me-407 midsem solution(s)

8/10/2019 ME-407 Midsem Solution(s)

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ME-407 (Mid-Sem. Solution)

1) Given,

P23: max { cx| Ax = 0 } where, A is 2*3 matrix and b, c have all real components.

a) To give an example where P23 has feasible solution and its dual (D) has infeasible solution. Since

D has infeasible solution so P23has either unbounded or infeasible solution. According to question P23has feasible solution so, it must has unbounded solution.

An example:-

P23: max { Z = 2x1+ x23x3 }

s.t. { - x1+x2x3= 5

(from (1))and x1 >= 6 + 2*

( from (2)). Hence {x1> = 6 + 2* } is the onlyconstraint on x1. 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*) + 03*)

= 18 + 3*.

Since x3>=0, we have >= 0. As Z and so max {Z} = . So P23has

unbounded solution.

Let D be the dual of P23

.

D: min { W = - 5y16y2}

s.t. { - y1y2>= 3 (3)

y1+ y2>= 1 (4)

- y16*y2>= -3 (5)

y1, y2 >= 0 }

y2 The diagram shows inequality (3) and non-

negativity constraints which cant be

satisfied together so D is infeasible.

y1

- y1y2= 3


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b) P23and Dual (D) both are infeasible.

Example

P23: max { Z = 3x1+ x2+ 3x3 }

s.t. { x1+x2 = 0 }. The inequality x1, + x2 = 3 (3)

y1+ 5y2>= 1 (4)

- y2 >= 1 (5)

y1, y2 >= 0 }

y2 for any value of y2 >= 0. The inequality - y2 >= 1 can never

be satisfied so D has infeasible solution.

y1

-y2= 1

2)

Given,

[p] : max { Z = x1- x2+ x3 - 2x4+ x5x6 }

s.t. { x1+ 2x2- 2x3 + x4- x5 = 13

2x1+ x2 + x6 = 12

x1+ - x4+ x5 = 3

2x2+ x3 - 2x4- x5x6= 5}

The above LP is in the form of max {Z = CX | AX = b, X>= 0 }

Let x1, x2,x3,x4be basic variables in the above LP. Then

X = [XB; XN] XB= { x1, x2,x3,x4}, XN= { x5, x6}

A = [B | N]


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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 x1 x2 x3 x4 x5 x6 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 x1, x2,x3and x4are the basic variables.

Step 4: Check values of XB (R.H.S. side), if all elements are non-negative then the basis is primal

feasible.

Step 5: Check the reduced costs (Zj- Cj), if all elements are non negative than the basis is dualfeasible.

Row no Z x1 x2 x3 x4 x5 x6 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 0R1 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


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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

me-407 midsem solution(s)

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