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Page 1
Edurev123
4. Simplex Method of Solution
4.1 Maximize :?????????????????????????????????????????? =?? ?? ?? +?? ?? ?? +?? ?? ??
Subject to :
?? ?? ?? +?? ?? ?? ?=?? ?? ?? ?? +?? ?? ?? +?? ?? ?? ?=????
?? ?? ?? +?? ?? ?? ?=????
?? ?? ?=??
(2009 : 30 Marks)
Solution:
Step 1 : Checking whether objective function is to be maximized and all ?? 's are non-
negative. ? These conditions are satisfied. So, proceed to next step.
Step 2 : Express the above problem in standard form. By introducing the slack variables
?? 1
,?? 2
and ?? 3
, the problem in standard form becomes :
Max. ?? =3?? 1
+5?? 2
+4?? 3
+0?? 1
+0?? 2
+0?? 3
subject to
2?? 1
+3?? 2
+0?? 3
+?? 1
+0?? 2
+0?? 3
=8
3?? 1
+2?? 2
+4?? 3
+0?? 1
+?? 2
+0?? 3
=15
0?? 1
+2?? 2
+5?? 3
+0?? 1
+0?? 2
+?? 3
=10
where ?? 1
,?? 2
,?? 3
,?? 1
,?? 2
,?? 3
=0
Step 3 : Find initial basic feasible solution. The basic (non-degenerate) feasible solution
is
?? 1
=?? 2
=?? 3
=0 (non-basic)
?? 1
=8,?? 2
=15,?? 3
=10 (basic)
? Initial basic feasible solution is given by the table below
?? ?? 3 5 4 0 0 0
?? ?? Basis ?? 1
?? 2
?? 3
?? 1
?? 2
?? 3
?? ??
0 ?? 1
2 (3) 0 1 0 0 8
8
3
Page 2
Edurev123
4. Simplex Method of Solution
4.1 Maximize :?????????????????????????????????????????? =?? ?? ?? +?? ?? ?? +?? ?? ??
Subject to :
?? ?? ?? +?? ?? ?? ?=?? ?? ?? ?? +?? ?? ?? +?? ?? ?? ?=????
?? ?? ?? +?? ?? ?? ?=????
?? ?? ?=??
(2009 : 30 Marks)
Solution:
Step 1 : Checking whether objective function is to be maximized and all ?? 's are non-
negative. ? These conditions are satisfied. So, proceed to next step.
Step 2 : Express the above problem in standard form. By introducing the slack variables
?? 1
,?? 2
and ?? 3
, the problem in standard form becomes :
Max. ?? =3?? 1
+5?? 2
+4?? 3
+0?? 1
+0?? 2
+0?? 3
subject to
2?? 1
+3?? 2
+0?? 3
+?? 1
+0?? 2
+0?? 3
=8
3?? 1
+2?? 2
+4?? 3
+0?? 1
+?? 2
+0?? 3
=15
0?? 1
+2?? 2
+5?? 3
+0?? 1
+0?? 2
+?? 3
=10
where ?? 1
,?? 2
,?? 3
,?? 1
,?? 2
,?? 3
=0
Step 3 : Find initial basic feasible solution. The basic (non-degenerate) feasible solution
is
?? 1
=?? 2
=?? 3
=0 (non-basic)
?? 1
=8,?? 2
=15,?? 3
=10 (basic)
? Initial basic feasible solution is given by the table below
?? ?? 3 5 4 0 0 0
?? ?? Basis ?? 1
?? 2
?? 3
?? 1
?? 2
?? 3
?? ??
0 ?? 1
2 (3) 0 1 0 0 8
8
3
0 ?? 2
3 2 4 0 1 0 15
15
2
0 ?? 3
0 2 5 0 0 1 10
10
2
?? =S(?? ?? ????
) 0 0 0 0 0 0
?? ?? =?? ?? -?? ?? 3 5 4 0 0 0
?
