Chapter 6.10, Problem 1P

a.

Explanation of Solution

Dual of the Glassco problem:

• The primal of the Glassco problem is as follows,

  ${\text{max z=6x}}_{\text{1}}{\text{+10x}}_2{\text{+9x}}_3{\text{+20x}}_4$

• The above function subject to the following constraints,

  $\begin{array}{l}{\text{4x}}_{\text{1}}{\text{+9x}}_2{\text{+7x}}_3{\text{+10x}}_4\le \text{200}\text{(Constraint}\text{on}\text{Molding)}\\ {\text{x}}_{\text{1}}{\text{+x}}_2{\text{+3x}}_3{\text{+40x}}_4\le 4\text{00}\text{(Constraint}\text{on}\text{Packaging)}\\ {\text{3x}}_{\text{1}}{\text{+4x}}_2{\text{+2x}}_3{\text{+x}}_4\le 5\text{00}\text{(Constraint}\text{on}\text{Glass)}\end{array}$

• The optimal solution to this problem is as follows,

  $\begin{array}{l}\text{z=}\frac{\text{2800}}{\text{3}}\\ {\text{x}}_{\text{1}}\text{=}\frac{\text{400}}{\text{3}}\\ {\text{x}}_{\text{2}}\text{=0}\\ {\text{x}}_{\text{3}}\text{=0}\\ {\text{x}}_{\text{4}}\text{=}\frac{\text{20}}{\text{3}}\\ {\text{s}}_{\text{1}}{\text{=s}}_2\text{=0}\\ {\text{s}}_{\text{3}}\text{=}\frac{\text{280}}{\text{3}}\end{array}$

• To find the dual of the problem consider the below table,

			max z
min w		${\text{x}}_{\text{1}}\le 0$	${\text{x}}_2\le 0$	${\text{x}}_3\le 0$	${\text{x}}_4\le 0$
		${\text{x}}_{\text{1}}$	${\text{x}}_2$	${\text{x}}_3$	${\text{x}}_4$
${\text{y}}_{\text{1}}\ge 0$

b.

Explanation of Solution

Optimal solution to the dual of the Glassco problem:

• “${\text{y}}_{\text{3}}=0$” because the value of “${\text{s}}_{\text{3}}$” is greater than “0”.
• The constraint (1) is binding because “${\text{x}}_1$” is greater than “0”.
• The constraint (4) is binding because “${\text{x}}_4$” is greater than “0”.
• Solving the constraint (1) and (4) simultaneously produces “${\text{y}}_1=\frac{22}{15}$”, , and “${\text{y}}_2=\frac{2}{15}$”.
• Substitute the values of “${\text{y}}_1$”,“${\text{y}}_2$”, and set the value of “${\text{y}}_3$” as “0” in equation “${\text{w=600y</}}_{}$

c.

Explanation of Solution

Example of complementary slackness condition:

Condition:

If “${\text{i}}^{\text{th}}\text{primal}\text{slack>0}$”, then “${\text{i}}^{\text{th}}\text{dual}\text{variable=0}$”.

• “${\text{s}}_3>0$” denotes “${\text{y}}_3=0$”.
• This is fair, because an additional ounce of glass will not increase “z”, if all available glass is not being used.
• Therefore, the constraint on glass should contain the shadow price of “0”.

Condition:

If “${\text{i}}^{\text{th}}\text{dual}\text{variable>0}$”, then “${\text{i}}^{\text{th}}\text{primal}\text{slack=0}$”.

• “${\text{y}}_2>0$” denotes “${\text{s}}_2=0$”.
• An extra minute of packaging time will increase “Z”, because “${\text{y}}_2$” is greater than “0”.
• Therefore, all currently available packaging time should be used.

Condition:

If “${\text{j}}^{\text{th}}\text{dual excess>0}$”, then “${\text{j}}^{\text{th}}\text{primal}\text{variable=0}$”.

• “${\text{e}}_2>0$” denotes “${\text{x}}_2=0$”.
• It is denoted by the following dual problem constraint,

${\text{9y}}_1{\text{+y}}_2+4{\text{y}}_3\ge 10$

• From this constraint, subtract the excess variable “${\text{e}}_{\text{2}}$” as follows,

$\begin{array}{c}{\text{9y}}_1{\text{+y}}_2+4{\text{y}}_3-{\text{e}}_{\text{2}}=10\\ {\text{e}}_{\text{2}}={\text{9y}}_1{\text{+y}}_2+4{\text{y}}_3-10\end{array}$