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ME-407 Midsem Solution(s)

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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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