What is a marketing problem in applications of linear progra
What is a marketing problem in applications of linear programming?
Briefly discuss the objective function and constraint requirements in a marketing problem.
Give a real world example of a marketing problem.
Solution
Marketing Applications
One application of linear programming in marketing is media selection.
LP can be used to help marketing managers allocate a fixed budget to various advertising media.
The objective is to maximize reach, frequency, and quality of exposure.
Media Selection
SMM Company recently developed a new instant salad machine, has $282,000 to spend on advertising. The product is to be initially test marketed in the Dallas area. The money is to be spent on a TV advertising blitz during one weekend (Friday, Saturday, and Sunday) in November.
The three options available are: daytime advertising, evening news advertising, and Sunday game-time advertising. A mixture of one-minute TV spots is desired.
Estimated Audience
Ad Type Reached With Each Ad Cost Per Ad
Daytime 3,000 $5,000
Evening News 4,000 $7,000
Sunday Game 75,000 $100,000
SMM wants to take out at least one ad of each type (daytime, evening-news, and game-time). Further, there are only two game-time ad spots available. There are ten daytime spots and six evening news spots available daily. SMM wants to have at least 5 ads per day, but spend no more than $50,000 on Friday and no more than $75,000 on Saturday.
Define the Decision Variables
DFR = number of daytime ads on Friday
DSA = number of daytime ads on Saturday
DSU = number of daytime ads on Sunday
EFR = number of evening ads on Friday
ESA = number of evening ads on Saturday
ESU = number of evening ads on Sunday
GSU = number of game-time ads on Sunday
Define the Objective Function
Maximize the total audience reached:
Max (audience reached per ad of each type) x (number of ads used of each type)
Max 3000DFR +3000DSA +3000DSU +4000EFR +4000ESA +4000ESU +75000GSU
Define the Constraints
Take out at least one ad of each type:
(1) DFR + DSA + DSU > 1
(2) EFR + ESA + ESU > 1
(3) GSU > 1
Ten daytime spots available:
(4) DFR < 10
(5) DSA < 10
(6) DSU < 10
Six evening news spots available:
(7) EFR < 6
(8) ESA < 6
(9) ESU < 6
Only two Sunday game-time ad spots available:
(10) GSU < 2
At least 5 ads per day:
(11) DFR + EFR > 5
(12) DSA + ESA > 5
(13) DSU + ESU + GSU > 5
Spend no more than $50,000 on Friday:
(14) 5000DFR + 7000EFR < 50000
Spend no more than $75,000 on Saturday:
(15) 5000DSA + 7000ESA < 75000
Spend no more than $282,000 in total:
(16) 5000DFR + 5000DSA + 5000DSU + 7000EFR + 7000ESA + 7000ESU + 100000GSU7 < 282000
Non-negativity:
DFR, DSA, DSU, EFR, ESA, ESU, GSU > 0
Objective Function Value = 199000.000
Variable Value Reduced Costs
DFR = 8.000 0.000
DSA = 5.000 0.000
DSU = 2.000 0.000
EFR = 0.000 0.000
ESA = 0.000 0.000
ESU = 1.000 0.000
GSU = 2.000 0.000



