Question

s, and maximum media usage limitations are as shown:
Constraints
Television
Radio
Online
Audience per advert
1000,000
18000
40,000
Cost per Advert
$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.
If the promotional budget is limited to $18,200, how many commercial messages should be run on each medium to maximize total audience contact? What is the al- location of the budget among the three media, and what is the total audience reached?
By how much would audience contact increase if an extra $100 were allocated to the promotional budget?

The New West 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 Constraints Audience per advert Television 1000,000 Radio 18000 Online 40,000 $300 $2000 10 S600 10 Cost per Advert Maximum media usage 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 a) If the promotional budget is limited to $18,200, how many commercial messages should be run on each medium to maximize total audience contact? What is the al location of the budget among the three media, and what is the total audience reached? b) By how much would audience contact increase if an extra $100 were allocated to the promotional budget?

Answer 1

Solution

Let x1, x2, x3 represent the number of advertisements in television, radio rand online respectively. The objective function to maximize is therefore z=100’000x1+18,000x2+40,000x3 under constraints:

20x1+3x2+6x3 <=182

x3<=10

-x1+x2-x3<=0

-9x1+x2+x3<=0

x1	x2	x3	s1	s2	s3	s4	b
20	3	6	1	0	0	0	182
0	0	1	0	1	0	0	0
-1	1	-1	0	0	1	0	0
-9	1	1	0	0	0	1	0
-100000	-18000	-40000	0	0	0	0	0

An optimal solution is as follows:

media	no. of adv.	cost	audience
televison	4	$8000	400000
radio	14	$4200	252000
online	10	$6000	400000
	28	$18200	1052000