Addition of a new constraint

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• Addition of a new constraint
• The addition of a new constraint to an existing
  model can lead to one of two cases
• The new constraint is redundant, meaning that it
  is satisfied by the current optimal solution and
  hence can be dropped altogether from the model
  altogether.
• The current solution violates the new constraint
  in which case the dual simplex method can be used
  to recover feasibility.

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We introduce this constraint in the final
simplex tableau as an additional row where the
usual additional (slack or surplus) variable is
taken as the basic variable for this new row.
Because this new row will probably have non-zero
coefficients for some of the other basic
variables, the conversion to proper form for
Gaussian elimination is done and then
reoptimization is done in the usual way. The
following example illustrates this.
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Example Consider the LPP Maximize
subject to
Applying the Simplex method, the optimal tableau
is given in the next slide. Now a new constraint

is added. Find the new optimal
solution.
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Basic z x1 x2 x3 s1
s2 s3 Sol
s4 0 0 0 0
z 1 0 0 3/2 0
3/2 1/2 25
s1 0 0 0 1 1
- 1 - 2 10
x1 0 1 0 1/2 0
1/2 1/2 15
x2 0 0 1 - 3/2 0
- 1/2 1/2 5
0 0 0 -7/2 0
- 5/2 -1/2 1 -7
s4 0 3 -2 1 0
0 0 1 28
z 1 0 0 0 0
3/7 2/7 3/7 22
s1 0 0 0 0 1
- 12/7 -15/7 2/7 8
x1 0 1 0 0 0
1/7 3/7 1/7 14
x2 0 0 1 0 0
4/7 5/7 -3/7 8
x3 0 0 0 1 0
5/7 1/7 -2/7 2
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• Changes affecting Optimality
• Changes in the objective coefficients cj
• The changes in the objective coefficients, cj
  , affect "zj - cj" and hence the optimality. We
  recompute the z-Row by replacing cj with c'j . We
  note that CB will also change to C'B.

In the previous problem, we replace the objective
function with
.
Find the new optimal solution.
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Thus c1 is changed from 2 to 3 c2 is changed
from -1 to -2 and c3 is changed from 1 to 3
We note that in z-Row, the coefficients of basic
variables x1 , x2 , and x4 will remain zero now
also. We have only to calculate the new
coefficients of the non-basic variables only and
the new z. Coefficient of x3 z3 c3
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Coefficient of x5 z5 c5
Coefficient of x6 z6 c6
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Since all zj cj are ? 0, the current solution
remains optimal. Of course the new z is
If in the previous problem, we replace the
objective function with
You can verify that z6 c6 -1/2 lt 0. Thus
optimality is spoiled. By applying the regular
Simplex method, we can show that the new optimum
solution is x1 10, x2 0, x3 0 with new z
20.
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Addition of a new activity ( Changes in the
coefficients of an existing activity) Suppose a
new activity n1 is added with coefficients
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To find the effect of this on the current optimal
solution, we pretend this activity was present
initially with all coefficients zero. Hence the
new coefficients are calculated using the
formulae In z-row, the coefficient of xn1 is
And in the constraint matrix its coefficients are
If zn1 cn1 satisfies the optimality
condition, the current solution is optimal. Else
we apply regular simplex method to restore
optimality.
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The same procedure is adopted when the
coefficients of an existing variable are changed.
If the variable ia basic variable, we should see
that the new tableau is in proper form (i.e.
coefficients of other basic variables in that
column should be made zero).
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In the previous problem the coefficients of the
(non-basic) variable x3 are changed from
to
Using sensitivity analysis, find the new optimal
solution and value.
The only change in the optimal tableau will be
the x3 column. We calculate the new x3 column and
the coefficient of x3 in the z-row.
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We first note that
Hence the new x3 column is
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New coefficient of x3 in z-row
Thus optimality is disturbed. We replace the x3
column in the original optimal tableau by the new
values and then find the new optimal solution and
optimal value by regular simplex method.
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Basic z x1 x2 x3 s1
s2 s3 Sol
z 1 0 0 0
3/2 1/2 25
-3/2 6 -1/2 -3/2
s1 0 0 0 1
- 1 - 2 10
x1 0 1 0 0
1/2 1/2 15
x2 0 0 1 0
- 1/2 1/2 5
z 1 0 0 0 1/4
5/4 0 55/2
x3 0 0 0 1 1/6
- 1/6 -1/3 5/3
x1 0 1 0 0 1/12
5/12 1/3 95/6
x2 0 0 1 0 1/4
-3/4 0 15/2
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In the previous problem a new variable x7 is
introduced with coefficients
Using sensitivity analysis, find the new optimal
solution and value.
This is like the previous case. We assume the
variable x7 was already present with all
coefficients 0.
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We first note that
Hence the new x7 column is
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New coefficient of x7 in z-row
Hence the original solution remains optimal with
the same objective value and it does not help by
introducing this new variable.
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In the previous problem the coefficients of the
(basic) variable x1 are changed from
to
Using sensitivity analysis, find the new optimal
solution and value.
The only change in the optimal tableau will be
the x1 column. We calculate the new x1 column and
the coefficient of x1 in the z-row.
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We first note that
Hence the new x1 column is
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New coefficient of x1 in z-row
We replace the x1 column in the original optimal
tableau by the new values.
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Basic z x1 x2 x3 s1
s2 s3 Sol
1 0 0 1/2 0
1/2 -1/2 -5
z 1 0 3/2 0
3/2 1/2 25
2 0 1 -1
s1 0 0 1 1
- 1 - 2 10
x1 0 0 1/2 0
1/2 1/2 15
x2 0 1 -3/2 0
- 1/2 1/2 5
0 0 1 -1 0
0 1 20
z 1 0 1/2 0 0
1/2 0 5
s1 0 0 2 -1 1
- 1 0 50
x1 0 1 -1/2 1 0
1/2 0 5
s3 0 0 1 -1 0
0 1 20