
x1
x2
S1
S2
S3
5
X2
-1
1
1
0
0
11
11/-1
= -11
0
S2
1-(1*-1)
=2
1-(1*1)
= 0
0-(1*1)
= - 1
1-(1*0)
= 1
0-(1*0)
= 0
27-(1*11)
16
16/2
= 8
0
S3
2-(5*-1)
= 7

6. 5-(5*1)
= 0
0-(5*1)
= -5
0-(5*0)
= 0
1-(5*0)
= 1
90-(5*11)
= 35
35/7
= 5
Zj
-5
5
5
0
0
55
j = Cj – Zj
9
0
-5
0
0
As the j is positive the optimal solution is not obtained.
X1 is having highest j, thus incoming variable is X1. It realizes highest profit for next iteration.
Pivot Column = X1, pivot elements are -1, 2, and 7.
The replacement ratio for basis variable S3 is lowest positive, this variable is outgoing variable.
S3 is completely utilized if 5 units of X1 are to be produced.
Pivot row = S3
Key element (KE) = 7
Cj
4
5
0

7. 0
0
B (Quantity)
RR
Cb
Xb
x1
x2
S1
S2
S3
5
X2
-1-(-1*1)
= 0
1-(-1*0)
= 1
1-(-1*-5/7)
= 2/7
0
0-(-1*1/7)
= 1/7
11-(-1*5)
= 16
16/(2/7)
= 56
0
S2
2-(2*1)
= 0
0-(2*0)
= 0
- 1-(2*-5/7)
= 3/7
1
0-(2*1/7)

8. = -2/7
16-(2*5)
= 6
16/(3/7)
= 37.33
4
X1
7/7
= 1
0/7
= 0
-5/7
0
1/7
35/7
= 5
5/(-5/7)
= -7
Zj
4
5
-10/7
0
9/7
(5*16+4*5)
= 100
j = Cj – Zj
0
0
10/7
0
-9/7
As the j is positive the optimal solution is not obtained.
S1 is having highest j, thus incoming variable is S1. It realizes highest profit for next iteration.
Pivot Column = S1, pivot elements are 2/7, 3/7, and -5/7.
The replacement ratio for basis variable S2 is lowest positive, this variable is outgoing variable.

9. S2 is completely utilized if 37.33 units of S2 are to be produced.
Pivot row = S2
Key element (KE) = 3/7
Cj
4
5
0
0
0
B (Quantity)
RR
Cb
Xb
x1
x2
S1
S2
S3
5
X2
0
1-(2/7*0)
= 1
2/7-(2/7*1)
= 0
0-(2/7*7/3)
= -2/3
1/7-(2/7*-2/3)
= 1/3
16-(2/7*14)
= 12
16/(2/7)
= 56
0
S1
0

10. 0
1
7/3
(-2/7)/(3/7)
= -2/3
6/(3/7)
= 14
16/(3/7)
= 37.33
4
X1
1
0
-5/7-(-5/7*1)
= 0
0-(-5/7*7/3)
= 5/3
1/7-(-5/7*-2/3)
= -1/3
5 –(-5/7*14)
= 15
5/(-5/7)
= -7
Zj
4
5
0
10/3
1/3
(5*12+4*15)
= 120
j = Cj – Zj
0
0
0
-10/0

11. -1/3
As all j is less than or equal to zero the optimal solution is obtained.
X1 = 15 and X2 = 12
Z = 120
Cj
4
5
0
0
0
B (Quantity)
Cb
Xb
x1
x2
S1
S2
S3
0
S1
-1
1
1
0
0
11
0
S2
1
1
0
1
0
27
0
S3

12. 2
5
0
0
1
90