Question: Consider the following LP and its optimal tableau max 2 = 2 X2 S1 0 s.t. 1 3x1 + 2x2 2x1 + 5.22 0 21

Consider the following LP and its optimal tableau

max 2 = 2 X2 S1 0 s.t. 1 3x1 + 2x2 2x1 + 5.22 0 21 0 0 S2 rhs 1 10 -2/3 4/3 1/3 10/3 0 0 1/3 7/3 0 1 0 to (a) The range of values of the right hand side by of the first constraint for which the current basis remains optimal is from (b) The range of values of the right hand side b2 of the second constraint for which the current basis remains optimal is from to (c) If the right hand side of the first constraint is changed to bi 7, the new optimal solution will be x1=0, x2=[ (d) If the right hand side of the second constraint is changed to b2=5, the new optimal solution will be X1 12

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