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| Some Exceptional Cases |
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| We may come across LPP which may have no feasible (infeasible) solution or may have unbounded solution. We conclude the chapter by giving one example for each of these exceptional cases. |
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| Example: |
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| Show graphically that the model |
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| Maximise Z = -5y |
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| Subject to |
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y 0 has no feasible solution. |
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| Suggested answer: |
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| Draw the graphs x + y = 1 |
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| - 0.5 -5y = - 10 |
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Shade the half planes of the constraints x + y  1 …(1) |
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-0.5x - 5y  -10 …(2) |
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| Note that the origin (0, 0) does not satisfy the inequation (2) hence the required region is the upper half plane. |
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| From the graph, it is clear that the intersection of the constraints is empty. Hence the given problem has no feasible solution. Therefore the given L.P.P has no solution. |
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| Example: Show graphically that the L.P.P |
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| Maximize |
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| Z = 6x + y |
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| Subject to the constraints |
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x  0, y  0 has an unbounded solution. |
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| Suggested answer: |
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The intersection of the half planes 2x + y  3 and x - y  0 is shown as shaded region in the figure. |
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| Feasible region is an unbounded convex region |
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| at A (0, 3), Z = 6 (0) + 3 = 3 |
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| At B (1, 1), Z = 6(1) + 1 = 7 |
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| Consider any point say (4, 5), the value of Z = 24 + 5 = 29. |
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| Hence the maximum flow is from (1, 1) to (4, 5) |
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| Hence the given problem has an unbounded solution. |
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