Definitions [8]
| Term | Definition |
| Feasible Solution | A feasible solution is any solution that satisfies all the constraints of the LPP, including non-negativity restrictions. |
| Feasible Region | The common region that satisfies all the constraints on the graph is called the feasible region. Every point inside or on this region represents a feasible solution. |
| Infeasible Solution | Any point that does not satisfy all the given constraints is an infeasible solution. |
| Optimal Solution | A feasible solution that gives the maximum or minimum value of the objective function is called the optimal solution. |
| Corner Point | A corner point is a vertex of the feasible region formed by the intersection of boundary lines. In the graphical method, these points are checked first to find the optimum value. |
| Bounded Region |
A feasible region that is enclosed within finite boundaries and does not extend indefinitely in any direction. |
| Unbounded Region |
A feasible region that extends indefinitely in one or more directions and is not completely enclosed by boundaries. |
A Linear Programming Problem (LPP) is a problem in which a linear objective function is to be maximised or minimised subject to a set of linear constraints and non-negative conditions on the variables.
An optimisation problem is a problem in which the value of one quantity has to be made as large as possible or as small as possible under given restrictions. If the quantity and restrictions are linear, the problem becomes a Linear Programming Problem (LPP).
A region is said to be convex if the line segment joining any two points in the region lies entirely within the region.
The linear function whose maximum or minimum value is to be determined is called the objective function.
The equation ax + by = c is called the associated equation of the inequality.
The common region satisfying all the given inequalities is called the solution set.
To optimise means to maximise or minimise.
Theorems and Laws [1]
Statement:
If a linear objective function has a maximum or minimum value over a feasible region, then the maximum or minimum occurs at one of the corner points of the feasible region.
Key Points
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An LPP is solved graphically when there are two variables.
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The feasible region is formed by the common solution of all constraints.
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The optimum value is found by evaluating the objective function at corner points.
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If two corner points give the same optimum value, then all points on the joining segment are also optimal.
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In an unbounded region, the required maximum or minimum may fail to exist.
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If no feasible region exists, the LPP has no feasible solution.
| Condition | Region represented |
|---|---|
| ( x > 0 ) | Right of the y-axis |
| ( x < 0 ) | Left of the y-axis |
| ( y > 0 ) | Above x-axis |
| ( y < 0 ) | Below x-axis |
| ( x ≥ 0 ) | Includes y-axis |
| ( y ≥ 0 ) | Includes x-axis |
Important Questions [10]
- Minimize z=4x+5y subject to 2x+y>=7, 2x+3y<=15, x<=3,x>=0, y>=0 solve using graphical method.
- Minimize: Z = 6x + 4y Subject to the conditions: 3x + 2y ≥ 12, x + y ≥ 5, 0 ≤ x ≤ 4, 0 ≤ y ≤ 4
- Solve the following LPP by using graphical method. Maximize : Z = 6x + 4y
- Solve the following L.P.P graphically: Maximize :Z = 10x + 25y Subject to : x ≤ 3, y ≤ 3, x + y ≤ 5, x ≥ 0, y ≥ 0
- Minimize :Z=6x+4y, Subject to : 3x+2y ≥12
- A company manufactures bicycles and tricycles each of which must be processed through machines A and B. Machine A has maximum of 120 hours available and machine B has maximum of 180 hours available.
- Solve the following LPP by graphical method: Maximize: z = 3x + 5ySubject to: x + 4y ≤ 24 3x + y ≤ 21 x + y ≤ 9 x ≥ 0, y ≥ 0 Also find maximum value of z.
- Solve the Following Linear Programming L. P. P. Graphically Minimize Z = 6x + 2y Subject to 5x + 9y ≤ 90
- Solve the following LPP by graphical method: Minimize Z = 7x + y subject to 5x + y ≥ 5, x + y ≥ 3, x ≥ 0, y ≥ 0
- Maximize: 3 5 Subjectto
