ProblemPrint Media Advertising (PMA) Contract
Problemprint Media Advertising Pma Has Been Given A Contract To Mark
Problem Print Media Advertising (PMA) has been given a contract to market Buzz Cola via newspaper ads in a major southern newspaper. Full-page ads in the weekday editions (Monday through Saturday) cost $2000, whereas on Sunday a full-page ad costs $8000. Daily circulation of newspaper is 30,000 on weekdays and 80,000 on Sundays. PMA has been given a $40,000 advertising budget for the month of August. The experienced advertising executives at PMA feel that both weekday and Sunday newspaper ads are important; hence they wish to run the equivalent of at least eight weekday and at least two Sunday ads during August. (Assume that a fractional ad would simply mean that a smaller ad is placed on one of the days; that is, 3.5 ads would mean three full-page ads and one half-page ad. Also, assume that smaller ads reduce exposure and costs proportionately.) This August has 26 weekdays and 5 Sundays. The objective is to determine the optimal placement of ads by PMA in the newspaper during August so as to maximize the cumulative total exposure (as measured by circulation) for the month of August.
Paper For Above instruction
Introduction
The advertising campaign for Buzz Cola by PMA involves strategic placement of full-page advertisements in a prominent southern newspaper throughout August. The primary goal is to maximize exposure, measured by circulation, while respecting budget constraints and minimum advertisement requirements (minimum number of ads for weekdays and Sundays). This presents a classic optimization challenge, suitable for linear programming, with the added nuance of fractional ad units representing smaller ads, which proportionally reduce costs and exposure.
Problem Formulation
Decision Variables:
- \( x_w \): number of full-page weekday ads (can be fractional) in August.
- \( x_s \): number of full-page Sunday ads (can be fractional) in August.
Objective Function:
Maximize total exposure:
\[
\text{Maximize } Z = 30,000 \times x_w + 80,000 \times x_s
\]
since circulation per ad is 30,000 for weekdays and 80,000 for Sundays, proportional to the fractional ad units.
Constraints:
1. Budget constraint:
\[
2000 \times x_w \times 26 + 8000 \times x_s \times 5 \leq 40,000
\]
where 26 weekdays and 5 Sundays in August.
2. Minimum ad constraints:
\[
x_w \geq \frac{8}{26} \quad \text{(minimum 8 weekday ads in total)}
\]
\[
x_s \geq \frac{2}{5} \quad \text{(minimum 2 Sunday ads in total)}
\]
Expressed relative to fractional units, these are:
\[
x_w \geq \frac{8}{26} \approx 0.3077
\]
\[
x_s \geq \frac{2}{5} = 0.4
\]
3. Non-negativity:
\[
x_w \geq 0, \quad x_s \geq 0
\]
Note: Because fractional ads are allowed, the problem is continuous, facilitating solution via graphical and solver methods.
Graphical Solution
Given the two decision variables, a graphical approach involves plotting the feasible region defined by the constraints and identifying the point that maximizes the objective function along the boundary.
Step 1: Plot the constraints
- Budget constraint:
\[
2000 \times 26 \times x_w + 8000 \times 5 \times x_s \leq 40,000
\]
which simplifies to:
\[
52,000 x_w + 40,000 x_s \leq 40,000
\]
Divide through by 40,000:
\[
1.3 x_w + x_s \leq 1
\]
- Minimum ad constraints:
\[
x_w \geq 0.3077
\]
\[
x_s \geq 0.4
\]
Step 2: Find the intersection points of constraints
- With the budget line:
When \( x_w = 0.3077 \):
\[
1.3 \times 0.3077 + x_s = 1 \implies 0.4 + x_s = 1 \implies x_s = 0.6
\]
which satisfies the minimum \( x_s \geq 0.4 \).
Similarly, for \( x_s = 0.4 \):
\[
1.3 x_w + 0.4 = 1 \implies 1.3 x_w = 0.6 \implies x_w \approx 0.4615
\]
- The intersection of the constraint lines with the minimum values.
Step 3: Identify feasible vertices
Feasible vertices are:
- (A) \( (x_w, x_s) = (0.3077, 0.4) \)
- (B) \( (x_w, x_s) = (0.4615, 0.4) \)
- (C) \( (x_w, x_s) = (0.3077, 0.6) \)
- Potential intersection point where the budget line meets the maximum allowable combination within constraints.
Step 4: Calculate the objective function at these vertices
At (A):
\[
Z = 30,000 \times 0.3077 + 80,000 \times 0.4 = 9,231 + 32,000 = 41,231
\]
At (B):
\[
Z = 30,000 \times 0.4615 + 80,000 \times 0.4 = 13,846 + 32,000 = 45,846
\]
At (C):
\[
Z = 30,000 \times 0.3077 + 80,000 \times 0.6 = 9,231 + 48,000 = 57,231
\]
The maximum occurs at point (C), with \( x_w = 0.3077 \), \( x_s=0.6 \), yielding the highest exposure.
Conclusion of Graphical Method:
Optimal fractional ad placement:
- Approximately 0.308 weekday ads (which is practical as fractional ads are allowed),
- 0.6 Sunday ads.
Total exposure is maximized under these values, respecting budget and minimum ad requirements.
Excel Solver Method
The problem is set up in Excel with cells representing \( x_w \) and \( x_s \), and formulas calculating total costs, exposure, and constraints. Solver is configured as follows:
- Set Objective: Maximize total exposure cell (\( Z = 30,000 \times x_w + 80,000 \times x_s \))
- Variables: Changing cells \( x_w \) and \( x_s \)
- Constraints:
- \( 52,000 x_w + 40,000 x_s \leq 40,000 \)
- \( x_w \geq 0.3077 \)
- \( x_s \geq 0.4 \)
Using Solver's Simplex LP method, the optimal solution replicates the graphical analysis, confirming:
- \( x_w \approx 0.308 \),
- \( x_s \approx 0.6 \).
Total exposure is approximately 57,231 exposure units, aligning with the graphical solution's maximum.
Final Practical Interpretation:
Because fractional ads are impractical, PMA may choose to place:
- 8 weekday ads (full ads),
- 3 Sunday ads (full ads),
which slightly exceeds the minimum required and stays within budget, giving a comparably high exposure (~\( 8 \times 30,000 + 3 \times 80,000 = 240,000 + 240,000 = 480,000 \)) and respecting the constraints, although actual calculation may suggest fractional placement for optimization.
Conclusion
Through linear programming modeling and the application of graphical and Excel Solver methods, PMA can optimally allocate their advertising budget in August to maximize exposure. The solution indicates fractional ad placements just below one full ad per day on average, emphasizing the benefit of smaller, cost-effective ads within constraints. Practical implementation would involve rounding these to integer values while maintaining a balance between cost and exposure, guided by the mathematical solution.
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