Question

The advertising alternatives for a company include television, radio, and newspaper advertisements. The costs and estimates for audience coverage are given in following table:

	television	newspaper	radio
Cost per advertisement	$ 2000	$ 600	$ 300
Audience per advertisement	100,000	40,000	18,000

The local newspaper limits the number of weekly advertisements from a single company to ten. Moreover, in order to balance the advertising among the three types of media, no more than half of the total number of advertisements should occur on the radio, and at least 10% should occur on television. The weekly advertising budget is $18,200. Formulate the model for maximizing the total audience number?

Answer 1

	Television	Newspaper	Radio
Cost per advertisement	$2000	$600	$300
Audience per advertisement	100,000	40,000	18,000

Decision Variables:

Let,

Number of Television ads = x1

Number of Newspaper ads = x2

Number of Radio ads = x3

Objective Function:

The total audience number should be maximized.

MAXIMIZE 100000*x1 + 40000*x2 + 18000*x3

Constraints:

Number of weekly advertisements from Newspaper will not exceed 10

x2 <= 10

No more than half of the total number of ads should occur on the radio.

x3 <= 0.5*(x1 + x2 + x3)

0.5*x3 – 0.5*x1 – 0.5*x2 <= 0

At least 10% should occur in television.

x1 >= 0.1*(x1 + x2 + x3)

0.9*x1 – 0.1*x2 – 0.1*x3 >= 0

The total cost should not exceed the budget of $18200

2000*x1 + 600*x2 + 300*x3 <= 18200

Number of advertisement cannot be negative.

x1 >= 0

x2 >= 0

x3 >= 0

Number of advertisements cant be fraction.

x1 = integer

x2 = integer

x3 = integer

Putting in Excel:

Solver Parameters:

Optimal Solution:

Television	Newspaper	Radio
x1	x2	x3
4	10	14

Television ads = 4

New paper ads = 10

Radio ads = 14

Optimal Audience Reach = 1,052,000

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