![a [a] f=a-[a]
3/2 1 ½
-7/3 -3 2/3
-1 -1 0
-2/5 -1 3/5](https://image.slidesharecdn.com/integer-programming-adopted-from-taha-and-other-sourcescompress-241016053327-9fe8968e/85/Integer-Programming-MNIT-Jaipur-Taha-et-al-25-320.jpg)
Integer Programming - MNIT Jaipur - Taha et al
- 1. Integer Programming Adopted fromTaha and other sources
- 3. No Model Type Decision variables 1 Linear Programming (LP) Can take continuous value. 2 Integer Linear Programming (ILP) At least one variable is integer valued. In integer optimization, at least one variable is restricted to integer values. But the decision variables can be linear and / or nonlinear. We will consider only linear variables. Linear Optimization Classification Within ILP, we can have either “all integer” or “mixed integer” model. A variable restricted to 0 or 1 values is called a binary variable.
- 4. Maximize x1+ 8x2 = Z Subject to x1 & x2 ≥ 0 x1 ≤ 2.0 x1 + 10x2 ≤ 20.5 What if x1and x2 must be integers? We can try four closest points. (1, 2), (2, 2) are infeasible. At (1, 1), Z = 9 . At (2, 1) , Z = 10 Optimal may not be a corner point 3 2 1 1 2 3 4 X1 LP optimal: Z = 16.80 x1= 2.00, x2 = 1.85 x1= 0.00, x2 = 2.05 Integer optimal:(0,2), Z = 16! LP example
- 5. A simple LP example Maximize 7x1+ 11x2 = Z Subject to x1 & x2 ≥ 0 x1 + x2 ≤ 6 18x1+ 34x2 ≤ 154 Optimal LP: x1= 3.125, x2 = 2.875 with Z = 53.5 Optimal ILP: x1= 1, x2 = 4 with Z = 51 9 8 7 6 5 4 3 2 1 1 2 3 4 5 6 7 8 9 There is no efficient procedure (like Simplex) available to find the optimal solution. Many approaches: Start with LP. 1.Cutting Plane: Add one constraint at a time till you get an optimal solution. 2.Branch and Bound: If the current solution is not integer, split the problem into two problems (with one constraint added to each) and solve again. Repeat till you get integer optimal solution. Solver uses this approach. 3.….
- 6. ILP: B&B Maximize 7x1+ 11x2 = Z Subject to x1 & x2 ≥ 0 x1 + x2 ≤ 6 18x1+ 34x2 ≤ 154 LP: Z = 53.5. (3.125, 2.875) 9 8 7 6 5 4 3 2 1 1 2 3 4 5 6 7 8 9 Start with the LP solution. Add two branches with extra constraints if there is a non- integer value and solve the two sub-problems. x2 ≤ 2 Z = 50. (4, 2) Z = 53.222 (2.889, 3) Feasible! No need to expand this branch. This is a lower bound. Add two branches
- 7. Root node LP: Z = 53.5 (3.125, 2.875) Branch and bound methodology Z = 50 (4, 2) Z = 52.222 (2.889, 3) x1 ≥ 3 Infeasible! Z = 52.176 (2, 3.47059) Z = 51. (1, 4) Z = 47. (2, 3) Optimum! B & B methodology can be applied to many other problems besides ILP.
- 8. Capital Budgeting Project P1 P2 P3 P4 P5 NPV 10 17 16 8 14 Expenditure 48 96 80 32 64 Assume budget (B) = 176 million rupees. Which projects should we select to maximize Net Present Value (NPV)? Maximize 10 y1 + 17 y2 + 16 y3 + 8 y4 + 14 y5 Subject to: 48 y1 + 96 y2 + 80 y3 + 32 y4 + 64 y5 ≤ 176 y1 , y2 , y3 , y4 , y5 all binary (0 or 1).
- 9. Capital Budgeting Any problems with the model?
- 10. Binary Variables & Logical Relationships In many models, one or more constraints involving binary variables can be added to satisfy desired logical relationship. Suppose we have several projects P1 , P2, P3, etc. and we define binary variables Y1, Y2, Y3, etc. Y1 = 0 means project P1 is not selected Y2 = 1 means project P2 is selected and so on. Suppose the logical relation is: Select P2 or P5 or both. We need to add constraint(s) such that when the relationship is satisfied, all constraints must be met and when the relationship is not satisfied, at least one of the constraint must fail. We need only one constraint. Y2+Y5 1.
