Module-III Varried Flow.pptx GVF Definition, Water Surface Profile Dynamic Eq...
simplex method-maths 4 mumbai university
1.
2. Simplex Method
•When decision variables are more than 2, we always use Simplex Method
•Slack Variable: Variable added to a constraint to convert it to an
equation (=).
• A slack variable represents unused resources
• A slack variable contributes nothing to the objective
function value.
•Surplus Variable: Variable subtracted from a constraint to convert it to
an equation (=).
• A surplus variable represents an excess above a constraint
requirement level.
• Surplus variables contribute nothing to the calculated value
of the objective function.
3. Cont….
Basic Solution(BS) : This solution is obtained by setting any n variables
(among m+n variables) equal to zero and solving for remaining m
variables, provided the determinant of the coefficients of these variables
is non-zero. Such m variables are called basic variables and remaining n
zero valued variables are called non basic variables.
Basic Feasible Solution(BFS) : It is a basic solution which also satisfies the
non negativity restrictions.
4. Cont…..
BFS are of two types:
◦ Degenerate BFS: If one or more basic variables are zero.
◦ Non-Degenerate BFS: All basic variables are non-zero.
Optimal BFS: BFS which optimizes the objective function.
5. Example
Max. Z = 13x1+11x2
Subject to constraints:
4x1+5x2 < 1500
5x1+3x2 < 1575
x1+2x2 < 420
x1, x2 > 0
6. Solution :
Step 1: Convert all the inequality constraints into equalities
by the use of slack variables.
Let S1, S2 , S3 be three slack variables.
Introducing these slack variables into the inequality constraints
and rewriting the objective function such that all variables are on
the left-hand side of the equation. Model can rewritten as:
Z - 13x1 -11x2 = 0
Subject to constraints:
4x1+5x2 + S1 = 1500
5x1+3x2 +S2= 1575
x1+2x2 +S3 = 420
x1, x2, S1, S2, S3 > 0
7. Cont…
Step II: Find the Initial BFS.
One Feasible solution that satisfies all the constraints is: x1= 0, x2= 0, S1=
1500,
S2= 1575, S3= 420 and Z=0.
Now, S1, S2, S3 are Basic variables.
Step III: Set up an initial table as:
8. Cont…
Row
NO.
Basic
Variable
Coefficients of: Sol. Rati
o
Z x1 x2 S1 S2 S3
A1 Z 1 -13 -11 0 0 0 0
B1 S1 0 4 5 1 0 0 1500 375
C1 S2 0 5 3 0 1 0 1575 315
D1 S3 0 1 2 0 0 1 420 420
Step IV: a) Choose the most negative number from row A1(i.e Z row). Therefore,
x1 is a entering variable.
b) Calculate Ratio = Sol col. / x1 col. (x1 > 0)
c) Choose minimum Ratio. That variable(i.e S2) is a departing
variable.
9. Cont….
Step V: x1 becomes basic variable and S2 becomes non basic
variable. New table is:
Row
NO.
Basic
Varia
ble
Coefficients of: Sol. Ratio
Z x1 x2 S1 S2 S3
A1 Z 1 0 -16/5 0 13/5 0 4095
B1 S1 0 0 13/5 1 -4/5 0 240 92.3
C1 x1 0 1 3/5 0 1/5 0 315 525
D1 S3 0 0 7/5 0 -1/5 1 105 75
23. Example
Max.. Z = 3x1 + 4x2
Subject to constraints:
x1 - x2 < 1
-x1 + x2 < 2
x1, x2 > 0
24. Cont…
Let S1 and S2 be two slack variables
.
Modified form is:
Z -3x1 - 4x2 = 0
x1 - x2 +S1 = 1
-x1 + x2 +S2 = 2
x1, x2, S1, S2 > 0
Initial BFS is : x1= 0, x2= 0, S1= 1, S2= 2
and Z=0.
25. Cont…
Basic
Variable
Coefficients of: Sol. Ratio
Z x1 x2 S1 S2
Z 1 -3 -4 0 0 0
S1 0 1 -1 1 0 1 -
S2 0 -1 1 0 1 2 2
Therefore, x2 is the entering variable and S2 is the
departing variable.
26. Cont…
Basic
Variable
Coefficients of: Sol. Ratio
Z x1 x2 S1 S2
Z 1 -7 0 0 4 8
S1 0 0 0 1 1 3 -
x2 0 -1 1 0 1 2 -
x1 is the entering variable, but as in x1 column every no. is less
than equal to zero, ratio cannot be calculated. Therefore given
problem is having a unbounded solution.
