Consider a linear programming problem in standard form, described in terms of th

Question

Consider a linear programming problem in standard form, described in terms of the following initial tableau: The entries alpha, beta, gamma, delta, eta, epsilon in the tableau are unknown parameters. Furthermore, let B be the basis matrix corresponding to having x_2, x_3, and x_1 (in that order) be the basic variables. For each one of the following statements, find the ranges of values of the various parameters that will make the statement to be true. Phase II of the simplex method can be applied using this as an initial tableau. The first row in the present tableau indicates that the problem is infeasible. The corresponding basic solution is feasible, but we do not have an optimal basis. The corresponding basic solution is feasible and the first simplex iteration indicates that the optimal cost is -infinity. The corresponding basic solution is feasible, x_6 is a candidate for entering the basis, and when x_6 is the entering variable, x_3 leaves the basis. The corresponding basic solution is feasible, x_7 is a candidate for entering the basis, but if it does, the solution and the objective value remain unchanged.

Explanation / Answer

a)

x1  x2  x3      S1    S2   S3       b

0    0   1      n       1    0       3

1    0   0     -2       2    n      -1

0   1    0      0     -1   2        1

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1    1    1

X1 = 0, x2 =0,x3=0, s1=3,s2=-1,s3=1

(x1,x2,x3,s1,s2,s3)=(0,0,0,3,-1,1)

Z=x1+x2+x3

=3/1=3, -1/0=0, 1/0=0.

0    0   n      -1       1    0       3

1    0   -2     0       2    n      -1

0    1    0      0     -1   2       1

_________________________

1     0   0      1     0    0      5

X1+x2+x3+s1+s2+s3 =               0,0,3, 0,-1,1

Z=x1+x2+x3=1(0)+1(0)+1(3)=0+0+3

0    0   n      -1       1    0       3

1    0   -2/-1     0       2    n      -1 /-2

0    1     0        0     -1     2         1

___________________________

0    0      2     0      0    0       4.50

Z= X1+x2+x3+s1+s2+s3 =               0,0,2, 0,0.50,1

0    n   0       -1       1    0       3

1    0   2       0       2    n      0.50

0    1   0       0    -1 /-1 2         1

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0        1        0              0                   4                0              4.50

Z= x1+x2+x3+s1+s2+s3=  (0,0,0.50,1,3)

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