# Quantitative Analysis

Topics: Optimization, Optimization problem, Operations research Pages: 4 (509 words) Published: December 24, 2010
CASE ONE

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)

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

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

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....