If an artificial variable is in an optimal solution of the equivalent model at a nonzero level, then no feasible solution for the original model exists. On the contrary, if the optimal solution of the equivalent model does not contain an artificial variable at a non-zero level, the solution is also optimal for the original model.
Properties of The Simplex
Method
1. The
Simplex method for maximizing the objective function starts at a basic feasible
solution for the equivalent model and moves to an adjacent basic feasible
solution that does not decrease the value of the objective function. If such a
solution does not exist, an optimal solution
for the equivalent model has been reached. That is, if all of the Coefficients
of the non-basic variables in the objective function equation are greater than
or equal to zero at some point, then an optimal solution for the equivalent
model has been reached.
2. If an
artificial variable is in an optimal solution of the equivalent model at a
nonzero level, then no feasible solution for the original model exists. On the
contrary, if the optimal solution of the equivalent model does not contain an
artificial variable at a non-zero level, the solution is also optimal for the
original model.
3.
If all the
slack, surplus, and artificial variables are zero when an optimal
solution of the equivalent model
is reached, then all of the
constraints in the original model are strict
“equalities” for the values of the variables that
optimize the objective function.
4. If a
non-basic variable has zero coefficients in the objective function equation
when an optimal solution is reached, there are multiple optimal solutions. In
fact, there is infinity of optimal solutions, the Simplex method finds only one
optimal solution and stops.
5. Once an
artificial variable leaves the set of basic variables (the basis), it will
never enter the basis again, so all calculations for that variable can be
ignored in future steps.
6.
When
selecting the variable to leave the current basis:
(a)
If two or
more ratios are smallest, choose one arbitrarily.
(b)
If a positive ratio does not exist, the objective
function in the original model is not bounded by the constraints. Thus a Finite
optimal solution for The original model does not exist.
7.
If a basis
has a variable at zero level, it is called a degenerate basis.
8. Although
cycling is possible, there have never been any practical problems for which the
Simplex method failed to converge.
Example Maximize z = X1+2X2 Subject to: -X1+2X2<=8, X1+2X2<=12, X1-X2<=3; X1>=0
and X2>=0. Solution Step 1 Introducing the slack Variable X3>=0, X4>=0
and X5>=0 to the first, second and third constraints respectively and
convert the problem into standard form. -X1+2X2 + X3 = 8, X1+2X2 +X4 =12, X1- X2 +X5 = 3; And the
modified objective function is
Z=X1+2X2+0X3+0X4+0X5 X1, X2,
X3, X4 & X5 >0 The constraints the given
L.P.P are converted into the system of equations:
Step 2 An
obvious initial basic feasible solution is given by XB=B-1b. Where
B=I3 and XB=[X3 X4 X5], & I3 stands for Identity matrix of order of 3 (that
is a 3X3 matrix). That is, [X3 X4 X5] = I3 [8 12 3] = [8 12 3] Step 3 We
compute yj and the net evaluations, zj-cj corresponding to the basic variables
X3, X4 and X5: y1=B-1a1 | = | I3 [-1 | 1 | 1] | = | [1 1 | 1] |
y2=B-1a1 | = | I3 [2 | 2 | -1] | = [2 2
-1] |
y3 =
B-1e1 = e1, y4 = B-1e2 = e2 | and | y5 =
B-1e3 = e3. |
Z1-C1 =
cB y1-c1 = (0 0 | 0) [-1 | 1 | 1]-1 =
-1. |
Z2-C2 =
cB y2-c2 = (0 0 0) [2 | 2 | | -1]-2 | = -2. |
Z3-C3 = cB y3-c3 = (0 0 0) e1-0 =
0, Z4-C4 = cB y4-c4 = (0 0 0) e2-0 =
0, Z5-C5 = cB y5-c5 = (0 0 0) e3-0 =
0. Step 4 -
Deciding the entering variable Making use of the above
information, the starting simplex tableau is written as follows:
From the
starting tableau, it is apparent that there are two Zj-Cj values, which are
having negative coefficients. We choose
the most negative of these, viz., -2. The corresponding column vector y2,
therefore, enters the basis. Step 5 - Deciding the leaving variable Now, we
will compute the ratios using the entering column elements and RHS of each
constraint. Each row of the table, the respective RHS
coefficient of the constraint is divided by entering column, non-zero element
and placed in the last column of the table. Then, the minimum among the value
is chosen as leaving variable. Min
{XBi/Yi2, Yi2>0} = Min. {8/2, 12/2, no ratio for third row} = 4. Since the
minimum ratio occurs for the first row, basis vector Y3 leaves the basis. The
common intersection element y12 (=2) become the leading element for updating.