27.
28. Introduction
LPP, in which constraints may also have > and = signs, we
introduce a new type of variable , called the artificial
variable. These variables are fictitious and cannot
have any physical meaning. Two Phase Simplex
Method is used to solve a problem in which some
artificial variables are involved. The solution is
obtained in two phases.
30. Cont…
Convert the objective function into the maximization form
Max. Z’ = -15/2 x1 + 3x2 where Z’= -Z
Subject to constraints:
3x1 - x2 - x3 > 3
x1 - x2 + x3 > 2
x1, x2, x3 > 0
31. Cont…
Modified form is :
Introduce surplus variables S1 and S2, and artificial variables
a1 and a2
Z’ + 15/2 x1 - 3x2 = 0
Subject to constraints:
3x1 - x2 - x3 –S1 + a1 = 3
x1 - x2 + x3 –S2 + a2 = 2
x1, x2, x3 , S1, S2, a1, a2 > 0
32. Cont…
Phase I : Simplex method is applied to a specially constructed
Auxiliary LPP leading to a final simplex table containing a BFS
to the original problem.
•Step 1: Assign a cost –1 to each artificial variable and a cost
0 to all other variables in the objective function.
•Step 2: Construct the auxiliary LPP in which the new
objective function Z* is to be maximized subject to the given
set of constraints.
34. Cont…
•Step 3: Solve the auxiliary problem by simplex method until either
of the following three possibilities arise:
•Max Z* < 0 and at least one artificial variable appear in the
optimum basis at a positive level. In this case given problem
does not have any feasible solution.
•Max Z* = 0 and at least one artificial variable appear in the
optimum basis at a zero level. In this case proceed to Phase II.
•Max Z* = 0 and no artificial variable appear in the optimum
basis. In this case also proceed to Phase II.
35. Cont…
Basic
Variable
Coefficients of: Sol.
Z* x1 x2 x3 S1 S2 a1 a2
Z* 1 0 0 0 0 0 1 1 0
a1 0 3 -1 -1 -1 0 1 0 3
a2 0 1 -1 1 0 -1 0 1 2
This table is not feasible as a1 and a2 has non zero coefficients in
Z* row. Therefore next step is to make the table feasible.
36. Cont…
Basic
Variable
Coefficients of: Sol. Ratio
Z* x1 x2 x3 S1 S2 a1 a2
Z* 1 -4 2 0 1 1 0 0 -5
a1 0 3 -1 -1 -1 0 1 0 3 1
a2 0 1 -1 1 0 -1 0 1 2 2
Therefore, x1 is the entering variable and a1 is the
departing variable.
37. Cont…
Basic
Variable
Coefficients of: Sol. Ratio
Z* x1 x2 x3 S1 S2 a2
Z* 1 0 2/3 -4/3 -1/3 1 0 -1
x1 0 1 -1/3 -1/3 -1/3 0 0 1 -
a2 0 0 -2/3 4/3 1/3 -1 1 1 3/4
Therefore, x3 is the entering variable and a2 is the
departing variable.
38. Cont…
Basic
Variable
Coefficients of: Sol.
Z* x1 x2 x3 S1 S2
Z* 1 0 0 0 0 0 0
x1 0 1 -1/2 0 -1/4 -1/4 5/4
x3 0 0 -1/2 1 1/4 -3/4 3/4
As there is no variable to be entered in the basis, this table is
optimum for Phase I. In this table Max. Z* = 0 and no artificial
variable appears in the optimum basis, therefore we can proceed
to Phase II.
39. Cont…
Phase II: The artificial variables which are non basic at the end of
Phase I are removed from the table and as well as from the
objective function and constraints. Now assign the actual costs to
the variables in the Objective function. That is, Simplex method is
applied to the modified simplex table obtained at the Phase I.
Basic
Variable
Coefficients of: Sol.
Z’ x1 x2 x3 S1 S2
Z’ 1 15/2 -3 0 0 0 0
x1 0 1 -1/2 0 -1/4 -1/4 5/4
x3 0 0 -1/2 1 1/4 -3/4 3/4
Again this table is not feasible as basic variable x1 has a non zero
coefficient in Z’ row. So make the table feasible.
43. Cont…
Basic
Variable
Coefficients of: Sol.
Z* x1 x2 x3 a1 a2
Z* 1 0 0 0 1 1 0
a1 0 -2 1 3 1 0 2
a2 0 2 3 4 0 1 1
This table is not feasible as a1 and a2 has non zero coefficients in
Z* row. Therefore next step is to make the table feasible.
44. Cont…
Basic
Variable
Coefficients of: Sol. Ratio
Z* x1 x2 x3 a1 a2
Z* 1 0 -4 -7 0 0 -3
a1 0 -2 1 3 1 0 2 2/3
a2 0 2 3 4 0 1 1 1/4
Therefore, x3 is the entering variable and a2 is the
departing variable.
45. Cont…
Basic
Variable
Coefficients of: Sol.
Z* x1 x2 x3 a1
Z* 1 7/4 5/4 0 0 -5/4
a1 0 -7/2 -5/4 0 1 5/4
x3 0 1/2 3/4 1 0 1/4
As there is no variable to be entered in the basis, therefore
Phase I ends here. But one artificial variable is present in the
basis and Z* < 0. Therefore we cannot proceed to Phase II.
Given problem is having a non-feasible solution.
46. Example
Min.. Z = 4x1 + x2
Subject to constraints:
3x1 + x2 = 3
4x1 + 3x2 > 6
x1 +2x2 < 4
x1, x2 > 0
51. Cont…
Basic
Variable
Coefficients of: Sol.
Z* x1 x2 S1 S2
Z* 1 0 0 0 0 0
x1 0 1 0 1/5 0 3/5
x2 0 0 1 -3/5 0 6/5
S2 0 0 0 1 1 1
As there is no variable to be entered in the basis, this table is
optimum for Phase I. In this table Max. Z* = 0 and no artificial
variable appears in the optimum basis, therefore we can proceed
to Phase II.
52. Cont…
Phase II:
Original Objective function is:
Min.. Z = 4x1 + x2
Convert the objective function into the maximization form
Max. Z’ = -4 x1 - x2 where Z’= -Z
Basic
Variable
Coefficients of: Sol.
Z* x1 x2 S1 S2
Z* 1 4 1 0 0 0
x1 0 1 0 1/5 0 3/5
x2 0 0 1 -3/5 0 6/5
S2 0 0 0 1 1 1
Again this table is not feasible as basic variable x1 and x2 has a non zero
coefficient in Z’ row. So make the table feasible.
53. Cont…
Basic
Variable
Coefficients of: Sol. Ratio
Z’ x1 x2 S1 S2
Z’ 1 0 0 -1/5 0 -18/5
x1 0 1 0 1/5 0 3/5 3
x2 0 0 1 -3/5 0 6/5 -
S2 0 0 0 1 1 1 1
Therefore, S1 is the entering variable and S2 is the
departing variable.
56. Introduction
At the stage of improving the solution during Simplex
procedure, if a tie for the minimum ratio occurs at least
one basic variable becomes equal to zero in the next
iteration and the new solution is said to be Degenerate.
57. Example
Max.. Z = 3x1 + 9x2
Subject to constraints:
x1 + 4x2 < 8
x1 + 2x2 < 4
x1, x2 > 0
58. Cont…
Let S1 and S2 be two slack variables.
Modified form is:
Z - 3x1 - 9x2 = 0
x1 + 4x2 + S1 = 8
x1 + 2x2 +S2 = 4
x1, x2, S1, S2 > 0
Initial BFS is : x1= 0, x2= 0, S1= 8, S2= 4 and Z=0.
59. Cont…
Basic
Variable
Coefficients of: Sol. Ratio
Z x1 x2 S1 S2
Z 1 -3 -9 0 0 0
S1 0 1 4 1 0 8 2
S2 0 1 2 0 1 4 2
In this table S1 and S2 tie for the leaving variable. So
any one can be considered as leaving variable.
Therefore, x2 is the entering variable and S1 is the
departing variable.
60. Cont…
Basic
Variable
Coefficients of: Sol. Ratio
Z x1 x2 S1 S2
Z 1 -3/4 0 9/4 0 18
x2 0 1/4 1 1/4 0 2 8
S2 0 1/2 0 -1/2 1 0 0
Therefore, x1 is the entering variable and S2 is the
departing variable.
61. Cont…
Basic
Variable
Coefficients of: Sol.
Z x1 x2 S1 S2
Z 1 0 0 3/2 3/2 18
x2 0 0 1 1/2 -1/2 2
x1 0 1 0 -1 2 0
Optimal Solution is : x1= 0, x2= 2
Z = 18
It results in a Degenerate Basic Solution.
