Problem 5 Problem Set 3'4B Pages 101102

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Problem 5 Problem Set 3.4B Pages 101-102
Maximize
Subject to the constraints
We shall solve this problem by two phase method.
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Phase I
Minimize
Subject to the constraints
Here s1 is a slack variable, s2 is a surplus
variable and R2 artificial variable.
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Basic r x1 x2 x3 s1 s2
R2 Sol.
3 4 2 0
-1 0 8
r 1 0 0 0 0 0
-1 0
s1 0 2 1 1 1 0
0 2
R2 0 3 4 2 0 -1
1 8
r 1 0 0 0 0 0
-1 0
s1 0 5/4 0 1/2 1 1/4
-1/4 0
x2 0 3/4 1 1/2 0 -1/4
1/4 2
Phase I has ended. We now start Phase II.
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Basic z x1 x2 x3 s1 s2
R2 Sol.
-1/2 0 -3 0
-1/2 4
z 1 -2 -2 -4 0 0
0
s1 0 5/4 0 1/2 1 1/4
0
x2 0 3/4 1 1/2 0 -1/4
2
z 1 7 0 0 6 1
4
x3 0 5/2 0 1 2 1/2
0
x2 0 -1/2 1 0 -1 -1/2
2
This is the optimal tableau.
Opt.Sol. x1 0, x2 2, x3 0. Opt.Value z 4
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We saw in Phase I, in the starting iteration,
there was a tie for the leaving variable namely
s1 and R2. We rightly chose R2 to leave the
basis. Now we shall see what happens if s1 was
allowed to leave the basis.
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Basic r x1 x2 x3 s1 s2
R2 Sol.
3 4 2 0
-1 0 8
r 1 0 0 0 0 0
-1 0
s1 0 2 1 1 1 0
0 2
R2 0 3 4 2 0 -1
1 8
r 1 -5 0 -2 -4 -1
0 0
x2 0 2 1 1 1 0
0 2
R2 0 -5 0 -2 -4 -1
1 0
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Phase I has ended. But we find the artificial
variable R2 is still in the basis. Since it is at
0 level, we now start Phase II having the
starting BFS as x2 and R2. (That is we have to
keep the R2 column in Phase II also.)
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Basic z x1 x2 x3 s1 s2
R2 Sol.
2 0 -2 2
0 0 4
z 1 -2 -2 -4 0 0
0 0
x2 0 2 1 1 1 0
0 2
R2 0 -5 0 -2 -4 -1
1 0
z 1 7 0 0 6 1
-1 4
x2 0 -1/2 1 0 -1 -1/2
1/2 2
x3 0 5/2 0 1 2 1/2
-1/2 0
We now make R2 leave the basis. Remember that
even though its coefficient in the pivot column
is negative, it does not spoil the feasibility as
RHS value is zero.
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As R2 has left the basis, we now forget the R2
column. The remaining tableau is optimal. Hence
the optimal solution is x1 0, x2 2, x3 0.
And the optimum value z 4 (as before.)
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• Now we shall look at four special cases that
  arise in the application of Simplex method.
• Degeneracy
• Alternative optimum solutions
• Unbounded solutions
• Infeasible solutions

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Degeneracy If in a basic feasible solution (BFS)
one of the basic variables is present at 0 level
(i.e. is zero) then we say we have a degenerate
BFS. This usually happens when there is a tie for
the leaving variable (as we saw in the previous
example). There is nothing alarming in having a
degenerate BFS except that in some cases it may
lead to
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Cycling a phenomenon in which the a leaving
variable again enters the basis and again leaves
repeatedly without optimality being reached. The
Optimization people claim that this is a rare
phenomenon and normally optimality is reached in
a finite number of steps. Degeneracy can be
permanent as in the previous example or can be
temporary as the following example shows.
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Problem 2 Problem Set 3.5A Page 105
Maximize
Subject to
We solve it by Simplex method.
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Basic z x1 x2 s1 s2
s3 Sol.
z 1 -3 -2 0
0 0 0
s1 0 4 -1 1
0 0 8
s2 0 4 3 0
1 0 12
s3 0 4 1 0
0 1 8
z 1 0 -11/4 3/4 0
0 6
x1 0 1 -1/4 1/4 0
0 2
s2 0 0 4 -1
1 0 4
s3 0 0 2 -1
0 1 0
Note that the BFS obtained is degenerate as s3
0. (We also observe that there was a tie for the
leaving variable.)
