Jonathan Geraillo MSC530M – Volunteer Exercises
Robin Miclat March 21, 2014

LINEAR PRGRAMMING
PROBLEM 7-16 A candidate for mayor in a small town has allocated $40,000 for last-minute advertising in the days preceding the election. Two types of ads will be used: radio and television. Each radio ad costs $200 and reaches an estimated 3,000 people. Each television ad costs $500 and reaches an estimated 7,000 people. In planning the advertising campaign, the campaign manager would like to reach as many people as possible, but she has stipulated that at least 10 ads of each type must be used. Also, the number of radio ads must be at least as the number of television ads. How many ads each type should be used? How many people will this reach?
SOLUTION:
Let
X1 = number of radio ads placed
X2= number of TV ads placed

Objective: Maximize audience reach 3000x1 + 7000x2
Subject to (Constraints): 200X1 + 500X2 < 40000 1X1 > 10 1X2 > 10 1X1- 1X2 > 0

OPTIMAL SOLUTION: X1 = 175 (number of radio ads that will be placed) and X2 = 10 (number of TV ads that will be placed); EXPOSURE = 595,000 people.
INTERPRETATION: The candidate for mayor is recommended to place 175 radio ads and 10 TV ads for last-minute advertising in the days preceding the election. The radio ads will reach 525,000 people and will cost 35,000. The TV ads, on the other hand, will reach 70,000 people and will cost 5,000. This combination of advertisement (175 radio ads and 10 TV ads) will reach 595,000 people and will cost 40,000. Type of Ad | Number of Ads | Number of people to reach per advertisement | Total Number of people to reach | Cost per advertisement | Total Cost for the Ads | Radio ads | 175 | 3000 | 525000 | 200 | 35000 | TV ads | 10 | 7000 | 70000 | 500 | 5000 | Total | | | 595000 | | 40000 |

SOLUTION USING POM-QM: