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ME-407 (Mid-Sem. Solution) 1)
Given, P
23
: max {
c’x
| Ax <= b , x >= 0 } where, A is 2*3 matrix and b, c have all real components. a) To give an example where P
23
has feasible solution and it
’
s dual (D) has infeasible solution. Since D has infeasible solution so P
23
has either unbounded or infeasible solution. According to question P
23
has feasible solution so, it must has unbounded solution. An example:- P
23
: max { Z = 2x
1
+ x
2
–
3x
3
} s.t. { - x
1
+x
2
–
x
3
<= -5 (1) - x
1
- x
2
+ 2x
3
<= -6 (2) x
1
,x
2
,x
3
>= 0 } In the above example Z = (l.h.s. of (1) ) * (-1) + (l.h.s 0f (2)) * (-2). This implies max Z =
.
To confirm that the above problem has unbounded solution let us take x
2
= 0 and x
3
=
( to see in (x
1
, x
3
) plane) then x
1
>= 5
–
(from (1))
and x
1
>= 6 + 2*
( from (2))
. Hence {x
1
> = 6 + 2* } is the only constraint on x
1
. We have a ray in the direction of (6 + 2*
, 0,
), starting at (6,0,0); and the entire ray is in the feasible region. Along this ray Z value becomes: Z = 3* (6 + 2*
)
+ 0
–
3*
) = 18 + 3*
.
Since x
3
>=0, we have
>= 0. As
⟶ Z ⟶
and so max {Z} = . So P
23
has unbounded solution.
Let D be the dual of P
23
. D: min { W = - 5y
1
–
6y
2
} s.t. { - y
1
–
y
2
>= 3 (3) y
1
+ y
2
>= 1 (4) - y
1
–
6*y
2
>= -3 (5) y
1
, y
2
>= 0 } y
2
The diagram shows inequality (3) and non-
negativity constraints which can’t be
satisfied together so D is infeasible. y
1
- y
1
–
y
2
= 3
b) P
23
and Dual (D) both are infeasible. Example P
23
: max { Z = 3x
1
+ x
2
+ 3x
3
}
s.t. { x
1
+x
2
<= -1 (1) -3x
1
+5x
2
+ 2x
3
<= 0 (2) x
1
, x
2
, x
3
>= 0 } For any value of {x
1
, x
2
>= 0 }. The inequality x
1
, + x
2
<= -
1 can’t be satisfied.
Hence P
23
has infeasible solution. Let D be the dual of P
23
D: min { W = - y
1
} s.t. {y
1
+ 3 y
2
>= 3 (3) y
1
+ 5y
2
>= 1 (4) - y
2
>= 1 (5) y
1
, y
2
>= 0 } y
2
for any value of y
2
>= 0. The inequality - y
2
>= 1 can never be satisfied so D has infeasible solution. y
1
-y
2
= 1 2) Given, [p] : max { Z = x
1
- x
2
+ x
3
- 2x
4
+ x
5
–
x
6
} s.t. { x
1
+ 2x
2
- 2x
3
+ x
4
- x
5
= 13 2x
1
+ x
2
+ x
6
= 12 x
1
+ - x
4
+ x
5
= 3 2x
2
+ x
3
- 2x
4
- x
5
–
x
6
= 5} The above LP is in the form of max {Z = C
’
X | AX = b, X>= 0 } Let x
1
, x
2,
x
3,
x
4
be basic variables in the above LP. Then X = [X
B
;
X
N
] X
B
= { x
1
, x
2,
x
3,
x
4
}, X
N
= { x
5
, x
6
} A = [B | N]
