Chapter 4.16, Problem 1P

Program Plan Intro

The advertising agency is trying to determine a TV advertising schedule for Priceler Auto company.

Let the following variables,

$x_1$=The number of minutes of ads shown during football games

$x_2$=The number of minutes of ads shown during soap operas

When the decision maker is not able to determine precisely the relative importance of goals, the pre-emptive goal programming is used. The decision maker must rank or order the goals from the most important to east important goal in order to apply preemptive goal programming.

The objective function coefficient for the variable representing goal $i$ will be $P_i$. Assume,

$P_1>>P_2>>\mathrm{...}>>P_n$

Therefore, the pre-emptive programming formulation for the Priceler problem is,

Minimize, ${\text{z=P}}_{\text{1}}{\text{S}}_{\text{1}}{}^{\text{-}}{\text{+P}}_{\text{2}}{\text{S}}_{\text{2}}{}^{\text{-}}{\text{+P}}_{\text{3}}{\text{S}}_{\text{3}}{}^{\text{-}}$

Subject to the constraints,

$\begin{array}{l}{\text{7x}}_{\text{1}}{\text{+3x}}_{\text{2}}{\text{+S}}_{\text{1}}{}^{\text{-}}{\text{-S}}_{\text{1}}{}^{\text{+}}\text{=40(HIM constraint)}\\ {\text{5x}}_{\text{1}}{\text{+10x}}_{\text{2}}{\text{+S}}_{\text{2}}{}^{\text{-}}{\text{-S}}_{\text{2}}{}^{\text{+}}\text{=60(LIP constraint)}\\ {\text{5x}}_{\text{1}}{\text{+4x}}_{\text{2}}{\text{+S}}_{\text{3}}{}^{\text{-}}{\text{-S}}_{\text{3}}{}^{\text{+}}\text{=35(HIW constraint)}\\ {\text{100x}}_{\text{1}}{\text{+60x}}_{\text{2}}\le \text{600(Budget constraint)}\end{array}$

In the above, the deviational variables are, ${\text{S}}_{\text{i}}{}^{\text{-}}{\text{,S}}_{\text{i}}{}^{+}$. All the variables are, non-negative.

The optimal solution can be found graphically when the preemptive goal programming involves only two decision variables. The graph of constraints for Priceler is shown below,

Explanation of Solution

b.

Determining HIM is highest priority goal, followed by LIP and then HIW:

• In this case, HIM has the highest priority goal followed by LIP and HIW. In the graph, it is found that $(6,0)$ is the only point that satisfies the HIM constraint, the budget constraint and the LIP constraint.
• Therefore, from all the points satisfying highest priority goals, the point ${\text{x}}_{\text{1}}$

Explanation of Solution

c.

Determining HIM is highest priority goal, followed by HIW and then LIP:

• In this case, HIM has the highest priority goal followed by HIW and LIP. The intersection between the HIM constraint and the budget line is the point that satisfies the HIM goal and the budget constraint and is closest to the HIW constraint.
• The above stated point can be found by solving the set of equations simultaneously.

  $\begin{array}{l}{\text{7x}}_{\text{1}}{\text{+3x}}_{\text{2}}\text{=40(HIM constraint)}\\ {\text{100x}}_{\text{1}}{\text{+60x}}_{\text{2}}\end{array}$

Explanation of Solution

d.

Determining HIM is highest priority goal, followed by HIM and then LIP:

• In this case, HIW has the highest priority goal followed by HIM and LIP. The HIW goal that is the closest to HIM is point C and is the point satisfying the budget constraint.
• The above stated point can be found by solving the set of equations simultaneously.