Q 4.14 pliz in details. there is no other information apart from those 3 4x1 3 6

Question

Q 4.14
pliz in details. there is no other information apart from those

3 4x1 3 60 xi + 2x2 's 40 The starting solution consists of artificial xs and xs for the first and second constraints and slack No for the third constraint. Using M optimal tableau is given as 100 for the artificial variables, the Basic2 x6 Solution 00-9861002 02 18 10 Write the associated dual problem, and determine its optimal solution in two ways 4-14. Consider the following LP: Maximize z = 2x1 + 4x2 + 4x3-3x4 subject to x+2+ x1 + 4x2 +x4 = 8 Using aj and x4 as starting variables,the optimal tableau is given as Basic x1 X2 x3 X4 Solution 16 X3 x2 75 .25 .25 Write the associated dual problem, and determine its optimal solution in two ways

Explanation / Answer

Dual

Min. W = 4y1 + 8y2
Subject to,
y1 + y2 >= 2 (corresponding to x1)
y1 + 4y2 >= 4 (corresponding to x2)
y1 >= 4 (corresponding to x3)
y2 >= -3 (corresponding to x4)
y1, y2 >= 0

Solution 1

According to the optimal tableau -

* Note Zj is negative of Z values of the optimal tableau as this is a maximization problem.

We will note the Cj - Zj values of the starting variables as the optimal solutions of Dual. So, at optimality, Y1=4 and Y2=0, and min.W = max.Z = 16.

Solution 2

Note that in the optimal solution, x3 has come up with a positive value of 2. So, the shadow price of the third constraint of dual (note: shadow price of dual constraint = optimal solution of the corresponding primal variable) is 2. A positive shadow price denotes a binding constraint and the inequality of that constraint can be replaced with equality. So, the third constraint of the dual can be written as -

y1 = 4

Since min.W = max.Z = 16,

y2 = (16 - 4 x 4) / 8 = 0

x1 x2 x3 x4 Zj * -2 0 0 -3 Cj 2 4 4 -3 Cj - Zj 4 4 4 0

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