Examples of LPP
Example 1:
The manager of an oil refinery must decide on the optimal mix of two possible blending processes of which the inputs and outputs per production run are as follows:
The maximum amounts available of crudes A & B are 225 units and 200 units respectively. Market demand shows that at least 150 units of gasoline X and 120 units of gasoline Y must be produced. The profits per production run from process 1 and process 2 are Rs. 200 and Rs. 300 respectively. Formulate the problem as a linear programming problem.
Solution:
Let x and y denote the number of production runs of the two processes, respectively. Then the appropriate mathematical formulation of the problem is:
Maximize Z=200x+300y
Subject to the constraints: 4x+5y≤225 3x+6y≤200 4x+5y≥150 7x+5y≥120 x,y≥0 The standard weight of a special purpose brick is equal to 5kg and it contains two basic ingredients B1 and B2. B1 costs Nu. 5 per kg and B2 costs Nu.8 per kg. Strength considerations dictate that the brick should contain not more than 4kg of B1 and minimum of 2kg of B2. Since the demand for the product is likely to be related to the price of the brick, find out graphically the minimum cost of the brick satisfying the above condition.
Let x be the ingredient B1 used in the brick
Let y be the ingredient B2 used in the brick
Minimize Z=5x+8y
Subject to the constraints: x+y=5 x≤4 y≥2 x,y≥0
We use the graphic method to find the solution of the above problem. To find the coordinates of the A, we solve the following equations: x+y=5 equation-1 x=4 equation-2 y=2 equation-3 let x=0
0+y=5
Y=5, coordinate A=(0,5)
Let y=0 x+0=5 x=5, coordinate B=(5,0)
Feasible regions lie on the line segment of A & E, which is shown in the graph with bold line
Thus feasible regions are A & E.
Take equation A and B to find the corners of E x+y=5 substitute x=3 in the following equation and find y y=2 x+y=5, 3+y=5, y=5-3,
The manager of an oil refinery must decide on the optimal mix of two possible blending processes of which the inputs and outputs per production run are as follows:
The maximum amounts available of crudes A & B are 225 units and 200 units respectively. Market demand shows that at least 150 units of gasoline X and 120 units of gasoline Y must be produced. The profits per production run from process 1 and process 2 are Rs. 200 and Rs. 300 respectively. Formulate the problem as a linear programming problem.
Solution:
Let x and y denote the number of production runs of the two processes, respectively. Then the appropriate mathematical formulation of the problem is:
Maximize Z=200x+300y
Subject to the constraints: 4x+5y≤225 3x+6y≤200 4x+5y≥150 7x+5y≥120 x,y≥0 The standard weight of a special purpose brick is equal to 5kg and it contains two basic ingredients B1 and B2. B1 costs Nu. 5 per kg and B2 costs Nu.8 per kg. Strength considerations dictate that the brick should contain not more than 4kg of B1 and minimum of 2kg of B2. Since the demand for the product is likely to be related to the price of the brick, find out graphically the minimum cost of the brick satisfying the above condition.
Let x be the ingredient B1 used in the brick
Let y be the ingredient B2 used in the brick
Minimize Z=5x+8y
Subject to the constraints: x+y=5 x≤4 y≥2 x,y≥0
We use the graphic method to find the solution of the above problem. To find the coordinates of the A, we solve the following equations: x+y=5 equation-1 x=4 equation-2 y=2 equation-3 let x=0
0+y=5
Y=5, coordinate A=(0,5)
Let y=0 x+0=5 x=5, coordinate B=(5,0)
Feasible regions lie on the line segment of A & E, which is shown in the graph with bold line
Thus feasible regions are A & E.
Take equation A and B to find the corners of E x+y=5 substitute x=3 in the following equation and find y y=2 x+y=5, 3+y=5, y=5-3,