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1. Graphical method example ( 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

9. Multiple optimal solution example
(Previous example)
11. Infeasible solution example
(Next example)

10. Unbounded solution example





Unbounded solution
In maximization problem, if shaded area is open-ended. This means that the maximization is not possible and the LPP has no finite solution. Hence the solution of the given problem is unbounded.
Example
Find solution using graphical method
MAX Z = 5X1 + 4X2
subject to
X1 - 2X2 <= 1
X1 + 2X2 >= 3
and X1,X2 >= 0


Solution:
Problem is
MAX `Z_x``=````5``X_1`` + ``4``X_2`
subject to
`````X_1`` - ``2``X_2``1`
`````X_1`` + ``2``X_2``3`
and `X_1,X_2 >= 0; `




Hint to draw constraints

1. To draw constraint `color{red}{X_1-2X_2<=1 ->(1)}`

Treat it as `color{red}{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




2. To draw constraint `color{green}{X_1+2X_2>=3 ->(2)}`

Treat it as `color{green}{X_1+2X_2=3}`

When `X_1=0` then `X_2=?`

`=>(0)+2X_2=3`

`=>2X_2=3`

`=>X_2=(3)/(2)=1.5`

When `X_2=0` then `X_1=?`

`=>X_1+2(0)=3`

`=>X_1=3`

`X_1`03
`X_2`1.50









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=5X_1 + 4X_2`
`color{red}{A(0,1.5)}``color{green}{2->X_1+2X_2>=3}`
`color{black}{3->X_1>=0}`
`5(0)+4(1.5)=6`
`color{green}{B(2,0.5)}``color{red}{1->X_1-2X_2<=1}`
`color{green}{2->X_1+2X_2>=3}`
`5(2)+4(0.5)=12`


Problem has an unbounded solution.



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9. Multiple optimal solution example
(Previous example)
11. Infeasible solution example
(Next example)





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