Chapter 6, Problem 5RP

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

  ${\text{Min w = 6y}}_{\text{1}}{\text{+ 4y}}_{\text{2}}\text{+3y3}$

  Subject to

  $\begin{array}{l}{\text{3y}}_{\text{1}}{\text{+ 2y}}_{\text{2}}\text{ + y3 >= 4}\hfill \\ \begin{array}{l}{\text{y}}_{\text{1}}{\text{+ y}}_{\text{2}}{\text{ + y}}_{\text{3}}\text{>= 4}\\ {\text{y}}_{\text{1}}{\text{+y}}_{\text{2}}\text{ + y3 >= 1}\\ {\text{y}}_{\text{1}}\text{<= 0, y2 <= 0}\end{array}\hfill \end{array}$

• The formulas for

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)...

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)...