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Max Z = x1 + x2 2 x1 + 3 x2  6 (1) x2  1.5 (2) x1 - x2  2 (3)

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Presentation on theme: "Max Z = x1 + x2 2 x1 + 3 x2  6 (1) x2  1.5 (2) x1 - x2  2 (3)"— Presentation transcript:

1 Max Z = x1 + x2 2 x1 + 3 x2  6 (1) x2  1.5 (2) x1 - x2  2 (3)
Graphical solution Max Z = x1 + x2 2 x1 + 3 x2  6 (1) x2  (2) x1 - x2  (3) x1 , x2  0

2 Graphical solution Max Z = x1 + x2 2 x1 + 3 x2  6 (1) x2  1.5 (2)
OABCD represents the feasible space x2 Max Z = x1 + x2 2 x1 + 3 x2  6 (1) x2  (2) x1 - x2  (3) x1 , x2  0 (3) A B (2) C x1 O D (1)

3 Graphical solution Max Z = x1 + x2 2 x1 + 3 x2  6 (1) x2  1.5 (2)
(3) A B (2) Z = 1 Optimal Point 2 x*1 + 3 x*2 = 6 x*1 – x*2 = 2 x2 * C Z = 0 x1 * x1 O D (1) x*1 = 2.4, x*2 = 0.4 Z* = 2.8 Direction of the maximization

4 Graphical solution: 2nd method
Theorem 1: If an optimal solution of a LP exists, then at least one of the corner points is optimal. x2 Corner points (3) O (0, 0) Z = 0 A (0, 1.5) Z = 1.5 B (0.75, 1.5 ) Z = 2.25 C (2.4, 0.4) Z = 2.8 D (2, 0) Z = 2 A B (2) C x1 O D (1)

5 Graphical solution: example 2
Max Z = x1 + 2 x2 2 x1 + 3 x2  6 (1) x2  (2) x1 - x2  (3) x1 , x2  0 x2 Optimal Point (3) A B (2) Z = 1 O (0, 0) Z = 0 A (0, 1.5) Z = 3 B (0.75, 1.5 ) Z = 3.75 C (2.4, 0.4) Z = 3.2 D (2, 0) Z = 2 C Z* Z = 0 x1 O D (1)

6 Graphical solution: example 3
Theorem 2: If a LP has more than one optimal solution, then there is an infinite number of optimal solutions. x2 Max Z = 4 x1 + 6 x2 2 x1 + 3 x2  6 (1) x2  (2) x1 - x2  (3) x1 , x2  0 All the points of [BC] are optimal (3) A B (2) Z = 4 O (0, 0) Z = 0 A (0, 1.5) Z = 9 B (0.75, 1.5 ) Z = 12 C (2.4, 0.4) Z = 12 D (2, 0) Z = 8 C Z* Z = 0 x1 O D (1)

7 Graphical solution: example 4
(2) Max Z = x1 + 3 x2 x1 + 2 x2  (1) - x1 + 2 x2  4 (2) x1 , x2  0 x2 Infeasible Linear Program: There is no solution that satisfies all the constraints x1 O (1)

8 Graphical solution: example 5
Unbounded feasible space Max Z = x1 + x2 -2 x1 + 3 x2  6 (1) x2  (2) x1 , x2  0 (1) x2 Unbounded Linear Program: The objective function can always increase (maximization) without violating any constraint. Z = 1 (2) Z = 0 No optimal solution x1 O


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