Quantitative Analysis

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CASE ONE
Two advertising media are being considered for promotion of a product. Radio ads cost $400 each, while newspaper ads cost $600 each. The total budget is $7,200 per week. The total number of ads should be at least 15, with at least 2 of each type, and there should be no more than 19 ads in total. The company does not want the number of newspaper ads to exceed the number of radio ads by more than 25 percent. Each newspaper ad reaches 6,000 people, 50 percent of whom will respond; while each radio ad reaches 2,000 people, 20 percent of whom will respond. The company wishes to reach as many respondents as possible while meeting all the constraints stated. 1. Develop the appropriate LP model for determining the number of ads of each type that should be placed.

2. Graph the constraints and identify the feasible region (clearly indicate the feasible region)

3. Solve the LP problem using e corner point method and comment on your findings.

CASE PROBLEM 1

1)

RadioNewspaperMin/Max
Cost4006007200 Max
Min1115 Min
Max1119 Max
At least
2R11 Min
At least
2N12 Min

N ≤ 1.25R
Objective: 20 (2000)5 (6000)

R = # Radio ads
N = # Newspaper ads

Therefore A= .20(2000) R + .5(6000) N

Constraints
400 R + 600N ≤ 7200
R + N ≥ 15
R + N ≤ 19
R ≥ 2
N ≥ 2
R, N ≥ 0
1.25 R – N ≥ 0

2)

400R+ 600N ≤ 7200 (4R + 6N ≤ 72)
R= (0, 18)N= (12, 0)

R + N ≥ 15
R= (0, 15)N= (15, 0)

R + N ≤ 19
R= (0, 19)N= (19, 0)

R ≥ 2
Vertical line

N ≥ 2
Horizontal line

1.25 R – N ≥ 0
R= (0, 10)N= (0, 12.5)

[pic]

3)

Corner Point 1: R+N = 15
4R+6N = 72
Therefore C (9,6) = 21,600

Corner Point 2: N = 2
4R+6N = 72
Therefore C (15,2) = 12,000

Corner Point 3: R = 2
R+N =15
Therefore C (13,2) = 11,200

Conclusion: The corner point 1 with the value of 21,600; is the value at which the objective function is at a maximum....
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