Question

Maximize:-x 3x2 subject to: + S8 (resource 1). -xx (resource 2). s6 (resource 3). xi20.x2 20. The optimal tableau determined by the simplex method is given below Current Basic variablesvax x x5 Using the above optimal tableau determined by the simplex method, determine i) the optimal solution. ii) the shadow prices on the three constraints ii the range on the objective coefficient ofeach variable, holding the coefficient of the other variable at its current value, for which the solution to part (i) remains optimal; iv) the range on the availability of each resource, holding the availability of the other resources at their current values, for which the shadow prices in part (i) remain optimal.

Answer 1

i)

Refer final values from the optimal tableau

The optimal solution is following:

x1 = 2

x2 = 6

ii)

Refer the values in the last row of optimal tableau,

Shadow price of first constraint is 2

Shadow price of second constraint is 1

iii)

Holding the coefficient of x2 constant, the optimality range of x1 is from: __-3__  (=1-2/(1/2)) to __3__ (=1-1/(-1/2))

Holding the coefficient of x1 constant, the optimality range of x2 is from: __1__  (=3-1/(1/2)) to __infinity__ (=3+0/0)

iv)

Holding the availabiility of other resources constant the feasibility range of constraints' Right Hand Sides (RHSs) are following:

Feasibility range of RHS of first constraint is from: __5__ (=8-3) to __17__ (=8+9)

Feasibility range of RHS of second constraint is from: __-4__ (=5-9) to __8__ (=5+3)

Feasibility range of RHS of third constraint is from: __1.5__ (=6-4.5) to __infinity__ (=6+infinity)