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Basic z x1 x2 s1 s2
s3 Sol.
z 1 0 0 -5/8 0
11/8 6
x1 0 1 0 1/8 0
1/8 2
s2 0 0 0 1
1 -2 4
x2 0 0 1 -1/2 0
1/2 0
z 1 0 0 0
5/8 1/8 17/2
x1 0 1 0 0
-1/8 3/8 3/2
s1 0 0 0 1
1 -2 4
x2 0 0 1 0
1/2 -3/2 2
This is the optimal tableau. Note that the
optimal BFS obtained is NOT degenerate.
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Alternative Optimal Solutions When the objective
function is parallel to a binding constraint
(i.e. a constraint that is satisfied as an
equation), then the objective function will
assume the same optimal value at more than one
corner point. We say the problem has alternative
optimal solutions. In this case there also will
be an infinity of (non-basic) feasible solutions
where the objective function has the same optimum
value.
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Criterion for the existence of alternative
optimal solutions from the optimal tableau If
,in the optimal tableau, the coefficient of a
non-basic variable is zero, then that can enter
the basis and the objective function will not
change (why?). Thus we shall have alternative
optimal solutions as the following example
illustrates.
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Maximize
Subject to the constraints
We shall solve this problem by the Simplex method.
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Basic z x1 x2 s1 s2
Sol.
z 1 -1 -1 0
0 0
s1 0 1 1 1
0 6
s2 0 1 2 0
1 10
z 1 0 0 1
0 6
This is the optimal tableau. Since coefficient of
the nonbasic variable x20, alt. opt. soln.
exists.
x1 0 1 1 1
0 6
s2 0 0 1 -1
1 4
z 1 0 0 1
0 6
x1 0 1 0 2
-1 2
x2 0 0 1 -1
1 4
Note that we have got an alt. opt. soln. We
understand this graphically also in the next
slide.
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Optimal BFSs
(0,6)
All points on this line segment (except the
endpoints) are nonbasic optimal feasible
solutions.
(0,5)
(2,4)
(6,0)
(10,0)
Note that the objective function line is parallel
to the first (binding) constraint line.
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Maximize
Subject to the constraints
We shall solve this problem by the Simplex method.
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Basic z x1 x2 s1 s2
Sol.
z 1 -2 1 0
0 0
s1 0 1 -1 1 0
10
s2 0 2 -1 0 1
40
z 1 0 -1 2
0 20
x1 0 1 -1 1 0
10
s2 0 0 1 -2
1 20
z 1 0 0 0
1 40
x1 0 1 0 -1
1 30
x2 0 0 1 -2
1 20
Note that we have got the optimal tableau.
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Since the coefficient of the nonbasic variable s1
is zero in the objective function row, we expect
an alternative optimal BFS to exist. We see that
the problem indeed has an infinite number of
alternative feasible solutions which are NOT
basic. Could you guess them? Answer
x1 30t x2 202t, where t is any nonnegative
number.
We can easily see this graphically also.
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(30,20)
All points on this line from (30,20) onwards are
Optimal nonbasic solutions.
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Problem 2 Problem Set 3.5B Page 108
Maximize
Subject to the constraints
Show that the above LPP has an infinite number of
Optimal (nonbasic) feasible solutions.
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Unbounded solutions In some LPPs, the values of
some decision variables can be increased
indefinitely without violating any of the
constraints, meaning that the solution space is
unbounded in at least one direction. As a result
the objective value may increase (maximization
case) or decrease (minimization case)
indefinitely. Thus both the feasible solution
space and the objective function are unbounded.
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The solution space is unbounded if in a Simplex
iteration, all the entries in a column
corresponding to a non-basic variable are
negative (or zero) in all the constraint rows and
in the objective function row, the entry is
(a) negative (or zero) in the case of a
maximization problem (b) positive (or zero) in
the case of a minimization problem We illustrate
by an example.
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Problem 2 Problem Set 3.5C Page 110
Maximize
Subject to the constraints
It is clear that the solution space is unbounded
in the x2 direction.
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Thus the variable x2 can be increased
indefinitely without violating any of the
constraints and thus the solution space and the
objective function are both unbounded. (You can
form the Simplex tableau and verify that in x2
column all the entries in the constraint rows are
negative or zero and that in the z-Row, it is
negative.)
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An example where cycling happens
Maximize
Subject to the constraints