Chapter 6, Problem 3RP

a.

Explanation of Solution

Dual linear problem and its optimal solution

• The dual of a given linear program is another linear program that is derived from the original linear program in following ways:
  • Each variable in the original linear program becomes a constraint in the dual linear program.
  • Each constraint in the original linear program becomes a variable in the dual linear program.
  • The objective direction is inversed which becomes maximum in the original and minimum in the dual.
• Hence the dual of the linear problem is

  ${\text{Max z = 5x}}_{\text{1}}{\text{+ x}}_{\text{2}}\text{+2x3}$

  Subject to

  $\begin{array}{l}{\text{x}}_{\text{1}}{\text{+ x}}_{\text{2}}\text{ + x3 <= 6}\hfill \\ \begin{array}{l}{\text{6x}}_{\text{1}}{\text{+ x}}_{\text{3}}\text{<= 8}\\ {\text{x}}_{\text{2}}\text{ + x3 <= 2}\\ {\text{x}}_{\text{1}}{\text{, x}}_{\text{2}}\text{ , x3 >= 0}\end{array}\hfill \end{array}$

• The formulas for computing the optimal table include:
  • ${\text{x}}_{}$

b.

Explanation of Solution

Range of values

• For a maximization problem, the new optimal value = (old optimal value) + (Constraint i’s shadow price).
• For a minimization problem, the new optimal value = (old optimal value) – (Constraint i’s shadow price).
• In this case, when ${\text{x}}_{\text{2}}$ enters the basis in the optimal table, then the value of ${}_{}$

c.

Explanation of Solution

Range of values

• For a maximization problem, the new optimal value = (old optimal value) + (Constraint i’s shadow price).
• For a minimization problem, the new optimal value = (old optimal value) – (Constraint i’s shadow price).
• Here ${\text{x}}_{\text{2}}$