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Home > Operation Research calculators > Graphical method example
1. Graphical method example ( Enter your problem )
( Enter your problem )
  1. Elements of LPP and Definition
  2. Algorithm & Example-1
  3. Maximization Example-2
  4. Maximization Example-3
  5. Minimization Example-4
  6. Minimization Example-5
  7. Mixed constraints Example-6
  8. Mixed constraints Example-7
  9. Multiple optimal solution example
  10. Unbounded solution example
  11. Infeasible solution example
 		Other related methods
  1. Formulate linear programming model
  2. Graphical method
  3. Simplex method (BigM method)
  4. Two-Phase method
  5. Primal to dual conversion
  6. Dual simplex method
  7. Integer simplex method
  8. Branch and Bound method
  9. 0-1 Integer programming problem
  10. Revised Simplex method
3. Maximization Example-2
(Previous example)		5. Minimization Example-4
(Next example)

4. Maximization Example-3


Find solution using graphical method
MAX z = 2x1 + x2
subject to
x1 + 2x2 <= 10
x1 + x2 <= 6
x1 - x2 <= 2
x1 - 2x2 <= 1
and x1,x2 >= 0

Solution:
Problem is
MAX `z_x``=````2``x_1`` + ````x_2`
subject to
`````x_1`` + ``2``x_2`≤`10`
`````x_1`` + ````x_2`≤`6`
`````x_1`` - ````x_2`≤`2`
`````x_1`` - ``2``x_2`≤`1`
and `x_1,x_2 >= 0; `



Hint to draw constraints

1. To draw constraint `color{red}{x_1+2x_2<=10 ->(1)}`

Treat it as `color{red}{x_1+2x_2=10}`

When `x_1=0` then `x_2=?`

`=>(0)+2x_2=10`

`=>2x_2=10`

`=>x_2=(10)/(2)=5`

When `x_2=0` then `x_1=?`

`=>x_1+2(0)=10`

`=>x_1=10`

`x_1`010
`x_2`50



2. To draw constraint `color{green}{x_1+x_2<=6 ->(2)}`

Treat it as `color{green}{x_1+x_2=6}`

When `x_1=0` then `x_2=?`

`=>(0)+x_2=6`

`=>x_2=6`

When `x_2=0` then `x_1=?`

`=>x_1+(0)=6`

`=>x_1=6`

`x_1`06
`x_2`60



3. To draw constraint `color{blue}{x_1-x_2<=2 ->(3)}`

Treat it as `color{blue}{x_1-x_2=2}`

When `x_1=0` then `x_2=?`

`=>(0)-x_2=2`

`=>-x_2=2`

`=>x_2=-2`

When `x_2=0` then `x_1=?`

`=>x_1-(0)=2`

`=>x_1=2`

`x_1`02
`x_2`-20



4. To draw constraint `color{brown}{x_1-2x_2<=1 ->(4)}`

Treat it as `color{brown}{x_1-2x_2=1}`

When `x_1=0` then `x_2=?`

`=>(0)-2x_2=1`

`=>-2x_2=1`

`=>x_2=(1)/(-2)=-0.5`

When `x_2=0` then `x_1=?`

`=>x_1-2(0)=1`

`=>x_1=1`

`x_1`01
`x_2`-0.50








The value of the objective function at each of these extreme points is as follows:
Extreme Point
Coordinates
(`x_1`,`x_2`)		Lines through Extreme PointObjective function value
`z=2x_1 + x_2`
`color{red}{O(0,0)}``color{black}{5->x_1>=0}`
`color{black}{6->x_2>=0}`		`2(0)+1(0)=0`
`color{green}{A(1,0)}``color{brown}{4->x_1-2x_2<=1}`
`color{black}{6->x_2>=0}`		`2(1)+1(0)=2`
`color{blue}{B(3,1)}``color{blue}{3->x_1-x_2<=2}`
`color{brown}{4->x_1-2x_2<=1}`		`2(3)+1(1)=7`
`color{brown}{C(4,2)}``color{green}{2->x_1+x_2<=6}`
`color{blue}{3->x_1-x_2<=2}`		`2(4)+1(2)=10`
`color{darkorange}{D(2,4)}``color{red}{1->x_1+2x_2<=10}`
`color{green}{2->x_1+x_2<=6}`		`2(2)+1(4)=8`
`color{darkviolet}{E(0,5)}``color{red}{1->x_1+2x_2<=10}`
`color{black}{5->x_1>=0}`		`2(0)+1(5)=5`


The maximum value of the objective function `z=10` occurs at the extreme point `(4,2)`.

Hence, the optimal solution to the given LP problem is : `x_1=4, x_2=2` and max `z=10`.

This material is intended as a summary. Use your textbook for detail explanation.
Any bug, improvement, feedback then Submit Here


3. Maximization Example-2
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