As ?? 1
are positive under the columns. But we choose the largest positive, i.e., ' 5 ' under
column of ' ?? 2
'. Now ?? =
8
3
is smallest. So, ?? 2
is the incoming variable and ?? 1
is the
outgoing variable and (3) is the key element. Again producing :
?? ?? 3 5 4 0 0 0
?? ?? Basis ?? 1
?? 2
?? 3
?? 1
?? 2
?? 3
?? ??
5 ?? 2
2
3
1 0
1
3
0 0
8
3
8
3(0)
=8
0 ?? 2
5
3
0 4 -
2
3
1 0
29
3
29
3×4
=
29
12
0 ?? 3
-
4
3
0 (5) -
2
3
0 1
14
3
14
3×5
=
14
15
?? =S(?? ?? ????
)
10
3
5 0
5
3
0 0
?? ?? =?? ?? -?? ?? -
1
3
0 4 -
5
3
0 0
?
Again largest positive ?? ?? =4 under column of ' ?? 3
' and now ?? =
14
15
is smallest. So, ?? 3
is
the incoming variable and ?? 3
is the outgoing variable and (5) is the key element. Again
proceeding:
?? ?? 3 5 4 0 0 0
Page 3
Edurev123
4. Simplex Method of Solution
4.1 Maximize :?????????????????????????????????????????? =?? ?? ?? +?? ?? ?? +?? ?? ??
Subject to :
?? ?? ?? +?? ?? ?? ?=?? ?? ?? ?? +?? ?? ?? +?? ?? ?? ?=????
?? ?? ?? +?? ?? ?? ?=????
?? ?? ?=??
(2009 : 30 Marks)
Solution:
Step 1 : Checking whether objective function is to be maximized and all ?? 's are non-
negative. ? These conditions are satisfied. So, proceed to next step.
Step 2 : Express the above problem in standard form. By introducing the slack variables
?? 1
,?? 2
and ?? 3
, the problem in standard form becomes :
Max. ?? =3?? 1
+5?? 2
+4?? 3
+0?? 1
+0?? 2
+0?? 3
subject to
2?? 1
+3?? 2
+0?? 3
+?? 1
+0?? 2
+0?? 3
=8
3?? 1
+2?? 2
+4?? 3
+0?? 1
+?? 2
+0?? 3
=15
0?? 1
+2?? 2
+5?? 3
+0?? 1
+0?? 2
+?? 3
=10
where ?? 1
,?? 2
,?? 3
,?? 1
,?? 2
,?? 3
=0
Step 3 : Find initial basic feasible solution. The basic (non-degenerate) feasible solution
is
?? 1
=?? 2
=?? 3
=0 (non-basic)
?? 1
=8,?? 2
=15,?? 3
=10 (basic)
? Initial basic feasible solution is given by the table below
?? ?? 3 5 4 0 0 0
?? ?? Basis ?? 1
?? 2
?? 3
?? 1
?? 2
?? 3
?? ??
0 ?? 1
2 (3) 0 1 0 0 8
8
3
0 ?? 2
3 2 4 0 1 0 15
15
2
0 ?? 3
0 2 5 0 0 1 10
10
2
?? =S(?? ?? ????
) 0 0 0 0 0 0
?? ?? =?? ?? -?? ?? 3 5 4 0 0 0
?
As ?? 1
are positive under the columns. But we choose the largest positive, i.e., ' 5 ' under
column of ' ?? 2
'. Now ?? =
8
3
is smallest. So, ?? 2
is the incoming variable and ?? 1
is the
outgoing variable and (3) is the key element. Again producing :
?? ?? 3 5 4 0 0 0
?? ?? Basis ?? 1
?? 2
?? 3
?? 1
?? 2
?? 3
?? ??
5 ?? 2
2
3
1 0
1
3
0 0
8
3
8
3(0)
=8
0 ?? 2
5
3
0 4 -
2
3
1 0
29
3
29
3×4
=
29
12
0 ?? 3
-
4
3
0 (5) -
2
3
0 1
14
3
14
3×5
=
14
15
?? =S(?? ?? ????
)
10
3
5 0
5
3
0 0
?? ?? =?? ?? -?? ?? -
1
3
0 4 -
5
3
0 0
?
