2. Capital Budgeting Problem - 1 example
1. Find solution using 0-1 Integer programming problem method MAX Z = 300x1 + 90x2 + 400x3 + 150x4 subject to 35000x1 + 10000x2 + 25000x3 + 90000x4 <= 120000 4x1 + 2x2 + 7x3 + 3x4 <= 12 x1 + x2 <= 1 and x1,x2,x3,x4 >= 0
Solution: To apply branch and bound method, the following 4 constraints have to be added to the LP model. `x_1<=1`
`x_2<=1`
`x_3<=1`
`x_4<=1`
Solution steps by BigM method
Max `Z` | `=` | `` | `300` | `x_1` | ` + ` | `90` | `x_2` | ` + ` | `400` | `x_3` | ` + ` | `150` | `x_4` |
| subject to | `` | `35000` | `x_1` | ` + ` | `10000` | `x_2` | ` + ` | `25000` | `x_3` | ` + ` | `90000` | `x_4` | ≤ | `120000` | `` | `4` | `x_1` | ` + ` | `2` | `x_2` | ` + ` | `7` | `x_3` | ` + ` | `3` | `x_4` | ≤ | `12` | `` | `` | `x_1` | ` + ` | `` | `x_2` | | | | | | | ≤ | `1` | `` | `` | `x_1` | | | | | | | | | | ≤ | `1` | | | | `` | `` | `x_2` | | | | | | | ≤ | `1` | | | | | | | `` | `` | `x_3` | | | | ≤ | `1` | | | | | | | | | | `` | `` | `x_4` | ≤ | `1` |
| and `x_1,x_2,x_3,x_4 >= 0; ` |
Solution is Max `Z_(A)=750` `(x_1=1,x_2=0,x_3=1,x_4=0.3333)`
and `Z_L=700` `(x_1=1,x_2=0,x_3=1,x_4=0)` obtainted by the rounded off solution values.
The branch and bound diagram
A `x_1=1,x_2=0,x_3=1,x_4=0.3333` `Z_A=750` `Z_L=700` Solution steps by BigM method |
In Sub-problem A, `x_4(=0.3333)` must be an integer value, so two new constraints are created, `x_4=0` and `x_4=1`
Sub-problem B : Solution is found by adding `x_4=0`. Solution steps by BigM method
Max `Z` | `=` | `` | `300` | `x_1` | ` + ` | `90` | `x_2` | ` + ` | `400` | `x_3` | ` + ` | `150` | `x_4` |
| subject to | `` | `35000` | `x_1` | ` + ` | `10000` | `x_2` | ` + ` | `25000` | `x_3` | ` + ` | `90000` | `x_4` | ≤ | `120000` | `` | `4` | `x_1` | ` + ` | `2` | `x_2` | ` + ` | `7` | `x_3` | ` + ` | `3` | `x_4` | ≤ | `12` | `` | `` | `x_1` | ` + ` | `` | `x_2` | | | | | | | ≤ | `1` | `` | `` | `x_1` | | | | | | | | | | ≤ | `1` | | | | `` | `` | `x_2` | | | | | | | ≤ | `1` | | | | | | | `` | `` | `x_3` | | | | ≤ | `1` | | | | | | | | | | `` | `` | `x_4` | ≤ | `1` | | | | | | | | | | `` | `` | `x_4` | = | `0` |
| and `x_1,x_2,x_3,x_4 >= 0; ` |
Solution is Max `Z_(B)=700` `(x_1=1,x_2=0,x_3=1,x_4=0)` and `Z_L=700` `(x_1=1,x_2=0,x_3=1,x_4=0)` obtainted by the rounded off solution values. This Problem has integer solution, so no further branching is required.
| Sub-problem C : Solution is found by adding `x_4=1`. Solution steps by BigM method
Max `Z` | `=` | `` | `300` | `x_1` | ` + ` | `90` | `x_2` | ` + ` | `400` | `x_3` | ` + ` | `150` | `x_4` |
| subject to | `` | `35000` | `x_1` | ` + ` | `10000` | `x_2` | ` + ` | `25000` | `x_3` | ` + ` | `90000` | `x_4` | ≤ | `120000` | `` | `4` | `x_1` | ` + ` | `2` | `x_2` | ` + ` | `7` | `x_3` | ` + ` | `3` | `x_4` | ≤ | `12` | `` | `` | `x_1` | ` + ` | `` | `x_2` | | | | | | | ≤ | `1` | `` | `` | `x_1` | | | | | | | | | | ≤ | `1` | | | | `` | `` | `x_2` | | | | | | | ≤ | `1` | | | | | | | `` | `` | `x_3` | | | | ≤ | `1` | | | | | | | | | | `` | `` | `x_4` | ≤ | `1` | | | | | | | | | | `` | `` | `x_4` | = | `1` |
| and `x_1,x_2,x_3,x_4 >= 0; ` |
Solution is Max `Z_(C)=595` `(x_1=0,x_2=0.5,x_3=1,x_4=1)` and `Z_L=550` `(x_1=0,x_2=0,x_3=1,x_4=1)` obtainted by the rounded off solution values. `Z_(C)=595 < Z_(B)=700`, so no further branching is required.
|
The branch and bound diagram
A `x_1=1,x_2=0,x_3=1,x_4=0.3333` `Z_A=750` `Z_L=700` Solution steps by BigM method | `x_4=0` | | `x_4=1` | B `x_1=1,x_2=0,x_3=1,x_4=0` `Z_B=700` `Z_L=700` Solution steps by BigM method | | C `x_1=0,x_2=0.5,x_3=1,x_4=1` `Z_C=595` `Z_L=550` Solution steps by BigM method |
The 0-1 Integer programming problem algorithm thus terminated and the optimal integer solution is : `Z_(B)=700` and `x_1=1,x_2=0,x_3=1,x_4=0`
This material is intended as a summary. Use your textbook for detail explanation. Any bug, improvement, feedback then
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