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 online. Audience estimates, costs, and maximum media usage limitations are as shown:
Constraint | Television | Radio | Online |
---|---|---|---|
Audience per advertisement | 100000 | 18000 | 20000 |
Cost per advertisement | $2000 | $300 | $600 |
Maximum media usage | 10 | 20 | 10 |
To ensure a balanced use of advertising media, radio advertisements must not exceed 50% of the total number of advertisements authorized. In addition, television should account for at least 10% of the total number of advertisements authorized.
Let | T = number of television spot advertisements |
R = number of radio advertisements | |
O = number of online advertisements |
Budget ($) | |
---|---|
T = | |
R = | |
O = | |
Total Budget = $ |
Objective function:
Objective function is to maximize the audience reach
Z = max 100000T + 18000R + 20000O
Constraints:
2000T + 300R + 600O <= 29,300 (Budget)
Subject to,
2000T+300R+600O <= 29300 (Promotional budget constraint)
T <= 10 (Max media - Television)
R<= 20 (Max media - Radio)
O <= 10 (Max media - Online)
R<=(T+R+O)*0.5
or, -0.5T+0.5R-0.5O <= 0 (Radio <= 50% of total ads)
T >= 0.1*(T+R+O) (TV >= 10% of total ads)
or, 0.9T-0.1R-0.1O >= 0
T,R,O >= 0 (non-negativity constraint)
Solving in solver we get,
T = number of television spot advertisements = 10
R = number of radio advertisements = 17
O = number of online advertisements = 7
Budget |
|
T |
20000 |
R |
5100 |
O |
4200 |
Maximum total audience can be reached = 1446000
b)
Shadow price = 42.22
So,
For each $100 increase in budget, the audience will increase by 100 * 42.22 = 4,222