Again largest positive ?? ?? =4 under column of ' ?? 3
' and now ?? =
14
15
is smallest. So, ?? 3
is
the incoming variable and ?? 3
is the outgoing variable and (5) is the key element. Again
proceeding:
?? ?? 3 5 4 0 0 0
?? ?? Basis ?? 1
?? 2
?? 3
?? 1
?? 2
?? 3
?? ??
5 ?? 2
2
3
1 0
1
3
0 0
8
3
8×3
3×2
=4
0 ?? 2
(
41
15
) 0 0 -
2
15
1 -
4
15
89
15
89×15
15×41
=
89
41
4 ?? 3
-
4
15
0 1 -
2
15
0
1
5
14
15
14×15
15×(-4)
?? =S?? ?? ?? ????
34
15
5 4
17
15
0
4
5
?? ?? =?? ?? -?? ??
11
15
0 0 -
17
15
0 -
4
5
?
Again
11
15
is the only positive ?? ?? under column of ' ?? 1
' and now ?? =
89
41
is the smallest
positive fraction. So,
?? 1
is the incoming variable and ?? 2
is the outgoing variabie and (
41
15
) is the key element.
Again proceeding :
?? ?? 3 5 4 0 0 0
?? ?? Basis ?? 1
?? 2
?? 3
?? 1
?? 2
?? 3
?? ??
5 ?? 2
0 1 0
15
41
-
10
41
8
41
50
41
3 ?? 1
1 0 0 -
2
41
15
41
-
12
41
89
41
4 ?? 3
0 0 1 -
6
41
4
41
5
41
62
41
?? =S?? ?? ?? ????
3 5 1
45
41
11
41
24
41
?? ?? =?? ?? -?? ?? 0 0 0
-
45
41
-
11
41
-
24
41
Page 4
Edurev123
4. Simplex Method of Solution
4.1 Maximize :?????????????????????????????????????????? =?? ?? ?? +?? ?? ?? +?? ?? ??
Subject to :
?? ?? ?? +?? ?? ?? ?=?? ?? ?? ?? +?? ?? ?? +?? ?? ?? ?=????
?? ?? ?? +?? ?? ?? ?=????
?? ?? ?=??
(2009 : 30 Marks)
Solution:
Step 1 : Checking whether objective function is to be maximized and all ?? 's are non-
negative. ? These conditions are satisfied. So, proceed to next step.
Step 2 : Express the above problem in standard form. By introducing the slack variables
?? 1
,?? 2
and ?? 3
, the problem in standard form becomes :
Max. ?? =3?? 1
+5?? 2
+4?? 3
+0?? 1
+0?? 2
+0?? 3
subject to
2?? 1
+3?? 2
+0?? 3
+?? 1
+0?? 2
+0?? 3
=8
3?? 1
+2?? 2
+4?? 3
+0?? 1
+?? 2
+0?? 3
=15
0?? 1
+2?? 2
+5?? 3
+0?? 1
+0?? 2
+?? 3
=10
where ?? 1
,?? 2
,?? 3
,?? 1
,?? 2
,?? 3
=0
Step 3 : Find initial basic feasible solution. The basic (non-degenerate) feasible solution
is
?? 1
=?? 2
=?? 3
=0 (non-basic)
?? 1
=8,?? 2
=15,?? 3
=10 (basic)
? Initial basic feasible solution is given by the table below
?? ?? 3 5 4 0 0 0
?? ?? Basis ?? 1
?? 2
?? 3
?? 1
?? 2
?? 3
?? ??
0 ?? 1
2 (3) 0 1 0 0 8
8
3
0 ?? 2
3 2 4 0 1 0 15
15
2
0 ?? 3
0 2 5 0 0 1 10
10
2
?? =S(?? ?? ????
) 0 0 0 0 0 0
?? ?? =?? ?? -?? ?? 3 5 4 0 0 0
?
As ?? 1
are positive under the columns. But we choose the largest positive, i.e., ' 5 ' under
column of ' ?? 2
'. Now ?? =
8
3
is smallest. So, ?? 2
is the incoming variable and ?? 1
is the
outgoing variable and (3) is the key element. Again producing :
?? ?? 3 5 4 0 0 0
?? ?? Basis ?? 1
?? 2
?? 3
?? 1
?? 2
?? 3
?? ??
