In: Statistics and Probability
The Westchester Chamber of Commerce periodically sponsors public service seminars and programs. Currently, promotional plans are under way for this year’s program. Advertising alternatives include television, radio, and newspaper. Audience estimates, costs, and maximum media usage limitations are as shown:
Constraint | Television | Radio | Newspaper |
---|---|---|---|
Audience per advertisement | 100000 | 18000 | 40000 |
Cost per advertisement | $1400 | $190 | $600 |
Maximum media usage | 10 | 10 | 20 |
To ensure a balanced use of advertising media, radio advertisements must not exceed 50% of the total number of advertisements authorized. In addition, television should account for at least 10% of the total number of advertisements authorized.
Let | T = number of television spot advertisements |
R = number of radio advertisements | |
N = number of newspaper advertisements |
Budget ($) | |
---|---|
T = | |
R = | |
N = | |
Total Budget = $ |
Solution :
Decision variables:
Let
T = No of television spot advertisements
R = No of radio advertisements
N = No of newspaper advertisements
Objective function:
Objective function is to maximize the total audience contact
Z =Max 100000 T + 18000 R + 40000 N
Constraints:
1400 T + 300 R + 600 N <= 22,400 (Budget limit)
T < =10 (Maximum Television usage)
R<= 20 (Maximum Radio usage)
N<= 10 (Maximum Newspaper usage)
R <= 0.5(T+R+N)
Or
-0.5T + 0.5R -0.5N <= 0 (radio advertisements must not exceed 50% of the total number of advertisements)
T>= 0.1(T+R+N)
Or
0.9T – 0.1R-0.1N >= 0
T,R,N >= 0
By using solver we can solve
Budget:
T = 14000
R= 2400
N= 6000
Total audience reached = 13,20,000
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