4. Problem 4: For each of the following simplex tableaux, find the basic feasibl

Question

4. Problem 4: For each of the following simplex tableaux, find the basic feasible solution (x1, r2, r3, r4, r5, T6) that is represented by rows 1 through 4 in the the tableau and write down which vari- ables are in the basis for the corresponding basic feasible solution. (Do not run the simplex algorithm on the tableaux) x1 12 13 14 2 01 0 4 0 0 1 -z27-20 5 0-10 0 3 1 1 0 3 0 T1 T2 T3 T4 270 0 0 3 -10 -4 6 0 0 1 03 10 7 1 0 0 03 2 T5 T6 z270 0 5 010 0 5 1 0 1 030 6 0 1 1 23 0 T1 T2 T3 4 7 0 0 0 1

Explanation / Answer

basic feasible solutions -

a) x1 = 0, x2 = 6, x3 = 0, x4 = 7, x5 = 0, x6 = 5, variables in the basis = x2.x4,x6

b) x1 = 7, x2 = 5, x3 = 6, x4 = 0, x5 = 0, x6 = 0, variables in the basis = x1,x2,x3

c) x1 = 5, x2 = 6, x3 = 0, x4 = 0, x5 = 0, x6 = 7, variables in the basis = x1,x2,x6

d) No,any of the basic feasible solution is not anoptimal solution.because at optimal solution condition all the elements present in negative Z row should be less than or equal to zero (<=0),which is not present in any of the three case.

variables that should brought in basis of optimal solutions are -

a) x3

b) x4

c) x3

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