5 ?? 2
2
3
1 0
1
3
0 0
8
3
8
3(0)
=8
0 ?? 2
5
3
0 4 -
2
3
1 0
29
3
29
3×4
=
29
12
0 ?? 3
-
4
3
0 (5) -
2
3
0 1
14
3
14
3×5
=
14
15
?? =S(?? ?? ????
)
10
3
5 0
5
3
0 0
?? ?? =?? ?? -?? ?? -
1
3
0 4 -
5
3
0 0
?
Again largest positive ?? ?? =4 under column of ' ?? 3
' and now ?? =
14
15
is smallest. So, ?? 3
is
the incoming variable and ?? 3
is the outgoing variable and (5) is the key element. Again
proceeding:
?? ?? 3 5 4 0 0 0
?? ?? Basis ?? 1
?? 2
?? 3
?? 1
?? 2
?? 3
?? ??
5 ?? 2
2
3
1 0
1
3
0 0
8
3
8×3
3×2
=4
0 ?? 2
(
41
15
) 0 0 -
2
15
1 -
4
15
89
15
89×15
15×41
=
89
41
4 ?? 3
-
4
15
0 1 -
2
15
0
1
5
14
15
14×15
15×(-4)
?? =S?? ?? ?? ????
34
15
5 4
17
15
0
4
5
?? ?? =?? ?? -?? ??
11
15
0 0 -
17
15
0 -
4
5
?
Again
11
15
is the only positive ?? ?? under column of ' ?? 1
' and now ?? =
89
41
is the smallest
positive fraction. So,
?? 1
is the incoming variable and ?? 2
is the outgoing variabie and (
41
15
) is the key element.
Again proceeding :
?? ?? 3 5 4 0 0 0
?? ?? Basis ?? 1
?? 2
?? 3
?? 1
?? 2
?? 3
?? ??
5 ?? 2
0 1 0
15
41
-
10
41
8
41
50
41
3 ?? 1
1 0 0 -
2
41
15
41
-
12
41
89
41
4 ?? 3
0 0 1 -
6
41
4
41
5
41
62
41
?? =S?? ?? ?? ????
3 5 1
45
41
11
41
24
41
?? ?? =?? ?? -?? ?? 0 0 0
-
45
41
-
11
41
-
24
41
As all ?? ?? are negative. So, the optimal solution is achieved.
Optimal solution is :
?? 1
=
89
41
?? 2
=
50
41
?? 3
=
62
41
and
Max. ?? ?=3×
89
41
+5×
50
41
+4×
62
41
?=
765
41
Max. ?? ?=
765
41
4.2 Solve by Simplex method, the following LP Problem:
Maximize????????????????????????????????????????????????????? =?? ?? ?? +?? ?? ??
Constraints.
?? ?? ?? +?? ?? ?? ?=????
?? ?? ?? +?? ?? ?? ?=????
?? ?? ,?? ?? ?=??
(2011 : 12 Marks)
Solution:
The give, : LPP is
Maximize????????????????????????????????????????????????????? =5?? 1
+3?? 2
Constraints.
3?? 1
+5?? 2
?=15
5?? 1
+2?? 2
?=10
?? 1
,?? 2
?=0
Convert the inequalities by the introduction of slack variables ?? 3
and ?? 4
as follows
3?? 1
+5?? 2
+?? 3
=15
5?? 1
+2?? 2
+?? 4
=10
Page 5
Edurev123
4. Simplex Method of Solution
4.1 Maximize :?????????????????????????????????????????? =?? ?? ?? +?? ?? ?? +?? ?? ??
Subject to :
?? ?? ?? +?? ?? ?? ?=?? ?? ?? ?? +?? ?? ?? +?? ?? ?? ?=????
?? ?? ?? +?? ?? ?? ?=????
?? ?? ?=??
(2009 : 30 Marks)
Solution:
Step 1 : Checking whether objective function is to be maximized and all ?? 's are non-
negative. ? These conditions are satisfied. So, proceed to next step.