The required procedure for solving above LP is given below: Step 1: the above LP can be written in tabular form as shown Row no Z x
1
x
2
x
3
x
4
x
5
x
6
RHS R0 1 -1 1 -1 2 -1 1 0 R1 0 1 2 -2 1 -1 0 13 R2 0 2 1 0 0 0 1 12 R3 0 1 0 0 -1 1 0 3 R4 0 0 2 1 -2 1 -1 5 Step 2: Convert B matrix into identity by applying elementary row operations only and operations must be applied to whole table. Step 3: if we get B as identity matrix that implies x
1
, x
2,
x
3
and x
4
are the basic variables. Step 4: Check values of X
B
(R.H.S. side), if all elements are non-negative then the basis is primal feasible. Step 5: Check the reduced costs (Z
j
- C
j
), if all elements are non negative than the basis is dual feasible. Row no Z x
1
x
2
x
3
x
4
x
5
x
6
RHS R0 1 -1 1 -1 2 -1 1 0 R1 0 1 2 -2 1 -1 0 13 R2 0 2 1 0 0 0 1 12 R3 0 1 0 0 -1 1 0 3 R4 0 0 2 1 -2 1 -1 5 Operation {Interchanging R1 and R3} (equivalent to pre-multiplication of the equation system AX=b by an invertible matrix). R0 1 -1 1 -1 2 -1 1 0 R1 0 1 0 0 -1 -1 0 3 R2 0 2 1 0 0 0 1 12 R3 0 1 2 -2 1 -1 0 13 R4 0 0 2 1 -2 1 -1 5 Operation {Interchanging R3 and R4} R0 1 -1 1 -1 2 -1 1 0 R1 0 1 0 0 -1 -1 0 3 R2 0 2 1 0 0 0 1 12 R3 0 0 2 1 -2 1 -1 5 R4 0 1 2 -2 1 -1 0 13 Operations {R0 <- R0 +R1, R2 <- R2
–
2*R1 and R4 <- R4 - R1} R0 1 0 1 -1 1 0 1 3 R1 0 1 0 0 -1 1 0 3
R2 0 0 1 0 2 -2 1 6 R3 0 0 2 1 -2 1 -1 5 R4 0 0 2 -2 2 -2 0 10 Operations {R0 <- R0 - R2, R3 <- R3- 2*R2 and R4 <- R4
–
2*R2} R0 1 0 0 -1 -2 2 0 --3 R1 0 1 0 0 -1 1 0 3 R2 0 0 1 0 2 -2 1 6 R3 0 0 0 1 -6 5 -3 -7 R4 0 0 0 -2 -2 2 2 2 Operations {R0 <- R0 + R3 and R4 <- R4
–
2*R3} R0 1 0 0 -7 7 3 -10 0 R1 0 1 0 0 -1 1 0 3 R2 0 0 1 0 2 -2 1 6 R3 0 0 0 1 -6 5 -3 -7 R4 0 0 0 0 -14 12 -8 -16 Operation {R4 <- R4*(-1/14) } R0 1 0 0 0 -7 7 3 -10 R1 0 1 0 0 -1 1 0 3 R2 0 0 1 0 2 -2 1 6 R3 0 0 0 1 -6 5 -3 -7 R4 0 0 0 0 1 -6/7 4/7 8/7 Operations {R0 <- R0 + 7* R4, R1 <- R1 + R4, R2 <- R2 -2*R2 and R3 <-R3 + 6*R4} R0 1 0 0 0 0 1 1 -2 R1 0 1 0 0 0 1/7 4/7 29/7 R2 0 0 1 0 0 -2/7 -1/7 26/7 R3 0 0 0 1 0 -1/7 3/7 -1/7 R4 0 0 0 0 1 -6/7 4/7 8/7 We have the identity matrix (4*4) in the table under the columns (x
1
, x
2,
x
3
and x
4
), thereby giving a bona fide simplex tabular We have x
1
= 29/7, x
2
= 26/7
,
x
3
= -1/7 and x
4
= 8/7. Since x
3
has negative value, this basic solution is not primal feasible. Since all reduced cost values (Zj
–
Cj) are non negative, this basic solution is dual feasible.

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