- 11. Relationship/Constraint(s) Explanation If only P2 is selected, Y2 = 1. If only P5 is selected, Y5 = 1. If P2, P5 selected, Y2+Y5 = 2 but when both are not selected, Y2+Y5 = 0 and the constraint is not satisfied. Select P2 or P5, or both. Y2 + Y5 ≥ 1 Select at most one project from P3 , P4 , P5. Y3 + Y4 + Y5 ≤ 1 If 0 or 1 projects are selected, the constraint is satisfied. If two are selected, Y3+Y4 +Y5 will be equal to 2 and the constraint is not satisfied. Same if you select all three. If P5 is selected, then P4 must be as well. Y4 - Y5 0 If P5 is selected, Y5 = 1 then Y4 must also be 1. If Y5=0, Y4 can be 0 or 1. If Y5 =1 and Y4 = 0, the constraint is not satisfied.
- 12. 12 Linking Constraints and Fixed Costs Suppose we purchase x units of a product at unit cost C. For sending the purchase order, there is a fixed cost F. So the total purchase cost will be (F + Cx). To correct this, we introduce a binary variable y and add a new constraint as follows. Minimize Fy + Cx Subject to: x ≤ My where M is a large number. y = 0 or 1 Note that when x = 0, y can be 0 or 1 according to the first constraint and minimization will force it to zero. When x > 0, y will be forced to a value of 1. If this value is included in the objective function, then we will end up with fixed cost F even when x = 0. This should not happen. There should be no fixed cost when we don’t buy.
- 13. Fixed-Charge Problem 𝐶𝑗 𝑥𝑗 = ቊ 𝑘𝑗 + 𝑐𝑗𝑥𝑗 𝑥𝑗 > 0 0 𝑥𝑗 = 0 Where, 𝑐𝑗 is variable cost per unit, 𝑘𝑗 is fixed cost. The objective criterion then becomes: 𝑀𝑖𝑛𝑖𝑚𝑖𝑧𝑒 𝑍 = 𝑗=1 𝑁 𝑐𝑗𝑥𝑗 Because of discontinuity at the origin, Z becomes untractable from the analytic stand point.
- 17.
- 18.
- 19.
- 20.
- 21.
- 22.
- 23. The concept of the cutting will first be illustrated by an example. Consider the integer linear programming problem Maximize z = 7x1+ 9x2 Subject to -x1 + 3x2 ≤ 6 7x1 + x2 ≤ 35 x1 ,x2 are nonnegative integers The optimal continuous solution (ignoring the integrality condition) is shown graphically. This is given by z = 63, x1=9/2, x2= 7/2 which is non-integer. CUTTING PLANE ALGORITHMS
- 24. Integer programming, MA-4020- Operational Research 24 Cutting plane algorithm: The general procedure for cutting plane method is as follows: Step1:find the original LP solution (using the simplex method) ignoring the integer constraint. Step 2:if the solution is integer stop. Otherwise , construct a “cut” derived from the row that has a non integer variable with the largest fractional value and add to the current final tableau. If there is a tie select any row arbitrarily. Step 3:solve the augmented LP problem ,and return to step 2.
- 25. a [a] f=a-[a] 3/2 1 ½ -7/3 -3 2/3 -1 -1 0 -2/5 -1 3/5
- 26. Integer programming, MA-4020- Operational Research 26 Consider the following problem . Maximize Z=7x1+ 9x2 subject to: 7x1+ x2 35 ◦ -x1+ 3x2 6 ◦ x1, x2 are non negative integers.
- 27. 0 1 2 3 4 5 6 7 8 0 1 2 3 4 5 6 7 fr secondary constraints primary constraints (9/2,7/2); Z = 63 (4,3); Z = 55 X2 X1
- 28. Basic x1 x2 x3 x4 S1 R.H.S. Z 0 0 28/11 15/11 0 63 x2 0 1 7/22 1/22 0 7/2 x1 1 0 -1/22 3/22 0 9/2 S1 0 0 -7/22 -1/22 1 -1/2
- 29. Basic x1 x2 x3 x4 S1 Solution Z 0 0 0 1 8 59 x2 0 1 0 0 1 3 x1 1 0 0 1/7 -1/7 32/7 x3 0 0 1 1/7 -22/7 11/7
- 30. Since the solution is still non-integer , a new cut is constructed.The x1 - equation is written as which gives the cut
- 31. Basic x1 x2 x3 x4 S1 S2 R.H.S. Z 0 0 0 1 8 0 59 x2 0 1 0 0 1 0 3 x1 1 0 0 1/7 -1/7 0 32/7 x3 0 0 1 1/7 -22/7 0 11/7 S2 0 0 0 -1/7 -6/7 1 -4/7
- 32. Basic x1 x2 x3 x4 S1 S2 Solution z 0 0 0 0 2 7 55 x2 0 1 0 0 1 0 3 x1 1 0 0 0 -1 1 4 x3 0 0 1 0 -4 1 1 x4 0 0 0 1 6 -7 4
- 33. Can be expressed in terms of x1 and x2 only by using the appropriate substitution as follows: or Which is equivalent to Similarly for the second cut, The equivalent terms in case of x1 & x2