Step 2 : Express the above problem in standard form. By introducing the slack variables
?? 1
,?? 2
and ?? 3
, the problem in standard form becomes :
Max. ?? =3?? 1
+5?? 2
+4?? 3
+0?? 1
+0?? 2
+0?? 3
subject to
2?? 1
+3?? 2
+0?? 3
+?? 1
+0?? 2
+0?? 3
=8
3?? 1
+2?? 2
+4?? 3
+0?? 1
+?? 2
+0?? 3
=15
0?? 1
+2?? 2
+5?? 3
+0?? 1
+0?? 2
+?? 3
=10
where ?? 1
,?? 2
,?? 3
,?? 1
,?? 2
,?? 3
=0
Step 3 : Find initial basic feasible solution. The basic (non-degenerate) feasible solution
is
?? 1
=?? 2
=?? 3
=0 (non-basic)
?? 1
=8,?? 2
=15,?? 3
=10 (basic)
? Initial basic feasible solution is given by the table below
?? ?? 3 5 4 0 0 0
?? ?? Basis ?? 1
?? 2
?? 3
?? 1
?? 2
?? 3
?? ??
0 ?? 1
2 (3) 0 1 0 0 8
8
3
0 ?? 2
3 2 4 0 1 0 15
15
2
0 ?? 3
0 2 5 0 0 1 10
10
2
?? =S(?? ?? ????
) 0 0 0 0 0 0
?? ?? =?? ?? -?? ?? 3 5 4 0 0 0
?
As ?? 1
are positive under the columns. But we choose the largest positive, i.e., ' 5 ' under
column of ' ?? 2
'. Now ?? =
8
3
is smallest. So, ?? 2
is the incoming variable and ?? 1
is the
outgoing variable and (3) is the key element. Again producing :
?? ?? 3 5 4 0 0 0
?? ?? Basis ?? 1
?? 2
?? 3
?? 1
?? 2
?? 3
?? ??
5 ?? 2
2
3
1 0
1
3
0 0
8
3
8
3(0)
=8
0 ?? 2
5
3
0 4 -
2
3
1 0
29
3
29
3×4
=
29
12
0 ?? 3
-
4
3
0 (5) -
2
3
0 1
14
3
14
3×5
=
14
15
?? =S(?? ?? ????
)
10
3
5 0
5
3
0 0
?? ?? =?? ?? -?? ?? -
1
3
0 4 -
5
3
0 0
?
Again largest positive ?? ?? =4 under column of ' ?? 3
' and now ?? =
14
15
is smallest. So, ?? 3
is
the incoming variable and ?? 3
is the outgoing variable and (5) is the key element. Again
proceeding:
?? ?? 3 5 4 0 0 0
?? ?? Basis ?? 1
?? 2
?? 3
?? 1
?? 2
?? 3
?? ??
5 ?? 2
2
3
1 0
1
3
0 0
8
3
8×3
3×2
=4
0 ?? 2
(
41
15
) 0 0 -
2
15
1 -
4
15
89
15
89×15
15×41
=
89
41
4 ?? 3
-
4
15
0 1 -
2
15
0
1
5
14
15
14×15
15×(-4)
?? =S?? ?? ?? ????
34
15
5 4
17
15
0
4
5
?? ?? =?? ?? -?? ??
11
15
0 0 -
17
15
0 -
4
5
?
Again
11
15
is the only positive ?? ?? under column of ' ?? 1
' and now ?? =
89
41
is the smallest
positive fraction. So,
?? 1
is the incoming variable and ?? 2
is the outgoing variabie and (
41
15
) is the key element.
Again proceeding :
?? ?? 3 5 4 0 0 0
?? ?? Basis ?? 1
?? 2
?? 3
?? 1
?? 2
?? 3
?? ??
5 ?? 2
0 1 0
15
41
-
10
41
8
41
50
41
3 ?? 1
1 0 0 -
2
41
15
41
-
12
41
89
41
4 ?? 3
0 0 1 -
6
41
4
41
5
41
62
41
?? =S?? ?? ?? ????