We indicate the leading element in bold type with a star *. Step 6 Convert
the leading element y12 to unity and all other elements in its column (i.e.y2)
to zero by the following transformations: Y11 =Y11/Y12 =1/2, Y10 = Y10/Y12 =8/2
or 4, so on,
Y20=y20-(y10/y11) y22=12-(8/2) (2) =4.
Y30=y30-(y10/y12)
y32=3-(8/2) (-2) =11.
Y21=y21-(y11/y12) y22=1-(-1/2) (2) =2. Y31=y31-(y11/y12) y32=1-(-1/2)
(-2) =0. And so on……… Step 7 Using the
above computations, the following iterated simplex tableau is obtained: The above
simplex tableau yields a new basic feasible solution with the increased value
of z. Now since
z1-c1 < 0, y1 enters the basis. Also,
since Min. {X Bi / yi >0} = Min {4/2, 7/ (1/2)} = 2, y4 leaves the basis. Thus the leading element will be
y21 (=2). Converting
the leadwing element to unity and all other elements of yi to zero by usual row
transformations the next iterated tableau is obtained.
Since,
all Zj-Cj>=0, we conclude that there is no more improvement possible and the
problem is in its optimum stage. Therefore, the optimal solution for the given
problem is X1= 2 X2= 5 Max
Z = 12 Example-2 Solve the
following problem by simplex method [the same problem is solved under graphical
method already] Maximize z = 15X1 + 10X2 Subject to constraints 4X1+6X2
<=360 3X1+0X2<=180 0X1+5X2
<=200 X1,
X2>=0 Solution The
problem is converted into standard form by adding slack variables X3, X4 &
X5 to the each of the constraint. 4X1+6X2 +X3=360 3X1+0X2+X4=180 0X1+5X2+X5=200 And the modified objective function is Maximize z = 15X1 + 10X2 +0X3 +
0X4 + 0X5
Then the first simplex table is
arrived. Then we
will compute the Zj-Cj (net evaluations) for each column as follows; For the
first column Y1, it is computed as follows; Z1-C1= cB y1-c1 [(0*4) + (0*1) + (0* 0)]-15 = -15 In similar lines, the evaluations are calculated for all the columns and
placed in the last row of the table. Then we will compute the
ratios, using RHS of each constraint and dividing it by the respective entering
column non-zero, non-negative quantities; this is placed in the last column of
the following table.
Then, the minimum among the value is chosen as leaving variable;
therefore, X1 entering the basis and X4 leaving the basis. Then, we need to do the simple ‘row’ operations or modified ‘Gauss-Jordon’
Elimination process. We have to have 1 at the row/column intersection [known as
pivotal element] and the rest of the entries in the pivotal column as zero. The first row is replaced
by dividing all the entries by 4, which is the ‘pivotal’ element of the table
[The number in the intersection of ‘entering column’ and ‘leaving row’] and
placed in the second iteration table in the corresponding position of the
original place. For example, consider the original first row
Now the X3=360 is replaced by the
following calculation New
element= Old element – [respective element in the pivotal row X respective element in the pivotal column/pivotal element] New
element for 360 = 360 – [(180 * 4) /3] = 360-240 = 120 Similarly,
for the Y2 column, of the first row, for the ‘6’, the new element is arrived in
similar way New element for 6 = 6 – [(4 * 0)
/3] = 6-0 = 6 The new table-2 is arrived as follows with the
following changes; •
X3 is
replaced by X1 •
Row
operations are carried out Table-2- which is showing the entering column vector
and leaving variable
In
the next step, X2 is entering the basis and X3 leaves the basis; the row
operations are carried out and the new table-3 is given below;
You can notice that in the end of second iteration,
all the net evaluations (Zj-Cj) are non-negative; this implies that the current
solution is in the optimal stage. Therefore, X1 = 60
X2= 20 Max Z=(15* 60) + (10 * 20) = 900+200=1100 You can recall that the same
answer was obtained through the graphical method. In the
next sections, we solve the special cases, attempted by graphical method.
Tags : Operations Management - Introduction to Operations Research
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