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7. Branch and Bound method example ( Enter your problem )
  1. Algorithm
  2. Example-1
  3. Example-2
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

2. Example-1
(Previous example)
9. 0-1 Integer programming problem
(Next method)

3. Example-2





Find solution using Branch and Bound method
MAX Z = 7x1 + 9x2
subject to
-x1 + 3x2 <= 6
7x1 + x2 <= 35
x2 <= 7
and x1,x2 >= 0


Solution:
Solution steps by BigM method, Graphical method

Max `Z``=````7``x_1`` + ``9``x_2`
subject to
` - ````x_1`` + ``3``x_2``6`
```7``x_1`` + ````x_2``35`
`````x_2``7`
and `x_1,x_2 >= 0; `


Solution is
Max `Z_(A)=63` `(x_1=4.5,x_2=3.5)`

and `Z_L=55` `(x_1=4,x_2=3)` obtainted by the rounded off solution values.

The branch and bound diagram
A
`x_1=4.5,x_2=3.5`
`Z_A=63`
`Z_L=55`
Solution steps by
BigM method,
Graphical method


In Sub-problem A, `x_1(=4.5)` must be an integer value, so two new constraints are created, `x_1<=4` and `x_1>=5`

Sub-problem B : Solution is found by adding `x_1<=4`.
Solution steps by BigM method, Graphical method
Max `Z``=````7``x_1`` + ``9``x_2`
subject to
` - ````x_1`` + ``3``x_2``6`
```7``x_1`` + ````x_2``35`
`````x_2``7`
`````x_1``4`
and `x_1,x_2 >= 0; `

Solution is
Max `Z_(B)=58` `(x_1=4,x_2=3.3333)`
and `Z_L=55` `(x_1=4,x_2=3)` obtainted by the rounded off solution values.
Sub-problem C : Solution is found by adding `x_1>=5`.
Solution steps by BigM method, Graphical method
Max `Z``=````7``x_1`` + ``9``x_2`
subject to
` - ````x_1`` + ``3``x_2``6`
```7``x_1`` + ````x_2``35`
`````x_2``7`
`````x_1``5`
and `x_1,x_2 >= 0; `

Solution is
Max `Z_(C)=35` `(x_1=5,x_2=0)`
and `Z_L=35` `(x_1=5,x_2=0)` obtainted by the rounded off solution values.
This Problem has integer solution, so no further branching is required.




The branch and bound diagram
A
`x_1=4.5,x_2=3.5`
`Z_A=63`
`Z_L=55`
Solution steps by
BigM method,
Graphical method
`x_1≤4``x_1≥5`
B
`x_1=4,x_2=3.3333`
`Z_B=58`
`Z_L=55`
Solution steps by
BigM method,
Graphical method
C
`x_1=5,x_2=0`
`Z_C=35`
`Z_L=35`
Solution steps by
BigM method,
Graphical method




In Sub-problem B, `x_2(=3.3333)` must be an integer value, so two new constraints are created, `x_2<=3` and `x_2>=4`

Sub-problem D : Solution is found by adding `x_2<=3`.
Solution steps by BigM method, Graphical method
Max `Z``=````7``x_1`` + ``9``x_2`
subject to
` - ````x_1`` + ``3``x_2``6`
```7``x_1`` + ````x_2``35`
`````x_2``7`
`````x_1``4`
`````x_2``3`
and `x_1,x_2 >= 0; `

Solution is
Max `Z_(D)=55` `(x_1=4,x_2=3)`
and `Z_L=55` `(x_1=4,x_2=3)` obtainted by the rounded off solution values.
This Problem has integer solution, so no further branching is required.
Sub-problem E : Solution is found by adding `x_2>=4`.
Solution steps by BigM method, Graphical method
Max `Z``=````7``x_1`` + ``9``x_2`
subject to
` - ````x_1`` + ``3``x_2``6`
```7``x_1`` + ````x_2``35`
`````x_2``7`
`````x_1``4`
`````x_2``4`
and `x_1,x_2 >= 0; `

Solution is
This Problem has an infeasible solution, so this branch is terminated.




The branch and bound diagram
A
`x_1=4.5,x_2=3.5`
`Z_A=63`
`Z_L=55`
Solution steps by
BigM method,
Graphical method
`x_1≤4``x_1≥5`
B
`x_1=4,x_2=3.3333`
`Z_B=58`
`Z_L=55`
Solution steps by
BigM method,
Graphical method
C
`x_1=5,x_2=0`
`Z_C=35`
`Z_L=35`
Solution steps by
BigM method,
Graphical method
`x_2≤3``x_2≥4`
D
`x_1=4,x_2=3`
`Z_D=55`
`Z_L=55`
Solution steps by
BigM method,
Graphical method
E

Infeasible Solution

Solution steps by
BigM method,
Graphical method




The branch and bound algorithm thus terminated and the optimal integer solution is :
`Z_(D)=55` and `x_1=4,x_2=3`




This material is intended as a summary. Use your textbook for detail explanation.
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2. Example-1
(Previous example)
9. 0-1 Integer programming problem
(Next method)





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