Sensitivity Recap
Optimal simplex tableau
 Optimal solution is the right hand column
 Shadow prices for the scarce resources are in the z row
 Basic variables are expressed in terms of non-basic
variables
Changing the quantity of a scarce resource while maintaining
the same optimum solution
 Write the new constraint as an equation using a new slack
variable
 Substitute for the original slack variable in terms of this
new slack variable into the equations for the optimal values
of the basic variables
 Calculate the upper bound of the change, such that the
values of all the basic variables remain non-negative when
the non-basic variables are set to zero
Changing the profit of one component while maintaining the
same optimum solution
 Write the new objective function in terms of the original
objective function and the basic variable corresponding to
the component for which the profit is changed
 Substitute for that basic variable in the new objective
function in terms of the original non-basic variables
 Calculate the lower and upper bounds for the profit change,
such that the coefficients of the non-basic variables in the
objective function are all non-positive
Changing the profit on two components
 As above, using the ratio of the two changes in profit
Adding an extra variable, finding the level of profit on the new
variable above which original optimum solution no longer
holds
 Express the new objective function in terms of the original
objective function and the new basic variable
 Express the new slack variables in terms of the original
slack variables and the new basic variable
 Hence express the new objective function in terms of the new
slack variables and the new basic variable
 Find the maximum level of profit which ensures that the
coefficient of the new basic variable in the new objective
function is non-positive
Finding a new optimum solution
corresponding to a new variable is fixed
when
the
profit
 Rewrite the original optimal equations in terms of the new
slack variables and enter into tableau form
 Iterate the tableau once to obtain the optimal solution
Adding an extra constraint
 Substitute the optimal solution values of the basic variables
into the constraint expressed as an equation. If the new slack
variable is non negative the new constraint is redundant.
 If the new slack variable is negative, write the new constraint
in terms of the non-basic variables of the original optimum
solution and the new slack variable and enter into tableau
form.
 Calculate the absolute value of the ratio of each negative
entry in the row corresponding to the new slack variable to
the entry below it in the z row. The non-basic variable
corresponding to the column with the smallest value of this
ratio is the entering variable, the new slack variable is the
departing variable.
 Iterate the tableau once to obtain the optimal solution