# Statistics_SL1 by ClassOf1

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```									               Statistics

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Sub: Statistics                                                                    Topic: Operation Research

Design a LPP Model

Question :

Consider the following LP problem:

Maximize profit = 1X1 + 1X2

Subject to 2X1 + 1X2 ≤ 100

1X1 + 2X2 ≤ 100

a) What is the optimal solution to this problem? Solve it graphically.

b) If a technical breakthrough occurred that raised the profit per unit of X1 to \$3, would this
affect the optimal solution?

c) Instead of an increase in the profit coefficient X1 to \$3, suppose the profit was overestimated
and should have only been \$1.25. Does this change the optimal solution?

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Sub: Statistics                                                                    Topic: Operation Research

Solution :

a. Replace the variable x1 by x and x2 by y.

2X1 + 1X2 ≤ 100

1X1 + 2X2 ≤ 100

Rewrite the inequality in explicit form.

y ≤ 100 - 2x

y ≤ (100 - x)/2

Enter the above inequality in the Graphing Calculator.

Solving the set of linear inequality constraints graphically, we obtain the feasible region.

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Sub: Statistics                                                                    Topic: Operation Research

x1                 x2                 Max

0                  0                            0

0                 50                          50

33.33              33.33                    66.66

50                  0                          50

So we conclude that the profit is maximum at x1 = 33.33 and x2 = 33.33

So the optimal solution is (33.33, 33.33).

The profit at optimal point is \$66.66.

b. Objective Function is z = 3X1 + 1X2.

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*The Homework solutions from Classof1 are intended to help students understand the approach to solving the problem and not for
Sub: Statistics                                                                    Topic: Operation Research
x1                x2              Max

0                 0                        0

0                50                      50

33.33            33.33               133.32

50                 0                    150

If we increase the objective function coefficient of x1 from 1 to 3 then the optimal solution is (50, 0).
Consequently, the profit at optimal point is \$150.

c. Objective Function is z = 1.25X1 + 1X2.

x1               x2               Max

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