3 5 1
45
41
11
41
24
41
?? ?? =?? ?? -?? ?? 0 0 0
-
45
41
-
11
41
-
24
41
As all ?? ?? are negative. So, the optimal solution is achieved.
Optimal solution is :
?? 1
=
89
41
?? 2
=
50
41
?? 3
=
62
41
and
Max. ?? ?=3×
89
41
+5×
50
41
+4×
62
41
?=
765
41
Max. ?? ?=
765
41
4.2 Solve by Simplex method, the following LP Problem:
Maximize????????????????????????????????????????????????????? =?? ?? ?? +?? ?? ??
Constraints.
?? ?? ?? +?? ?? ?? ?=????
?? ?? ?? +?? ?? ?? ?=????
?? ?? ,?? ?? ?=??
(2011 : 12 Marks)
Solution:
The give, : LPP is
Maximize????????????????????????????????????????????????????? =5?? 1
+3?? 2
Constraints.
3?? 1
+5?? 2
?=15
5?? 1
+2?? 2
?=10
?? 1
,?? 2
?=0
Convert the inequalities by the introduction of slack variables ?? 3
and ?? 4
as follows
3?? 1
+5?? 2
+?? 3
=15
5?? 1
+2?? 2
+?? 4
=10
Taking ?? 1
=0,?? 2
=0, we get, ?? 3
=15,?? 4
=10 which is the starting basic feasible
solution. Construct the starting simplex table as follows :
?? ?? 5 3 0 0
Minimum
Ratio
?? ?? ?? 1
?? ?? ?? ?? ?? ?? 1
?? 2
?? 3
?? 4
1/3 0 15 3 5 1 0
15
3
=5
1/4 0 10 5 2 0 1
10
5
=2(Min)
?
??
?
Incoming Vector
?
?? ?=?? ?? -?? ?? ?? ?? ?
?? ?=?? 1
-?? ?? ?? 1
?=5-[(0,0)(3,5)]=5-(0×3+0×5)
?=5-0=5
?
2
?=?? 2
-?? ?? ?? 2
?=3-[(0,0)(5,2)]=3
The values of ?
3
and ?
4
must be zero as they correspond to unit column vectors.
Since all ?
?? are not less than or equal to zero, therefore the solution is not optimal.
Since ?
1
=5 is maximum, ??? 1
is the incoming vector.
Since,
?? ?? 1/1
ratio is minimum against ?? 4
in the table, therefore ?? 4
is the outgoing vector.
The element at the intersection of the minimum ratio arrow and incoming vector arrow is
the pivot element.
We mark this element in ?.
Divide the second row containing the key element (=5) by 5 to get unity at this position
and add 'minus 3 times' the second row to first row to zero at all other positions of the
column containing the pivot element.
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FAQs on Simplex Method of Solution - Mathematics Optional Notes for UPSC
| 1. What is the Simplex Method of solution in linear programming? |  |
Ans. The Simplex Method is a mathematical technique used for solving linear programming problems. It involves iteratively moving from one feasible solution to another along the edges of the feasible region until the optimal solution is reached.
| 2. How does the Simplex Method work in solving linear programming problems? | |
Ans. The Simplex Method works by starting at a basic feasible solution and then moving to adjacent basic feasible solutions that improve the objective function value. This process continues until the optimal solution is found.
| 3. What are the key components of the Simplex Method algorithm? | |
Ans. The key components of the Simplex Method algorithm include identifying initial basic feasible solution, determining entering and leaving variables, calculating pivot values, and updating the basic feasible solution.
| 4. When is the Simplex Method preferred over other optimization techniques? | |
Ans. The Simplex Method is preferred over other optimization techniques when dealing with linear programming problems with multiple variables and constraints. It is particularly useful when there are many feasible solutions.
| 5. Can the Simplex Method be used for non-linear programming problems? | |
Ans. No, the Simplex Method is specifically designed for solving linear programming problems. For non-linear programming problems, other optimization techniques such as gradient descent or genetic algorithms are more appropriate.
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Nov 24, 2024 Last updated |
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Additional Information about Simplex Method of Solution for UPSC Preparation
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The content of Simplex Method of Solution is prepared as per the latest UPSC syllabus.
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