Unit.4.integer programming
- 1. Chapter Four Integer Programming By;-Dagnaygebaw Goshme (MSc.)
- 2. 4.1.Need For Integer Programming Integer:-Values such as 0, 1, 2, 3, etc. are perfectly valid for these variables as long as these values satisfy all model constraints.) In LPP all the decision variables were allowed to take any non-negative values as it is quite possible and appropriate to have fractional values in many situations. However, there are several conditions in business and industry that lead to planning models involving integer valued variables. That is, many practical problems involve decision variables that must be integer valued.
- 3. There are cases that the fractional values of variables may be meaningless in the context of the actual decision problem. For example: In production, manufacturing is frequently scheduled interims of discrete quantities like labor, tables, chairs, machines … Production of livestock involves discrete number of animals Integer programming models require all of the assumptions implicit in linear programming models except that certain specific variables must be integer valued.
- 4. When all the variables are constrained to be integers, it is called a pure-integer programming model, and In case only some of the variables are restricted to have integer values, the problem is said to be a mixed-integer programming problem. There are two methods used to solve IPP, namely 1.Gomory’s fractional cut method 2. Branch and bound method (search method)
- 5. 1. Gomory’s Fractional Cut Method This method consists of first solving the IPP as an ordinary LPP by ignoring the restriction of integer values introducing a new constraint to the problem such that the new set of feasible solution includes all the original feasible integer solution, but does not include the optimum non-integer solution initially found. This new constraint is called Fractional Cut or Gomorian constraint. Then the revised problem is solved using simplex till an optimal integer solution is obtained.
- 6. Gomory’s Fractional Cut Method involves the following steps: 1st. Solve the IPP as an ordinary LPP by ignoring the restriction of integer values 2nd .Test the integrality of the optimum solution. If all Xbi ≥ 0 and are integers, an optimum integer solution is obtained If at least one bi is not an integer then go to the next step
- 7. 3rd .Select the source row If only one Xbi is non-integer, the row corresponding to non-integer solution will be the source row If more than one Xbi is non-integer, choose the row having the largest fractional part (fi) as a source row. In this case write each Xbi as (Xbi = xbi +fi): Where xbi is integer part of Xbi (solution) and fi is the positive fractional part of Xbi. 0< fi<1 If there is a tie when two or more rows have the same positive larger fi), select arbitrarily.
- 8. 4th.Find the new constraint (Gomorian constraint) from the source row. i. ∑aijXj = Xbi ii.∑((aij)+fi)Xj = xbi +fi iii. ∑fiXj ≥ fi iv.∑(-fi)Xj ≤ -fi v. ∑ (-fi)Xj + Gi = -fi Gomorian constraint and G is Gomorian slack. 5th .Add the new (Gomorian) constraint at the bottom of the simplex table obtained in step 1 and find the new feasible solution. Note that Gomorian slack enter in to the objective function with zero coefficient 6th. Test the integrality of the new solution Again if at least one Xbi (solution) is not an integer, go to step 3 and repeat the procedure until an optimal integer solution is obtained.
- 9. Example: Find the optimal integer solution for the following LPP Max Z = x1+x2 Subject to: 3x1 + 2x2 ≤ 5, x2≤ 2, x1, x2 ≥ 0 and are integers Solution 1st. Solve the IPP as an ordinary LPP by ignoring the restriction of integer values Max Z = x1+x2 + 0s1+0s2 Subject to: 3x1 + 2x2 +s1= 5, x2 +s2= 2, x1, x2 , s1, s2 ≥ 0 and are integers
- 11. 2nd.Test the integrality of the optimum solution. The solution is x1 =1/3, x2=2 and Z=7/3. Since x1 =1/3 is non-integer solution so that the current feasible solution is not optimal integer solution. 3rd. Select the source row. The row corresponding to non-integer solution. x1 + 1/3s1- 2/3s2 = 1/3,
- 12. 4th.Find the new constraint (Gomorian constraint) from the source row. x1 + 1/3s1- 2/3s2 = 1/3, (1+0) x1 + (0+1/3) s1 + (-1+ 1/3) s2 = 0+1/3, and eliminate the integer part 1/3s1+1/3s2 ≥ 1/3 -1/3s1-1/3s2 ≤ -1/3 -1/3s1-1/3s2 +G1= -1/3Gomorian constraint and G is Gomorian slack. 5th.Add the new (Gomorian) constraint at the bottom of the simplex table obtained in step 1 and find the new feasible solution.
- 13. The solution is not feasible because the solution (G1=-1/3) is –ve. Therefore, G1 will leave the base. Select the new constraint as a pivot row. In order to select the entering variable, use: ) 0 , ( ij ij j a a Max Entering variable is known by dividing Zj-Cj /aij in G raw and fimally sellect maximum value variable
- 14. Row operation is applied . Since all ∆j ≥ 0 and all Xbi ≥ 0 and are integers, the current feasible solution is optimal integer solution. Therefore, the solution is x1=0, x2=2 and Z=2
- 15. 2. Branch and bound method (search method) Reading assignment ………………..